Therefore, R's=0.057/26.5^2=0.000078 Ω2% of 415 V is 8.3 V

[andyskin]

Homework Statement

A 415V to 11 kV transformer has a rating of 200 kVA. The winding resistance
and leakage reactance when referred to the primary are 0.014 Ω and 0.057 Ω
respectively.

(a) Determine the % regulation of the transformer at 0.8 power factor lagging.

(b) In designing a particular 415V to 11 kV, 200 kVA transformer, the primary
winding resistance is to be 10 mΩ. Find the maximum winding resistance of the
secondary winding if the transformer is to have 2% regulation at unity power
factor.

Relevant Equations

.

Primary winding voltage v1 = 415V
Secondary winding voltage v2 = 11kv
Rating of transformer is 200KVA
winding resistance Rieq = 0.14 Ω
winding leaking reactance Xieq = 0.057 Ω

A) % regulation at 0.8 power factor lagging

Primary winding current I1 = 200/415 = 482a
Power factor CosΦ = 0.8 lag
SinΦ = 0.6

% Voltage regulation = I1Rieq/V1 cosΦ + I1Xieq/V1 SinΦ
= (482x0.014/415)x 0.8 + (482x0.057/415)x0.6

= 0.013 + 0.038972
= 0.05272
= 5.272%

B) Primary winding resistance R1 = 10mΩ = 0.01Ω

Power factor = cos Φ = 1
sin Φ = 0

% Voltage regulation = I1Rieq/v1 cosΦ + I1Xieq/v1 sinΦ

0.02 = 482xRieq/415 x1 + 482xXieq/415 x 0

482x0.014/415 x1 + 482x0.057/415 x 0

= 0.0162

Where k is transformation ration

K = N2/N1 = V2/V1 = 100/415 = 26.5

0.0162 = 0.01 + r2/26.5√

r2/25.5√ = 0.0072

r2 = 5.0585 ΩNot sure if I'm correct , the course notes are not much good

 

[Babadag]

a) it is correct.

b) How much is 2% of 415 V?

Req=Rp+R's

R's=Rs/k^2 k=11/.415=26.5 indeed.