(a)
To Find:
The resistance and inductive reactance of the plant’s total load.
(a)
Answer to Problem 18P
Resistance
Explanation of Solution
Given:
Operated power of plant is
Voltage supply to the plant is
Frequency supply to the plant is
Resistance of the transmission line in
Phase
Formula used:
The relation between
Impedance in terms of
Average power
Calculation:
From equation (3) rms current can be rewritten as
Substitute the value
Substitute the values:
=
Conclusion:
Hence, theResistance
(b)
To Find:
RMS current and RMS voltage of the power line.
(b)
Answer to Problem 18P
Explanation of Solution
Given:
Operated power of plant is
Voltage supply to the plant is
Frequency supply to the plant is
Resistance of the transmission line in
Phase
Formula use:
RMS current
Calculation:
Substituting the values of
Apply Kirchhoff’s loop principle to the circuit:
Substituting the value of
Conclusion:
Hence,RMS current and RMS voltage are
(c)
To Find:
Amount of power lost in transmission.
(c)
Answer to Problem 18P
Explanation of Solution
Given:
Resistance of the transmission line in
Transmission current
Formula use:
Transmission power
Calculation:
Substituting the value of
Conclusion:
Hence,the power loss in the transmission line is
(d)
To Find:
The amount of money that would be saved by the electric utility during one month of operation.
(d)
Answer to Problem 18P
$128
Explanation of Solution
Given:
Phase angle between voltage and current
Operated power of plant is
Voltage supply to the plant is
Frequency supply to the plant is
Resistance of the transmission line in
Formula used:
Cost saving equation
Transmission power
RMS current
Calculation:
Substituting the value of
Substituting the value of
Cost saving is
Conclusion:
Hence, the cost is
(e)
To Find:
The amount ofcapacitance required to achieve the charge.
(e)
Answer to Problem 18P
Explanation of Solution
Given:
Phase angle between voltage and current
Operated power of plant is
Voltage supply to the plant is
Frequency supply to the plant is
Resistance of the transmission line in
Saved cost per kilowatt-hour is $128.
Formula used:
Capacitance
Relation between
Conclusion:
From equation (2)
Substituting the value of
Conclusion:
Hence, the required capacitance is
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Chapter 29 Solutions
Physics for Scientists and Engineers, Vol. 3
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