Transformer calculation. Help!

[Krist Midbrod]

I have a question that I need some clarification on. Its in Mike Holt's exam prep book based on 2011 NEC. Unit 12 page 402

Question:
a 4160 to 240 v, three phase transformer is connected delta/ delta. the secondary PHASE current is 300A and the secondary LINE voltage is 240V. How much current will be flowing in the primary conductors supplying the transformer (Line current)?

Can someone please tell me why the answer book says 17A. This is a mistake right?????? Please check my work below.



So naturally... delta phase I x 1.73 = Line I 300A x 1.732= 519.6A Secondary Line current

Find VA 240v x 1.732 x 519.6A = 21,5987VA

Find primary line current 21,5987VA / 4160V x 1.732 = 29.9A Line current

Primary phase current is 17.3A NOT LINE CURRENT right????? Please tell me Im not still messed up on this concept!

 

[wsbeih]

I have a question that I need some clarification on. Its in Mike Holt's exam prep book based on 2011 NEC. Unit 12 page 402

Question:
a 4160 to 240 v, three phase transformer is connected delta/ delta. the secondary PHASE current is 300A and the secondary LINE voltage is 240V. How much current will be flowing in the primary conductors supplying the transformer (Line current)?

Can someone please tell me why the answer book says 17A. This is a mistake right?????? Please check my work below.



So naturally... delta phase I x 1.73 = Line I 300A x 1.732= 519.6A Secondary Line current

Find VA 240v x 1.732 x 519.6A = 21,5987VA

Find primary line current 21,5987VA / 4160V x 1.732 = 29.9A Line current

Primary phase current is 17.3A NOT LINE CURRENT right????? Please tell me Im not still messed up on this concept!

You are multiplying by SQRT (3) twice. Think of it this way. KVA primary = KVA secondary.

 

[GoldDigger]

When you are given a primary and secondary voltage and then told that the secondary voltage is a line voltage are we justified in assuming that the primary voltage is line too.
And just what is the line voltage in a delta?

Anyway, the ratio of the voltages must be the inverse of the current ratio. And one of the line currents is given.

 

[augie47]

The way I see it the ratio 17.33:1
You have the secondary current of 300 and applying the ratio you find a primary current of 17 amps.
If you add a 3 phase factor the number come out the same.

 

[charlie b]

I agree with the OP's answer of 29.9. This presumes that the original problem was correctly described. I understand the problem to have given the following:

• Line-to-line voltage on the primary side is 4160.
• Line-to-line voltage on the secondary side is 240.
• Current in the secondary phases of the transformer (not the current in the secondary conductors leading away from the transformer) is 300.

The OP's calculation method is correct, though it took a bit more work than was necessary. Here is a quicker way to the same answer:

• The transformer's turns ratio is 4160/240, or 17.3.
• The transformer's primary current will be lower than its secondary current by this same factor of 17.3.
• Therefore, the primary phase current is 300/17.3, or 17.3.
• The line current in the primary feeders will be this value of 17.3 amps times the square root of 3.
• 17.3 x 1.732 is 29.9.

 

[charlie b]

You are multiplying by SQRT (3) twice.

We always do. For example, given a delta-wye transformer with secondary line-to-neutral voltage of 120 volts, and a line current of 50 amps, calculate the load's VA.
Answer:

• Secondary line-to-line voltage is 120 times the square root of 3, or 208V.
• Power is that same 208V times 50 amps times the square root of 3.

You see that we multiplied by the square root of 3 twice. It does not matter whether we are dealing with a delta or wye secondary. The calculations use line-to-line voltages and line currents.

 

[augie47]

charlieb:
You are much sharper than I so I should just accept your answer but I tried another approach and still came up with 17.3. Please correct my thinking .
A 300 amp load at 240v 3 phase would be 124.704 kw { 240 X 1.732 X 300/1000 }
124.704 kw load at 4160 3 phase = 17.3 {124704/4160 X 1.732)

 

[wsbeih]

We always do. For example, given a delta-wye transformer with secondary line-to-neutral voltage of 120 volts, and a line current of 50 amps, calculate the load's VA.
Answer:

• Secondary line-to-line voltage is 120 times the square root of 3, or 208V.
• Power is that same 208V times 50 amps times the square root of 3.

You see that we multiplied by the square root of 3 twice. It does not matter whether we are dealing with a delta or wye secondary. The calculations use line-to-line voltages and line currents.

I respectfully disagree, you contradicts yourself. 240 is the L-L voltage. and the primary is also 4160V L-L. your calculation is correct if the given voltage is 240 L-N. Always approach this problem as KVA primary = KVA secondary (this requires you get the V-LL at primary and secondary). As a result, the line current can be obtained
.

