https://wiki.eddyn.net/api.php?action=feedcontributions&user=Eddynetweb&feedformat=atomeddynetweb's cesspit - User contributions [en]2024-03-28T19:32:57ZUser contributionsMediaWiki 1.40.0https://wiki.eddyn.net/index.php?title=User:BotCollection&diff=233User:BotCollection2023-08-29T20:57:20Z<p>Eddynetweb: Just a small change. :o</p>
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<div>Bot dumping ground.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=User:AshelyWilkie590&diff=232User:AshelyWilkie5902023-08-29T17:32:13Z<p>Eddynetweb: Eddynetweb moved page User:AshelyWilkie590 to User:BotCollection: Automatically moved page while renaming the user "AshelyWilkie590" to "BotCollection"</p>
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<div>#REDIRECT [[User:BotCollection]]</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=User:BotCollection&diff=231User:BotCollection2023-08-29T17:32:13Z<p>Eddynetweb: Eddynetweb moved page User:AshelyWilkie590 to User:BotCollection: Automatically moved page while renaming the user "AshelyWilkie590" to "BotCollection"</p>
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<div>Exploring the Outdo Online Meter reading Choices: Crown Internet site Recommendations, Majuscule Blog Suggestions, and Premiere Subject Platforms<br><br>With the wage increase of the internet, we at present birth accession to an tremendous amount of substance at our fingertips, including blogs, websites, and early online sources. However, the abundance of options fanny be overwhelming, and it nates be intriguing to bed which sites provide the better subject matter. This article explores respective exceed website recommendations, heavy web log suggestions, and chancellor message platforms to assistance you uncovering the apotheosis locate for you.<br><br>Teetotum Internet site Recommendations<br><br>When it comes to finding excellent online content, close to websites rack out above the pillow. [http://www.godmart.com.ve Unrivaled] so much place is Medium, which is a hub for thoughtful, well-written articles on a broad kind of subjects, including technology, politics, and modus vivendi. The situation has a sleek, easy-to-exercise port and allows writers to print their cultivate straightaway on the internet site or to consortium their depicted object from other platforms.<br><br>Some other crown internet site passport is The Atlantic, which is an fantabulous generator for in-depth, long-shape articles on stream events and refinement. The site has a immense archive of articles geological dating rearward to the mid-1800s and features an telling roll of writers and journalists. The Atlantic Ocean as well offers a gainful subscription service, which provides extra admittance to scoop contented and features.<br><br>log Suggestions<br><br>In accession to accomplished websites, blogs nates besides be an [https://ant.com excellent] informant of online content. Unity standout web log is Mastermind Pickings, which offers thoughtful, piquant comment on a mountain chain of topics, from literature to philosophical system to science. The author, Maria Popova, has a discrete vocalization and committal to writing flair that ready her separate from former bloggers.<br><br>Some other swell blog mesmerism is Hold But Why, which offers in-depth analytic thinking and vox populi pieces on a astray kind of subjects. The site's tagline, "A popular long-form, stick-figure-illustrated blog about almost everything," sums up its eclecticist and piquant approaching. The site's author, Tim Urban, is known for his elaborated and often humourous authorship mode.<br><br>Prime Minister Capacity Platforms<br><br>Finally, in that location are several chancellor message platforms that bid curated substance or provide users to plowshare their have act. Unrivaled such weapons platform is Medium, which we mentioned in the beginning as a pass internet site testimonial. In the event you beloved this post and you want to acquire details concerning news - [https://blog.chickabug.com/patriotic-crafts/ click hyperlink] - - [https://blog.chickabug.com/patriotic-crafts/ click hyperlink] - generously pay a visit to our website. Medium's editor-curated contentedness feeds and algorithms come on the better subject matter on the land site founded on a user's interests, making it slowly to find new writers and blogs.<br><br>Another prime minister contented platform is Quora, which is a question-and-resolve residential area that allows users to pass on questions and find responses from early users. The website has a full straddle of topics, from science to business organisation to personal finance, and its user-generated message tush ofttimes allow for singular insights and perspectives.<br><br>With so many online options available, it terminate be challenging to witness the C. H. Best depicted object for your interests and tastes. However, by exploring exceed site recommendations, corking blog suggestions, and premiere substance platforms, you arse pick up approximately of the outdo online recitation choices useable today. Whether you're looking for in-profoundness analysis, engaging commentary, or simply a just read, thither are flock of options to explore in the online human beings.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=CS410:_Programming_Paradigms_Notes&diff=150CS410: Programming Paradigms Notes2022-08-22T17:45:40Z<p>Eddynetweb: Create course page.</p>
<hr />
<div>Welcome! Notes for Fall 2022 Programming Paradigms course. Will make this more pretty as things evolve. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
We first need to consider the following materials: <br />
* AAA<br />
* BBB</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE492_Section_3_Notes&diff=149ECE492 Section 3 Notes2022-04-12T00:28:21Z<p>Eddynetweb: Starting 3.1 notes on MOSFET.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 3.1 == <br />
<br />
What is a '''MOSFET'''?<br />
* Stands for: Metal Oxide Semiconductor Field Effect Insulator.<br />
* Yay!<br />
<br />
See Figure 3a.<br />
<br />
Has 3 sides:<br />
* Gate, Drain, and Source<br />
* The gate current is always 0 (Ig = 0). <br />
* It is a symmetrical device (drain as source and vice versa). <br />
* Length of channel is typically 0.03um to 1um<br />
* Width of channel is typically 1um to 100um.<br />
* Gate current is typically 0mA (10^-15mA)<br />
* Drain voltage in position...<br />
<br />
Threshold Voltage (Vt)<br />
* At which a sufficient number of mobile electrons accumulate in the channel region to form a conducting channel. <br />
<br />
Effective Voltage (or Overdrive Voltage) <br />
* Vov = Vgs - Vt<br />
<br />
Additional formulas to consider: <br />
* Kn' = UnCox = Process Transconductance Parameter<br />
** Un = Electron mobility<br />
** Cox = Oxide capacitance<br />
** Kn = MOSFET transconductance parameter = Kn'[w/L]<br />
* r_DS = Channel Resistance = [1/g_DS] = [1/[UnCox][w/L][Vgs-Vt]]<br />
* At V_DS, the channel is going to be deepest at the source and shallower at the drain. <br />
* Channel resistance is inversely proportional to the gate voltage. <br />
<br />
* Channel pinch off: <br />
* The zero depth of the channel at the drain ends because of V_DS = V_GS - V_E<br />
** If V_GS < Vtn then no channel, so transistor is working in cut off region (I_D = 0)<br />
<br />
* MOSFET operation<br />
** Tniode region<br />
** Saturation region<br />
<br />
<br />
<br />
Now we'll cover (B) Common Emitter Current gain: <br />
* <math>B=\frac{I_{c}}{I_{b}}</math><br />
* 50 < B < 200<br />
<br />
...and (Alpha) Common Base Current Gain: <br />
* <math>Alpha = \frac{B}{B+1} = \frac{I_{c}}{I_{e}}</math><br />
<br />
=== Header 2 ===<br />
<br />
It sometimes allow current, and sometimes it doesn't.<br />
<br />
<br />
===Assignment 1.1===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE492_Section_2_Notes&diff=148ECE492 Section 2 Notes2022-03-07T20:59:11Z<p>Eddynetweb: Added additional day content.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 2.1 == <br />
<br />
We'll begin by covering a '''Bi-polar Junction Transistor''', also known as BJT.<br />
<br />
What is it exactly? Well, it's a transistor. <br />
* It is a 3 terminal devices.<br />
* It is a unsymmetrical device.<br />
* It amplifies the signal. <br />
<br />
There are two types of BJTs:<br />
* NPN, in which...<br />
* PNP, in which...<br />
<br />
Now we'll cover (B) Common Emitter Current gain: <br />
* <math>B=\frac{I_{c}}{I_{b}}</math><br />
* 50 < B < 200<br />
<br />
...and (Alpha) Common Base Current Gain: <br />
* <math>Alpha = \frac{B}{B+1} = \frac{I_{c}}{I_{e}}</math><br />
<br />
V-I Characteristics of Transistors:<br />
* AAAA<br />
<br />
NPN Transistor Example: <br />
* Cut Off: Both junction are reverse biased.<br />
* Saturation: Both junction are forward biased. <br />
* Active: One junction is forward biased, another one is reverse biased.<br />
<br />
Understanding cut off regions for NPN transistors: <br />
* Cut off region (open switch - both junctions one reverse biased)<br />
* Saturation region (closed switch)<br />
* Active region (amplification - one junction forward biased another is reverse biased)<br />
* Breakdown region (Damaged - acts like open circuit essentially) <br />
<br />
How do you know when one of these three (or four counting breakdown) apply? <br />
* Voltage<br />
** V<sub>E</sub> < V<sub>B</sub> < V<sub>c</sub><br />
** V<sub>E</sub> < V<sub>B</sub> > V<sub>c</sub><br />
** V<sub>E</sub> > V<sub>B</sub> < V<sub>c</sub><br />
** V<sub>E</sub> > V<sub>B</sub> > V<sub>c</sub><br />
* NPN<br />
** Active <br />
** Saturation<br />
** Cut off<br />
** Reverse active<br />
* PNP<br />
** Reverse active<br />
** Cut off<br />
** Saturation<br />
** Active<br />
<br />
Understanding common emitter (CE) configuration<br />
* Consider the following configurations...<br />
<br />
Consider the following example: <br />
* We'll assume that V<sub>B</sub> is equal to 0 (since it is grounded). <br />
* Start solving the example by assuming active mode of operation. <br />
* In active mode of operation<br />
** B<sub>BE</sub> = 0.7 (NPN), B<sub>CE</sub> >= 0.3V<br />
** B<sub>EB</sub> = 0.7 (PNP), B<sub>EB</sub> >= 0.3V<br />
Therefore, V<sub>EB</sub> = 0.7V = V<sub>E</sub> - V<sub>B</sub> such that V<sub>B</sub> = 0<br />
<br />
<br />
<br />
=== Header 2 ===<br />
<br />
It sometimes allow current, and sometimes it doesn't.<br />
<br />
<br />
===Assignment 1.1===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE492_Section_2_Notes&diff=147ECE492 Section 2 Notes2022-03-02T19:42:22Z<p>Eddynetweb: Section 2.1 additional notes</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 2.1 == <br />
<br />
We'll begin by covering a '''Bi-polar Junction Transistor''', also known as BJT.<br />
<br />
What is it exactly? Well, it's a transistor. <br />
* It is a 3 terminal devices.<br />
* It is a unsymmetrical device.<br />
* It amplifies the signal. <br />
<br />
There are two types of BJTs:<br />
* NPN, in which...<br />
* PNP, in which...<br />
<br />
Now we'll cover (B) Common Emitter Current gain: <br />
* <math>B=\frac{I_{c}}{I_{b}}</math><br />
* 50 < B < 200<br />
<br />
...and (Alpha) Common Base Current Gain: <br />
* <math>Alpha = \frac{B}{B+1} = \frac{I_{c}}{I_{e}}</math><br />
<br />
V-I Characteristics of Transistors:<br />
* AAAA<br />
<br />
NPN Transistor Example: <br />
* Cut Off: Both junction are reverse biased.<br />
* Saturation: Both junction are forward biased. <br />
* Active: One junction is forward biased, another one is reverse biased.<br />
<br />
=== Header 2 ===<br />
<br />
It sometimes allow current, and sometimes it doesn't.<br />
<br />
<br />
===Assignment 1.1===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE492_Section_2_Notes&diff=146ECE492 Section 2 Notes2022-03-02T19:15:40Z<p>Eddynetweb: Section 2.1 created for ECE492!</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 2.1 == <br />
<br />
We'll begin by covering a '''Bi-polar Junction Transistor''', also known as BJT.<br />
<br />
What is it exactly? Well, it's a transistor. <br />
* It is a 3 terminal devices.<br />
* It is a unsymmetrical device.<br />
* It amplifies the signal. <br />
<br />
There are two types of BJTs:<br />
* NPN, in which <br />
* PNP<br />
<br />
Now we'll cover (B) Common Emitter Current gain: <br />
* <math>B=\frac{I_{c}}{I_{b}}</math><br />
* 50 < B < 200<br />
<br />
...and (Alpha) Common Base Current Gain: <br />
* <math>Alpha = \frac{B}{B+1} = \frac{I_{c}}{I_{e}}</math><br />
<br />
=== Header 2 ===<br />
<br />
It sometimes allow current, and sometimes it doesn't.<br />
<br />
<br />
===Assignment 1.1===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=145ECE488 Homework2022-03-01T05:53:20Z<p>Eddynetweb: Homework 8</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''<br />
<br />
== Homework 4 ==<br />
<br />
A load impedance consists of 25 ohm resistance in series with 38 ohm inductive reactance. The load is connected across a 60 Hz, 240 V rms source. Use the voltage source as the reference (zero phase angle).<br />
<br />
# Draw the circuit diagram.<br />
# Calculate the value of inductance.<br />
# Calculate the phasor value of current through the load impedance.<br />
# Which leads, current or voltage?<br />
# What is the phase angle between current and voltage?<br />
# Calculate the phasor value of voltage across the resistor.<br />
# Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
=== Breakdown ===<br />
<br />
1. Draw the circuit diagram. <br />
<br />
https://wiki.eddyn.net/images/thumb/2/25/ECE488_HW4_CircuitDiagram.png/600px-ECE488_HW4_CircuitDiagram.png<br />
<br />
2. Calculate the value of inductance. <br />
<br />
<math> <br />
\begin{align}<br />
X_{L} = 38 \\<br />
wL = 38 \\<br />
L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H<br />
\end{align}<br />
</math><br />
<br />
3. Calculate the phasor value of current through the load impedance.<br />
<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\<br />
\sqrt{25^2+38^2} = 45.49\\<br />
tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\<br />
= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps<br />
\end{align}<br />
</math><br />
<br />
4. Which leads, current or voltage?<br />
