- #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 = 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.