# Calculate the power loss in electrical cable

## Main Question or Discussion Point

Hi

I'm trying to calculate the power loss over an electrical cable.

The information of the system is:
Maximum current of 20.92 A per unit at 34.5kV

The cable has the following properties:
DC Resistance at 25 deg C (0.1672 ohms/1000 ft)
AC Resistance at 25 deg C (0.1672 ohms/1000ft)
The length of the cable is 47370.1ft

The technical report paper I got the calculation from gives the maximum losses in the cable as 10398.8W

So far I haven't been able to get the same answer using P=I^2*R where R is the impedance of the cable R=sqrt(0.1672^2+0.1672^2)=0.2365 ohms/1000ft then (0.2365/1000)*47370.1=11.2 ohms as the impedance of the length of the cable. P=I^2*R=4902.1W

Could someone please tell me where I've gone wrong here?

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PhanthomJay
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Hi

I'm trying to calculate the power loss over an electrical cable.

The information of the system is:
Maximum current of 20.92 A per unit at 34.5kV

The cable has the following properties:
DC Resistance at 25 deg C (0.1672 ohms/1000 ft)
AC Resistance at 25 deg C (0.1672 ohms/1000ft)
The length of the cable is 47370.1ft

The technical report paper I got the calculation from gives the maximum losses in the cable as 10398.8W

So far I haven't been able to get the same answer using P=I^2*R where R is the impedance of the cable R=sqrt(0.1672^2+0.1672^2)=0.2365 ohms/1000ft then (0.2365/1000)*47370.1=11.2 ohms as the impedance of the length of the cable. P=I^2*R=4902.1W

Could someone please tell me where I've gone wrong here?

Looks like you left out some important information about whether this is single phase or a 3 phase circuit. Is it an AC or DC voltage?

I think the problem will be that you take the forward and return resistance as parallel forward resistance. But it's series instead.

You already have the resistance, you have to multiply it with the length: 0.1672*47.370 => 7.92Ohm for full length.
For the forward line it gives 3466W loss. If the return path is similar, it'll be doubled: 6932W

The paper most likely calculates with a three phase system, where the loss will be tripled: 3*3466 = 10398W

Last edited:
Xeno1221
Apologies it's a three phase circuit with AC voltage

PhanthomJay