Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 29, Problem 18P

(a)

To determine

To Find:

The resistance and inductive reactance of the plant’s total load.

(a)

Expert Solution
Check Mark

Answer to Problem 18P

Resistance R=571Ω and inductive impedance XL = 266.4Ω .

Explanation of Solution

Given:

Operated power of plant is P=2.3MW

Voltage supply to the plant is ε=40kV

Frequency supply to the plant is F=60Hz

Resistance of the transmission line in R=5.2Ω

Phase δ = 25°

Formula used:

The relation between Z and δ are given as

  R=Zcosδ and XL=Zsinδ.......... (1)

Impedance in terms of Irms and εrms ,

  Z=εrmsIrms.......... (2)

Average power Pav=εrmsIrmscosδ.......... (3)

Calculation:

From equation (3) rms current can be rewritten as

  Irms=Pavεrmscosδ.......... (4)

Substitute the value Irms from equation (4) in equation (2)

  Z=ε2rmscosδPav.......... (5)

Substitute the values:

  Z=( 40kV)2cos25°2.3MW

= =630.3Ω

  R=630.3cos25°=571ΩXL=630.3Ωsin25°=266.4Ω

Conclusion:

Hence, theResistance R=571Ω and inductive impedance XL = 266.4Ω .

(b)

To determine

To Find:

RMS current and RMS voltage of the power line.

(b)

Expert Solution
Check Mark

Answer to Problem 18P

  63.4 A; 40.3kV

Explanation of Solution

Given:

Operated power of plant is P=2.3MW

Voltage supply to the plant is ε=40kV

Frequency supply to the plant is F=60Hz

Resistance of the transmission line in R=5.2Ω

Phase δ = 25°

Formula use:

RMS current

  Irms=Pavεrmscosδ.......... (1)

Calculation:

Substituting the values of Pav , εrms and δ in equation (1)

  Irms=2.3 MW40 kV cos 25°=63.4A

Apply Kirchhoff’s loop principle to the circuit:

  εrmsIrmsRIrmsZ=0

  εrms=Irms(R+Z).......... (2)

Substituting the value of Irms , R and Z in equation (2)

  εrms=63.4×(5.2+630.3)=40.3kV

Conclusion:

Hence,RMS current and RMS voltage are 63.4 A and 40.3kV .

(c)

To determine

To Find:

Amount of power lost in transmission.

(c)

Expert Solution
Check Mark

Answer to Problem 18P

  20.9kW

Explanation of Solution

Given:

Resistance of the transmission line in R=5.2Ω

Transmission current Irms = 63.4 A

Formula use:

Transmission power

  Ptrans=I2rmsR .......... (1)

Calculation:

Substituting the value of Irms , R in equation (1)

  Ptrans=I2rmsR = (63.4A) 2 ×5.2Ω=20.9kW

Conclusion:

Hence,the power loss in the transmission line is 20.9kW .

(d)

To determine

To Find:

The amount of money that would be saved by the electric utility during one month of operation.

(d)

Expert Solution
Check Mark

Answer to Problem 18P

$128

Explanation of Solution

Given:

Phase angle between voltage and current δ = 18°

Operated power of plant is P=2.3MW

Voltage supply to the plant is ε=40kV

Frequency supply to the plant is F=60Hz

Resistance of the transmission line in R=5.2Ω

Formula used:

Cost saving equation ΔC=(P25°P18°)Δt.......... (1)

Transmission power Ptrans=I2rmsR.......... (2)

RMS current Irms=Pavεrmscosδ.......... (3)

Calculation:

Substituting the value of Pav , εrms and δ in equation (3)

  Irms=2.3 MW40 kV cos18°=60.5A

Substituting the value of Irms , R in equation (2)

  Ptrans=I2rmsR = (60.5A) 2 ×5.2Ω=19kW

Cost saving is ΔC=(20.9kW19kW)(16h/d)(30d/month)($0.14kWh)

  =$128

Conclusion:

Hence, the cost is $128 .

(e)

To determine

To Find:

The amount ofcapacitance required to achieve the charge.

(e)

Expert Solution
Check Mark

Answer to Problem 18P

  33μF

Explanation of Solution

Given:

Phase angle between voltage and current δ = 18°

Operated power of plant is P=2.3MW

Voltage supply to the plant is ε=40kV

Frequency supply to the plant is F=60Hz

Resistance of the transmission line in R=5.2Ω

Saved cost per kilowatt-hour is $128.

Formula used:

Capacitance C=12πfXC .......... (1)

Relation between δ , XC , XL and R is

  tanδ=XLXCR.......... (2)

Conclusion:

From equation (2) XC can be rewritten as

  XC=XLRtanδ

Substituting the value of XL , R and δ in equation (1)

  C=16.28×60s 1×( 264.7571tan18°)=33μF

Conclusion:

Hence, the required capacitance is 33μF .

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Students have asked these similar questions
When a certain resistor is connected to anac generator with a maximum voltage of 15 V, the average powerdissipated in the resistor is 22 W. (a) What is the resistance of theresistor? (b) What is the rms current in the circuit? (c) We knowthat Pav = I2rmsR, and hence it seems that reducing the resistanceshould reduce the average power. On the other hand, we alsoknow that Pav = V2rms>R, which suggests that reducing R increasesPav. Which conclusion is correct? Explain.
A power plant generator produces 100 A at 15 kV(rms). A transformer is used to step up the transmissionline voltage to 150 kV (rms). (a) What is rms current in thetransmission line? (b) If the resistance per unit length of theline is 8.6 × 10−8 Ω/m, what is the power loss per meterin the line? (c) What would the power loss per meter be ifthe line voltage were 15 kV (rms)?
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