* With voltage source as reference, '''voltage leads current'''.<br />
<br />
5. What is the phase angle between current and voltage?<br />
* The phase angle is ϕ = <math>56.66^\circ</math><br />
<br />
6. Calculate the phasor value of voltage across the resistor.<br />
<br />
<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math><br />
<br />
7. Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math><br />
<br />
== Homework 5 ==<br />
<br />
An electric load consists of 5 Ω resistance, 20 Ω capacitive reactance, and 10 Ω inductive reactance, all connected in parallel. A voltage source of 100 V is applied across the load. Compute the following: <br />
<br />
# The current of the load<br />
# The real power of the load<br />
# The reactive power of the load<br />
<br />
=== Breakdown ===<br />
<br />
https://wiki.eddyn.net/images/thumb/e/ef/ECE488_Homework_5_Circuit_Diagram.png/600px-ECE488_Homework_5_Circuit_Diagram.png<br />
<br />
1) The current of the load<br />
<br />
<math> <br />
\begin{align}<br />
Z = \frac{1}{\frac{1}{5}+j(\frac{1}{20}-\frac{1}{10})} = \\ 4.71+j1.18 = 4.85\angle 14.04^\circ Ohms \\<br />
I = \frac{V}{Z} = \frac{100\angle 0^\circ}{4.85\angle 14.04^\circ} = 20.62\angle -14.04^\circ Amps\\<br />
\end{align}<br />
</math><br />
<br />
2) The real power of the load<br />
<br />
<math> <br />
\begin{align}<br />
S = V * I = (100\angle 0^\circ)(20.62\angle -14.04^\circ) = \\ 2062\angle 14.04^\circ VA = 2000 + j500 VA<br />
\end{align}<br />
</math><br />
<br />
Real power is <math>P=2000W = 2kW</math>.<br />
<br />
3) The reactive power of the load <br />
<br />
Reactive power is <math>Q=500VAr = 0.5kVAr</math> from above.<br />
<br />
== Homework 8 ==<br />
<br />
In the system shown, The voltages V<sub>ab</sub>, V<sub>bc</sub>, and V<sub>ca</sub> are balanced three-phase with magnitude 208 V. The resistors R are 18 ohms.<br />
<br />
# Calculate the currents I<sub>ab</sub>, I<sub>bc</sub>, and I<sub>ca</sub> (Ohm's law)<br />
# Calculate the power consumed by each resistor, and the total 3-phase power.<br />
# Calculate the currents I<sub>a</sub>, I<sub>b</sub>, and I<sub>c</sub>. (KCL)<br />
<br />
Convert the delta source to a Y source.<br />
Convert the delta load to a Y load.<br />
Draw the single-phase equivalent for the circuit.<br />
Use the single-phase equivalent to calculation Ia. Its magnitude should be the same as part c.<br />
<br />
https://wiki.eddyn.net/images/thumb/b/b2/ECE488_HW8_Figure.jpg/600px-ECE488_HW8_Figure.jpg<br />
<br />
=== Breakdown ===<br />
<br />
1) Calculate the currents I<sub>ab</sub>, I<sub>bc</sub>, and I<sub>ca</sub> (Ohm's law)<br />
<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{R} = \frac{208V}{18 Ohm} = 11.56A \\<br />
I_{ab} = 11.56\angle 0^\circ \\<br />
I_{bc} = 11.56\angle -120^\circ \\<br />
I_{ca} = 11.56\angle 120^\circ \\<br />
\end{align}<br />
</math><br />
<br />
2) Calculate the power consumed by each resistor, and the total 3-phase power.<br />
<br />
<math> <br />
\begin{align}<br />
P = V * I = 208V * 11.56A = 2404.48W or 2.4kW<br />
\end{align}<br />
</math><br />
<br />
Power consumed by each of the shown resistors: 2.4kW<br />
Total power with the 3-phase: <math>2.4kW * 3 = 7.2kW</math><br />
<br />
3) Calculate the currents I<sub>a</sub>, I<sub>b</sub>, and I<sub>c</sub>. (KCL)<br />
<br />
<math> <br />
\begin{align}<br />
I_{a} = I_{ab} - I_{ac} = 11.56A - 11.56\angle 120^\circ \\ <br />
= 11.56A - (-5.78 + j10.01)A = (11.56 + 5.78 - j10.01)A \\<br />
= (17.34 - j10.01) = 20.02\angle -30^\circ \\<br />
----- \\<br />
I_{b} = I_{bc} - I_{ab} = 11.56\angle - 120^\circ - 11.56\angle 0^\circ \\ <br />
= (-5.78 + j10.01)A - 11.56 = (17.34-j10.01)A \\<br />
= 20.02\angle -150^\circ \\<br />
----- \\<br />
I_{c} = I_{ac} - I_{bc} = 11.56\angle 120^\circ - 11.56\angle -120^\circ \\ <br />
= (-5.78 + j10.01)A - (-5.78 - j10.01)A = (0+j20.02) \\<br />
= 20.02\angle 90^\circ \\<br />
\end{align}<br />
</math><br />
<br />
4) Convert the delta source to a Y source. Convert the delta load to a Y load. Draw the single-phase equivalent for the circuit. Use the single-phase equivalent to calculation I<sub>a</sub>. Its magnitude should be the same as part c.<br />
<br />
a)<br />
<br />
<math> <br />
\begin{align}<br />
I_{an} = \frac{|V_{ab}|}{\sqrt{3}} =120.09V = 120V\angle 0^\circ -30^\circ = 120V\angle -30^\circ \\<br />
I_{bn} = \frac{|V_{bc}|}{\sqrt{3}} =120.09V = 120V\angle -120^\circ -30^\circ = 120V\angle -150^\circ \\<br />
I_{cn} = \frac{|V_{ca}|}{\sqrt{3}} =120.09V = 120V\angle 120^\circ -30^\circ = 120V\angle 90^\circ<br />
\end{align}<br />
</math><br />
<br />
b) <br />
<br />
<math> <br />
\begin{align}<br />
Z_{y} = \frac{Z}{3} = \frac{18 Ohms}{3} = 6 Ohms<br />
\end{align}<br />
</math><br />
<br />
c) <br />
<br />
<br />
d) <br />
<br />
<math> <br />
\begin{align}<br />
I_{an} = \frac{|V_{ab}|}{\sqrt{3}} =120.09V = 120V\angle 0^\circ -30^\circ = 120V\angle -30^\circ \\<br />
\end{align}<br />
</math><br />
<br />
e) Verification:<br />
<br />
<math> <br />
\begin{align}<br />
P = \sqrt{3}(208V)(20.02A)(1) = 7.2kW \\<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=144ECE488 Homework2022-03-01T04:16:20Z<p>Eddynetweb: Added Homework 8.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''<br />
<br />
== Homework 4 ==<br />
<br />
A load impedance consists of 25 ohm resistance in series with 38 ohm inductive reactance. The load is connected across a 60 Hz, 240 V rms source. Use the voltage source as the reference (zero phase angle).<br />
<br />
# Draw the circuit diagram.<br />
# Calculate the value of inductance.<br />
# Calculate the phasor value of current through the load impedance.<br />
# Which leads, current or voltage?<br />
# What is the phase angle between current and voltage?<br />
# Calculate the phasor value of voltage across the resistor.<br />
# Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
=== Breakdown ===<br />
<br />
1. Draw the circuit diagram. <br />
<br />
https://wiki.eddyn.net/images/thumb/2/25/ECE488_HW4_CircuitDiagram.png/600px-ECE488_HW4_CircuitDiagram.png<br />
<br />
2. Calculate the value of inductance. <br />
<br />
<math> <br />
\begin{align}<br />
X_{L} = 38 \\<br />
wL = 38 \\<br />
L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H<br />
\end{align}<br />
</math><br />
<br />
3. Calculate the phasor value of current through the load impedance.<br />
<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\<br />
\sqrt{25^2+38^2} = 45.49\\<br />
tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\<br />
= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps<br />
\end{align}<br />
</math><br />
<br />
4. Which leads, current or voltage?<br />
* With voltage source as reference, '''voltage leads current'''.<br />
<br />
5. What is the phase angle between current and voltage?<br />
* The phase angle is ϕ = <math>56.66^\circ</math><br />
<br />
6. Calculate the phasor value of voltage across the resistor.<br />
<br />
<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math><br />
<br />
7. Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math><br />
<br />
== Homework 5 ==<br />
<br />
An electric load consists of 5 Ω resistance, 20 Ω capacitive reactance, and 10 Ω inductive reactance, all connected in parallel. A voltage source of 100 V is applied across the load. Compute the following: <br />
<br />
# The current of the load<br />
# The real power of the load<br />
# The reactive power of the load<br />
<br />
=== Breakdown ===<br />
<br />
https://wiki.eddyn.net/images/thumb/e/ef/ECE488_Homework_5_Circuit_Diagram.png/600px-ECE488_Homework_5_Circuit_Diagram.png<br />
<br />
1) The current of the load<br />
<br />
<math> <br />
\begin{align}<br />
Z = \frac{1}{\frac{1}{5}+j(\frac{1}{20}-\frac{1}{10})} = \\ 4.71+j1.18 = 4.85\angle 14.04^\circ Ohms \\<br />
I = \frac{V}{Z} = \frac{100\angle 0^\circ}{4.85\angle 14.04^\circ} = 20.62\angle -14.04^\circ Amps\\<br />
\end{align}<br />
</math><br />
<br />
2) The real power of the load<br />
<br />
<math> <br />
\begin{align}<br />
S = V * I = (100\angle 0^\circ)(20.62\angle -14.04^\circ) = \\ 2062\angle 14.04^\circ VA = 2000 + j500 VA<br />
\end{align}<br />
</math><br />
<br />
Real power is <math>P=2000W = 2kW</math>.<br />
<br />
3) The reactive power of the load <br />
<br />
Reactive power is <math>Q=500VAr = 0.5kVAr</math> from above.<br />
<br />
== Homework 8 ==<br />
<br />
In the system shown, The voltages V<sub>ab</sub>, V<sub>bc</sub>, and V<sub>ca</sub> are balanced three-phase with magnitude 208 V. The resistors R are 18 ohms.<br />
<br />
# Calculate the currents I<sub>ab</sub>, I<sub>bc</sub>, and I<sub>ca</sub> (Ohm's law)<br />
# Calculate the power consumed by each resistor, and the total 3-phase power.<br />
# Calculate the currents I<sub>a</sub>, I<sub>b</sub>, and I<sub>c</sub>. (KCL)<br />
<br />
Convert the delta source to a Y source.<br />
Convert the delta load to a Y load.<br />
Draw the single-phase equivalent for the circuit.<br />
Use the single-phase equivalent to calculation Ia. Its magnitude should be the same as part c.<br />
<br />
[[File:ECE488 HW8 Figure.jpg|thumb]]<br />
<br />
=== Breakdown ===<br />
<br />
Hello.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=File:ECE488_HW8_Figure.jpg&diff=143File:ECE488 HW8 Figure.jpg2022-03-01T04:15:46Z<p>Eddynetweb: </p>
<hr />
<div>ECE488 Homework 8 Figure</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=File:ECE488_HW6_Circuit_Diagram.jpg&diff=142File:ECE488 HW6 Circuit Diagram.jpg2022-02-17T05:42:25Z<p>Eddynetweb: </p>
<hr />
<div>ECE488 HW6 Circuit Diagram</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=PHIL385_Weekly_Discussions&diff=141PHIL385 Weekly Discussions2022-02-15T18:38:16Z<p>Eddynetweb: PHIL385 Discussion Post Page</p>
<hr />
<div>This is meant to allow me to draft my thoughts. I'll put more information here eventually.<br />
<br />
== Week 1 Discussion Board ==<br />
<br />
The following topics were covered in week 1: <br />
<br />
# Lin (2016) The Ethical Dilemma of Self-driving Cars, 4m<br />
# Norman (1990) Design Thinking, 217-257<br />
# Roy (2015) When We Design for Disability, We All Benefit, 13m<br />
# Blockley (2012) From Idea to Reality, 1-17<br />
<br />
Discussion: <br />
<br />
Blah blah. We'll get there. :)<br />
<br />
== Week 2 Discussion Board ==<br />
<br />
# NSPE (2007) Code of Ethics and Reference Guide, 3-4<br />
# Heffernon (2012) Dare to Disagree, 13m<br />
# Satris (1986) Student Relativism, 51-54<br />
# Timmons (2013) An Introduction to Moral Theory, 1-12<br />
<br />
Discussion: <br />
<br />
Blah blah. We'll get there. :)<br />
<br />
== Week 3 Discussion Board ==<br />
<br />
# Timmons (2013) Moral Relativism, 41-67<br />
# Skakoon and King (2004) What the Beginner Needs to Learn at Once, 5-19<br />
<br />
Discussion: <br />
<br />
Blah blah. We'll get there. :)<br />
<br />
== Week 4 Discussion Board ==<br />
<br />
# Davis (1998) Codes of Ethics and the Challenger, 43-60<br />
# Timmons (2013) Evaluation of Ethical Egoism [Prisoner’s Dilemma], 194-201<br />
<br />
Discussion: <br />
<br />
Blah blah. We'll get there. :)<br />
<br />
== Week 5 Discussion Board ==<br />
<br />
# Science Buddies (2011) Decision Making Matrix<br />
# Harris et al. (1997) Analyzing and Resolving Ethical Problems, 21-37<br />
# NSPE Ethics Board (2006) 5-4 Failure to Disclose Full Impact of Development<br />
# SEP (2013) Moral Reasoning 2.3: Sorting Out Which Considerations Are Most Relevant<br />
<br />
<br />
Discussion: <br />
<br />
Blah blah. We'll get there. :)</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=140ECE488 Homework2022-02-15T05:35:38Z<p>Eddynetweb: Finish Homework 5</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''<br />
<br />
== Homework 4 - Eduardo Santillan ==<br />
<br />
A load impedance consists of 25 ohm resistance in series with 38 ohm inductive reactance. The load is connected across a 60 Hz, 240 V rms source. Use the voltage source as the reference (zero phase angle).<br />
<br />
# Draw the circuit diagram.<br />
# Calculate the value of inductance.<br />
# Calculate the phasor value of current through the load impedance.<br />
# Which leads, current or voltage?<br />
# What is the phase angle between current and voltage?<br />
# Calculate the phasor value of voltage across the resistor.<br />
# Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
=== Breakdown ===<br />
<br />
1. Draw the circuit diagram. <br />
<br />
https://wiki.eddyn.net/images/thumb/2/25/ECE488_HW4_CircuitDiagram.png/600px-ECE488_HW4_CircuitDiagram.png<br />
<br />
2. Calculate the value of inductance. <br />
<br />
<math> <br />
\begin{align}<br />
X_{L} = 38 \\<br />
wL = 38 \\<br />
L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H<br />
\end{align}<br />
</math><br />
<br />
3. Calculate the phasor value of current through the load impedance.<br />
<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\<br />
\sqrt{25^2+38^2} = 45.49\\<br />
tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\<br />
= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps<br />
\end{align}<br />
</math><br />
<br />
4. Which leads, current or voltage?<br />
* With voltage source as reference, '''voltage leads current'''.<br />
<br />
5. What is the phase angle between current and voltage?<br />
* The phase angle is ϕ = <math>56.66^\circ</math><br />
<br />
6. Calculate the phasor value of voltage across the resistor.<br />
<br />
<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math><br />
<br />
7. Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math><br />
<br />
== Homework 5 - Eduardo Santillan ==<br />
<br />
An electric load consists of 5 Ω resistance, 20 Ω capacitive reactance, and 10 Ω inductive reactance, all connected in parallel. A voltage source of 100 V is applied across the load. Compute the following: <br />
<br />
# The current of the load<br />
# The real power of the load<br />
# The reactive power of the load<br />
<br />
=== Breakdown ===<br />
<br />
https://wiki.eddyn.net/images/thumb/e/ef/ECE488_Homework_5_Circuit_Diagram.png/600px-ECE488_Homework_5_Circuit_Diagram.png<br />
<br />
1) The current of the load<br />
<br />
<math> <br />
\begin{align}<br />
