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Archived Transformer physics help

  • Thread starter graygoh
  • Start date
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1. 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

2. Homework Equations

efficiency = output power / input power

3. The Attempt at a Solution
I totally don't know how to start. help pls. thanks
 

cnh1995

Homework Helper
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open circuit (LV winding supplied): 200V, 3.2A, 450W
This gives the core loss of the transformer.
Pi=450W
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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