THREE-PHASE TRANSFORMER (Examples) Example 1 A 500-kVA, 3-phase, 50 Hz transformer has a voltage ratio (line voltages) of 33/11 kV and is delta/star connected. The resistances per phase are: high voltage 35 Ω, low voltage 0.876 Ω and the iron loss is 3050 W. Calculate the value of efficiency at full-load and one-half of full-load respectively (a) at unity pf and (b) 0.8 pf Solution Transformer ratio, a = 33,000/(11,000/√3) = 3√3 Per phase resistance referred to secondary side R02 = 35/(3√3)2 + 0.876 = 2.172 Ω Secondary phase current, Is-p = 500,000 / (√3 x11,000) = 26.24 A Full-load condition Full-load total Cu loss, Pcu = 3 (Is-p)2 x R02 = 3 x 26.242 x 2.172 = 4,490 W Iron loss, Pcore = 3,050 W Total full-load loss = 3.050 + 4.490 = 7,540 W Output Power at unity pf, Pout = 500 kVA x 1 = 500 kW Full load efficiency at unity pf, =
+
=
500 = 0.9854 500 + 7.54
.
%
=
400 = 0.9815 400 + 7.54
.
%
Output power at 0.8 pf, Pout = 500 kVA x 0.8 = 400 kW
Full load efficiency at 0.8 pf,
Half-load condition
=
+
Half-load output Power at unity pf, Pout = (1/2) x 500 kVA x 1 = 250 kW Half-load total Cu loss, Pcu = (1/2)2 x 4,490 = 1,222 W Total half-load loss = 3.050 + 1,222 = 4,172 W
Half-load efficiency at unity pf, =
+
=
250 = 0.9835 250 + 4.172
.
%
=
200 = 0.9796 200 + 4.172
.
%
Half-load output power at 0.8 pf, Pout = (1/2) x 500 kVA x 0.8 = 200 kW
Half-load efficiency at 0.8 pf, =
+
Example 2 A 100 kVA,3-phase, 50 Hz, 3,300/400 V transformer is ∆-connected on HV side and Yconnected on LV side. The resistance of the HV winding is 3.5 Ω per phase and that of the LV winding 0.02 Ω per phase. Calculate the iron losses of the transformer at normal voltage and frequency if its full-load efficiency be 95.8 % at 0.8 pf lagging. Solution Full-load output, Pout = 100 x 0.8 = 80kW Input power,
Pin = 80/0.958 = 83.5 kW
Total loss, Ploss = Pin – Pout = 83.5 – 80 = 3.5 kW Let us find full-load Cu losses for which purpose, we would first calculate R02. / /
=
=
/√
=
√
R02 = R2 + R1/a2 = 0.02 + 3.5/(33/√3)2 = 0.037 Ω Full-load secondary current =
√
= 144.1
Total Cu loss, Pcu = I22xR02 = 144.12 x 0.037 = 2,305 W Iron loss, Pcore = Ploss – Pcu = 3500 – 2305 = 1,195 W
Example 3 A 2,500 kVA,3-phase, 6,600/400 V, 3-phase transformer is Y-connected on HV side and ∆-connected on LV side. The test results are as follow: SC test:
HV side:
400 V 219 A and
24 kW
OC test:
LV side:
400 V 175 A and
18 kW
Determine i). The parameter equivalent circuit referred to high voltage (HV) side ii). The percentage of voltage regulation on full-load at power factor of 0.8 lagging iii). The efficiency of the transformer on full-load at power factor of 0.9 lagging. Solution i) From SC test data on HV side with Y-connection Primary voltage/phase = primary phase voltage = 400/√3 = 231 V Primary current/phase = primary phase current = 219 A Equivalent impedance/phase Z01 = 231 / 219 = 1.055 Ω Equivalent resistance/phase I12R01 = Psc / 3 = 24,000 / 3 (219)2 R01-= 8,000 or R01 = 0.167 Ω and equivalent reactance / phase =
−
=
1.055 − 0.167 = .
From OC test data on LV side with ∆-connected Secondary voltage/phase
= 400 V
Secondary phase current/phase = 175/√3 = 101 A Po = √3 VoL IoL cos φo with
IoL = secondary line current
and 18,000 = √3 x 400 x 175 cos φo cos φo = 0.148 and φo = 81.5o sin φo = 0.99 =
= =
= =
400 = 26.76 101 0.148
=
Transformer turn ratio
400 =4 101 0.99
/ /
=
/√
=
Core parameter referred to HV (primary)
= 9.53
Rc = a2Rc’ = (9.53)2 x 26.75 = 2430 Ω and
Xm = a2Xm’ = (9.53)2 x 4 = 363 Ω
ii) For pf or cos φ = 0.8 lag, sinφ = 0.6 Percentage voltage regulation %
=
%
=
%
=
+
100
219x0.167x0.8 + 219x1.042x0.6 6600 √3
166.18 3810
100% = .
100%
%
iii) Full-load primary line current can be found from √3 x 6,600 x I1 = 2,500,000 I1 = 219 A It shows that SC test has been carried out under full load condition. Therefore, PcuFL = 24 kW
Total loss = Ploss = Pcore + Pcu = 18 + 24 = 42 kW
Full-load output at pf =0.9 is Pout = 2,500 x 0.9 = 2,250 kW The full-load efficiency
Note
=
+
=
2,250 = 0.9817 2,250 + 42
.
%
Please TRY for ∆ / Y connection of example 3.