Z = \frac{1}{\frac{1}{5}+j(\frac{1}{20}-\frac{1}{10})} = \\ 4.71+j1.18 = 4.85\angle 14.04^\circ Ohms \\<br />
I = \frac{V}{Z} = \frac{100\angle 0^\circ}{4.85\angle 14.04^\circ} = 20.62\angle -14.04^\circ Amps\\<br />
\end{align}<br />
</math><br />
<br />
2) The real power of the load<br />
<br />
<math> <br />
\begin{align}<br />
S = V * I = (100\angle 0^\circ)(20.62\angle -14.04^\circ) = \\ 2062\angle 14.04^\circ VA = 2000 + j500 VA<br />
\end{align}<br />
</math><br />
<br />
Real power is <math>P=2000W = 2kW</math>.<br />
<br />
3) The reactive power of the load <br />
<br />
Reactive power is <math>Q=500VAr = 0.5kVAr</math> from above.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=139ECE488 Homework2022-02-15T05:07:03Z<p>Eddynetweb: Homework 5 - add circuit diagram.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''<br />
<br />
== Homework 4 - Eduardo Santillan ==<br />
<br />
A load impedance consists of 25 ohm resistance in series with 38 ohm inductive reactance. The load is connected across a 60 Hz, 240 V rms source. Use the voltage source as the reference (zero phase angle).<br />
<br />
# Draw the circuit diagram.<br />
# Calculate the value of inductance.<br />
# Calculate the phasor value of current through the load impedance.<br />
# Which leads, current or voltage?<br />
# What is the phase angle between current and voltage?<br />
# Calculate the phasor value of voltage across the resistor.<br />
# Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
=== Breakdown ===<br />
<br />
1. Draw the circuit diagram. <br />
<br />
https://wiki.eddyn.net/images/thumb/2/25/ECE488_HW4_CircuitDiagram.png/600px-ECE488_HW4_CircuitDiagram.png<br />
<br />
2. Calculate the value of inductance. <br />
<br />
<math> <br />
\begin{align}<br />
X_{L} = 38 \\<br />
wL = 38 \\<br />
L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H<br />
\end{align}<br />
</math><br />
<br />
3. Calculate the phasor value of current through the load impedance.<br />
<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\<br />
\sqrt{25^2+38^2} = 45.49\\<br />
tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\<br />
= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps<br />
\end{align}<br />
</math><br />
<br />
4. Which leads, current or voltage?<br />
* With voltage source as reference, '''voltage leads current'''.<br />
<br />
5. What is the phase angle between current and voltage?<br />
* The phase angle is ϕ = <math>56.66^\circ</math><br />
<br />
6. Calculate the phasor value of voltage across the resistor.<br />
<br />
<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math><br />
<br />
7. Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math><br />
<br />
== Homework 5 - Eduardo Santillan ==<br />
<br />
An electric load consists of 5 Ω resistance, 20 Ω capacitive reactance, and 10 Ω inductive reactance, all connected in parallel. A voltage source of 100 V is applied across the load. Compute the following: <br />
<br />
# The current of the load<br />
# The real power of the load<br />
# The reactive power of the load<br />
<br />
=== Breakdown ===<br />
<br />
https://wiki.eddyn.net/images/thumb/e/ef/ECE488_Homework_5_Circuit_Diagram.png/600px-ECE488_Homework_5_Circuit_Diagram.png<br />
<br />
1) The current of the load</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=138ECE488 Homework2022-02-15T05:04:39Z<p>Eddynetweb: Homework 5 start</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''<br />
<br />
== Homework 4 - Eduardo Santillan ==<br />
<br />
A load impedance consists of 25 ohm resistance in series with 38 ohm inductive reactance. The load is connected across a 60 Hz, 240 V rms source. Use the voltage source as the reference (zero phase angle).<br />
<br />
# Draw the circuit diagram.<br />
# Calculate the value of inductance.<br />
# Calculate the phasor value of current through the load impedance.<br />
# Which leads, current or voltage?<br />
# What is the phase angle between current and voltage?<br />
# Calculate the phasor value of voltage across the resistor.<br />
# Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
=== Breakdown ===<br />
<br />
1. Draw the circuit diagram. <br />
<br />
https://wiki.eddyn.net/images/thumb/2/25/ECE488_HW4_CircuitDiagram.png/600px-ECE488_HW4_CircuitDiagram.png<br />
<br />
2. Calculate the value of inductance. <br />
<br />
<math> <br />
\begin{align}<br />
X_{L} = 38 \\<br />
wL = 38 \\<br />
L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H<br />
\end{align}<br />
</math><br />
<br />
3. Calculate the phasor value of current through the load impedance.<br />
<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\<br />
\sqrt{25^2+38^2} = 45.49\\<br />
tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\<br />
= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps<br />
\end{align}<br />
</math><br />
<br />
4. Which leads, current or voltage?<br />
* With voltage source as reference, '''voltage leads current'''.<br />
<br />
5. What is the phase angle between current and voltage?<br />
* The phase angle is ϕ = <math>56.66^\circ</math><br />
<br />
6. Calculate the phasor value of voltage across the resistor.<br />
<br />
<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math><br />
<br />
7. Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math><br />
<br />
== Homework 5 - Eduardo Santillan ==<br />
<br />
An electric load consists of 5 Ω resistance, 20 Ω capacitive reactance, and 10 Ω inductive reactance, all connected in parallel. A voltage source of 100 V is applied across the load. Compute the following: <br />
<br />
# The real power of the load<br />
# The reactive power of the load<br />
# The current of the load<br />
<br />
=== Breakdown ===<br />
<br />
[[File:ECE488 Homework 5 Circuit Diagram.png|thumb]]</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=File:ECE488_Homework_5_Circuit_Diagram.png&diff=137File:ECE488 Homework 5 Circuit Diagram.png2022-02-15T05:04:22Z<p>Eddynetweb: </p>
<hr />
<div>ECE488 Homework 5 Circuit Diagram</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=136ECE488 Homework2022-02-10T05:30:54Z<p>Eddynetweb: Homework 4 changed sub-question 4.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''<br />
<br />
== Homework 4 - Eduardo Santillan ==<br />
<br />
A load impedance consists of 25 ohm resistance in series with 38 ohm inductive reactance. The load is connected across a 60 Hz, 240 V rms source. Use the voltage source as the reference (zero phase angle).<br />
<br />
# Draw the circuit diagram.<br />
# Calculate the value of inductance.<br />
# Calculate the phasor value of current through the load impedance.<br />
# Which leads, current or voltage?<br />
# What is the phase angle between current and voltage?<br />
# Calculate the phasor value of voltage across the resistor.<br />
# Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
=== Breakdown ===<br />
<br />
1. Draw the circuit diagram. <br />
<br />
https://wiki.eddyn.net/images/thumb/2/25/ECE488_HW4_CircuitDiagram.png/600px-ECE488_HW4_CircuitDiagram.png<br />
<br />
2. Calculate the value of inductance. <br />
<br />
<math> <br />
\begin{align}<br />
X_{L} = 38 \\<br />
wL = 38 \\<br />
L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H<br />
\end{align}<br />
</math><br />
<br />
3. Calculate the phasor value of current through the load impedance.<br />
<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\<br />
\sqrt{25^2+38^2} = 45.49\\<br />
tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\<br />
= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps<br />
\end{align}<br />
</math><br />
<br />
4. Which leads, current or voltage?<br />
* With voltage source as reference, '''voltage leads current'''.<br />
<br />
5. What is the phase angle between current and voltage?<br />
* The phase angle is ϕ = <math>56.66^\circ</math><br />
<br />
6. Calculate the phasor value of voltage across the resistor.<br />
<br />
<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math><br />
<br />
7. Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=135ECE488 Homework2022-02-10T05:27:37Z<p>Eddynetweb: Homework 4 added circuit diagram</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''<br />
<br />
== Homework 4 - Eduardo Santillan ==<br />
<br />
A load impedance consists of 25 ohm resistance in series with 38 ohm inductive reactance. The load is connected across a 60 Hz, 240 V rms source. Use the voltage source as the reference (zero phase angle).<br />
<br />
# Draw the circuit diagram.<br />
# Calculate the value of inductance.<br />
# Calculate the phasor value of current through the load impedance.<br />
# Which leads, current or voltage?<br />
# What is the phase angle between current and voltage?<br />
# Calculate the phasor value of voltage across the resistor.<br />
# Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
=== Breakdown ===<br />
<br />
1. Draw the circuit diagram. <br />
<br />
https://wiki.eddyn.net/images/thumb/2/25/ECE488_HW4_CircuitDiagram.png/600px-ECE488_HW4_CircuitDiagram.png<br />
<br />
2. Calculate the value of inductance. <br />
<br />
<math> <br />
\begin{align}<br />
X_{L} = 38 \\<br />
wL = 38 \\<br />
L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H<br />
\end{align}<br />
</math><br />
<br />
3. Calculate the phasor value of current through the load impedance.<br />
<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\<br />
\sqrt{25^2+38^2} = 45.49\\<br />
tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\<br />
= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps<br />
\end{align}<br />
</math><br />
<br />
4. Which leads, current or voltage?<br />
* With voltage source as reference, voltage leads current by <math>56.66^\circ</math><br />
<br />
5. What is the phase angle between current and voltage?<br />
* The phase angle is ϕ = <math>56.66^\circ</math><br />
<br />
6. Calculate the phasor value of voltage across the resistor.<br />
<br />
<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math><br />
<br />
7. Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=File:ECE488_HW4_CircuitDiagram.png&diff=134File:ECE488 HW4 CircuitDiagram.png2022-02-10T05:26:31Z<p>Eddynetweb: </p>
<hr />
<div>ECE388 Homework 4 Circuit Diagram</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=133ECE488 Homework2022-02-10T05:25:02Z<p>Eddynetweb: Homework 4 formatting changes.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''<br />
<br />
== Homework 4 - Eduardo Santillan ==<br />
<br />
A load impedance consists of 25 ohm resistance in series with 38 ohm inductive reactance. The load is connected across a 60 Hz, 240 V rms source. Use the voltage source as the reference (zero phase angle).<br />
<br />
# Draw the circuit diagram.<br />
# Calculate the value of inductance.<br />
# Calculate the phasor value of current through the load impedance.<br />
# Which leads, current or voltage?<br />
# What is the phase angle between current and voltage?<br />
# Calculate the phasor value of voltage across the resistor.<br />
# Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
=== Breakdown ===<br />
<br />
1. Draw the circuit diagram. <br />
<br />
2. Calculate the value of inductance. <br />
<br />
<math> <br />
\begin{align}<br />
X_{L} = 38 \\<br />
wL = 38 \\<br />
L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H<br />
\end{align}<br />
</math><br />
<br />
3. Calculate the phasor value of current through the load impedance.<br />
<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\<br />
\sqrt{25^2+38^2} = 45.49\\<br />
tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\<br />
= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps<br />
\end{align}<br />
</math><br />
<br />
4. Which leads, current or voltage?<br />
* With voltage source as reference, voltage leads current by <math>56.66^\circ</math><br />
<br />
5. What is the phase angle between current and voltage?<br />
* The phase angle is ϕ = <math>56.66^\circ</math><br />
<br />
6. Calculate the phasor value of voltage across the resistor.<br />
<br />
<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math><br />
<br />
7. Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=132ECE488 Homework2022-02-10T05:23:56Z<p>Eddynetweb: Homework 4 added</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''<br />
<br />
== Homework 4 - Eduardo Santillan ==<br />
<br />
A load impedance consists of 25 ohm resistance in series with 38 ohm inductive reactance. The load is connected across a 60 Hz, 240 V rms source. Use the voltage source as the reference (zero phase angle).<br />
<br />
# Draw the circuit diagram.<br />
# Calculate the value of inductance.<br />
# Calculate the phasor value of current through the load impedance.<br />
# Which leads, current or voltage?<br />
# What is the phase angle between current and voltage?<br />
# Calculate the phasor value of voltage across the resistor.<br />
# Calculate the phasor value of voltage across the inductive reactance.<br />
<br />
=== Breakdown ===<br />
<br />
1. Draw the circuit diagram. <br />
<br />
2. Calculate the value of inductance. <br />
<br />
<math> <br />
\begin{align}<br />
X_{L} = 38 \\<br />
wL = 38 \\<br />
L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H<br />
\end{align}<br />
</math><br />
<br />
3. Calculate the phasor value of current through the load impedance.<br />
<math> <br />
\begin{align}<br />
I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\<br />
\sqrt{25^2+38^2} = 45.49\\<br />
tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\<br />
= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps<br />
\end{align}<br />
</math><br />
<br />
4. Which leads, current or voltage?<br />
* With voltage source as reference, voltage leads current by <math>56.66^\circ</math><br />
<br />
5. What is the phase angle between current and voltage?<br />
* The phase angle is ϕ = <math>56.66^\circ</math><br />
<br />
6. Calculate the phasor value of voltage across the resistor.<br />
<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math><br />
<br />
7. Calculate the phasor value of voltage across the inductive reactance.<br />
<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=CloudAtCost_Custom_OS&diff=131CloudAtCost Custom OS2022-02-09T14:14:42Z<p>Eddynetweb: Clarification of procedures.</p>
