ECE488 Homework - Revision history

Eddynetweb: Homework 8

2022-03-01T05:53:20Z

Homework 8

@@ Line 297: / Line 297: @@
 Use the single-phase equivalent to calculation Ia. Its magnitude should be the same as part c.
-[[File:ECE488 HW8 Figure.jpg|thumb]]
+https://wiki.eddyn.net/images/thumb/b/b2/ECE488_HW8_Figure.jpg/600px-ECE488_HW8_Figure.jpg
 === Breakdown ===
-Hello.
+) Calculate the currents I<sub>ab</sub>, I<sub>bc</sub>, and I<sub>ca</sub> (Ohm's law)

Eddynetweb: Added Homework 8.

2022-03-01T04:16:20Z

Added Homework 8.

← Older revision		Revision as of 04:16, 1 March 2022
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	* 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).		* 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).

	== Homework 2 ~~- Eduardo Santillan~~ ==		== Homework 2 ==

	Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.		Assume the total electric generation for a region in 2025 will be 11,000 GWh/day.
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	* '''Total cost per day''': $596,490,000		* '''Total cost per day''': $596,490,000

	== Homework 3 ~~- Eduardo Santillan~~ ==		== Homework 3 ==

	A power system is supplying 2000 MW of power for one hour from the following generating units:		A power system is supplying 2000 MW of power for one hour from the following generating units:
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	'''Total Fuel Costs: $17,825.75'''		'''Total Fuel Costs: $17,825.75'''

	== Homework 4 ~~- Eduardo Santillan~~ ==		== Homework 4 ==

	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).		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).
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	<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math>		<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math>

	== Homework 5 ~~- Eduardo Santillan~~ ==		== Homework 5 ==

	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:		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:
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	Reactive power is <math>Q=500VAr = 0.5kVAr</math> from above.		Reactive power is <math>Q=500VAr = 0.5kVAr</math> from above.

			== Homework 8 ==

			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.

			# Calculate the currents I<sub>ab</sub>, I<sub>bc</sub>, and I<sub>ca</sub> (Ohm's law)
			# Calculate the power consumed by each resistor, and the total 3-phase power.
			# Calculate the currents I<sub>a</sub>, I<sub>b</sub>, and I<sub>c</sub>. (KCL)

			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 Ia. Its magnitude should be the same as part c.

			[[File:ECE488 HW8 Figure.jpg\|thumb]]

			=== Breakdown ===

			Hello.

Eddynetweb: Finish Homework 5

2022-02-15T05:35:38Z

Finish Homework 5

← Older revision		Revision as of 05:35, 15 February 2022
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	1) The current of the load		1) The current of the load

			<math>
			\begin{align}
			Z = \frac{1}{\frac{1}{5}+j(\frac{1}{20}-\frac{1}{10})} = \\ 4.71+j1.18 = 4.85\angle 14.04^\circ Ohms \\
			I = \frac{V}{Z} = \frac{100\angle 0^\circ}{4.85\angle 14.04^\circ} = 20.62\angle -14.04^\circ Amps\\
			\end{align}
			</math>

			2) The real power of the load

			<math>
			\begin{align}
			S = V * I = (100\angle 0^\circ)(20.62\angle -14.04^\circ) = \\ 2062\angle 14.04^\circ VA = 2000 + j500 VA
			\end{align}
			</math>

			Real power is <math>P=2000W = 2kW</math>.

			3) The reactive power of the load

			Reactive power is <math>Q=500VAr = 0.5kVAr</math> from above.

Eddynetweb: Homework 5 - add circuit diagram.

2022-02-15T05:07:03Z

Homework 5 - add circuit diagram.

← Older revision		Revision as of 05:07, 15 February 2022
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	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:		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:

			# The current of the load
	# The real power of the load		# The real power of the load
	# The reactive power of the load		# The reactive power of the load
	~~# The current of the load~~

	=== Breakdown ===		=== Breakdown ===

	~~[[File~~:~~ECE488 Homework 5 Circuit Diagram~~.png~~\|thumb]]~~		https://wiki.eddyn.net/images/thumb/e/ef/ECE488_Homework_5_Circuit_Diagram.png/600px-ECE488_Homework_5_Circuit_Diagram.png

			1) The current of the load

Eddynetweb: Homework 5 start

2022-02-15T05:04:39Z

Homework 5 start

← Older revision		Revision as of 05:04, 15 February 2022
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	<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math>		<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math>

			== Homework 5 - Eduardo Santillan ==

			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:

			# The real power of the load
			# The reactive power of the load
			# The current of the load

			=== Breakdown ===

			[[File:ECE488 Homework 5 Circuit Diagram.png\|thumb]]

Eddynetweb: Homework 4 changed sub-question 4.

2022-02-10T05:30:54Z

Homework 4 changed sub-question 4.

← Older revision		Revision as of 05:30, 10 February 2022
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	4. Which leads, current or voltage?		4. Which leads, current or voltage?
	* With voltage source as reference, voltage leads current ~~by <math>56~~.~~66^\circ</math>~~		* With voltage source as reference, '''voltage leads current'''.

