Calculate the power loss in electrical cable

AI Thread Summary
The discussion revolves around calculating power loss in an electrical cable with a maximum current of 20.92 A at 34.5 kV. The user initially calculated power loss using the formula P=I^2*R but arrived at a lower figure of 4902.1W, while a technical report indicated losses of 10398.8W. Key feedback highlighted the need to clarify whether the system is single-phase or three-phase, noting that the resistance should be treated as series rather than parallel. For a three-phase AC circuit, the resistance should be multiplied by three to account for losses across all phases. The correct approach leads to a total loss of 10398W, aligning with the technical report's findings.
Xeno1221
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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?

Thanks in advance
 
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Xeno1221 said:
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?

Thanks in advance
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
 
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Apologies it's a three phase circuit with AC voltage
 
Xeno1221 said:
Apologies it's a three phase circuit with AC voltage
Ok. Resistance is either AC or DC, not both. For a balanced AC 3 phase circuit with the same conductor on each phase, each conductor of each phase carries the same current
 

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