<hr />
<div>This tutorial attempts to remake the original one present on the original CloudAtCost wiki. The original wiki had become rather overrun with spam, so the owner shut it down. <br />
<br />
== Supported OS ==<br />
<br />
The following OS's are supported, according to [https://netboot.xyz netboot.xyz]: <br />
<br />
* Antergos<br />
* Arch Linux<br />
* Architect Linux<br />
* CentOS<br />
* CoreOS<br />
* Debian<br />
* Fedora<br />
* FreeBSD<br />
* Gentoo<br />
* Mageia<br />
* Manjaro Linux<br />
* Microsoft Windows<br />
* MirOS<br />
* OpenBSD<br />
* OpenSUSE<br />
* RancherOS<br />
* Scientific<br />
* Tiny Core Linux<br />
* Ubuntu<br />
* Many more (check the site)! <br />
<br />
== Script (Broken, will fix later) ==<br />
<br />
Simply create and execute this script to boot directly into the generic iPXE kernel without hassle.<br />
<br />
Special thanks to [https://www.reddit.com/user/rschulze /u/rschulze] for the script. <br />
<br />
'''Update:''' People are reporting that the script is broken again! I'm keeping it here for when I have time to edit it, but just be aware. Feel free to fix it if you have time (I think the lkrn file just needs to be updated). <br />
<br />
<source lang="bash"><br />
#!/usr/bin/env bash<br />
<br />
cdr2mask ()<br />
{<br />
# Number of args to shift, 255..255, first non-255 byte, zeroes<br />
set -- $(( 5 - ($1 / 8) )) 255 255 255 255 $(( (255 << (8 - ($1 % 8))) & 255 )) 0 0 0<br />
[[ $1 -gt 1 ]] && shift ${1} || shift<br />
echo "${1-0}.${2-0}.${3-0}.${4-0}"<br />
}<br />
<br />
cd /boot || exit 1<br />
wget --continue --output-document=/boot/generic-ipxe.lkrn https://boot.netboot.xyz/ipxe/netboot.xyz.lkrn<br />
<br />
_Dev="$(ip route list | awk '/^default/ {print $5}')"<br />
_Gateway="$(ip route list | awk '/^default/ {print $3}')"<br />
_IP="$(ip -o -4 addr show "${_Dev}" | awk '{print $4}')"<br />
_Netmask="$(cdr2mask "${_IP#*/}")" <br />
_IP=${_IP%/*}<br />
<br />
cat <<_EOF_ > /boot/netboot.xyz-initrd<br />
#!ipxe<br />
#/boot/netboot.xyz-initrd<br />
imgfree<br />
set net0/ip ${_IP%/*}<br />
set net0/netmask ${_Netmask}<br />
set net0/gateway ${_Gateway}<br />
set dns 208.67.222.222<br />
ifopen net0<br />
chain --autofree https://boot.netboot.xyz<br />
_EOF_<br />
<br />
cat <<_EOF_ > /etc/grub.d/09_custom<br />
echo "<br />
menuentry 'netboot.xyz' {<br />
set root='hd0,msdos1'<br />
linux16 /generic-ipxe.lkrn<br />
initrd16 /netboot.xyz-initrd<br />
}<br />
"<br />
_EOF_<br />
chmod +x /etc/grub.d/09_custom<br />
<br />
update-grub2 || { grub2-mkconfig -o /boot/grub2/grub.cfg; grub2-set-default 0; }<br />
<br />
reboot<br />
</source><br />
<br />
== Prerequisites ==<br />
<br />
I would advise doing this on a fresh CloudAtCost server, preferably using the Debian 8 Jessie image. Make sure your console works, as you'll need it throughout the entire installation. <br />
<br />
It appears that the console works much more consistently on DC-2 as opposed to DC-3, so I'd give it a shot there for the most successful installation (plus, DC-2 is probably as stable you're gonna get at the moment).<br />
<br />
== Steps ==<br />
<br />
1. Login into your fresh server, preferably using the console (SSH will work too). <br />
<br />
2. Run the following command: <br />
<br />
<source lang="bash"><br />
cd /boot && wget -c "https://boot.netboot.xyz/ipxe/netboot.xyz.lkrn" <br />
</source> <br />
<br />
'''Note''': This is where you'll need the console. If you're not already on it, go to https://panel.cloudatcost.com/ and open the console. <br />
<br />
3. Reboot your server using Ctrl+Alt+Del (or clicking the blue square in the corner of the console), or using the "reboot" command. <br />
<br />
4. Right after the VMWare splash screen, the GRUB boot menu will appear. ''Quickly'' press the letter "'''C'''". <br />
<br />
'''Note''': You should now see a "'''grub>'''" prompt. If not, you'll need to reboot and try step 4 again. <br />
<br />
5. In the grub prompt, type the following command: <br />
<br />
<source lang="bash">linux16 /boot/netboot.xyz.lkrn</source> <br />
<br />
'''Note''': Be prepared to ''quickly'' press "'''M'''" after this next command: <br />
<br />
6. Then boot into it using the following command: <br />
<br />
<source lang="bash">boot</source> <br />
<br />
7. ''Quickly'' press "'''M'''" to bring up the menu. <br />
<br />
8. Select "Manual network configuration".<br />
<br />
9. Now you'll need to provide the following information in order for the script to download the image. This can all be found on the information icon in the CloudAtCost panel: <br />
<source lang="text">IP address (found in panel)<br />
Subnet mask (if you have issues relating to networking, try leaving this blank)<br />
Default gateway (found in panel)<br />
Set DNS to 8.8.8.8<br />
</source><br />
<br />
'''Note''': To avoid HTTP/S timeouts when fetching iPXE bootfiles, avoid inputting netmask while performing the previous steps. If iPXE times out HTTP/s connections, reboot and perform the steps above starting from the grub prompt. This is a handshake bug between CloudAtCost VMs and our iPXE host server that we are currently working on addressing. <br />
<br />
10. If everything is successful, you should now get a list of OS's which you can choose to your hearts content. Pick the OS you'd like and then follow the instructions. The script will lead you to installing the operating system of your choice. <br />
<br />
https://netboot.xyz/images/netboot.xyz.gif<br />
<br />
That's it!<br />
<br />
----<br />
<br />
If you have any issues, check the ##CloudAtCost channel on Freenode or /r/CloudAtCost on reddit. Plenty of people there available to help.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=CloudAtCost_Custom_OS&diff=130CloudAtCost Custom OS2022-02-09T14:12:26Z<p>Eddynetweb: Script is broken, will fix later.</p>
<hr />
<div>This tutorial attempts to remake the original one present on the original CloudAtCost wiki. The original wiki had become rather overrun with spam, so the owner shut it down. <br />
<br />
== Supported OS ==<br />
<br />
The following OS's are supported, according to [https://netboot.xyz netboot.xyz]: <br />
<br />
* Antergos<br />
* Arch Linux<br />
* Architect Linux<br />
* CentOS<br />
* CoreOS<br />
* Debian<br />
* Fedora<br />
* FreeBSD<br />
* Gentoo<br />
* Mageia<br />
* Manjaro Linux<br />
* Microsoft Windows<br />
* MirOS<br />
* OpenBSD<br />
* OpenSUSE<br />
* RancherOS<br />
* Scientific<br />
* Tiny Core Linux<br />
* Ubuntu<br />
* Many more (check the site)! <br />
<br />
== Script (Broken, will fix later) ==<br />
<br />
Simply create and execute this script to boot directly into the generic iPXE kernel without hassle.<br />
<br />
Special thanks to [https://www.reddit.com/user/rschulze /u/rschulze] for the script. <br />
<br />
'''Update:''' People are reporting that the script is broken again! I'm keeping it here for when I have time to edit it, but just be aware. Feel free to fix it if you have time (I think the lkrn file just needs to be updated). <br />
<br />
<source lang="bash"><br />
#!/usr/bin/env bash<br />
<br />
cdr2mask ()<br />
{<br />
# Number of args to shift, 255..255, first non-255 byte, zeroes<br />
set -- $(( 5 - ($1 / 8) )) 255 255 255 255 $(( (255 << (8 - ($1 % 8))) & 255 )) 0 0 0<br />
[[ $1 -gt 1 ]] && shift ${1} || shift<br />
echo "${1-0}.${2-0}.${3-0}.${4-0}"<br />
}<br />
<br />
cd /boot || exit 1<br />
wget --continue --output-document=/boot/generic-ipxe.lkrn https://boot.netboot.xyz/ipxe/netboot.xyz.lkrn<br />
<br />
_Dev="$(ip route list | awk '/^default/ {print $5}')"<br />
_Gateway="$(ip route list | awk '/^default/ {print $3}')"<br />
_IP="$(ip -o -4 addr show "${_Dev}" | awk '{print $4}')"<br />
_Netmask="$(cdr2mask "${_IP#*/}")" <br />
_IP=${_IP%/*}<br />
<br />
cat <<_EOF_ > /boot/netboot.xyz-initrd<br />
#!ipxe<br />
#/boot/netboot.xyz-initrd<br />
imgfree<br />
set net0/ip ${_IP%/*}<br />
set net0/netmask ${_Netmask}<br />
set net0/gateway ${_Gateway}<br />
set dns 208.67.222.222<br />
ifopen net0<br />
chain --autofree https://boot.netboot.xyz<br />
_EOF_<br />
<br />
cat <<_EOF_ > /etc/grub.d/09_custom<br />
echo "<br />
menuentry 'netboot.xyz' {<br />
set root='hd0,msdos1'<br />
linux16 /generic-ipxe.lkrn<br />
initrd16 /netboot.xyz-initrd<br />
}<br />
"<br />
_EOF_<br />
chmod +x /etc/grub.d/09_custom<br />
<br />
update-grub2 || { grub2-mkconfig -o /boot/grub2/grub.cfg; grub2-set-default 0; }<br />
<br />
reboot<br />
</source><br />
<br />
== Prerequisites ==<br />
<br />
I would advise doing this on a fresh CloudAtCost server, preferably using the Debian 8 Jessie image. Make sure your console works, as you'll need it throughout the entire installation. <br />
<br />
It appears that the console works much more consistently on DC-2 as opposed to DC-3, so I'd give it a shot there for the most successful installation (plus, DC-2 is probably as stable you're gonna get at the moment).<br />
<br />
== Steps ==<br />
<br />
1. Login into your fresh server, preferably using the console (SSH will work too). <br />
<br />
2. Run the following command: <br />
<br />
<source lang="bash"><br />
cd /boot && wget -c "https://boot.netboot.xyz/ipxe/netboot.xyz.lkrn" <br />
</source> <br />
<br />
'''Note''': This is where you'll need the console. If you're not already on it, do so now. <br />
<br />
3. Reboot your server using Ctrl+Alt+Del (or clicking the blue square in the corner of the console), or using the "reboot" command. <br />
<br />
4. Right after the VMWare splash screen, the GRUB boot menu will appear. ''Quickly'' press the letter "'''C'''". <br />
<br />
'''Note''': You should now see a "'''grub>'''" prompt. If not, you'll need to reboot and try step 4 again. <br />
<br />
5. In the grub prompt, type the following command: <br />
<br />
<source lang="bash">linux16 /boot/netboot.xyz.lkrn</source> <br />
<br />
'''Note''': Be prepared to ''quickly'' press "'''M'''" after this next command: <br />
<br />
6. Then boot into it using the following command: <br />
<br />
<source lang="bash">boot</source> <br />
<br />
7. ''Quickly'' press "'''M'''" to bring up the menu. <br />
<br />
8. Select "Manual network configuration".<br />
<br />
9. Now you'll need to provide the following information in order for the script to download the image. This can all be found on the information icon in the CloudAtCost panel: <br />
<source lang="text">IP address (found in panel)<br />
Subnet mask (if you have issues relating to networking, try leaving this blank)<br />
Default gateway (found in panel)<br />
Set DNS to 8.8.8.8<br />
</source><br />
<br />
'''Note''': To avoid HTTP/S timeouts when fetching iPXE bootfiles, avoid inputting netmask while performing the previous steps. If iPXE times out HTTP/s connections, reboot and perform the steps above starting from the grub prompt. This is a handshake bug between CloudAtCost VMs and our iPXE host server that we are currently working on addressing. <br />
<br />
10. If everything is successful, you should now get a list of OS's which you can choose to your hearts content. Pick the OS you'd like and then follow the instructions. The script will lead you to installing the operating system of your choice. <br />
<br />
https://netboot.xyz/images/netboot.xyz.gif<br />
<br />
That's it!<br />
<br />
----<br />
<br />
If you have any issues, check the ##CloudAtCost channel on Freenode or /r/CloudAtCost on reddit. Plenty of people there available to help.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=129ECE488 Homework2022-02-08T03:27:02Z<p>Eddynetweb: Finish Homework 3</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Subbituminous Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
Combined-cycle gas turbine Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
Open-cycle gas turbine Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=128ECE488 Homework2022-02-08T02:59:04Z<p>Eddynetweb: Update Homework 3.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
* 500 MW nuclear (33% efficiency)<br />
* 500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
* 200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
* 100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
* 600 MW wind<br />
* 100 MW solar photovoltaic<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The CO<sub>2</sub> emissions from electric generators, by fuel type, are shown on the right. <br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu (mmBtu is 106 Btu)<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO<sub>2</sub> emissions and the total fuel cost for the hour.<br />
<br />
=== Breakdown ===<br />
<br />
We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.<br />
<br />
Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\<br />
\frac{2202.92 lbs}{MWh} * 500 MWh = 1101460 lbs<br />
\end{align}<br />
</math><br />
<br />
CC Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\<br />
\frac{726.03 lbs}{MWh} * 200 MWh = 145206.33 lbs<br />
\end{align}<br />
</math><br />
<br />
OC Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = 1140.91 \\<br />
\frac{1140.91 lbs}{MWh} * 100 MWh = 114091 lbs<br />
\end{align}<br />
</math><br />
<br />
'''Total CO<sub>2</sub> Emissions: 1,360,757.33 lbs'''<br />
<br />
Now we'll calculate the '''fuel costs for the hour''' for the dataset.<br />
<br />
Nuclear: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$0.44}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $4.55 \\<br />
\frac{$4.55}{MWh} * 500 MWh = $2275.31<br />
\end{align}<br />
</math><br />
<br />
Coal: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$1.55}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = $16.03 \\<br />
\frac{$16.03}{MWh} * 500 MWh = $8015.31<br />
\end{align}<br />
</math><br />
<br />
CC Gas:<br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = $21.10 \\<br />
\frac{$21.10}{MWh} * 200 MWh = $4219.67<br />
\end{align}<br />
</math><br />
<br />
OC Gas: <br />
<br />
<math> <br />
\begin{align}<br />
\frac{$}{MWh} = \frac{$3.40}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.35} = $33.15 \\<br />
\frac{$33.15}{MWh} * 100 MWh = $3315.46<br />
\end{align}<br />
</math><br />
<br />
