Efficiency Calculation of 10-kVA Transformer at 0.85 Power Factor Lagging

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SUMMARY

The efficiency of a 10-kVA transformer operating at a 0.85 power factor lagging is calculated to be 89%. The transformer was tested under open circuit conditions, yielding a core loss of 450W, and under short circuit conditions, resulting in a copper loss of 600W. The total losses amount to 1.050 kW, which is used to determine the efficiency using the formula η = Output power / Input power.

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Homework Statement


A 10-kVA, 200:400 V transformer gave these test results:
open circuit (LV winding supplied): 200V, 3.2A, 450W
short circuit (HV winding supplied) : 38 V, 25A, 600W
calculate the efficiency when the transformer delivers its rated kVA at 0.85 power factor lagging

Homework Equations



efficiency = output power / input power

The Attempt at a Solution


I totally don't know how to start. help pls. thanks
 
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graygoh said:
open circuit (LV winding supplied): 200V, 3.2A, 450W
This gives the core loss of the transformer.
Pi=450W
graygoh said:
short circuit (HV winding supplied) : 38 V, 25A, 600W
This gives the copper loss of the transformer at full load.
Pcu=600W
Iron loss is fixed loss and doesn't depend on the load. Copper loss varies proportional to the square of the current. In general, copper loss=x2Pcu, where x is the fraction of full load with which the transformer is loaded. Since the transformer is operated at full load, x=1 and the losses will be,
P=Pi+Pcu
=450+600=1.050kW.
Efficiency η= Output power/input power
=Output kW/(Output kW+losses)
∴ η=10×0.85/(10×0.85+1.050)
∴ η=0.89
The transformer is 89% efficient.
 
Last edited:

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