LINE LOSS CALCULATION: SAMPLE PROBLEM 4

PROBLEM:  A 230kV transmission line has impedance of 50 angle 78 ohms and a capacitive reactance of 1200ohms. It transmits the power of a base load plant. On a certain dry season the sending end power is 100MW at 235kV and 95% power factor continuously for a period of one month. If cost of generation is 1.3dollars per kW-hr, what is the cost of the line losses for the one month period?

SOLUTION:

In analyzing this problem, we assume a nominal pi equivalent circuit. Also, assume the sending voltage as the reference and with a lagging power factor.


Nominal pi equivalent circuit

 First step is to solve for the sending current by using the equation below.

CURRENT = (POWER GENERATED) / (SQRT(3)*LINE VOLTAGE*POWER FACTOR)

                   = (100X10^6) / (SQRT(3)*(235X10^3)(0.95))

CURRENT = 258.6 A

θS = arccos (0.95) = 18.194 degrees

Solving forY/2:

Y = (1/Xc) = (1/(-j1200)) = (1/1200)angle 90degrees

Y/2 = (1/(2(1200))) = (1/2400) angle 90degrees

Solving for I1 :

 I1 = (voltage sending)*(Y/2) = ((235)/(sqrt(3))) angle 90 degrees = 56.532 angle 90 degrees A

 I2   = IL - I1

      = 258.6 angle (-18.194)degrees  -  56.532 angle 90 degrees

      = 245.67 - j80.744 - (j56.532)

 I2   = 245.67 - j137.276 = 281.422 angle (-29.195) A

ZT  =  R + jXT = 50 angle 78 degrees = 10.395 + j48.9 ohms

Ploss = 3 I squared RL = 3(281.422)squared(10.395) = 2469.8 kW

Wloss = Ploss x (24hrs / day) x (30 days / month)

          = (2469.8)(24)(30)

Wloss = 1778256 kW-hr

Total Cost = Wloss x cost per kW-hr

                 = 1778256 x 1.3

TOTAL COST = $ 2.312 MILLION


Reference: 1001 Solved Problems in Electrical Engineering by Romeo Rojas Jr.