A transformer:

-50KVA
-11000/220V 50Hz
-NO load 0.3Amps
-No Load losses = 650W

determine;
a)
i)iron loss current. = 59.1mA
ii)magnetizing current = 294mA
iii)no load pf. = 0.201

b) a s.c. test gives Zeq =125Ω ref. to primary
and full load Power losses = 850W

determine:

i) full load sec current = 227.275Amps
ii)efficiency with 70% full load and 0.8 pf lagging = 97.41% (not sure)
iii) the regulation for the condition of b ii)

My problem is that I'm not sure if i should consider Inot(no load component) in part a) to workout b ii) and where to apply the 0.8 pf lag.

 

I used the Power loss at full load and worked it out at 70% load to find the regulation at b ii) to be 0.9885

Can anyone confirm it for me pls?

 

The worst case efficiency would occur at rated load of 50kVA.

The stated full load loss is 850 Watts.

So the full load efficiency would be 100* (50,000-850)/50,000 expressed as a %age - i.e 98.3%. So your answer of 97.41% at 70% of rated load is not consistent.

I did some rough calcs as follows.

Primary side Full load current = 50,000/11,000=4.545A

The primary series resistance loss is approximately the full load loss minus the no load loss - i.e. 850W-650W=200W

The series resistance is then Rps=200W/(4.545^2)=9.68Ω

The stated primary series impedance is Zeq=125Ω (complex)

Hence Xps=√(125^2-9.68^2)=124.6Ω

At 70% rated load and 0.8pf the current will be 0.7*4.545 or 3.182A @ -36.87° [2.546-j1.91].

The primary series power loss would then be 3.182^2*9.68=98W.

Assume the iron losses are about the same @ 650W.

Total losses at 70% rated power would be 748W.

Giving an efficiency of 100*(50,000-748)/50,000=98.5%

The primary voltage drop at 70% rated load and 0.8pf would be

(9.68+j124.6)*(2.546-j1.91)=262.5+j298.7=397.7 @48.68° Volts

The regulation would then be

100*(11,000-397.7@46.68°)/11,000=97.65%.

So I also get different answer on that score.