A 5 KVA, 400/200, 50 Hz single phase transformers gave the follow results during open and short circuits test. - O.C test: 400 V, 1 A, 60 W (H.V.) - S.C test: 15 V, 12.5 A, 50 W (H.V.) - Calculate No laod parameters Ro and Xo Equivalent resistance and reactance referred to high voltage side Regulation at full load and 0.8 pf lagging Iron and copper losses at full load Efficiency at half load 0.8 pt lagging -

Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering

Marks: 10 M

Year: May 2012

Given:

KVA = 5

V1= 400 V

V2= 200 V, K=0.5

F= 50

Wo = 60W

Io = 1 A

VSC = 15 V, ISC = 12.5 Amp, WSC = 50 W

No load parameters Ro and Xo:

From the OC test results, we get,

Cos ф= Wo / V1Io = 60/ 400 * 1= 0.15

∴ фo = sin (cos-1 (фo)

∴ Ro = V1/ Iocos фo = 400/ 110.15 = 2.667 π

∴ Vo =V1/Iosin фo = 400/ 1* 0.9886 = 404.61 Ω

Equivalent resistance and reactance

From SC test

$R1T = WSC/ISC = 50/ (12.5)2 = 0.32 Ω \\ Z1T = VSC/ISC = 15/12.5 = 1.2 Ω \\ ∴ \sqrt{(I2 – R2)}= [1.22- 0.322] ½ \\ X1T = 1.16-2$

Regulation of full load and 0.8 pf lagging.

Cos ф = 0.8

Therefore sin ф = 0.6

$ƪ Reg = [I1R1T cos ф + I1 X1Tsin ф]/ V1 * 100 \\ = [(12.5 * 0.32*0.8) + (12.5 * 1.16 * 0.6)]/400 * 100 \\ = 2.975 ƪ$

Iron and copper losses art full load

Iron losses P1 = Wo = 60W

Full load current = 5 * 103 / 400 = 12.5 A

∴ I1(FL) = ISC

∴ PCU(FL) = WSC = 50 W

Efficiency at half load and cos ф = 0.8

$ƪ ƞ HL = 0.5 *KVA * 103 * cos ф/ (0.5) *KVA * 103 * cos ф) + P1 + Pu (HL) * 100$

But $PCU (HL) = ¼ * PCU(HL) \\ = ¼ * 50 = 12.5 W \\ ƪ ƞ HL= [0.5 * 5000 * 0.8]/ [(0.5 * 5000 * 0.8) + 60+ 12.5] \\ = 96.5 ƪ$