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

AI Thread Summary
The efficiency of a 10-kVA transformer operating at a 0.85 power factor lagging is calculated using the given test results for open circuit and short circuit conditions. The core loss is determined to be 450W, while the copper loss at full load is 600W. Total losses amount to 1.050 kW, combining both core and copper losses. The efficiency formula yields a result of 89% when substituting the output power and total losses. Thus, the transformer operates at an efficiency of 89%.
graygoh
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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.
 
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