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

In summary, the conversation discusses the testing of a 50 kVA, 6360 V/240 V transformer on open circuit and short circuit to determine its efficiency. The open circuit test resulted in a primary voltage of 6360 V, primary current of 1.0 A, and input power of 2 kW. The short circuit test showed a voltage of 280 V, current in the secondary of 175 A, and input power of 1 kW. The power factor was found to be 0.8 lagging. The conversation also involves calculations of impedances and power equations, with some confusion and discrepancies in the results.
  • #1
that.kid
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.
 
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  • #2
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
I get η>90%. What did you calculate the impedances to be?
 
  • #4
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
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
that.kid said:
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
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
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 to I'm currently stuck on a transformer question and require a little clarification

1. What is a transformer and how does it work?

A transformer is an electrical device that is used to transfer electrical energy from one circuit to another through electromagnetic induction. It consists of two or more coils of wire wound around a core, which is typically made of iron. When an alternating current flows through the primary coil, it creates a changing magnetic field that induces a current in the secondary coil, allowing for the transfer of energy.

2. What is the purpose of a transformer?

The main purpose of a transformer is to change the voltage level of an electrical circuit. It can step up or step down the input voltage, depending on the number of turns in the primary and secondary coils. This allows for efficient transmission of electricity over long distances and also allows for the use of different voltage levels in various electronic devices.

3. How is a transformer different from other electrical devices?

A transformer is different from other electrical devices because it does not contain any moving parts and does not generate or consume electricity. Instead, it relies on the principles of electromagnetic induction to transfer energy between circuits. It is also more efficient than other devices, with minimal energy losses during the transformation process.

4. What factors affect the performance of a transformer?

The performance of a transformer is affected by several factors, including the number of turns in the primary and secondary coils, the material of the core, the frequency of the input current, and the amount of load on the secondary circuit. The quality of the insulation between the coils and the core also plays a significant role in the performance of a transformer.

5. What are some common issues with transformers?

Some common issues with transformers include overheating, insulation breakdown, and short circuits. These can be caused by factors such as excessive load on the secondary circuit, poor quality insulation, or damage to the core. Regular maintenance and proper use can help prevent these issues and ensure the efficient performance of a transformer.

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