Transformer hypothetical question

[mickys86]

A transformer is rated at 10kw but is only delivering 9.8kw. Copper losses are 0.33kw and iron losses are 0.2kw. Calculate the efficiency of the transformer
A 88.6%
B 98%
C 90.4%
D 99%

I make the answer 91.33%

What am I missing??

 

[telectrix]

i read it as a trick question. the 9.8kW takes into account the copper loss and the iron loss. answer 98%

 

[topquark]

^^^^ ditto.

 

[mickys86]

Hmmmm the answer is actually c. Apparently. According to the Swindon massive site

 

[topquark]

Guess it depends what they mean by delivering. I asssumed that was to the point of usage and not the output of the transformer

 

[mickys86]

Either way i think they're wrong haha

 

[Silly Sausage]

According to my Hughes Electrical book,

eff=output power/(output power+losses) = 9.8/(9.8+0.53) = 94.87%

or more accurately...

eff= (input power-losses)/input power = (10-0.53)/10 = 94.7%

so I don't know.

Where's Tony?

 

[mickys86]

Putting this to the top of the thread..... Is mr Tony online yet???

 

[Moses]

According to my Hughes Electrical book,

eff=output power/(output power+losses) = 9.8/(9.8+0.53) = 94.87%

or more accurately...

eff= (input power-losses)/input power = (10-0.53)/10 = 94.7%

so I don't know.

Where's Tony?

Wrong values/answers in the question!!!

 

[mickys86]

Wrong values/answers in the question!!!

So what do ya think it is??

 

[Moses]

So what do ya think it is??

I think the copper losses have been incorrectly listed as 0.33KW, they should be 0.83KW

 

[MarkieSparkie]

According to my Hughes Electrical book,

eff=output power/(output power+losses) = 9.8/(9.8+0.53) = 94.87%

or more accurately...

eff= (input power-losses)/input power = (10-0.53)/10 = 94.7%

so I don't know.

Where's Tony?

The two equations are derived from first principles and are equivalent ways of calculating the efficiency of a transformer, neither is more accurate than the other.
The first answer is in error because the copper lossses are proportional to the load VI[SUP]2[/SUP], the copper loss at 9.8kW is therefore [0.33kW/(10/9.8)[SUP]2[/SUP]]= 0.317kW, the iron losses are approximately constant over the working load range, giving total losses at 9.8kW of 0.517kW.
So plugging these into equation 1:
eff = [9.8/(9.8+0.517)] x 100/1
eff = 95.0% (to 1DP)
The second answer is in error because the rated power 10kW is not the input power at 9.8kW.
From above the Input Power = Output Power + Total Losses
Input Power = 9.8kW + 0.517kW
Input Power = 10.317kW
Checking using equation 2
eff = (10.317-0.517)/10.317
eff= 95.0% (to 1DP) as was expected from equation 1
The above workings are based on the data provided in the OPs original question and as can be seen does not yield any of the multiple choice answers.
To achieve answer c) 90.4% the total losses would obviously need to be higher.
Total Losses = [(Output Power/eff)- output power]
Total Losses = [(9.8/0.904)-9.8]kW
Total Losses = 1.04kW
If we assume iron losses in the original question were correct at 0.2kW
Copper Losses = Total Losses-Iron Losses
Copper Losses = 1.04-0.2
Copper Losses = 0.84kW
Suggesting that the Copper Losses in the original question of 0.33kW might be a transcription error of the intended 0.83kW which would yield a rounded answer of c) 90.4% efficiency as proposed by Moses.

 

[RobWansbeck]

MarkieSparkie, I think you meant to say the copper loss @10kW = ... 0.344kW.
Moses has the best guess, the copper loss should be 0.83kW.
This gives 90.38% at 10kW.