Electrical Engineering Transmission Line Question

1. Sep 18, 2008

BlueForce

1. The problem statement, all variables and given/known data

First id like to say that if after reading this if anyone has any other forums to suggest to where i can post this and get some help please let me know! I cant figure this problem out for the life of me! Here goes.

1. Provide electrical network analysis based on the data(Diagram #1):
a)calc of line reactance
c)draw equivalent diagram
d)determine voltage Vs at the sending end

2. Based on the results of the previous question calc the savings from the improvement of the pf from .75 to .95
Note that the load factor is .7, there are 8760 hours in a year and energy costs are 9 cents per KW-Hr.
What is the new sending voltage?
There is a shunt capacitor that has been in to improve the power factor to .95
Draw the network model for this case. Diagram #2 that i have drawn.

This is Diagram #1:

This is Diagram #2:

Diameter of each solid cylindrincal conductor is d = 17.1mm
i converted this to .0171 meters then divided it by 2 to get the radius of .00855

Distance between conductors D = 4m

The length of the line is l = 20km

Resistance per unit length is r = 19.8 ohms per 100km
i converted this to 3.96 ohms per 20km(divide by 5) because the length of my line is 20km

Voltage at receiving end is Vr = 115 KV

Power Factor of the load is pf = 0.75

Real power of the load P = 3500 KW

2. Relevant equations
Well i have these equations and im not sure but i may be missing some or doing too many calculations. Im trying to verify if im doing this problem the correct way. I churned through many calculations and im fairly sure im not doing this right but im trying.

L is inductance per unit of the line i believe
L = (2 * 10^7) ln((cuberoot(d12*d23*d13))/(r*e^(-.25)))

Xl is the inductive reactance(Ohms)
Xl = omega * L * l

Z is impedance of the line
Z = R + jXl

S is the Apparent power
S = P / pf

I is current
I*(conjugate)= S / (sqrt(3) * Vr)

Vs is sending voltage
Vs = Vr + (I * Z)

Loss factor Lsf
Lsf = .15(Ldf) + .85(Ldf)^2

Avg loss = I^2 * R * Lsf

Total annual loss = 3(3 phase system) * 8760(hours in a year) * Avg loss

Cost per year = total annual loss * cost per KW-HR($0.09) Again, im not even sure that i have these calculations right or that im even using the right equations omega = 2 * pi * f assumed frequency is 60Hz 3. The attempt at a solution Ok here's what i have so far after 3 hours. Haha. L = (2 * 10^7) ln ((cuberoot(4*4*8)/(.00855*e^(-.25) = 1.32583 * 10^-6 H/m a) Xl = omega * L * l = (376.99) * (1.32583 * 10^-6) * (20 Km) = 9.996 Ohms Z = R + jXl = 3.96 + j9.996 = 10.752 ohms at 68.39 deg S = P / pf = 3500 Kw / .75 = 4666 KV-A cos^-1 (.75) = 41.41 deg I* = S / (sqrt(3) * 115 KV) = 23.425 A at 41.41 deg I = 23.425 A at -41.41 deg Vs = Vr + (I * Z) = 115 KV + (23.425 at -41.41) * (10.752 at 68.39 deg) d)Vs = 115.224 KV at .0568 deg b) Zload = Vr^2 / S = 115 KV ^ 2 / 4666KV-A at -41.41 deg = 2.834 Ohms at 41.41 deg 2. pf from .75 to .95 cos^-1 (.95) = 18.19 deg S = P / pf = 3500 KW / .95 = 3685 KV-A at 18.19 deg I* = S / sqrt(3) * Vr = 3684 KV-A / sqrt(3) * 115 KV = 18.495 A at 18.19 deg I = 18.495 at -18.19 deg Lsf = .15(Ldf) + .85(Ldf)^2 = .15(.7) + .85(.7)^2 = .522 Avg loss = I^2 * R * Lsf = (18.425)^2 * (3.96) (.522) = 707 W Total annual loss = 3 * 8760 * 707 W = 18.579 KW Cost per year = 18.579 KW * 0.09 =$1,672 per year

This is as far as i got in 3 hours and im fairly sure its not right. I dont really have much of a clue of what ive done wrong or how to do it properly. I would really appreciate some light on the situation. If someone cannot help me please direct me to somewhere, where i can get help on this heh. Thank you very much for reading this and helping me out. It took me over an hour to type this out haha. Thanks again :)