- #1
Jason-Li
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1. 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.
Hi, I have read through other threads re. this question and can't seem to follow what was being discussed, this I my attempt would appreciate if someone could tell me if I'm going wrong.
For reference I use R'p & X'p to represent secondary reflected onto primary.
a)
Cos^-10.8= 36.86989765°
Voltage regulation = VA/V1^2*(R'p*cosΦ+X'p*sinΦ)
= 200000 / 415^2 *(0.014*0.8+0.057*sin 36.86989765)
= 0.0527217303
= 5.27%
b)
n=E1/E2 = 415/11000 = 0.03772727273
as cos^-1 *1 = 0
Reg. = VA/V1^2*(R'p*cosΦ+X'p*sinΦ)
0.02 = 200000/415^2*(R'p*1+X'p*0)
R'p = 0.02 / 1.161271592
R'p = 0.017222501
R'p = Rp + Rs * n^2
Rs = (R'p-Rp) / n^2
Rs = (0.0172225001-0.01) / 0.03772727273^2
Rs = 5.074Ω
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.
The Attempt at a Solution
Hi, I have read through other threads re. this question and can't seem to follow what was being discussed, this I my attempt would appreciate if someone could tell me if I'm going wrong.
For reference I use R'p & X'p to represent secondary reflected onto primary.
a)
Cos^-10.8= 36.86989765°
Voltage regulation = VA/V1^2*(R'p*cosΦ+X'p*sinΦ)
= 200000 / 415^2 *(0.014*0.8+0.057*sin 36.86989765)
= 0.0527217303
= 5.27%
b)
n=E1/E2 = 415/11000 = 0.03772727273
as cos^-1 *1 = 0
Reg. = VA/V1^2*(R'p*cosΦ+X'p*sinΦ)
0.02 = 200000/415^2*(R'p*1+X'p*0)
R'p = 0.02 / 1.161271592
R'p = 0.017222501
R'p = Rp + Rs * n^2
Rs = (R'p-Rp) / n^2
Rs = (0.0172225001-0.01) / 0.03772727273^2
Rs = 5.074Ω