Transformer -- find equivalent circuit from efficiency and V.R

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SUMMARY

This discussion focuses on calculating the equivalent circuit parameters of a 200 KVA, 4000/1000 V transformer, specifically addressing efficiency and voltage regulation. The transformer operates at an efficiency of 0.97 at both full load and 60% load with a unity power factor. The calculated copper loss (Pcu) at full load is 3868.98 VA, and the iron loss (Pi) is 2319.59 VA. The equivalent resistance referred to the low voltage side is determined to be 0.0967 ohm, closely matching the book's answer of 0.0961 ohm.

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jaus tail
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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 = 502 * 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:
E2 = V2 + (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 X2.
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.
 
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jaus tail said:
Full load Voltage regulation at 0.8 pf lag is 5
You can use the formula for approximate voltage regulation here.
Regulation= (IRcosΦ+IXsinΦ)/V2, where R and X are referred to lv and I is the secondary full load current.

(Note IR/V2= Rpu and IX/V2= Xpu).
 
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Got it. Thanks. I was using R as only R secondary and not the R equivalent.
 

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