# Confirmation of transformer efficiency

1. Mar 7, 2013

### sandy.bridge

1. The problem statement, all variables and given/known data
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,
$$\theta_{OC}=arccos((186W)/((240V)(5.41A)))=81.7639°$$

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

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

Short circuit test:

$$\theta_{SC}=arccos(617W/((48V)(20.8A)))=51.3°[tex] [tex]R_{eq,p}=1.42615Ω, X_{eq,p}=1.81426Ω$$

I put everything in terms of the LV side and hence $R_{G,S},X_{M,S}$ remain the same. The primary equivalents are divided by a factor of a=2400/240-100.

Full load on secondary side at 0.8 PF:
$$I_S=(50 000VA)/(240V)e^{-j36.9°}$$

hence
$$V_p/a=(0.0142262+j0.0181426)(208.33e^{-j36.9°})+240V=484.641e^{j0.146956}$$

$$P_{Cu}=(208.33A)^2(0.0142262Ω)=617W$$
$$P_{Core}=(484.641V)^2/(309.7W)=758W$$
$$P_{Supplied}=(240V)(208.33A)cos(36.9°)=39984W$$

The efficiency turns out to be 96.7%.

#### Attached Files:

• ###### Soln Transf Example.pdf
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Last edited: Mar 7, 2013
2. Mar 8, 2013

### rude man

How can there be a voltage reading if the transformer is short-circuited at the output? The problem makes no sense to me.

3. Mar 8, 2013

### sandy.bridge

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