Question

A 50 KVA, 2200/220 V, 50 Hz transformer has
an iron loss of 300 Watts. The resistances of low and high
voltage windings are 0.005 ohm and 0.5 ohm respectively. If the
load power factor is 0.8 lagging, calculate its efficiency on full
load and half load.
​

Answer 1

Answer:

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Answer 2

Answer:

Explanation:-

Given:-

R1 =  0.005 ohm

R2=0.5 ohm

Wi = 300W

pf= 0.8

To find :-

its efficiency on full

load and half load.

Solution:

The full load primary and secondary and secondary current.

   $(l1)FL = \frac{kVA rating *1000}{E1} = \frac{50*1000}{2200} =22.72A\\(l2)FL = \frac{kVArating *1000}{E1} = \frac{50*1000}{220} = 227.272A$

1) Efficiency at full load 0.8 pf =

First , we will calculate the copper  lost at full load.

$[Wcu]FL = (l)^2 FL R1 +(l2)^2FLR2\\ = (22.72)^2 *0.005 + (227.272)^2*0.5\\ = 2461692.200992\\ = 246kW$

We know that efficiency at any load can be calculated as

%η

$\frac{(x *full -loadk VA*pf)}{(x*full-loadkVA*pf)+Wi+x^2[Wcu]FL} *100$

where

x = ratio of given load to full load

Wi = iron loss in kW

[Wcu]FL = copper loss at full load in kW

Now, efficiency at full load 0.8 pf is required

we have X= 1 full load kvA= = rated kVA = 50 , pF = 0.8 Wi= 0.142kW

Hence

% η = (0.05*50*0.8)/(0.05*50*0.8)+(300)+(50)²*246*100

= 3.25044kw

1) Efficiency at half full load 0.8 pf =

 $\frac{(x *full -loadk VA*pf)}{(x*full-loadkVA*pf)+Wi+x^2[Wcu]FL} *100$

= (0.5*50*0.8)/(0.5*50*0.8)+300+(50)²*246 *100

= 61500kw