• Support PF! Buy your school textbooks, materials and every day products Here!

I'm currently stuck on a transformer question and require a little clarification

  • Thread starter that.kid
  • Start date
  • #1
6
0

Homework Statement



A 50 kVA 6360 V/240 V, 50 Hz, transformer is tested on open circuit and short circuit to obtain its efficiency. The results of the tests are as follows:
Open Circuit: Primary Voltage 6360 V, Primary current 1.0 A, Input power 2 kW.
Short Circuit: Voltage across primary 280 V, Current in secondary 175 A, Input power 1kW

power factor at 0.8 lagging

The Attempt at a Solution



So far I've calculate the impedances using the open circuit and short circuit equations, and I'm having trouble getting the power equations.
I've used P(open circuit) = P(input)*I(open circuit)*cos(theta)
where cos(theta) = PF
I've used P(short circuit) = P(input)*I(short circuit)*cos(theta)

And then P(output)/(P(output) + P(open circuit)+P(short circuit)), and I'm getting an efficiency of ~61%. I'm pretty sure this is wrong.
 

Answers and Replies

  • #2
6
0
I think it's wrong because transformers should usually have an efficiency of ~90% and above. To be getting such an inefficient transformer with those values just doesn't seem right. Can anybody please confirm?
 
  • #3
NascentOxygen
Staff Emeritus
Science Advisor
9,243
1,068
I get η>90%. What did you calculate the impedances to be?
 
  • #4
6
0
From the open circuit test, I got Rc = 7950 ohm, and Xm = 10600 ohm

From the short circuit test, I got z = 1.6 angle 36.86 degrees using the formula z(s/c) = (short circuit voltage/short circuit current) angle theta

I found the angle (theta) from PF = cos theta

I'm not sure what I've done wrong from there, is it possible that you could show me your working out? I followed the notes on open circuit/short circuit test that my lecturer gave me, and I re-watched the video of him lecturing this particular topic, but I found that it hasn't helped much at all, and it's only gotten me more confused.
 
  • #5
6
0
I know that if I do

(50 * 0.8)/(50 * 0.8 +(2+1)) I get an efficiency of about 93%,

where the 50 is the 50 KVA input voltage, the 0.8 is the power factor, the 2 comes from the 2KW input on the open circuit test, and the 1 comes from the 1KW input on the short circuit test. I think I'm just confusing myself more by doing that though. I saw my lecturer do it this way, and I don't really understand why or what exactly it achieves. Electricity just doesn't like me :P
 
  • #6
NascentOxygen
Staff Emeritus
Science Advisor
9,243
1,068
From the short circuit test, I got z = 1.6 angle 36.86 degrees using the formula z(s/c) = (short circuit voltage/short circuit current) angle theta
Have you referred the S/C current back to the primary side for this calculation?
 
  • #7
6
0
Yes, I have referred it back to the primary and drawn an equivalent circuit. Referring it to the primary I got R(E)=1123.6 and X(Z)=674.16. What equation did you use to calculate the copper/core losses? I've tried P=I^R and P(open circuit)=P(input)*I(open circuit)*power factor and P(short circuit)=V(short circuit)*I(short circuit)*power factor using the values given/calculated for each.
 
  • #8
NascentOxygen
Staff Emeritus
Science Advisor
9,243
1,068
From From the short circuit test, I got z = 1.6 angle 36.86 degrees using the formula z(s/c) = (short circuit voltage/short circuit current) angle thetathe short circuit test, I got z = 1.6 angle 36.86 degrees using the formula z(s/c) = (short circuit voltage/short circuit current) angle theta
How do I relate that to this ?
Yes, I have referred it back to the primary and drawn an equivalent circuit. Referring it to the primary I got R(E)=1123.6 and X(Z)=674.16.
 

Related Threads for: I'm currently stuck on a transformer question and require a little clarification

  • Last Post
Replies
2
Views
2K
Replies
0
Views
1K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
3
Views
2K
Replies
2
Views
986
Replies
4
Views
856
Top