Question

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.

Answer 1

(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)
𝑊_𝑆𝐶=175𝑤 𝑉_𝑆𝐶=20𝑉 𝐼_𝑆𝐶=12.5𝐴

`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.

𝑊_𝐶𝑈= 𝐼₂²𝑅₀₂

`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:-
𝑊_𝑖=𝑊_𝑐𝑢=150𝑊=0.150𝐾𝑊
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%`