# Calculating the reactive power of a transmission line

I have a single machine connected to an infinite bus with the following parameters:

- Sending end power = 3.75MW
- RL=5.18 ohms
- XL=130 ohms
- VS (Generator bus)=161kV
- VR (Infinite bus)=161kV
- Sending end voltage angle (delta) = 0.01877 (Angle which allows 3.75MW to be transferred)

Using these I want to calculate the amount of reactive power sent by the sending end as well.

I calculated it first without factoring in the line resistance (RL=0) Then got the result for Q as : **35.24kVAR**

Next I calculated it with the line resistance factored in Then got the result for Q as: **-114.77kVAR**

When the line resistance is factored in, the answers for reactive power vary immensely. And when the resistance is factored in, the sign of Q is negative indicating the generator bus is absorbing VARS which doesn't seem to make sense.

Why are they so different? What am I doing wrong?

The equations I used are here

I derived these equations using Matlab to do the algebra, so it hopefully should be correct :)

I did the calculation using Microsoft.Excel 2010 and the result is close to your.

If you require abs(VS)=abs(VR) the reactive power has to be 114.7 kVAr only.

In my opinion it is capacitive.

In order to preserve P=3.75 MW delta has to be 0.018831

Did you include the resistance in your calculation or no?

berkeman
Mentor
@anon6912 -- is this for schoolwork? If so, I can move your thread to the schoolwork forums.

@anon6912 -- is this for schoolwork? If so, I can move your thread to the schoolwork forums.
Hey yeah, could you move it if you dont mind :)
Thanks

Of course, I did it. The formulas in complex are as following:

VS=COMPLEX(161*cos(delta),161*sin(delta))

VR=COMPLEX(161,0)

Z=COMPLEX(5.18,130)

IS=IMDIV(IMSUB(VS,VR),Z)

S=VS*IMCONJUGATE(IS)

P=IMREAL(S)

Q=IMAGINARY(S)

Of course, I did it. The formulas in complex are as following:

VS=COMPLEX(161*cos(delta),161*sin(delta))

VR=COMPLEX(161,0)

Z=COMPLEX(5.18,130)

IS=IMDIV(IMSUB(VS,VR),Z)

S=VS*IMCONJUGATE(IS)

P=IMREAL(S)

Q=IMAGINARY(S)

Ok but im still confused:

a. Its an inductive line how can it be capacitive? (is it capacitive because the terminal voltage magnitudes are the same?)
b. and how/why does the effect of the resistance have such a huge impact on the value of Q?
b.1 Does the incusion of the resistance make it "capacitive" in a sense somehow?

anorlunda
Staff Emeritus
If you are looking at the equations, then the answers to your questions are implicit in the equations.

If you are taking the data from #1 and putting them into a computer, then I recommend that you write and solve the equations sybolicly on paper instead.

Once you have the expression for Q, ask "what conditions other than line reactance can make Q negative?"

jim hardy
My mistake! It is not good to say capacitive. Since it is referred to the generator delivered power if it is negative that means the generator is underexcitated and gets reactive power from the Grid[Infinite].
The transmission line reactive power it is :
If Is=23 A then line reactive power it is [X*IS^2]=130*23^2=68.77 kVAr.

jim mcnamara and jim hardy
My mistake! It is not good to say capacitive. Since it is referred to the generator delivered power if it is negative that means the generator is underexcitated and gets reactive power from the Grid[Infinite].
The transmission line reactive power it is :
If Is=23 A then line reactive power it is [X*IS^2]=130*23^2=68.77 kVAr.
View attachment 205900
Thanks a lot. That vector diagram makes makes sense :)