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An ac generator provides emf to a resistive load in a remote factory over a two-cable transmission line. At the factory a stepdown transformer reduces the voltage from its (rms) transmission value Vtto a much lower value that is safe and convenient for use in the factory. The transmission line resistance is R=0.30/cable, and the power of the generator is P=250kW. If Vt=80kV, what are (a) the voltage decreases V along the transmission line and (b) the rate V at which energy is dissipated in the line as thermal energy? If Vt=80kV, what are (c) V and (d)V ? If Vt=80kV, what are (e) V and (f) Pd?

Short Answer

Expert verified
  1. The voltage V decreases along the transmission lineis 1.9 V.
  2. The ratePdis5.9W
  3. The voltage VatV1=80KVis 19 V.
  4. The power PdatVt=8.0kVis 590 W.
  5. The voltage V atVt=0.80kVis 190 V.
  6. The power Pd at 0.8kV is 59000W

Step by step solution

01

Listing the given quantities:

The resistance of transmission line is, R=0.30/cable

The power of the generator is,

P=250kW=250103W

02

Understanding the concepts of power:

Ohm's law states that the current flowing through a conductor between two points is directly proportional to the voltage across both points.

Calculate the total resistance over two lines first. Then use the expression relating power, voltage, and current to determine the desired values.

Formula:

Write the equation for voltage as below.

V=IR

Here, V is the voltage, I is the current, R is the resistance.

Write the equation for power as below.

P=VI

The resistance R for two cables; hence, the total resistance is as follows:

R=20.3=0.6

03

(a) Calculations of the ΔV along transmission line:

Irms=PVt=25010380103=3.125A

V=IrmsR=3.1250.6=1.9V

The voltage V along transmission line is 1.9 V.

04

(b) Calculations of the rate of  Pd:

Pd=Irms2R=3.12520.6=5.9W

The rate Pd is 5.9 W.

05

(c) Calculations of the  ΔV at Vt=8.00 kV:

Irms=PVt=2501038103=31.25A

V=IrmsR=31.250.6=19V

The voltage Vat Vt=8.0kVis 19 V.

06

(d) Calculations of the  Pd at  Vt=8.00  kV

Pd=Irms2R=31.2520.6=590W

The power Pd at Vt=8.0kV is 590 W .

07

(e) Calculations of the  ΔV at  Vt=0.80 kV:

Irms=PVt=2501030.8103=312.5A

V=IrmsR=312.50.6=190V

The voltage V at Vt=0.80kV is 190 V.

08

(f) Calculations of the  Pd at  0.8 kV

Pd=Irms2R=1.87520.6=59000W

The power Pd at 0.8 kV is 59000 W.

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