Question

A 3-phase load of 2,000 kVA, 0.8 p.f. is supplied at 6.6 kV, 50-Hz by means of a 33 kV transmission line 20 km long and a 5 : 1 transformer. The resistance per km of each conductor is 0.4 Ω and reactance 0.5Ω. The resistance and reactance of the transformer primary are 7.5 Ω and 13.2 Ω, whilst the resistanceof the secondary is 0.35 Ω and reactance 0.65 Ω. Find the voltage necessary at the sending end of transformission line when 6.6 kV is maintained at the load-end and find the sending-end power factor. Determine also the efficiency of transmission.

Answer 1

One phase of the system is shown in Fig.Impedance per phase of high voltage line = (8 + j 10)

Impedance of the transformer primary 

= (7.5 + j 13.2) ohm

Total impedance on the high-tension side

= (8 + j 10) + (7.5 + j 13.2) = 15.5 + j 23.2

This impedance can be transferred to secondary side by using the relation given in Hence, impedance as referred to secondary side is

= (15.5 + j 23.2)/52 = 0.62 + j 0.928

Adding to them the impedance of the transformer secondary, we get the total impedance as referred to low-voltage side

= (0.62 + j 0.928) + (0.35 + j 0.65) = 0.97 + j 1.578

Now, kVA load per phase = 2,000/3.0 = 667

Receiving-end voltage/phase = 6.6/ √3 = 3.81 kV

∴ current in the line = 667/3.81 = 175 A

Drop per conductor = I(R cos φ + X sin φ) = 175(0.97 × 0.8 + 1.578 × 0.6) = 302 V

Now, E_S = E_R + I(R cos φ + X sin φ)

Hence, sending-end voltage (phase to neutral as referred to the lower voltage side) is 3,810 + 302 = 4,112 V. As referred to high-voltage side, its value = 4,112 × 5 = 20,560 V

Line voltage = 20,560 × √3 /1000 = 35.6 kV

If φ_S is the power factor angle at the sending-end, then

 ∴ φ_S = tan–1(0.796) = 38°31′ 

∴ cos φ_S = cos 38°31′ = 0.782

Power loss/phase = 1752 × 0.97/1000 = 29.7 kW

Power at the receiving end/phase = 2000 × 0.8/3 = 533.3 kW

∴ transmission efficiency