The OC and SC test data are given below for a single phase, 5KVA, 200V/400V, 50Hz transformer

OC test from LV side | 200V | 1.25A | 150w |

SC test from HV side | 20V | 12.5A | 175w |

Determine the following :-

1) Draw the equivalent circuit of the transformer referred to LV side.

2) At what load or KVA the transformer is to be operated for maximum efficiency?

3) Calculate the value of maximum efficiency.

4) Regulation of the transformer at full load 0.8power factor lagging.

#### Solution

(1) Approximate equivalent circuit.

From OC test (meters are connected on LV side i.e. secondary)

`cosvarphi_0^'=W_0/(V_0I_0^')=150/(200xx1.25)=0.6 sinvarphi_0^'`

`I_W^'=I_0^'cosvarphi_0^'=1.25xx0.6=0.75A` `R_0^'=V_0/I_W^'=200/0.75`=266.6Ω

`I_mu^'=I_0^'sinvarphi_0^'`=1.25×0.8=1A `X_0^'=V_0/I_mu^'=200/1`=200Ω

From SC test (meters are connected on the HV side i.e. primary)

π_{ππΆ}=175π€ π_{ππΆ}=20π πΌ_{ππΆ}=12.5π΄

`Z(01)=V_(SC)/I_(SC)=20/12.5=1.6Omega` `R(01)=W_(SC)/I_(SC)^2=175/12.5^2=1.12Omega`

`X_(01)=sqrt(Z_(01)^2-R_(01)^2)=sqrt(1.6^2-1.12^2)=1.1426Omega`

`K=400/200=2`

`R_(02)=K^2R_(01)=2^2xx1.12=4.48Omega`

`X_(02)=K^2R_(01)=2^2xx1.1426=4.57Omega`

(2) Maximum efficiency and load at which it occurs.

π_{πΆπ}= πΌ_{2}^{2}π
_{02}

`I_2=(5xx1000)/400-12.5`

`W_(CU)=12.5^2xx4.48=700`

Load KVA = Full-load KVA×`sqrt(W_1/W_(cu))=5xxsqrt(150/700)=2.314KVA`

FOR MAXIMUM EFFICIENCY:-

π_{π}=π_{ππ’}=150π=0.150πΎπ

Pf =1

`%eta_(max)=("load KVA" xx "pf")/("load KVA" xx "pf"+W_i+W_i)xx100`

`%eta_(max)=(5xx1)/(5xx1+0.15+0.15)xx100`

`%eta_(max)=94.33%`

(3) Regulation of transform at full load 0.8power factor lag.

`% "regulation"=(I_2(R_(02)cosvarphi+X_(02)sinvarphi))/E_2xx100`

`% "regulation"=(12.5(4.48xx0.8+4.57xx0.6))/E_2xx100=19.76%`

`% "regulation"=19.76%`