'''Total Fuel Costs: $17,825.75'''</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=127ECE488 Homework2022-02-07T23:11:22Z<p>Eddynetweb: Homework 3 upload</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000<br />
<br />
== Homework 3 - Eduardo Santillan ==<br />
<br />
A power system is supplying 2000 MW of power for one hour from the following generating units:<br />
500 MW nuclear (33% efficiency)<br />
500 MW coal (Subbituminous coal in pulverized fuel steam plant; 33% efficiency)<br />
200 MW natural gas (Combined-cycle gas turbine (CCGT) plant; 55% efficiency)<br />
100 MW natural gas, (Open-cycle gas turbine (OCGT) plant; 35% efficiency)<br />
600 MW wind<br />
100 MW solar photovoltaic<br />
<br />
The CO2 emissions from electric generators, by fuel type, are:<br />
<br />
[[File:CO2Emissions.png|thumb]]<br />
<br />
The cost of fuel for the three types of generators are:<br />
Nuclear fuel: 0.44 $/mmBtu mmBtu is 106 Btu<br />
Coal: $1.50/mmBtu<br />
Gas: $3.40/mmBtu<br />
<br />
This is the cost of fuel going into the plant, so you must use the efficiency of the plant to calculate cost of electricity out.<br />
<br />
Calculate the total CO2 emissions and the total fuel cost for the hour.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=File:CO2Emissions.png&diff=126File:CO2Emissions.png2022-02-07T23:08:30Z<p>Eddynetweb: </p>
<hr />
<div>CO2 emissions tracking.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=125ECE488 Homework2022-02-01T05:30:44Z<p>Eddynetweb: Homework 2 formatting update</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=124ECE488 Homework2022-02-01T05:28:52Z<p>Eddynetweb: /* Homework 2 - Eduardo Santillan */</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV || <br />
b. 50% ultra-supercritical coal, 50% combined cycle ||<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=123ECE488 Homework2022-02-01T05:27:49Z<p>Eddynetweb: Homework 2 formatting update</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV || <br />
b. 50% ultra-supercritical coal, 50% combined cycle ||<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=122ECE488 Homework2022-02-01T05:25:42Z<p>Eddynetweb: Homework 2 submission</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).<br />
<br />
== Homework 2 - Eduardo Santillan ==<br />
<br />
Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.<br />
Hydroelectric generation will provide 1,000 GWh/day.<br />
Nuclear will provide 2,000 GWh/day.<br />
Use the costs of generation for generators that start service in 2025 (table of data is below), calculate the total daily cost to generate if the rest of the energy is provided by:<br />
<br />
a. 50% onshore wind, 50% solar PV<br />
<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
<br />
<br />
Total System Levelized Cost of Energy (2019 $/MWh)<br />
* Solar photovoltaic 35.74<br />
* Geothermal 37.47 <br />
* Combined cycle 38.07 <br />
* Wind, onshore 39.95 <br />
* Hydroelectric 52.79 <br />
* Combustion turbine 66.62 <br />
* Ultra-supercritical coal 76.44 <br />
* Advanced nuclear 81.65 <br />
* Biomass 94.83 <br />
* Wind, offshore 122.25<br />
<br />
=== Breakdown ===<br />
<br />
We know that the total electric generation for a region in 2025 will be 11,000 GWh/day, and Hydroelectric/Nuclear will always provide a set amount (1,000/2,000 GWh/day respectively). <br />
<br />
This means we only need to split the remaining sources evenly factoring the guaranteed sources. Also convert GWh to MWh for consistency of calculations. <br />
<br />
a. 50% onshore wind, 50% solar PV<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 4000 GWh x $39.95/MWh = $159800000<br />
* Solar PV: 4000 GWh x $35.74/MWh = $142960000<br />
* '''Total cost per day''': $518,850,000<br />
b. 50% ultra-supercritical coal, 50% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Ultra-supercritical Coal: 4000 GWh x $76.44/MWh = $305760000<br />
* Combined Cycle: 4000 GWh x $38.07/MWh = $152280000<br />
* '''Total cost per day''': $674,130,000<br />
c. 25% onshore wind, 25% solar PV, 25% ultra-supercritical coal, 25% combined cycle<br />
* Hydroelectric: 1000 GWh x $52.79/MWh = $52790000<br />
* Nuclear: 2000 GWh x $81.65/MWh = $163300000<br />
* Onshore Wind: 2000 GWh x $39.95/MWh = $79900000<br />
* Solar PV: 2000 GWh x $35.74/MWh = $71480000<br />
* Ultra-supercritical Coal: 2000 GWh x $76.44/MWh = $152880000<br />
* Combined Cycle: 2000 GWh x $38.07/MWh = $76140000<br />
* '''Total cost per day''': $596,490,000</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=121ECE488 Homework2022-01-27T05:31:29Z<p>Eddynetweb: Publish assignment one.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
# Calculate the total energy supplied by SPP on this date.<br />
# Calculate the peak load for SPP on this date.<br />
# Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
The numbers have been extrapolated from the <br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8: <br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br />
===Breakdown===<br />
<br />
* 1. In finding the total energy supplied by SPP on April 8, simply take the below listed values and multiply them by 1 hour, and then sum them together. The total energy supplied by SPP on this date is '''666827 MWh'''.<br />
<br><br />
{|<br />
|+<br />
|-<br />
26739 || 27247 || 28610 || 28654 || 28039 || 27333<br />
|-<br />
26849 || 26637 || 26505 || 26512 || 26652 || 27157<br />
|-<br />
27950 || 28499 || 29186 || 29395 || 29096 || 28768<br />
|-<br />
28376 || 28005 || 27617 || 27528 || 27642 || 27831<br />
|}<br />
<br><br />
<br />
* 2. The peak load listed on the chart is '''29395 MW''' listed at hour 15 or so.<br />
<br />
* 3. Peak load for the SPP is significantly lower in April 8 as opposed to August 8 example because it is typically colder during the April month throughout the region. It is known that heating/cooling is an electrically expensive process, and most midwestern homes are heated using natural gas burners (https://www.ncdc.noaa.gov/temp-and-precip/us-maps/1/202004#us-maps-select).</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE492_Section_1_Notes&diff=120ECE492 Section 1 Notes2022-01-27T02:03:52Z<p>Eddynetweb: Jan 26 class publishing.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
We first need to consider the following materials: <br />
* Conductors<br />
* Insulators<br />
* Semiconductors<br />
<br />
Consider the concept of the charge for Oxygen (O):<br />
* 8 protons around the nucleus<br />
* 8 electrons (two on the inner orbit, 6 on the outer orbit). <br />
<br />
Let's look at the outer most orbit in context with a conductor:<br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is approximately 0 (meaning electrons can move freely between valency band and conduction band). <br />
* That is the reason why conductive materials act as they do. <br />
<br />
Let's look at the outer most orbit in the context of insulators: <br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is very high (meaning electrons have a very hard time moving between the valency band and conduction band). <br />
* That is the reason insulators, are well, insulators (or their property is insulative). <br />
<br />
Let's look at the outer most orbit in the context of semi-conductors: <br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is exactly 1.12eV (1eV = 1.602 * 10<sup>-19</sup>J). <br />
* Once electrons are in the conduction band, the semi-conductor will act as a conductor. <br />
<br />
This entire class is based on semi-conductor materials. <br />
<br />
=== What is a semiconductor? ===<br />
<br />
It sometimes allow current, and sometimes it doesn't.<br />
<br />
Semi-conductors only have '''4''' electrons in its outermost orbit. <br />
* Carbon<br />
* Silicon <br />
* Germanium<br />
* Lead<br />
<br />
What's the best two?<br />
* Silicon and Germanium<br />
<br />
Consideration of Silicon: <br />
* Silicon will acquire other silicon atoms to form a covalent bond. <br />
<br />
Now let's take a Silicon atom: <br />
* Take a pure silicon atoms and put energy to them: go from insulator to conductive properties, and electrons from the furthest orbit detach and jump to another position. <br />
* When electrons detach, it leaves a hole where it once was, that is viewed as a positive charge for clarity. <br />
<br />
Doping: Adding impurity to a pure semiconductor material. <br />
* Tri-Valent Impurity: majority of "holes" (positive charge), also know as P-type semiconductor. <br />
* Penta-Valent Impurity: majority of electrons, also known as N-type semiconductor. <br />
<br />
Let's take Aluminum for Tri-Valent example:<br />
* Pair the Aluminum with Silicon atoms, knowing that Aluminum has an outer electron count of 3. <br />
* Silicon will be bounded to the three outer Aluminum electrons, but 1 Silicon will be left.<br />
<br />
Let's try Phosphorous for Penta-Valent example:<br />
* Pair Phosphorous with Silicon atoms, knowing that Phosphorous has an outer electron count of 5. <br />
<br />
Now what is P-N Junction? It's a diode. <br />
* A P-type semiconductor connected to an N-type semiconductor.<br />
* Recall that P-type just has a lot of "holes" (positive). <br />
* Recall that N-type has a lot of electrons (negative).<br />
* Shouldn't these holes and electrons combine? Electric field is present due to electrons on the N-type side of the P-N junction. This electric field does not allow for combining. <br />
* There is an anode and a cathode on the P-type and N-type side respectively. A diode can be represented with A -> C. <br />
* Now let's connect a battery on each side of the junction, and we supplement 0.5V.<br />
* The electric field barrier becomes thinner and thinner, but not completely gone. <br />
* Now apply 1.0V, the electrons and holes will push against the junction. <br />
<br />
Now consider a reverse biased diode: <br />
* A diode where the energy source charge is reverse from the P-N junction. <br />
* Holes to negative and electrons to positive. <br />
<br />
Important consideration: A P-N Junction (diode, or biased/reversed biased diode) allows current in only one direction. <br />
<br />
What are the V-I characteristics of the P-N Junction/Diode:<br />
* What is PIV? Peak Inverse Voltage. <br />
* Consider the graph below for a typical V-I characteristic of a diode. <br />
<br />
https://media.geeksforgeeks.org/wp-content/uploads/20211025120251/VIcharacteristics.jpg<br />
<br />
Therefore, important considerations added for P-N Junction diode:<br />
* A P-N Junction (diode, or biased/reversed biased diode) allows current in only one direction.<br />
* Acts like a closed switch in forward biased condition. <br />
* Acts like an open switch in reversed biased condition. <br />
<br />
Now consider the following example: Find the current passing through each diode. <br />
* We'll need to break the problem apart. <br />
* Current always flows from higher potential to lower potential.<br />
* Assume ideal diode. <br />
<br />
Let's break the problem down: <br />
* Step 1: Fill me in. :)<br />
<br />
'''Insert example 1A'''<br />
<br />
Now let's consider the following example:<br />
<br />
* Step 1: Find all possible current flow directions (current always flows from high potential to low potential) and make a determination of whether a diode is open/closed/don't know. <br />
* Step 2: We'll take the assumption that D<sub>2</sub> and D<sub>3</sub> are closed because we don't know what they are. <br />
** Note: We can't have 3 different voltages at the same node, so our assumption is wrong. <br />
* Step 3: Now we'll verify that our assumption is correct.<br />
** Is D<sub>2</sub> and D<sub>3</sub> open? Since anode voltage is higher than cathode, then diode acts like a closed switch [in ideal case]. <br />
** Therefore, both D<sub>2</sub> and D<sub>3</sub> are open. <br />
<br />
'''Insert example 1B'''<br />
<br />
Now let's consider this next example.<br />
<br />
* Step 1: Find I<sub>1</sub>, I<sub>2</sub>, and I<sub>3</sub> assuming ideal diodes. <br />
** Find all possible current flow directions (current always flows from high potential to low potential). <br />
* Step 2: We'll assume that <br />
<br />
<br />
===Assignment 1.1===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE488_Homework&diff=119ECE488 Homework2022-01-25T22:43:27Z<p>Eddynetweb: Publish changes.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Machines & Transformers course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Homework 1 ==<br />
<br />
Average hourly load data for Southwest Power Pool (SPP) on April 8 is given in this document.<br />
* a. Calculate the total energy supplied by SPP on this date.<br />
* b. Calculate the peak load for SPP on this date.<br />
* c. Why is the peak load significantly lower than in the class example (posted above)?<br />
<br />
[[File:Southwest Power Pool April 8 Data.jpg|thumb]]<br />
<br />
===Breakdown===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=File:Southwest_Power_Pool_April_8_Data.jpg&diff=118File:Southwest Power Pool April 8 Data.jpg2022-01-25T22:42:04Z<p>Eddynetweb: </p>
<hr />
<div>This data shows the average hourly load data for the Southwest Power Pool on April 8.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE492_Section_1_Notes&diff=117ECE492 Section 1 Notes2022-01-24T19:47:35Z<p>Eddynetweb: Additional lecture content.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
We first need to consider the following materials: <br />
* Conductors<br />
* Insulators<br />
* Semiconductors<br />
<br />
Consider the concept of the charge for Oxygen (O):<br />
* 8 protons around the nucleus<br />
* 8 electrons (two on the inner orbit, 6 on the outer orbit). <br />
<br />
Let's look at the outer most orbit in context with a conductor:<br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is approximately 0 (meaning electrons can move freely between valency band and conduction band). <br />
* That is the reason why conductive materials act as they do. <br />
<br />
Let's look at the outer most orbit in the context of insulators: <br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is very high (meaning electrons have a very hard time moving between the valency band and conduction band). <br />
* That is the reason insulators, are well, insulators (or their property is insulative). <br />
