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8 kVA 400/120 V tests, half-load efficiency

Short answer

Referred to HV side: Ro=2133 Ω, Xo=337.5 Ω, R01=0.275 Ω, X01=0.387 Ω. At half load and 0.8 pf lagging, efficiency = 96.9%.

The following test results were obtained for open circuit and short circuit tests on a 8 kVA, 400/120 V, 50 Hz transformer: Open-circuit Test (LV Side): 120 V, 4 A, 75 W Short-circuit Test (HV Side): 9.5 V, 20 A, 110 W Calculate the equivalent circuit parameters referred to high voltage side. Also calculate the efficiency at half full load and 0.8 power factor lagging load.

Given: 8 kVA, 400/120 V, 50 Hz. OC test (LV): 120 V, 4 A, 75 W. SC test (HV): 9.5 V, 20 A, 110 W. HV side is 400 V; K = 120/400 = 0.3.

Shunt branch

On the LV side:

cos φ0 = 75/(120 × 4) = 0.15625
Iw = 4 × 0.15625 = 0.625 A ;  Iμ = √(4² − 0.625²) = 3.951 A
Ro = 120/0.625 = 192 Ω ;  Xo = 120/3.951 = 30.37 Ω

Refer to the HV side (× 1/K² = 11.11):

Ro(HV) = 192 × 11.11 = 2133 Ω
Xo(HV) = 30.37 × 11.11 = 337.5 Ω

Series branch (SC test is on the HV side)

Z01 = 9.5/20 = 0.475 Ω
R01 = 110/20² = 0.275 Ω
X01 = √(0.475² − 0.275²) = 0.387 Ω

Equivalent circuit (HV side)

        R=0.275Ω    X=0.387Ω
 o──┬────┬───/\/\/───────(((────────o
    │    │                          
 2133Ω  337.5Ω                  load
    │    │                       V2'
 o──┴────┴───────────────────────────o

Efficiency at half load, 0.8 pf lagging

Rated HV current = 8000/400 = 20 A, the same as the SC test current, so the full-load copper loss is 110 W.

Output   = 0.5 × 8000 × 0.8 = 3200 W
Iron loss = 75 W
Cu loss  = 0.5² × 110 = 27.5 W
η = 3200/(3200 + 75 + 27.5) = 96.9 %

Answer: Ro = 2133 Ω, Xo = 337.5 Ω, R01 = 0.275 Ω, X01 = 0.387 Ω (HV side); η at half load, 0.8 pf = 96.9 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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