current transformation through delta-wye xmer

[trainee]

Here are the details of what I've been thinking about...

Primary side data:

-delta connection

-13.8kV

-impedance assumed negligible

Secondary side data:

-wye ground connection

-4160V

-impedance assumed negligible

Xmer data:

-let's say for argument's sake 5MVA

Load data:

-let's say a total of 3000hp load connected to secondary side....single bus system. from 1 switchgear to 1 mcc

-assume 0.9 PF

-assume 1.0 diversity factor

-------------

thought process...of trying to find current on primary side.

P=3000hp*746*1.0=2237kW (real power)

S=2237kW/0.9=2486kVA (apparent power)

Q= sqrt(2486^2 - 2237^2)= 1084kVAR(reactive power)

I (secondary)=2486kVA/4160V/sqrt3=345A (current on secondary)

transformation from wye secondary to delta primary:

use turns ratio...we know:

Vp/Vs=Is/Ip

13800V/4160V=345A/Ip(delta)

Answer: Ip(delta)=104A

Does that look OK to you?

Thank you.

 

[trainee]

Anyone be kind enough to give me a "yah" or a "nah" on this one?

i know it's basic but it'll be greatly appreciated.

 

[Wolverine]

OW!!! You sprained my eyeballs by making me look at this problem. I thought I passed and could just slum it from now on, never having to do math ever again. Okay, since the damage is done, here's my input:

A simpler solution to the problem:

You identified that 2486kVA3ph is the power supplied.

Well, that's true for either side of the transformer.

So simply solve the problem using the high side values.

S3ph=2486kVA

V3ph=13.8kV

2486kVA3ph/3 = 828kVA1ph

13.8kV3ph = 13.8kV1ph (for the delta)

S1ph = V1ph*I1ph or I=V/S

828k/13.8k = 60A per phase in the delta

You don't have to, but you can also look at the low side Y:

S1ph = 828kVA

V1ph = ( 4.16k/sqrt3) = 2401V (remember, sqrt3 for the Y-G)

I=S/V or 828k/2.4k = 345A per phase

Check my work for me, Biomudartbengrovpilot?

 

[Flyer_PE]

Given my understanding of the problem:

High Side Voltage = 13.8kVL-L

Load = 3000hp at 0.9pf.

There are no transformer losses.

The configuration of the transformer doesn't matter if all you are interested in is line current.

Wolverine is correct in that you don't even have to bother with line current on the 4kV side.

For the load: S=2487kVA

ILine-13.8=S/(sprt3*13.8kv) = 104 Amps

If I've over-simplified the problem, let me know. I'm typing this while watching Blue's Clues with the boy.

Jim

 

[trainee]

Hi Wolverine:

Thanks for bearing with the sprained eyeballs. get well soon Your calculation is making use of the fact that

S[SIZE=8pt]delta[/SIZE]=S[SIZE=8pt]wye[/SIZE]=sqrt3*V[SIZE=8pt]L[/SIZE]*I[SIZE=8pt]L[/SIZE] correct?

So the 60A you calculated on the delta side is a phase current...I[SIZE=8pt]line[/SIZE]=sqrt3*I[SIZE=8pt]phase[/SIZE]...which gives me I[SIZE=8pt]line[/SIZE]=104A (the solution I got and IFR_pilot got)

My question now, the method that I went about solving 104A...is that the same as the method IFR_Pilot used to solve 104A? I know I took some shortcuts but it looks the same to me and the principles are alike. But why did you solve for phase currents?

Hi IFR_pilot:

Thanks for your response. A lead-off question is....why do I not need to be concerned with line currents on the primary side? Don't I have to be conscious of conductor sizes, upstream protection, etc...

Thanks.

 

[Wolverine]

Oops. I solved for the per-phase current in the delta of the transformer. [note to self: always read the question]

I stand corrected:

Iline = sqrt3*Iphase for a delta winding.

IP = 60A

IL = 104A

Did I get that right?

 

[Flyer_PE]

Hi Wolverine:Thanks for bearing with the sprained eyeballs. get well soon Your calculation is making use of the fact that

S[SIZE=8pt]delta[/SIZE]=S[SIZE=8pt]wye[/SIZE]=sqrt3*V[SIZE=8pt]L[/SIZE]*I[SIZE=8pt]L[/SIZE] correct?

So the 60A you calculated on the delta side is a phase current...I[SIZE=8pt]line[/SIZE]=sqrt3*I[SIZE=8pt]phase[/SIZE]...which gives me I[SIZE=8pt]line[/SIZE]=104A (the solution I got and IFR_pilot got)

My question now, the method that I went about solving 104A...is that the same as the method IFR_Pilot used to solve 104A? I know I took some shortcuts but it looks the same to me and the principles are alike. But why did you solve for phase currents?

Hi IFR_pilot:

Thanks for your response. A lead-off question is....why do I not need to be concerned with line currents on the primary side? Don't I have to be conscious of conductor sizes, upstream protection, etc...

Thanks.

The short answer is that yes, in a "real world" installation, you will have to concern yourself with conductor sizes and protection schemes for both the supplied loads as well as the transformer itself. The way you phrased the question indicated to me that the only parameter you were interested in finding was the line current on the 13.8kV side of the transformer. Given that power in = power out for the theoretical no-loss transformer, the voltage level where the load is applied doesn't matter.

As an example, suppose the 3000hp at 0.9pf load is a 480 Volt load instead of a 4kV load. The transformer is now a 13kV:480V transformer.

The line current on the 13.8kV side will be 104 amps since power in = power out.

Since the power was a given for the problem, you can solve directly for the current on any voltage level you wish.

 

[trainee]

Thanks a lot guys. I really appreciated your help on this one!