Transmission Line having length of approximately up to 50 km and line-to-line voltage of less than 20kV is considered Short Transmission Line. In this case, capacitance effects are small and therefore neglected in calculation of line performance.

Let us see an example of calculation of performance of a short transmission line

An overhead line delivers 2MW at 11kV at a power factor 0.8 lagging. Total line impedance consists of Resistance of 8 Ohm and Reactance of 10 Ohm.

Find (i) Sending end voltage, V_R (ii) Sending end power factor (iii) voltage drop and (iv) Efficiency of transmission line

Let receiving end voltage of Red phase to neutral , V_R= 11,000/square root (3) =11,000/1.732 = 6,350.9V
V_R= (6,350.9 + j0 ) V = 6,350.9/0° V

Power per phase = 2MW/ 3 = 0.67 MW

Receiving end current per phase, I_R =0.67 MW / (6,350.9V x 0.8) = 131.9 A
I_R = 131.9 /- cos ^-1 0.8 A = 131.9 /-36.9° A

Line impedance, Z = R + jX = 8 + j10 Ohm = 12.8 /51.3° (i) Sending end voltage, V_s = IZ + V_R
= (131.9 /-36.9° ) ( 12.8 /51.3°) + 6,350.9/0°
= 1688.3 /14.4° + 6350.9/0°'
= 7986.2 + j419.9
= 7997.2/3.0° V

V_s (line-line) = 1.732 x 7997.2V =13.85 kV

(ii) Sending end power factor

cos (Ø_s + Ø_R) = cos (3.0° + 39.9°) = cos 39.4° = 0.77

(iii) voltage drop (or regulation)

Voltage drop = ( V_s - V_R ) / V_R = (7997.2 - 6,350.9) / 6,350.9 =25.9% (iv) Efficiency of transmission line, eff
Efficiency of transmission line = Power (delivered) / { Power (delivered) + Line Losses}

a) Using single phase circuit
Line Losses per single phase =I² R = 131.9² x 8 /10⁶ = 0.139 MW
eff =Power (delivered) / { Power (delivered) + Line Losses}
= 0.67 MW /(0.67MW + 0.139 MW) = 82.8%

b) Using three phase circuit
Total Line Losses per three phase =3 x I² R = 3 x 131.9 ² x 8 /10⁶ = 0.417MW
eff = Power (delivered) / { Power (delivered) + Line Losses}
= 2 MW /(2 MW + 0.417 MW) = 82.7%

Note: Line efficiency of either single phase circuit or three phase circuit is the same !