	5. What is the phase angle between current and voltage?		5. What is the phase angle between current and voltage?

Eddynetweb: Homework 4 added circuit diagram

2022-02-10T05:27:37Z

Homework 4 added circuit diagram

← Older revision		Revision as of 05:27, 10 February 2022
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	1. Draw the circuit diagram.		1. Draw the circuit diagram.

			https://wiki.eddyn.net/images/thumb/2/25/ECE488_HW4_CircuitDiagram.png/600px-ECE488_HW4_CircuitDiagram.png

	2. Calculate the value of inductance.		2. Calculate the value of inductance.

Eddynetweb: Homework 4 formatting changes.

2022-02-10T05:25:02Z

Homework 4 formatting changes.

← Older revision		Revision as of 05:25, 10 February 2022
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	3. Calculate the phasor value of current through the load impedance.		3. Calculate the phasor value of current through the load impedance.

	<math>		<math>
	\begin{align}		\begin{align}
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	6. Calculate the phasor value of voltage across the resistor.		6. Calculate the phasor value of voltage across the resistor.

	<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math>		<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math>

	7. Calculate the phasor value of voltage across the inductive reactance.		7. Calculate the phasor value of voltage across the inductive reactance.

	<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math>		<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math>

Eddynetweb: Homework 4 added

2022-02-10T05:23:56Z

Homework 4 added

← Older revision		Revision as of 05:23, 10 February 2022
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	'''Total Fuel Costs: $17,825.75'''		'''Total Fuel Costs: $17,825.75'''

			== Homework 4 - Eduardo Santillan ==

			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).

			# Draw the circuit diagram.
			# Calculate the value of inductance.
			# Calculate the phasor value of current through the load impedance.
			# Which leads, current or voltage?
			# What is the phase angle between current and voltage?
			# Calculate the phasor value of voltage across the resistor.
			# Calculate the phasor value of voltage across the inductive reactance.

			=== Breakdown ===

			1. Draw the circuit diagram.

			2. Calculate the value of inductance.

			<math>
			\begin{align}
			X_{L} = 38 \\
			wL = 38 \\
			L = \frac{38}{2\pi f} = \frac{38}{2\pi 60} = 0.101 H
			\end{align}
			</math>

			3. Calculate the phasor value of current through the load impedance.
			<math>
			\begin{align}
			I = \frac{V}{Z_{Load}} = \frac{240\angle 0^\circ}{25+j38} = 0.101 H \\
			\sqrt{25^2+38^2} = 45.49\\
			tan{}^{-1}(\frac{38}{25}) = 56.66^\circ \\
			= \frac{240\angle 0^\circ}{45.49\angle 56.66^\circ} = 5.28\angle -56.66^\circ Amps
			\end{align}
			</math>

			4. Which leads, current or voltage?
			* With voltage source as reference, voltage leads current by <math>56.66^\circ</math>

			5. What is the phase angle between current and voltage?
			* The phase angle is ϕ = <math>56.66^\circ</math>

			6. Calculate the phasor value of voltage across the resistor.
			<math>V_{R} = I * R = 5.276\angle -56.66^\circ * 25 = 131.9\angle -56.66^\circ V</math>

			7. Calculate the phasor value of voltage across the inductive reactance.
			<math>V_{L} = I * X_{L} = 5.276\angle -56.66^\circ * 38\angle 90^\circ = 200.49\angle 33.34^\circ V</math>

Eddynetweb: Finish Homework 3

2022-02-08T03:27:02Z

Finish Homework 3

← Older revision		Revision as of 03:27, 8 February 2022
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	We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.		We'll first calculate the '''CO<sub>2</sub> emissions for the hour''' for the dataset.

	Coal:		Subbituminous Coal:

	<math>		<math>
	\begin{align}		\begin{align}
	\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\		\frac{lbs}{MWh} = \frac{213 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.33} = 2202.92 \\
	\frac{2202.92 lbs}{MWh} * 500 MWh = ~~1101460~~ lbs		\frac{2202.92 lbs}{MWh} * 500 MWh = 1101458 lbs
	\end{align}		\end{align}
	</math>		</math>

	CC Gas:		Combined-cycle gas turbine Gas:

	<math>		<math>
	\begin{align}		\begin{align}
	\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\		\frac{lbs}{MWh} = \frac{117 lbs}{mmBtu} * \frac{mmBtu}{293 KWh} * \frac{1000 KWh}{MWh} * \frac{1}{0.55} = 726.03 \\
	\frac{726.03 lbs}{MWh} * 200 MWh = 145206~~.33~~ lbs		\frac{726.03 lbs}{MWh} * 200 MWh = 145206 lbs
	\end{align}		\end{align}
	</math>		</math>

	OC Gas:		Open-cycle gas turbine Gas:

	<math>		<math>
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	</math>		</math>

	'''Total CO<sub>2</sub> Emissions: 1,360,~~757.33~~ lbs'''		'''Total CO<sub>2</sub> Emissions: 1,360,755 lbs'''

	Now we'll calculate the '''fuel costs for the hour''' for the dataset.		Now we'll calculate the '''fuel costs for the hour''' for the dataset.
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	</math>		</math>

	Coal:		Subbituminous Coal:

	<math>		<math>
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	</math>		</math>

	CC Gas:		Combined-cycle gas turbine Gas:

	<math>		<math>
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	</math>		</math>

	OC Gas:		Open-cycle gas turbine Gas:

	<math>		<math>