 

[charlie b]

I respectfully disagree, you contradicts yourself. 240 is the L-L voltage. and the primary is also 4160V L-L. your calculation is correct if the given voltage is 240 L-N. Always approach this problem as KVA primary = KVA secondary (this requires you get the V-LL at primary and secondary). As a result, the line current can be obtained.

I did not contradict myself. The L-L voltage is 240, so that is the right number to use in the power calculation. But the current value to use in the power calculation is the line current. For a delta secondary, the line current is 1.732 times the phase current. So we multiple the 300 amps times 1.732 to get the line current, then we get power from 1.732 times line current times L-L voltage. That brings the 1.732 into the math problem a second time.

 

[charlie b]

Please correct my thinking .

Where you went off course, Gus, is using the value of 300 for the current. See my post just above this one for details.

 

[ggunn]

We always do. For example, given a delta-wye transformer with secondary line-to-neutral voltage of 120 volts, and a line current of 50 amps, calculate the load's VA.
Answer:

• Secondary line-to-line voltage is 120 times the square root of 3, or 208V.
• Power is that same 208V times 50 amps times the square root of 3.

You see that we multiplied by the square root of 3 twice. It does not matter whether we are dealing with a delta or wye secondary. The calculations use line-to-line voltages and line currents.

Or more simply put, line power is three times phase power.

 

[MD84]

I agree with the OP.

An interesting twist would be if I changed one detail in the question.

Question:
a 4160 to 240 v, three phase transformer is connected delta/ wye. the secondary PHASE current is 300A and the secondary LINE voltage is 240V. How much current will be flowing in the primary conductors supplying the transformer (Line current)?

 

[don_resqcapt19]

Where you went off course, Gus, is using the value of 300 for the current. See my post just above this one for details.

I don't understand the difference between phase and line currents on a delta system.

 

[Ingenieur]

I don't understand the difference between phase and line currents on a delta system.

Phase is through one leg/branch of the delta load (source)
line is the feed/supply to one corner
line = sqrt3 x phase

the line divides into 2 currents, one per leg/branch
not quite 2 when phase is factored but 1.732

 

[MD84]

I don't understand the difference between phase and line currents on a delta system.

Phase current will be the current through one winding. Line current will be the current through the lines attached to the winding nodes. Line current is equal to the vector sum of the two adjoining phase windings or phase current times the square root of three.

It is similar to voltage properties of a wye connection.

 

[Krist Midbrod]

I agree with the OP's answer of 29.9. This presumes that the original problem was correctly described. I understand the problem to have given the following:

• Line-to-line voltage on the primary side is 4160.
• Line-to-line voltage on the secondary side is 240.
• Current in the secondary phases of the transformer (not the current in the secondary conductors leading away from the transformer) is 300.

The OP's calculation method is correct, though it took a bit more work than was necessary. Here is a quicker way to the same answer:

• The transformer's turns ratio is 4160/240, or 17.3.
• The transformer's primary current will be lower than its secondary current by this same factor of 17.3.
• Therefore, the primary phase current is 300/17.3, or 17.3.
• The line current in the primary feeders will be this value of 17.3 amps times the square root of 3.
• 17.3 x 1.732 is 29.9.

I agree with you here and thank you for the confirmation. The question is quoted correctly and the answer given is wrong.... The key to this question is that it states the 300A as PHASE current in a delta transformer, there for you must multiply by 1.732 to get the LINE current.... Also I apreciate the short cut to the same answer using the turns ratio. I was aware of the but wanted to write out all of the details for discussion.... Thanks again!

 

[Krist Midbrod]

I agree with the OP.

An interesting twist would be if I changed one detail in the question.

Question:
a 4160 to 240 v, three phase transformer is connected delta/ wye. the secondary PHASE current is 300A and the secondary LINE voltage is 240V. How much current will be flowing in the primary conductors supplying the transformer (Line current)?

17.3A Line current in the primary

9.9 Phase current in the primary

300A Line current in secondary

300A Phase current in secondary

 

[MD84]

17.3A Line current in the primary

9.9 Phase current in the primary

300A Line current in secondary

300A Phase current in secondary

:thumbsup:

 

[don_resqcapt19]

17.3A Line current in the primary

9.9 Phase current in the primary

300A Line current in secondary

300A Phase current in secondary

And that goes to my question of the difference between line and phase current in a delta system. I understand it to be the same in a delta system as stated in this post.

 

[MD84]

Hi Don. Sorry to muddle the topic. This answer was in reply to a delta/wye connection.

If you notice the delta primary has different phase/line currents. The wye secondary has the same phase and line currents.

Delta:
I line = I phase x sqrt3
V line = V phase

Wye:
I line = I phase
V line = V phase x sqrt3