Calculating the reactive power of a transmission line

In summary, the sending end is transferring 3.75MW of power to the generator bus. The sending end voltage angle is 0.01877 degrees which allows 3.75MW to be transferred. The reactive power sent by the sending end is **35.24kVAR** when the line resistance is not factored in, and **-114.77kVAR** when the line resistance is factored in.
  • #1
anon6912
21
3
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 :)
 
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  • #2
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
 
  • #3
Did you include the resistance in your calculation or no?
 
  • #4
@anon6912 -- is this for schoolwork? If so, I can move your thread to the schoolwork forums.
 
  • #5
berkeman said:
@anon6912 -- is this for schoolwork? If so, I can move your thread to the schoolwork forums.
Hey yeah, could you move it if you don't mind :)
Thanks
 
  • #6
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)
 
  • #7
Babadag said:
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 I am 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?
 
  • #8
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?"
 
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  • #9
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.
upload_2017-6-22_17-22-53.png
 
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  • #10
Babadag said:
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 :)
 

1. How is reactive power calculated in a transmission line?

Reactive power is calculated using the equation Q = V * I * sin(θ), where Q is the reactive power in volt-amperes reactive (VAR), V is the voltage in volts, I is the current in amperes, and θ is the phase angle difference between the voltage and current.

2. Why is reactive power important in a transmission line?

Reactive power is important because it represents the portion of power that is not used to perform work, but is instead used to maintain the magnetic field in the transmission line. This helps regulate voltage levels and ensure efficient power delivery.

3. How does reactive power affect the efficiency of a transmission line?

Reactive power can cause a phenomenon known as "reactive power loss," where excess power is needed to maintain the magnetic field in the transmission line. This can result in a decrease in the overall efficiency of the line, as more power is required to deliver the same amount of usable power.

4. What factors can affect the reactive power of a transmission line?

The reactive power of a transmission line can be affected by factors such as the length and type of the line, the amount of inductive and capacitive components in the line, and the amount of power being transmitted.

5. How can the reactive power of a transmission line be reduced?

The reactive power of a transmission line can be reduced by using devices such as capacitors and inductors to balance out the inductive and capacitive components, or by using power factor correction techniques to improve the power factor of the line. Proper design and maintenance of the line can also help reduce reactive power loss.

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