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A small town with a demand of 800kW of electric power at 220v is situated 15km away from an electric plant generating power at 440v. The resistance of the two-wire line carrying power is 0.5ohm/km. The town gets power from the line through a 4000-220volt. Step-down transformer at a sub-station in the town. (a) Estimate the line power loss in the form of heat. (b) How much power must the plant supply, assuming there is negligible power loss due to leakage? (c) Characterize the step-up transformer at the plant.


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Solution

Step1: Given data

  1. The input voltage of the plant is Vi=4000volt.
  2. The output voltage is Vo=220volt.
  3. The total required electric power is Pd=800KW.
  4. Resistance of the wire is 15×0.5ohm.

Step2: Concepts and formulae

  1. The RMS voltage of an ac circuit is Irms=PowerVrms, where, VrmsandIrmsare the RMS voltage of the circuit current.
  2. Power loss on an ac circuit is P=I2R, where, I is the current through the circuit and R is the resistance of the circuit.
  3. From, Ohm's law V=IR, where, V is the voltage across the circuit and I is the current through the circuit.

Step3: Finding the RMS current

The RMS current in the coil is Irms=PowerVrms.

So,

Irms=PowerVrms=800×1034000=200orIrms=200A.

Step4(a): Finding the power loss

The power loss due to the production of heat is PL=I2R.

So,

PL=I2R=2002×30×0.5orPL=40000×15orPL=600000orPL=600KW.

Therefore, the power loss in the circuit line is 600KW.

Step4(b): Finding the total power supply

We know, that the total power supply is the sum of demand power and power loss.

So,

P=Pd+PLorP=800+600orP=1400KW

The total power supply by the plant is 1400KW.

Step4(c): Characterization of the transformer

The voltage drop by the circuit is Vd=IR

So,

Vd=IR=200×15=3000orVd=3000volt.

Now, the secondary voltage is

Vs=Vi+Vd=4000+3000orVs=7000volt.

According to the question, the power generation takes place at the primary voltage 440volt.

Therefore, the transformer rating is 440-7000volt.


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