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.