20 kVA 2200/220 V tests, efficiency

Given: 20 kVA, 2200/220 V. OC (LV side): 220 V, 1.1 A, 125 W. SC (HV side): 52.7 V, 8.4 A, 287 W. Primary = HV side (2200 V); K = 220/2200 = 0.1.

(i) Parameters referred to the primary

Shunt branch (the OC test is on the LV side, so transfer it to the HV side with 1/K² = 100):

cos φ0 = 125/(220 × 1.1) = 0.5165
Iw = 1.1 × 0.5165 = 0.568 A ;  Iμ = √(1.1² − 0.568²) = 0.942 A
On LV:  Ro = 220/0.568 = 387.2 Ω ;  Xo = 220/0.942 = 233.6 Ω
On HV:  Ro = 387.2 × 100 = 38 720 Ω ;  Xo = 233.6 × 100 = 23 357 Ω

Series branch (the SC test is on the HV side, so the values are used directly):

Z01 = 52.7/8.4 = 6.274 Ω ;  R01 = 287/8.4² = 4.067 Ω
X01 = √(6.274² − 4.067²) = 4.777 Ω

Equivalent circuit (referred to primary)

 o──┬──────┬──────/\/\/────(((────────o
    │      │     4.067 Ω   4.777 Ω
  38 720  23 357                  load
    Ω      Ω                      V2'
 o──┴──────┴────────────────────────o

This is the approximate equivalent circuit: the shunt branch is moved to the input terminals, and the series branch follows it.

(ii) Maximum efficiency

Rated HV current I1 = 20 000/2200 = 9.091 A
Full-load Cu loss  = 287 × (9.091/8.4)² = 336.2 W
Iron loss Pi = 125 W
x (max η) = √(Pi/Pcu) = √(125/336.2) = 0.610
Output at max η, unity pf = 0.610 × 20 000 = 12 196 W
η_max = 12 196/(12 196 + 2 × 125) = 97.99 %

(Full-load efficiency at unity pf: 20 000/(20 000 + 125 + 336.2) = 97.75 %.)

(iii) Half load, 0.8 lagging

Output = 0.5 × 20 000 × 0.8 = 8000 W
Cu loss = 0.5² × 336.2 = 84.0 W
η = 8000/(8000 + 125 + 84.0) = 97.45 %

Answer: (i) Ro = 38.72 kΩ, Xo = 23.36 kΩ, R01 = 4.07 Ω, X01 = 4.78 Ω; (ii) η_max ≈ 98.0 % (at unity pf, about 61 % load); (iii) η ≈ 97.45 %.