- #1

jaus tail

- 615

- 48

## Homework Statement

For 200 KVA, 4000/1000 V transformer. 1 phase.

Efficiency is 0.97 at full load and 60% load at unity power factor.

No load pf = 0.25

Full load Voltage regulation at 0.8 pf lag is 5%

Find transformer ckt parameters for referred to lv side

## Homework Equations

efficiency is output / (output + copper loss + iron loss)

cu loss varies as square of current

or as square of percent load

voltage regulation is (|E2| - |V2|)/|V2|

## The Attempt at a Solution

Well efficiency is 0.97 at both times:

full load and 60% load. both times unity pf.

so

.97 = output/(output + losses_

0.97 = 200000/(200000 + Pi + Pcu)

and

for 60% load

0.97 = 200000*0.6/(200000*0.6 + Pi + .36Pcu)

I got Pcu at full load = 3868.98 VA

Pi = 2319.59 VA

Full load current at unity pf = 200000/4000 = 50A

So Pcu = 50

^{2}* R(eq)

So R(eq) = 1.545 on HV Side

R

_{(2)}= 0.7725 ohm

Referring to LV side by multiplying by (4000/1000)

^{2}I get 0.0967 ohm. (This matches book answer of 0.0961)

I don't know how to find out X parameter of leakage impedance.

The formula for voltage regulation is

(|E| - |V||)/|V| = 0.05 (given 5% regulation at 0.8 pf lag and full load)

V is 4000 V

So I get |E| = 4200 V.

Now:

E

_{2}= V

_{2}+ (IZ) --- All vectors

V2 = 4000 V and is reference so angle is 0

I is 200000/4000 = 50A at angle 36.87 lagging V2 (given pf is 0.8). ---> I'm not sure whether I should be 50 or 50/0.8 A as pf is 0.8 at full load.

Z = R2 + jX2

R2 = 0.7725ohm.

I have to find X2.

So

4200 (at unknown angle) = 4000 + 50(lagging angle is 36.87) * Z(also some angle)

I don't know how to proceed to find X

_{2}.

Do I have to split the above equation in imaginary and real terms?

Book has used per unit method, but I want to try without per unit method to see if it works.