<br />
Let's look at the outer most orbit in the context of semi-conductors: <br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is exactly 1.12eV (1eV = 1.602 * 10<sup>-19</sup>J). <br />
* Once electrons are in the conduction band, the semi-conductor will act as a conductor. <br />
<br />
This entire class is based on semi-conductor materials. <br />
<br />
=== What is a semiconductor? ===<br />
<br />
It sometimes allow current, and sometimes it doesn't.<br />
<br />
Semi-conductors only have '''4''' electrons in its outermost orbit. <br />
* Carbon<br />
* Silicon <br />
* Germanium<br />
* Lead<br />
<br />
What's the best two?<br />
* Silicon and Germanium<br />
<br />
Consideration of Silicon: <br />
* Silicon will acquire other silicon atoms to form a covalent bond. <br />
<br />
Now let's take a Silicon atom: <br />
* Take a pure silicon atoms and put energy to them: go from insulator to conductive properties, and electrons from the furthest orbit detach and jump to another position. <br />
* When electrons detach, it leaves a hole where it once was, that is viewed as a positive charge for clarity. <br />
<br />
Doping: Adding impurity to a pure semiconductor material. <br />
* Tri-Valent Impurity: majority of "holes" (positive charge), also know as P-type semiconductor. <br />
* Penta-Valent Impurity: majority of electrons, also known as N-type semiconductor. <br />
<br />
Let's take Aluminum for Tri-Valent example:<br />
* Pair the Aluminum with Silicon atoms, knowing that Aluminum has an outer electron count of 3. <br />
* Silicon will be bounded to the three outer Aluminum electrons, but 1 Silicon will be left.<br />
<br />
Let's try Phosphorous for Penta-Valent example:<br />
* Pair Phosphorous with Silicon atoms, knowing that Phosphorous has an outer electron count of 5. <br />
<br />
Now what is P-N Junction? It's a diode. <br />
* A P-type semiconductor connected to an N-type semiconductor.<br />
* Recall that P-type just has a lot of "holes" (positive). <br />
* Recall that N-type has a lot of electrons (negative).<br />
* Shouldn't these holes and electrons combine? Electric field is present due to electrons on the N-type side of the P-N junction. This electric field does not allow for combining. <br />
* There is an anode and a cathode on the P-type and N-type side respectively. A diode can be represented with A -> C. <br />
* Now let's connect a battery on each side of the junction, and we supplement 0.5V.<br />
* The electric field barrier becomes thinner and thinner, but not completely gone. <br />
* Now apply 1.0V, the electrons and holes will push against the junction. <br />
<br />
Now consider a reverse biased diode: <br />
* A diode where the energy source charge is reverse from the P-N junction. <br />
* Holes to negative and electrons to positive. <br />
<br />
Important consideration: A P-N Junction (diode, or biased/reversed biased diode) allows current in only one direction. <br />
<br />
What are the V-I characteristics of the P-N Junction/Diode:<br />
* What is PIV? Peak Inverse Voltage. <br />
* Consider the graph below for a typical V-I characteristic of a diode. <br />
<br />
https://media.geeksforgeeks.org/wp-content/uploads/20211025120251/VIcharacteristics.jpg<br />
<br />
Therefore, important considerations added for P-N Junction diode:<br />
* A P-N Junction (diode, or biased/reversed biased diode) allows current in only one direction.<br />
* Acts like a closed switch in forward biased condition. <br />
* Acts like an open switch in reversed biased condition. <br />
<br />
Now consider the following example: Find the current passing through each diode. <br />
* We'll need to break the problem apart. <br />
* Current always flows from higher potential to lower potential.<br />
* Assume ideal diode. <br />
<br />
Let's break the problem down: <br />
* Step 1: Fill me in. :)<br />
<br />
===Assignment 1.1===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE492_Section_1_Notes&diff=116ECE492 Section 1 Notes2022-01-24T19:17:06Z<p>Eddynetweb: </p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
We first need to consider the following materials: <br />
* Conductors<br />
* Insulators<br />
* Semiconductors<br />
<br />
Consider the concept of the charge for Oxygen (O):<br />
* 8 protons around the nucleus<br />
* 8 electrons (two on the inner orbit, 6 on the outer orbit). <br />
<br />
Let's look at the outer most orbit in context with a conductor:<br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is approximately 0 (meaning electrons can move freely between valency band and conduction band). <br />
* That is the reason why conductive materials act as they do. <br />
<br />
Let's look at the outer most orbit in the context of insulators: <br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is very high (meaning electrons have a very hard time moving between the valency band and conduction band). <br />
* That is the reason insulators, are well, insulators (or their property is insulative). <br />
<br />
Let's look at the outer most orbit in the context of semi-conductors: <br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is exactly 1.12eV (1eV = 1.602 * 10<sup>-19</sup>J). <br />
* Once electrons are in the conduction band, the semi-conductor will act as a conductor. <br />
<br />
This entire class is based on semi-conductor materials. <br />
<br />
=== What is a semiconductor? ===<br />
<br />
It sometimes allow current, and sometimes it doesn't.<br />
<br />
Semi-conductors only have '''4''' electrons in its outermost orbit. <br />
* Carbon<br />
* Silicon <br />
* Germanium<br />
* Lead<br />
<br />
What's the best two?<br />
* Silicon and Germanium<br />
<br />
Consideration of Silicon: <br />
* Silicon will acquire other silicon atoms to form a covalent bond. <br />
<br />
Now let's take a Silicon atom: <br />
* Take a pure silicon atoms and put energy to them: go from insulator to conductive properties, and electrons from the furthest orbit detach and jump to another position. <br />
* When electrons detach, it leaves a hole where it once was, that is viewed as a positive charge for clarity. <br />
<br />
Doping: Adding impurity to a pure semiconductor material. <br />
* Tri-Valent Impurity: majority of "holes" (positive charge), also know as P-type semiconductor. <br />
* Penta-Valent Impurity: majority of electrons, also known as N-type semiconductor. <br />
<br />
Let's take Aluminum for Tri-Valent example:<br />
* Pair the Aluminum with Silicon atoms, knowing that Aluminum has an outer electron count of 3. <br />
* Silicon will be bounded to the three outer Aluminum electrons, but 1 Silicon will be left.<br />
<br />
Let's try Phosphorous for Penta-Valent example:<br />
* Pair Phosphorous with Silicon atoms, knowing that Phosphorous has an outer electron count of 5. <br />
<br />
<br />
===Assignment 1.1===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE492_Section_1_Notes&diff=115ECE492 Section 1 Notes2022-01-24T18:56:51Z<p>Eddynetweb: Update lecture work with content.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
We first need to consider the following materials: <br />
* Conductors<br />
* Insulators<br />
* Semiconductors<br />
<br />
Consider the concept of the charge for Oxygen (O):<br />
* 8 protons around the nucleus<br />
* 8 electrons (two on the inner orbit, 6 on the outer orbit). <br />
<br />
Let's look at the outer most orbit in context with a conductor:<br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is approximately 0 (meaning electrons can move freely between valency band and conduction band). <br />
* That is the reason why conductive materials act as they do. <br />
<br />
Let's look at the outer most orbit in the context of insulators: <br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is very high (meaning electrons have a very hard time moving between the valency band and conduction band). <br />
* That is the reason insulators, are well, insulators (or their property is insulative). <br />
<br />
Let's look at the outer most orbit in the context of semi-conductors: <br />
* There are two bands: valency and conduction. <br />
* Electrons are always in the valency band. <br />
* In between, there is an energy gap, which is exactly 1.12eV (1eV = 1.602 * 10<sup>-19</sup>J). <br />
<br />
===Assignment 1.1===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=ECE492_Section_1_Notes&diff=114ECE492 Section 1 Notes2022-01-24T18:28:54Z<p>Eddynetweb: Create Electronic Circuits wiki for note taking.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Electronic Circuits course. Will make this more pretty as things evolve. <br />
<br />
My goal is to make Electronic Circuits so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
Insert lecture notes here!<br />
<br />
===Assignment 1.1===<br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=MATH511_Linear_Algebra_Chapter_1&diff=113MATH511 Linear Algebra Chapter 12022-01-24T15:36:50Z<p>Eddynetweb: Added 1.1 problem 1(b).</p>
<hr />
<div>Welcome! Notes for Spring 2022 Linear Algebra course. Will make this more pretty as things evolve. <br />
<br />
Wichita State University has standardized their course work for Linear Algebra. Almost all professors will likely adapt to what is being shown below, basically. <br />
<br />
My goal is to make Linear Algebra so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
Insert lecture notes here!<br />
<br />
===Assignment 1.1===<br />
<br />
We'll be doing 1(b), 5(c), 6(e), 8*, 10 this round from the source textbook. <br />
<br />
For '''1(b)''', consider the following: <br />
<br />
We'll need to use back substitution to solve this system of linear equation. <br />
<math display=block> <br />
\begin{align}<br />
x_{1}+x_{2}+x_{3}=8 \\<br />
2x_{2}+x_{3}=5 \\<br />
3x_{3}=9<br />
\end{align}<br />
</math> <br />
<br />
We'll first need to use substitution to simply and find the values x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>.<br />
A good rule of thumb is to find the easiest value to simplify that can then be used to find the other values. In this example, x<sub>3</sub> can be used to help find x<sub>1</sub>, x<sub>2</sub>:<br />
<br />
<math> <br />
\begin{align}<br />
3x_{3}=9 \\<br />
x_{3}=\frac{9}{3}<br />
\end{align}<br />
</math> <br />
<br />
We'll take this <math>x_{3}=\frac{9}{3}</math> and plug it into <math>2x_{2}+x_{3}=5</math> to find x<sub>2</sub>: <br />
<br />
<math> <br />
\begin{align}<br />
2x_{2}+x_{3}=5 \\<br />
x_{3}=\frac{9}{3} \\<br />
\text{Plug in } x_{3}=\frac{9}{3} \text{ into } 2x_{2}+x_{3}=5 \text{:} \\<br />
2x_{2}+\frac{9}{3}=5 \\<br />
2x_{2}=2 \\<br />
x_{2}=1<br />
\end{align}<br />
</math></div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=MATH511_Linear_Algebra_Chapter_1&diff=112MATH511 Linear Algebra Chapter 12022-01-24T14:07:22Z<p>Eddynetweb: Adding 1.1 Assignment.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Linear Algebra course. Will make this more pretty as things evolve. <br />
<br />
Wichita State University has standardized their course work for Linear Algebra. Almost all professors will likely adapt to what is being shown below, basically. <br />
<br />
My goal is to make Linear Algebra so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
Here is a LaTeX equation. <br />
<br />
<math><br />
\begin{bmatrix}<br />
1 & 2 & 3 \\<br />
4 & 5 & 6 \\<br />
7 & 8 & 9\\<br />
\end{bmatrix}<br />
</math><br />
<br />
===Assignment 1.1===<br />
<br />
We'll be doing 1(b), 5(c), 6(e), 8*, 10 this round. <br />
<br />
For 1(b), consider the following: <br />
<br />
<math display=block> <br />
\begin{align}<br />
x_{1}+x_{2}+x_{3}=8 \\<br />
2x_{2}+x_{3}=5 \\<br />
3x_{3}=9<br />
\end{align}<br />
</math> <br />
<br />
We'll first need to</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=MATH511_Linear_Algebra_Chapter_1&diff=111MATH511 Linear Algebra Chapter 12022-01-24T02:54:23Z<p>Eddynetweb: Testing TeX support.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Linear Algebra course. Will make this more pretty as things evolve. <br />
<br />
Wichita State University has standardized their course work for Linear Algebra. Almost all professors will likely adapt to what is being shown below, basically. <br />
<br />
My goal is to make Linear Algebra so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
Here is a LaTeX equation. <br />
<br />
<math><br />
\begin{bmatrix}<br />
1 & 2 & 3 \\<br />
4 & 5 & 6 \\<br />
7 & 8 & 9\\<br />
\end{bmatrix}<br />
</math><br />
<br />
https://latex.codecogs.com/png.image?\dpi{110}%20\begin{bmatrix}1%20&%202%20&%203\\a%20&%20b%20&%20c\end{bmatrix}</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=MATH511_Linear_Algebra_Chapter_1&diff=110MATH511 Linear Algebra Chapter 12022-01-24T02:48:44Z<p>Eddynetweb: Testing TeX support.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Linear Algebra course. Will make this more pretty as things evolve. <br />
<br />
Wichita State University has standardized their course work for Linear Algebra. Almost all professors will likely adapt to what is being shown below, basically. <br />
<br />
My goal is to make Linear Algebra so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
Here is a LaTeX equation. <br />
<br />
<math>\alpha</math> <br />
<br />
https://latex.codecogs.com/png.image?\dpi{110}%20\begin{bmatrix}1%20&%202%20&%203\\a%20&%20b%20&%20c\end{bmatrix}</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=MATH511_Linear_Algebra_Chapter_1&diff=109MATH511 Linear Algebra Chapter 12022-01-24T02:44:24Z<p>Eddynetweb: Testing TeX support.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Linear Algebra course. Will make this more pretty as things evolve. <br />
<br />
Wichita State University has standardized their course work for Linear Algebra. Almost all professors will likely adapt to what is being shown below, basically. <br />
<br />
My goal is to make Linear Algebra so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
Here is a LaTeX equation. <br />
<br />
{{math|<var>&alpha;</var>}}<br />
<br />
https://latex.codecogs.com/png.image?\dpi{110}%20\begin{bmatrix}1%20&%202%20&%203\\a%20&%20b%20&%20c\end{bmatrix}</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=MATH511_Linear_Algebra_Chapter_1&diff=108MATH511 Linear Algebra Chapter 12022-01-24T01:22:01Z<p>Eddynetweb: </p>
