Help with Voltage losses question

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(edited 9 years ago)

Hi, I need help with this question;
If the output from a generator with output line voltage of 25kV is 500 MVA and it is feeding onto a 200km long, 4 wire (Squab) cable circuit via a step up transformer (Ratio 1:16) calculate the voltage loss in the cable run.

If anyone could guide me in the right direction with where to start with this, it would be helpful.

Thanks

Not enough information to answer.

You need to post the whole question.

Reply 2

dc2209

Original post by uberteknik

Not enough information to answer.

You need to post the whole question.

That's the whole question. I think I have to assume that the generator like a lot of generators, has an output of 25kV and then the output of the transformer is 400kV. Then use the ratio to calculate a voltage loss maybe?

Reply 3

dc2209

Original post by uberteknik

Not enough information to answer.

You need to post the whole question.

I'm thinking there is no voltage loss in the cable run but this doesn't make sense?

Like I said, assuming the output voltage from the generator is 25kV and the output from the step up transformer is 400kV, this is 1:16 - which the transformer is rated. So does this mean there is no voltage loss in the cable run?

Reply 4

dc2209

by the way, the table may be relevant but it is part of a different question. http://gyazo.com/863ec916d04ca83341ec2f1a1d9feacb

Thanks

Reply 5

uberteknik

Original post by dc2209

by the way, the table may be relevant but it is part of a different question. http://gyazo.com/863ec916d04ca83341ec2f1a1d9feacb

Thanks

That makes a lot more sense.

Do you know what the transmission configuration is? i.e. how the squabs are wired?
Live and neutral meaning 2 squabs parallel? 3-phase? etc.

I have assumed 2 squabs used for live, 2 squabs used for return. If you know the transmission configuration you will need to modify accordingly:

Aluminium alloy cable with steel core Al/St or ALSC

Cable length = 200 km = 200x10³ metres

Cable resistance is 0.113 ohms/km (single squab)

Squab resistance = 200 x 0.113 = 22.6 ohms

2 squab parallel resistance = 22.6 || 22.6 = 11.3 ohms

Maximum current per squab = 765A

Generator output = 25kV @ 500MVA

Output current = 500x10⁶/25x10³ = 20 kAmps

Step up transformer ration = 1:16

Line voltage = 25x10³ x 16 = 400 kV

Line current = 20x10³/16 = 1250 Amps (two squab = 1530 Amps max current so this is within specification for the cable).

Voltage drop per run = I*R = 1250x11.3 = 14125 Volts

Out and return voltage drop = 2x14125 = 28250 Volts.

i.e voltage at step down transformer = 400kV - 28250 = 371750 Volts.

16:1 step down ratio

Output voltage from step down transformer = 371750/16 = 23.234 kV

Power loss = I²R = (1250)² x 22.6 = 35.3125 MW

(edited 9 years ago)

Reply 6

dc2209

Original post by uberteknik

All I know is that it is a 3-phase, 4 wire cable circuit.

Your calculations make sense to me. Thank you.

Your final answer, would that be divided by 4 to find the voltage loss in each of the cables since it is a 4 wire circuit?

Thanks

Reply 7

uberteknik

Original post by dc2209

All I know is that it is a 3-phase, 4 wire cable circuit.

In which case the 4 squab will be star connected (3-phase + return)

So the resistance calculation will mean each squab is 22.6 ohms

My example calculation assumed a purely resistive load with no power factor correction etc.

Volt drops are dependent on resistance AND reactance so unless you know the impedance characteristics of the load (i.e. is it balanced pf corrected etc) the volt drops and power losses will not be correct.

Also for a 3-phase system:

Power (MVA) = sqrt(3)*V*I

VI = 500x10⁶/1.73 = 288.68x⁶ Watts

I = 288.68x10⁶/25x10³ = 11547 Amps available from the generator not 20kA as previously stated.

Reply 8

dc2209

Yes, it would be star connected. So are there 4 squabs? I don't understand. If there are 4 then:

4 squab parallel resistance = 22.6 || 22.6 || 22.6 || 22.6 = 5.65 ohms?

I see what you're saying, but nothing more is given. The question is very badly worded :angry:

If O/P current from the generator = 11547 amps then divided by 16 this gives 721.69. If there are 4 squabs this would mean a max current of 3060A?

Thank You...