Confirmation of transformer efficiency

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SUMMARY

The discussion centers on confirming the efficiency calculation of a transformer, which was determined to be 96.7%. The calculations involved open circuit and short circuit tests, yielding parameters such as R_{G,S}, X_{M,S}, and equivalent resistance and reactance values. A critical error was identified in the voltage calculation during the full load condition, where the correct voltage should have been V_p/a=244.643e^{j0.29112} instead of 484.641e^{j0.146956}. This correction did not affect the overall efficiency percentage at 0.8 power factor.

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sandy.bridge
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Homework Statement


The problem at hand involved determining the equivalent circuit model of the transformer and then determining its efficiency. I for the life of me cannot get the results that he has provided us with, so perhaps someone here can either confirm my results or indeed confirm his.

Here are my steps:
For open circuit,
[tex]\theta_{OC}=arccos((186W)/((240V)(5.41A)))=81.7639°[/tex]

[tex]Y_E=(5.41A)/(240V)e^{-j81.76°}=0.003229-j0.02231[/tex]

[tex]R_{G,S}=1/0.003229=310Ω,X_{M,S}=1/0.02231=44.825Ω[/tex]

Short circuit test:

[tex]\theta_{SC}=arccos(617W/((48V)(20.8A)))=51.3°[tex] <br /> [tex]R_{eq,p}=1.42615Ω, X_{eq,p}=1.81426Ω[/tex]<br /> <br /> I put everything in terms of the LV side and hence [itex]R_{G,S},X_{M,S}[/itex] remain the same. The primary equivalents are divided by a factor of a=2400/240-100.<br /> <br /> Full load on secondary side at 0.8 PF:<br /> [tex]I_S=(50 000VA)/(240V)e^{-j36.9°}[/tex]<br /> <br /> hence<br /> [tex]V_p/a=(0.0142262+j0.0181426)(208.33e^{-j36.9°})+240V=484.641e^{j0.146956}[/tex]<br /> <br /> [tex]P_{Cu}=(208.33A)^2(0.0142262Ω)=617W[/tex]<br /> [tex]P_{Core}=(484.641V)^2/(309.7W)=758W[/tex]<br /> [tex]P_{Supplied}=(240V)(208.33A)cos(36.9°)=39984W[/tex]<br /> <br /> The efficiency turns out to be 96.7%.[/tex][/tex]
 

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How can there be a voltage reading if the transformer is short-circuited at the output? The problem makes no sense to me.
 
The voltage reading is not on the short circuit. This is a standard problem in the textbook we are using.
Short circuit comes with a voltage reading, current reading and a power reading.
Open circuit comes with a voltage reading, current reading and a power reading.

I already know where my mistake was which was at this point:
[tex]V_p/a=(0.0142262+j0.0181426)(208.33e^{-j36.9°})+240V=484.641e^{j0.146956}[/tex]
Somehow I managed to calculate this step wrong three times. The values are right, but the end result should have been:
[tex]V_p/a=(0.0142262+j0.0181426)(208.33e^{-j36.9°})+240V=244.643e^{j0.29112}[/tex]
Using this result I ended up with the same percentage at 0.8 PF.
 

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