<hr />
<div>Welcome! Notes for Spring 2022 Linear Algebra course. Will make this more pretty as things evolve. <br />
<br />
Wichita State University has standardized their course work for Linear Algebra. Almost all professors will likely adapt to what is being shown below, basically. <br />
<br />
My goal is to make Linear Algebra so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
Here is a LaTeX equation. <br />
<br />
https://latex.codecogs.com/png.image?\dpi{110}%20\begin{bmatrix}1%20&%202%20&%203\\a%20&%20b%20&%20c\end{bmatrix}</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=MATH511_Linear_Algebra_Chapter_1&diff=107MATH511 Linear Algebra Chapter 12022-01-24T01:20:34Z<p>Eddynetweb: Replaced content with "a"</p>
<hr />
<div>a</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=MATH511_Linear_Algebra_Chapter_1&diff=106MATH511 Linear Algebra Chapter 12022-01-24T01:16:43Z<p>Eddynetweb: Create course work.</p>
<hr />
<div>Welcome! Notes for Spring 2022 Linear Algebra course. Will make this more pretty as things evolve. <br />
<br />
Wichita State University has standardized their course work for Linear Algebra. Almost all professors will likely adapt to what is being shown below, basically. <br />
<br />
My goal is to make Linear Algebra so easily digestible, you could teach a middle schooler. We'll see if this works. <br />
<br />
== Prologue ==<br />
<br />
Hey how are you. I will insert something here later. :)<br />
<br />
== Section 1.1 == <br />
<br />
Here is a LaTeX equation. <br />
<br />
https://latex.codecogs.com/png.image?\dpi{110}%20\begin{bmatrix}1%20&%202%20&%203\\a%20&%20b%20&%20c\end{bmatrix}</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=CloudAtCost_Custom_OS&diff=104CloudAtCost Custom OS2021-01-30T14:02:22Z<p>Eddynetweb: Fixed issue with generic kernel download wget not working - causing the script to fail.</p>
<hr />
<div>This tutorial attempts to remake the original one present on the original CloudAtCost wiki. The original wiki had become rather overrun with spam, so the owner shut it down. <br />
<br />
== Supported OS ==<br />
<br />
The following OS's are supported, according to [https://netboot.xyz netboot.xyz]: <br />
<br />
* Antergos<br />
* Arch Linux<br />
* Architect Linux<br />
* CentOS<br />
* CoreOS<br />
* Debian<br />
* Fedora<br />
* FreeBSD<br />
* Gentoo<br />
* Mageia<br />
* Manjaro Linux<br />
* Microsoft Windows<br />
* MirOS<br />
* OpenBSD<br />
* OpenSUSE<br />
* RancherOS<br />
* Scientific<br />
* Tiny Core Linux<br />
* Ubuntu<br />
* Many more (check the site)! <br />
<br />
== Script ==<br />
<br />
Simply create and execute this script to boot directly into the generic iPXE kernel without hassle.<br />
<br />
Special thanks to [https://www.reddit.com/user/rschulze /u/rschulze] for the script. <br />
<br />
'''Update:''' Script has been fixed! The other generic kernel was moved to a different directory from boot.netboot.xyz. <br />
<br />
<source lang="bash"><br />
#!/usr/bin/env bash<br />
<br />
cdr2mask ()<br />
{<br />
# Number of args to shift, 255..255, first non-255 byte, zeroes<br />
set -- $(( 5 - ($1 / 8) )) 255 255 255 255 $(( (255 << (8 - ($1 % 8))) & 255 )) 0 0 0<br />
[[ $1 -gt 1 ]] && shift ${1} || shift<br />
echo "${1-0}.${2-0}.${3-0}.${4-0}"<br />
}<br />
<br />
cd /boot || exit 1<br />
wget --continue --output-document=/boot/generic-ipxe.lkrn https://boot.netboot.xyz/ipxe/netboot.xyz.lkrn<br />
<br />
_Dev="$(ip route list | awk '/^default/ {print $5}')"<br />
_Gateway="$(ip route list | awk '/^default/ {print $3}')"<br />
_IP="$(ip -o -4 addr show "${_Dev}" | awk '{print $4}')"<br />
_Netmask="$(cdr2mask "${_IP#*/}")" <br />
_IP=${_IP%/*}<br />
<br />
cat <<_EOF_ > /boot/netboot.xyz-initrd<br />
#!ipxe<br />
#/boot/netboot.xyz-initrd<br />
imgfree<br />
set net0/ip ${_IP%/*}<br />
set net0/netmask ${_Netmask}<br />
set net0/gateway ${_Gateway}<br />
set dns 208.67.222.222<br />
ifopen net0<br />
chain --autofree https://boot.netboot.xyz<br />
_EOF_<br />
<br />
cat <<_EOF_ > /etc/grub.d/09_custom<br />
echo "<br />
menuentry 'netboot.xyz' {<br />
set root='hd0,msdos1'<br />
linux16 /generic-ipxe.lkrn<br />
initrd16 /netboot.xyz-initrd<br />
}<br />
"<br />
_EOF_<br />
chmod +x /etc/grub.d/09_custom<br />
<br />
update-grub2 || { grub2-mkconfig -o /boot/grub2/grub.cfg; grub2-set-default 0; }<br />
<br />
reboot<br />
</source> <br />
<br />
== Prerequisites ==<br />
<br />
I would advise doing this on a fresh CloudAtCost server, preferably using the Debian 8 Jessie image. Make sure your console works, as you'll need it throughout the entire installation. <br />
<br />
It appears that the console works much more consistently on DC-2 as opposed to DC-3, so I'd give it a shot there for the most successful installation (plus, DC-2 is probably as stable you're gonna get at the moment).<br />
<br />
== Steps ==<br />
<br />
1. Login into your fresh server, preferably using the console (SSH will work too). <br />
<br />
2. Run the following command: <br />
<br />
<source lang="bash"><br />
cd /boot && wget -c "https://boot.netboot.xyz/ipxe/netboot.xyz.lkrn" <br />
</source> <br />
<br />
'''Note''': This is where you'll need the console. If you're not already on it, do so now. <br />
<br />
3. Reboot your server using Ctrl+Alt+Del (or clicking the blue square in the corner of the console), or using the "reboot" command. <br />
<br />
4. Right after the VMWare splash screen, the GRUB boot menu will appear. ''Quickly'' press the letter "'''C'''". <br />
<br />
'''Note''': You should now see a "'''grub>'''" prompt. If not, you'll need to reboot and try step 4 again. <br />
<br />
5. In the grub prompt, type the following command: <br />
<br />
<source lang="bash">linux16 /boot/netboot.xyz.lkrn</source> <br />
<br />
'''Note''': Be prepared to ''quickly'' press "'''M'''" after this next command: <br />
<br />
6. Then boot into it using the following command: <br />
<br />
<source lang="bash">boot</source> <br />
<br />
7. ''Quickly'' press "'''M'''" to bring up the menu. <br />
<br />
8. Select "Manual network configuration".<br />
<br />
9. Now you'll need to provide the following information in order for the script to download the image. This can all be found on the information icon in the CloudAtCost panel: <br />
<source lang="text">IP address (found in panel)<br />
Subnet mask (if you have issues relating to networking, try leaving this blank)<br />
Default gateway (found in panel)<br />
Set DNS to 8.8.8.8<br />
</source><br />
<br />
'''Note''': To avoid HTTP/S timeouts when fetching iPXE bootfiles, avoid inputting netmask while performing the previous steps. If iPXE times out HTTP/s connections, reboot and perform the steps above starting from the grub prompt. This is a handshake bug between CloudAtCost VMs and our iPXE host server that we are currently working on addressing. <br />
<br />
10. If everything is successful, you should now get a list of OS's which you can choose to your hearts content. Pick the OS you'd like and then follow the instructions. The script will lead you to installing the operating system of your choice. <br />
<br />
https://netboot.xyz/images/netboot.xyz.gif<br />
<br />
That's it!<br />
<br />
----<br />
<br />
If you have any issues, check the ##CloudAtCost channel on Freenode or /r/CloudAtCost on reddit. Plenty of people there available to help.</div>Eddynetwebhttps://wiki.eddyn.net/index.php?title=Abusive_Cop_Tracker&diff=102Abusive Cop Tracker2020-06-09T13:16:10Z<p>Eddynetweb: Added another story about MN.</p>
<hr />
<div>The 2020 protests show that cops do big oopsie when they think nobody is watching. <br />
<br />
This is a compilation of those "oopsis" to help with research efforts on a project I am working on. <br />
<br />
== Video Compilation == <br />
<br />
'''Note''': These are just links to content on different sites that typically go to their original author. All credit is given to those that recorded the works. If you are the owner of a video and do not want the item listed on the wiki, feel free to remove it and simply add the reason in the subject line. <br />
<br />
@greg_doucette on Twitter did an amazing job of cataloging a lot of content. <br />
<br />
# [https://twitter.com/greg_doucette/status/1268235074580987906 Des Moines, IA: police use pepper-spray in an elevator on apartment residents trying to go home, including on one woman carrying a baby]<br />
# [https://twitter.com/greg_doucette/status/1268234782204362752 Kalamazoo, MI: protestors lay on the ground, so cops start shooting teargas canisters at their heads for sport]<br />
# [https://twitter.com/greg_doucette/status/1268231604461342720 Columbus, OH: more police destroying private property without due process]<br />
# [https://twitter.com/greg_doucette/status/1268225612478590977 San Diego, CA: peaceful protest – downright festive really – interrupted by tear gas and bullets]<br />
# [https://twitter.com/greg_doucette/status/1268203414598684672 Boston, MA: cop destroying private property without due process]<br />
# [https://twitter.com/greg_doucette/status/1268202974498762757 Cincinnati, OH: the casual banality of evil]<br />
# [https://twitter.com/greg_doucette/status/1268201894490656769 Multiple: mashup of several videos at once, locations are at the bottom of each panel]<br />
# [https://twitter.com/greg_doucette/status/1268195207713038336 Originally I thought police were just shooting and it happened the 16yo boy got hit. Turns out the Austin cops deliberately sniper-shot him in the head with nobody else around him.]<br />
# [https://twitter.com/greg_doucette/status/1268193746329485312 Walnut Creek, CA: cops casually threatening to murder citizens 👮🏼♂️: "IF YOU DO NOT MOVE, YOU WILL BE DEAD."]<br />
# [https://twitter.com/greg_doucette/status/1268193208200318976 Note the person on a motorized scooter getting caught in a cloud of gas]<br />
# [https://twitter.com/greg_doucette/status/1268192571043524610 Seattle, WA: another night of indiscriminate flash-bang and teargas use on peaceful protestors]<br />
# [https://twitter.com/greg_doucette/status/1268192220529778688 San Jose, CA: cop casually hitting a guy with his motorcycle and sending him to asphalt as 3 more cops pile on]<br />
# [https://twitter.com/greg_doucette/status/1268190300209586177 Portland, OR: casual war-zoning as police shoot into a crowd for fun]<br />
# [https://twitter.com/greg_doucette/status/1268189755143053323 Portland, OR: cops defacing private property with spray paint]<br />
# [https://twitter.com/greg_doucette/status/1268189319002537989 New York City, NY: asked this before, but how many NYPD cops does it take to beat a woman? Bicyclist abruptly arrested without cause, as at least two cops beat her for sport]<br />
# [https://twitter.com/greg_doucette/status/1268053040524918786 Washington, DC: police leaving unexploded ordnance laying around. This one is a Stinger (not flash-bang): it's a glorified IED that shoots rubber bullets]<br />
# [https://twitter.com/greg_doucette/status/1268052132252602368 Los Angeles, CA: protestors shot in the head at 3rd & Fairfax by the Grove. Per the journalist who got the video, police shot first, *then* gave order to disperse]<br />
# [https://twitter.com/greg_doucette/status/1268051560489193473 Portland, OR: indiscriminate teargas and flash-bang grenade use]<br />
# [https://twitter.com/greg_doucette/status/1268049145912348673 Los Angeles, CA: at least 6 cops destroy a car, breaking out its windows to drag out the occupants]<br />
# [https://twitter.com/greg_doucette/status/1268047092079112198 Wilmington, NC: police kneel "in solidarity"...so protestors will get closer...for the police to shoot them.]<br />
# [https://twitter.com/greg_doucette/status/1268046296016969730 Washington, DC: cops pepper-spray a CNN crew as they're broadcasting live on-air]<br />
# [https://twitter.com/greg_doucette/status/1268044563828129792 Los Angeles, CA: police doing casual drive-by shootings from the comfort of their SUV]<br />
# [https://twitter.com/greg_doucette/status/1268044305677070337 Seattle, WA: police pre-beatings huddle. 👮🏼♂️: "Don't kill 'em. But hit 'em HARD!"]<br />
# [https://twitter.com/greg_doucette/status/1268043104822665218 Austin, TX: police shoot a 16-year-old boy in the head and leave him for dead on the side of the road]<br />
# [https://twitter.com/greg_doucette/status/1268041199077728256 Dallas, TX: police shoot out the left eye of 25yo Brandon Saenz]<br />
# [https://twitter.com/greg_doucette/status/1268037398845427713 San Antonio, TX: police abruptly open fire into crowd of peaceful protestors. 🗣: "Everybody put your hands up!" 👮🏼♂️👮🏼♀️👮🏼: ::starts shooting::]<br />
# [https://twitter.com/greg_doucette/status/1268034065120276480 Police looking like invaders]<br />
# [https://twitter.com/greg_doucette/status/1268033284166074370 San Diego, CA: no video, but the woman is an employee of Child Protective Services, shot as she was leaving work]<br />
# [https://twitter.com/greg_doucette/status/1268031453956321280 Orlando, FL: overhead view of when bike cops started beating protestors with their bikes for sport]<br />
# [https://twitter.com/greg_doucette/status/1268031102100426753 Orlando, FL: 7-8 police officers pile on to one guy who's already on the ground not resisting]<br />
# [https://twitter.com/greg_doucette/status/1268030671706099717 Charlotte, NC: police kettle protestors in between two parking decks, then open fire on them]<br />
# [https://twitter.com/greg_doucette/status/1268029918992109568 Orlando, FL: you see a protestor in gray hoodie walking away with his hands up. A cop rushes into the crowd to grab him so some officers can beat him for sport, then other officers indiscriminately pepper-spray everyone else]<br />
# [https://twitter.com/greg_doucette/status/1268028872404217857 San Diego, CA: One set of cops kettle protestors toward another set of cops, so the second set of cops can shoot them]<br />
# [https://twitter.com/greg_doucette/status/1268028575850147848 Seattle, WA: not a brutality video, but instead showing the scale of the protestors. *Years* of data – decades likely – have repeatedly proven that crackdowns radicalize people and lead to more protestors]<br />
# [https://twitter.com/greg_doucette/status/1268025470764888066 New York City, NY: how many NYPD cops does it take to beat one woman? Seems like at least 4-5 here]<br />
# [https://twitter.com/greg_doucette/status/1267984839254454278 Asheville, NC: police destroy medical supplies and cut open all the water bottles creating a tremendous plastic mess outside of Farm Burger]<br />
# [https://twitter.com/greg_doucette/status/1267981687289532421 Walnut Creek, CA: police using teargas and bullets on a peaceful protest in broad daylight]<br />
# [https://twitter.com/greg_doucette/status/1267978682951491585 Athens, GA: more indiscriminate teargas use]<br />
# [https://twitter.com/greg_doucette/status/1267978372249989124 Athens, GA: policing firing teargas and bullets near the UGA campus]<br />
# [https://twitter.com/greg_doucette/status/1267977008115572736 Hoover, AL: when people say "defund the police," this is the type of overreactive Pile of poo they're talking about. 