# voltage regulation



## cabby (Sep 19, 2008)

This is out of Kaplan's Review, Prob 2.8. Can anyone shed some light? I keep going around in circles with this and never get the correct voltage regulation in the book. Would be nice to see in pu if anyone has it.

Problem states to find the voltage regulation at .8 PF lagging.

transformer-60hz, 50KVA, 2300/230, PF=.8 lagging

&gt;

&gt; open ckt test

&gt; I=6.5A

&gt; V=230V

&gt; W=187W

&gt;

&gt; short ckt test

&gt; I=21.7A

&gt; V=115V

&gt; W=570W

&gt;

thanks,

cabby


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## Flyer_PE (Sep 20, 2008)

Here's my shot at it:

For the SC test, the primary voltage is 115Vac.

115/2300=0.05 The transformer impedance is 5%.

On a pu basis, assuming you have primary voltage of 1 pu and a 5% impedance. Rated current of 1 pu will result in a secondary voltage of 0.95pu.

%Voltage regulation = 100*(VNL - VFL)/VFL

= 100*(1-0.95)/0.95 = 5.26%


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## mudpuppy (Sep 25, 2008)

I haven't had time to sit down and calculate a solution to this, but it's been in the back of my mind for a while. I like Flyer's solution and I think that is the extent of the difficulty you're likely to see on the actual exam. However, I'm not sure it is preciscely correct for the Kaplan problem. The thing that is bugging me is that question give the loss Watts and also asks for the voltage regulation at 0.8 power factor. I'm thinking this implies they want you to calculate the complex impedance of the transformer and then find the voltage regulation with the lagging current. But I'm not going to try to solve it at 4 am with insomnia. Any thoughts?

By the way, cabby, do you have a copy of Chapman's _Electric Machinery Fundamentals_? The book has a good explanation of calculating transformer parameters from SC and OC tests.


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## NVRSTOP (Sep 27, 2008)

Here's a bit more lengthy solution:

Transformer-60hz, 50KVA, 2300/230, PF=.8 lagging

Open ckt test

I = 5A

V = 230V

W = 187W

Short ckt test

I = 21.7A

V = 115V

W = 570W

First calculate the following:

Resistance referred to primary (Rp) - 570W/(21.7A)^2 = 1.2 Ohms

Impedance referred to Primary (Zp) - 115V/21.7A = 5.3 Ohms

Inductive Reactance referred to Primary (Xp) - sq rt [(5.3)^2 - (1.21)^2] = 5.2 Ohms

Rated Primary Current (Ip)- 50,000/2,300 = 21.7 A

E = sq rt [((2300 * 0.8) + (21.7 * 1.2))^2 + ((2300 * 0.6) + (21.7 * 5.2))^2] = 2,390

% VR = (Vnl - Vfl) / Vfl * 100 = (2390-2300) / 2300 * 100 = 3.91%

The first term (2300 * 0.8) is the in-phase voltage

then (21.7 * 1.2) is the IR drop, (2300 * 0.6) is the out of phase voltage (at 80% pf - sin theta = 0.6),

and then (21.7 * 5.2) is the IX drop.

Let me know if this is close to Kaplan's answer.

23 days left - time to bear down!

Good Luck


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