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The following results are obtained on a 50 KVA; 1000/200V 50Hz transformers. OC test: 200V, 1.2 A, 90 W L.S SC test: 50 V; 5A and 110 W H.V Determine deficiency as half load at 0.8 pf lagging
written 8.3 years ago by gravatar for teamques10 teamques10 ★ 68k •   modified 4.6 years ago
basic electrical engineering
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written 8.3 years ago by gravatar for teamques10 teamques10 ★ 68k

From OC test (meters are connected on secondary side)

W1= 90 W

W1= 0.09 KW

From SC test (meters are connected on H.V side)

WSC = 110 W

Full load current=

KVA rating *1000/V1

∴ I1= 3 * 1000/1000= 5A

WCU= WSC = 110 W = 0.11 KW

Efficiency at half load and 0.8 pf lagging

X= 0.5

Pf= 0.8

$Ƞ = x* full-load KVA * pf/x* full-load KVA*pf + wt+ x2WCu \\ = (0.6*6*0.8/ 0.5 * 5 * 0.8* 0.09 + (0.5)2 +0.11 )* 100 \\ = 94.45 \%$
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