Download Practical Problems

Example 3.6: A transformer has 600 primary turns and 150 secondary turns. The
primary and secondary resistances are 0.25Ω and 0.01Ω respectively and the
corresponding leakage reactances are 1.0Ω and 0.04Ω respectively. Determine (a) the
equivalent resistance referred to the primary winding, (b) the equivalent reactance
referred to the primary winding, (c) the equivalent impedance referred to the primary
winding, and (d) the phase angle of the impedance.
Solution:
(a) From equation (3.20), equivalent resistance
Req1  R1  2 R2
2
 600 
Req1  0.25  
 0.01 = 0.41 Ω
 150 
(b) X eq1  X1   2 X 2
2
 600 
Req1  1.0  
 0.04 = 1.64 Ω
 150 
(c) Z eq1  Req2 1  X eq2 1 =
(d) cos  
Req1
Z eq1

Hence   cos 1
0.412  1.64 2 = 1.69 Ω
0.41
1.69
0.41
 75.960
1.69
Example 3.6: A 15 KVA, 2200/110 V transformer has R1 = 1.75Ω, R2 = 0.0045 Ω
the leakage reactances are X1 = 2.6 Ω and X2 = 0.0075 Ω Calculate,
(a) equivalent resistance referred to primary.
(b) equivalent resistance referred to secondary.
(c) equivalent reactance referred to primary.
(d) equivalent reactance referred to secondary.
(e) equivalent impedance referred to primary.
(f) equivalent impedance referred to secondary.
(g) total copper loss.
Solution:
The given values are, R1 = 1.75 Ω, R2 = 0.0045Ω, X1 = 2.6 Ω, X2 = 0.0075 Ω
α = 110/2200 = 1/20 = 0.05
(a) Req1 = R1 + R2' = R1 + R2 / α 2 = 1.75 + 0.0045/0.052 = 3.55 Ω
(b) R eq2 = R2 + R1' = R2 + α 2 R1 =
= 0.0045 + (0.05)2 x 1.75 = 0.00887 Ω
(c) X eq1 = X1 + X2' = X1 + X2/ α 2 = 2.6 + 0.0075/(0.05)2 = 5.6 Ω
(d) X eq2 = X2 + X1' = X2 + α 2 X1
= 0.0075 + (0.05)2 x 2.6 = 0.014 Ω
(e) Zeq1 = Req1 + j Xeq1 = 3.55 + j 5.6 Ω
Zeq1 = √(3.552 + 5.62) = 6.6304 Ω
(f) Zeq2 = Req2 + j X eq2 = 0.00887 + j 0.014 Ω
Zeq2 = √(0.008872 + 0.014 2) = 0.01657 Ω
(g) To find the load copper loss, calculate full load current.
I1F.L. = (KVA x 1000)/V1 = (25 x 1000)/2200 = 11.3636 A
total copper loss = (I1F.L.)2 Req1 = (11.3636)2 x 355 = 458.4194 W
This can be checked as,
I1F.L.= (KVA x 1000)/V2 = (25 x 1000/110 = 227.272 A
total copper loss = I12 R1 + I22 R2
= (11.3636)2 x 1.75 + (227.373)2 x 0.0045
= 225.98 + 232.4365 = 458.419 W
Practical Problems
Solve the following problems:
1. A 6600/440 V single-phase 600 kVA transformer has 1200 primary turns. Find :
1.
2.
3.
4.
Transformation ratio ;
Secondary turns ;
Voltage/turn; and
Secondary current when it supplies a load of 400 kW at 0.8 power factor
lagging.
2. Find the primary and secondary turns of a 3300/300 V, single-phase, 50-H2, 30
kVA transformer if the flux in the core is to be about 0.06 Wb. Also determine the
primary and secondary currents if the losses are to be neglected.
3. The voltage/turn of a single-phase transformer is 1.1 V, when the primary winding
is connected to a 220 V, 50-Hz A.C. supply, the secondary voltage is found to be 600
V. Find : Primary and secondary turns ; and
Core area if the maximum flux density is 1.2 T
4. A 2000/200 V single-phase transformer gives 0.5 A and 40 W as ammeter and
wattmeter readings when supply is given to the low voltage winding and high voltage
winding is kept open. Find :
a. The magnetizing component,
b. The iron loss component, and
c. The power factor of no-load current,
8. Find: (i) active component and reactive components of no-load current and ;
(ii) no-load current of a 230 V/115 V single-phase transformer if the power
input on no-load to the high-voltage winding is 70 W and power factor of no-load
current is 0.25 lagging.
5. A single-phase transformer has 500 turns on the primary and 40 turns on the
secondary winding. The mean length of the magnetic path in the core is 150 cm and
the joints are equivalent to an air-gap of 0.1 mm. When a potential difference of 3000
V is applied to the primary, maximum flux density is 1.2 T. Calculate :
1.
2.
3.
4.
The cross-sectional are of the core,
No-load secondary voltage,
The no-load current drawn by the primary, and
Power factor on no-load.
Given that AT/cm for a flux density of 1.2 T in the iron to be 5, the corresponding
iron loss to be 2 W/kg at 50 Hz and density of iron as 7.8 g/cm3.
6. A 230 V/115 V single phase transformer takes a no-load current of 1.7 A at a
power factor of 0.18 lagging with low voltage winding kept open. If the low voltage
winding is now loaded to take a current of 13 A at 0.8 power factor lagging find the
current taken by high voltage winding.
7. A transformer has a primary winding of 800 turns and a secondary winding of 200
turns. When the load current on the secondary is 80 A at 0.8 power factor lagging, the
primary current is 25 A at 0.707 power factor lagging. Determine graphically or
otherwise the no-load current of the transformer and its phase with respect to the
voltage.
8. A 20 kVA, 2000/200 V single phase, 50 Hz transformer has a primary resistance
of 2.5 Ω and reactance of 4.8 Ω. The secondary resistance and reactance are 0.01 Ω
and 0.018 Ω respectively. Find
1.
2.
3.
4.
Equivalent resistance referred to primary
Equivalent impedance referred to primary
Equivalent resistance, reactance and impedance referred to secondary, and
Total copper loss of the transformer.
9. A 50 kVA, 4400/220 V transformer has R1 = 3.45 Ω, R2 = 0.009 Ω. The values of
reactances are X1 = 5.2 Ω and X2= 0.015 Ω. Calculate for the transformer :
1.
2.
3.
4.
5.
Equivalent resistance as referred to primary
Equivalent resistance as referred to secondary
Equivalent reactance as referred to both primary and secondary,
Equivalent impedance as referred to both primary and secondary, and
Total copper loss, first using individual resistances of the two windings and
secondly, using equivalent resistance as referred to each side.