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Unformatted text preview: kVA ≡ kW
Fp := 0.83 Problem 5-11: Solution
Srated := 100⋅ kVA kVAR ≡ kW Vprated := 7.97⋅ kV Vsrated := 240⋅ volt Voc := 240⋅ V Ioc := 12⋅ A Poc := 1200⋅ W Vsc := 500⋅ V Isc := 12.5⋅ A Psc := 1500⋅ W Solution
The expected efficiency can be found by recognizing that the losses of the transformer at rated
voltage and current are given by the short and open circuit tests
Pcu := Psc copper loss at rated current Pcu = 1.5 kW Pcore := Poc core loss at rated voltage Pcore = 1.2 kW The power out can be found from the rated load and given power factor
Pout := Fp ⋅ Srated
η := Pout = 83 kW Pout
Pout + Pcu + Pcore η = 0.968 or η = 96.849 % ...
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University-West Lafayette.
- Spring '12