14 high schoolers. 50 cops]<br />
# [https://twitter.com/greg_doucette/status/1267976565532614658 Portland, ME: Too far away to tell if this is police or a vigilante, but either way they've got snipers casually aiming from hotel rooftops]<br />
# [https://twitter.com/greg_doucette/status/1267975706258128899 Denver, CO: y'all have seen me tag @elisabeth a number of times over the years, who's been working with the city on police use of force. So Denver police shot her 2x. From behind.]<br />
# [https://twitter.com/greg_doucette/status/1267964835263852546 Police casually aiming their guns at kids recording on their phones from the roof.]<br />
# [https://twitter.com/greg_doucette/status/1267963097316233219 Washington, DC: no identifiable insignia, so people don't know who to sue when they get brutalized]<br />
# [https://twitter.com/greg_doucette/status/1267918152635203585 San Luis Obispo, CA: police open fire on peaceful protestors with their hands raised. The juxtaposition is striking... 🗣: "THIS IS WHAT DEMOCRACY LOOKS LIKE!" 👮🏼♂️👮🏼👮🏼♀️: ::starts shooting::]<br />
# [https://twitter.com/greg_doucette/status/1267916734364860421 Los Angeles, CA: extended montage (15+ minutes) of police violence footage out of LA]<br />
# [https://twitter.com/greg_doucette/status/1267913962697474048 New York City, NY: police beat the everlasting sh*t out of a hospital worker as he walks to his hotel. Video in the article; he ended up getting stitches and 2 CT scans at the hospital where he works]<br />
# [https://twitter.com/greg_doucette/status/1267911437596450817 Hollywood, CA: guy is live-streaming his walk home from work when police show up and start arresting everyone. Listen close to the officed's radio. "You should not be driving past anyone. Stop where you are, and take someone into custody."]<br />
# [https://twitter.com/greg_doucette/status/1267908849446653956 Philadelphia, PA: cop pulls down a protestor's mask to pepper-spray her in the face as she's kneeling with her hands up, then the cop sprays another one, then he pushes over a third to make sure he can pepper-spray them in the face too]<br />
# [https://twitter.com/greg_doucette/status/1267898837940809731 Los Angeles, CA: police open fire for sport on two guys standing still yards away having a conversation with each other]<br />
# [https://twitter.com/greg_doucette/status/1267895665285267458 Denver, CO: Colorado State Patrol shoot a Denver reporter (who didn't realize at the time that the camera was still recording)]<br />
# [https://twitter.com/greg_doucette/status/1267869387916271617 Richmond, VA: police fire teargas on peaceful protestors for sport, a half-hour before curfew]<br />
# [https://twitter.com/greg_doucette/status/1267863914290741251 Columbus, OH: police pepper-spray members of the news media so they'll stop reporting. News media are exempt from the city's curfew]<br />
# [https://twitter.com/greg_doucette/status/1267861989017161731 Washington, DC: video elsewhere in the thread, but here's the @DCist write-up on police pepper-spraying inside a private residence that they didn't have a warrant to enter]<br />
# [https://twitter.com/greg_doucette/status/1267861168384880650 Joliet, IL: Protestors arrested as they're leaving, including one grabbed by the neck and dogpiled by at least 3 cops]<br />
# [https://twitter.com/greg_doucette/status/1267859272639164416 West Philadelphia, PA: police shooting teargas down empty streets for sport? I assume there's a purpose for this]<br />
# [https://twitter.com/greg_doucette/status/1267845555654402049 Atlanta, GA: four police team up to arrest a photographer, as the nightly news broadcasts it]<br />
# [https://twitter.com/greg_doucette/status/1267835616823255046 Seattle, WA: overhead view of Seattle PD pepper-spraying protestors, then indiscriminately firing flash bang grenades deep into the crowd]<br />
# [https://twitter.com/greg_doucette/status/1267694239028334592 Seattle, WA: police shoot a teargas canister at a reporter – live on-air as she's reporting]<br />
# [https://twitter.com/greg_doucette/status/1267670735645966338 Riverside, CA: police destroying private property while being filmed live on the nightly news]<br />
# [https://twitter.com/greg_doucette/status/1267666729699807232 Washington, DC: by now you've probably seen this one on the news, but they decided to gas clergy and other protestors so the Yapping Yam could get a photo op at a church]<br />
# [https://twitter.com/greg_doucette/status/1267659749195165697 Sacramento, CA: police open fire on a peaceful crowd – *before* issuing a dispersal order (you'll hear that near the end) – and one of the first things they do is shoot a non-protesting legal observer (and US Navy veteran) in the face]<br />
# [https://twitter.com/greg_doucette/status/1267654043482181640 Washington, DC: borrowing a page from the best authoritarian regimes across the globe – didn't Hosni Mubarak do this in Egypt a few years ago? – the military hovers a helicopter low over a crowd of peaceful protestors as a "show of strengt]<br />
# [https://twitter.com/greg_doucette/status/1267653338545491968 Los Angeles, CA: black store owners call the police to get help with looters. So the police arrest the black store owners who asked for help]<br />
# [https://twitter.com/greg_doucette/status/1267652666227855361 Buffalo, NY: police and National Guard swarm in like Keystone Kops to beat and arrest some people, but then run away as an SUV drives through them and hits at least one officer]<br />
# [https://twitter.com/greg_doucette/status/1267646365557035010 Pittsburgh, PA: police pepperspray and shoot peaceful unarmed protestors as they chant]<br />
# [https://twitter.com/greg_doucette/status/1267645954100023296 Omaha, NE: police shoot protestors with pepperballs, kick several repeatedly after tackling them]<br />
# [https://twitter.com/greg_doucette/status/1267645048155496449 Washington, DC: peaceful protestors try to go home to comply with curfew. Police block them in until after curfew so they can arrest everyone]<br />
# [https://twitter.com/greg_doucette/status/1267638295166582789 Richmond, VA: no video for this, but a long thread on Richmond police flagrantly violating citizens' constitutional rights. Going to RT the whole thing separate from this thread]<br />
# [https://twitter.com/greg_doucette/status/1267628992175226885 Little Rock, AR: peaceful protestors, kneeling with hands raised, get blown away by LRPD]<br />
# [https://twitter.com/greg_doucette/status/1267625961115222016 The DC cop punches the camera man first, before hitting him with his shield]<br />
# [https://twitter.com/greg_doucette/status/1267610821984751618 Richmond, VA: another video where everything was peaceful, it wasn't curfew yet, and police started gassing everyone for sport]<br />
# [https://twitter.com/greg_doucette/status/1267606765769809920 Philadelphia, PA: all-white vigilante groups now roaming the streets]<br />
# [https://twitter.com/greg_doucette/status/1267603917480906753 Los Angeles, CA: LAPD officers beat the sh*t out of several unarmed peaceful protestors for sport, while other cops in the pack shoot them at close range]<br />
# [https://twitter.com/greg_doucette/status/1267603264901648386 Richmond, VA: police start using tear gas at peaceful protest near Monument Avenue]<br />
# [https://twitter.com/greg_doucette/status/1267602766412820480 Atlanta, GA: police beat the sh*t out of one person, then a male officer breaks away to beat, tackle, and mount the unarmed woman behind him]<br />
# [https://twitter.com/greg_doucette/status/1267602175997468672 Washington, DC: one cop repeatedly beats the sh*t out of a cameraman with his riot shield, as another beats the reporter next to him with a baton while they flee]<br />
# [https://twitter.com/greg_doucette/status/1267600596523266049 Los Angeles, CA: LAPD beat the sh*t out of well-known actor Kendrick Sampson (@kendrick38). Then once they're done beating him, the baton cops back away so another cop can shoot him at close range]<br />
# [https://twitter.com/greg_doucette/status/1267598308392009735 Los Angeles, CA: LAPD scanner chatter. 👮🏼♂️: "$100 for a n*gger! $50 for a Mexican! Kill 'em all!"]<br />
# [https://twitter.com/greg_doucette/status/1267597416938733576 Denver, CO: DPD no video for this one. DPD conducting an investigation after one of its officers posts an admission on Instagram that police are deliberating instigating riots]<br />
# [https://twitter.com/greg_doucette/status/1267597140819415040 Philadelphia, PA: four cops brutalize and arrest a man. Notice how one cop deliberately put the man's hand on the baton / stick / whatever, just so he could then beat the guy for touching it]<br />
# [https://twitter.com/greg_doucette/status/1267596048513540103 Denver, CO: casual conversation between a protestor and riot cop: "Can I ask you what's going to happen at 8? Sir?" 👮🏼♂️: "What's gonna happen is we're gonna start beating the fuck out of you."]<br />
# [https://twitter.com/greg_doucette/status/1267594960955748352 Chicago, IL: nearly a dozen police officers team up to destroy a car and pull at least one of the occupants out by their head. Original poster indicates police stole the car soon thereafter, leaving the driver and another passenger in the lot]<br />
# [https://twitter.com/greg_doucette/status/1267594159776501761 Philadelphia, PA: police deliberately herd protestors up an embankment against an unclimbable fence – and then repeatedly gas them all for sport]<br />
# [https://twitter.com/greg_doucette/status/1267591652304134153 Des Moines, IA: police teargas and arrest a Des Moines Register reporter so she'll stop tweeting updates]<br />
# [https://twitter.com/greg_doucette/status/1267588463001513984 Indianapolis, IN: a half-dozen police repeatedly shoot and club a protestor for sport]<br />
# [https://twitter.com/greg_doucette/status/1267587659771392001 Columbus, OH: police officer peppersprays 2 people in the face as they silently kneel. Then another officer takes their water]<br />
# [https://twitter.com/greg_doucette/status/1267586228045701120 Long Beach, CA: police point a gas canister launcher at folks as a young child sits atop dad's shoulders]<br />
# [https://twitter.com/greg_doucette/status/1267585724058087440 Los Angeles, CA: police shoot a USMC veteran in the head for sport]<br />
# [https://twitter.com/greg_doucette/status/1267585281223471104 Austin, TX: Police shoot a pregnant black woman in the abdomen for sport]<br />
# [https://twitter.com/greg_doucette/status/1267582414878105600 Los Angeles, CA: LAPD sharing a jovial chuckle with a white woman as she spraypaints a building to make it look like it was done by protestors. You can even hear police suggest she write "Floyd" too]<br />
# [https://twitter.com/greg_doucette/status/1267528084393283584 Dallas, TX: police shoot out the eye of an unarmed black man for sport]<br />
# [https://twitter.com/greg_doucette/status/1267526684976328705 Boston, MA: multiple police officers beat the everlasting sh*t out of two people for sport, apparently for crossing the street?]<br />
# [https://twitter.com/greg_doucette/status/1267525235806527488 Raleigh, NC: audio from police communication, admitting to using expired tear gas for crowd control]<br />
# [https://twitter.com/greg_doucette/status/1267523455827750912 Seattle, WA: police appear to be trying to break the glass window of whatever business this is at 2nd and Pine?]<br />
# [https://twitter.com/greg_doucette/status/1267520159557787649 Washington, DC: police pin a CNN reporter against a wall while another one clubs him in the knee with a baton]<br />
# [https://twitter.com/greg_doucette/status/1267519238224445442 Miami, FL: police burst out of their SUV to tackle and arrest a peaceful unarmed protestors talking to them. The man was charged with violating the 8:00pm curfew; original poster says the arrest happened ~5:45pm]<br />
# [https://twitter.com/greg_doucette/status/1267513121234653203 Louisville, KY: police shoot – and kill – renowned local chef David McAtee]<br />
<br />
== News Compilation ==<br />
<br />
# [https://www.washingtonpost.com/outlook/2020/06/03/priest-stjohns-church-trump/ I’m a priest. The police forced me off church grounds for Trump’s photo op.]<br />
# [https://www.usatoday.com/story/news/nation/2020/06/03/asheville-north-carolina-police-seen-destroying-protesters-supplies/3135539001/ Video shows police destroying medical station at North Carolina protest; mayor looks for answers]<br />
# [https://www.nydailynews.com/news/national/ny-vallejo-ca-police-officer-kills-unarmed-22-year-old-on-his-knees-20200604-ib27psiyvnckvcrphk4fu64nai-story.html California police officer kills man on his knees, mistaking hammer for gun]<br />
# [https://reason.com/2020/06/08/video-shows-cops-slashing-tires-across-minneapolis-during-george-floyd-protests/ Video Shows Cops Slashing Tires Across Minneapolis During Protests]<br />
<br />
== Useful Links == <br />
<br />
* [https://www.niemanlab.org/2020/06/well-try-to-help-you-follow-the-police-attacks-on-journalists-across-the-country/ Attacks against reporters]<br />
* Here's a [https://docs.google.com/spreadsheets/d/1YmZeSxpz52qT-10tkCjWOwOGkQqle7Wd1P7ZM1wMW0E/edit?usp=sharing Google Sheets document] that is being updated more.</div>Eddynetweb