9.14. Transformer Efficiency. The efficiency of a transformer at a particular land and power factor is defined as the ratio of power output to power input .
It may be noted that the efficiency is based on power output in watts and not in volt-amperes, although losses are proportional to volt-amperes. Hence at any volt-ampere load, the efficiency depends on power factor, being maximum at unity power factor.
Efficiency can be calculated by determining core losses from open-circuit test and copper losses from short-circuit test.
Condition for maximum efficiency:
Iron losses, Pi = hysteresis loss + eddy current loss
=Ph + Pe
Copper losses, Pc = I12R01 or I22R02
Considering primary side:
Input to primary = V1I1 cos ɸ1
By proper design it is possible to make the maximum efficiency occur at any desired load.
Variation of efficiency with power factor. We know that transformers efficiency,
The variations of efficiency with power factor at different loading on a typical transformer are shown in Fig. 41.
Fig. 41. Variations of efficiency with power factor at different loadings.
Transformer on Load
Example 10. A 230 VI 115 V single-phase transformer takes a no-load current of 2 A at a power factor of 0.2 lagging with low voltage winding kept open. If the low voltage winding is now loaded to take a current of 15 A at 0.8 power factor lagging find the current taken by high voltage winding.
Solutlon. Primary, E1 = V1 = 230 V
Secondary, E2 = V2 = 115 V
No-load current 10 = 2 A
No-load power factor, cos ɸ0 = 0.2 or ɸ0 = 78.46° or 78° 29′
Load power factor, cos ɸ2 = 0.8 or ɸ2 = 36.9° or 36° 54′
Current taken by h.v, winding, I1:
Now, transformation ratio,
= 9.09 A
Hence, current taken by h.v., winding
I1 = 9.09 A. (Ans.)
Example 11. The number of turns on the primary and secondary windings of a transformer are 1000 and 200 respectively. When the load current on the secondary is 100 A at 0.8 power factor lagging, the primary current is 30 A at 0.707 power factor lagging. Determine the no-load current of the transformer and its phase with respect to the voltage.
Solution. Primary turns N1 = 1000
Secondary turns, N2 = 200
Secondary current I2 = 100
Power factor, cos ɸ2 = 0.8 or ɸ2 = 36.9° or 36°54’
Primary current, I1 = 30 A
Power factor, cos ɸ1 = 0.707 or ɸ1 = 45°
No-load current, I0, ɸ0:
Refer Fig. 43. I1 is the vector sum of I0 and I2’. Let I0 lag behind V1 by an angle ɸ0.
Resolving currents into their X and Y components, we get
I0 cos ɸ0 + 20 cos 36.9° = 30 cos 45°
I0 cos ɸ0 = 20 cos 45° – 20 cos 36.9°
= 21.21 – 16 = 5.21 A (i)
I0 sin ɸ0 + 20 sin 36.9° = 30 sin 45°
I0 sin ɸ0 = 20 sin 45° – 20 sin 36.9°
= 21.21 – 12 = 9.21 A (ii)
From (i) and (ii), we get
Hence, no-load current = 10.58 A
and ɸ0 = 60.5°. (Ans.)
Example 12.A 30 kVA, 2000/200V, single-phase, 50 Hz transformer has a primary resistance of 3.5
5 Ω. The secondary resistance and reactance are 0.015 Ω and 0.02 Ωrespectively. Find:
(i) Equivalent resistance, reactance and impedance referred to primary.
(ii) Equivalent resistance, reactance and impedance referred to secondary.
(iii) Total copper loss of the transformer.
Solution. Primary resistance, R1 = 3.5 Ω
Primary reactance, X1 = 4.5 Ω
Secondary resistance, R2 = 0.015 Ω
Secondary reactance, X2 = 0.02 Ω
Total copper loss = I22 R02 = 1502 × 0.05 = 1125 W. (Ans.)
Example 13. A single phase transformer has the following data:
Turn ratio 20 : 1 ; R1= 20Ω
X1 = 80 Ω
R2 = 0.04 Ω
X2 = 0.2 Ω
No-load current = 1.2 A leading the flux by 300.
The secondary delivers 180 A at a terminal voltage of 400 V and at a power factor of 0.8 lagging.
Determine by the aid of a vector diagram:
(i) The primary applied voltage.
(ii) The primary power factor
(iii) The efficiency.
Solution. Refer Fig. 44:
(i) Primary applied voltage, V1 :
Taking V2 as the reference vector
(ii) Primary power factor, cos ɸ1 :
Phase angle between V1 and I1,
ɸ1= – =
Primary power factor = =7.303 (lag). (Ans.)
No-load primary input power
= V1I1 cos ɸ0 = 9269 × 1.2 × cos = 5561.4 W
Total copper losses as referred to secondary
= I22R02 = (180)2 × 0.09 = 2916 w
Output = V2I2 cos ɸ2 = 400 × 180 × 0.8 = 57600 W
Total losses = 5561.4 + 2916 = 8477.4 W
Input = Output + Losses = 57600 + 8477.4 = 66077.4 W
EXAMPLE 14. The high voltage and low voltage windings of a 2200/220 V single-phase 50 Hz transformer has resistances of 4.8 Ω and 0.04 Ω and reactance 2 Ω and 0.018 Ω respectively. The low voltage winding is connected to a load having an impedance of (6+j4) Ω. Determine:
(i) Current in l.v. winding,
(ii) Current in h.v. winding
(iii) Load voltage, and
(iv) Power consumed by the load.
Solution. Primary resistance. R1 = 4.8 Ω
Primary reactance, X1 = 2 Ω
Secondary resistance, R2 = 0.04 Ω
Secondary reactance, X2 = 0.018 Ω
Impedance of load, ZL = (6+j4)
Example 15. A single phase transformer has Z1=1.4 + j5.2 Ω and Z2=0.0117+j0.0465 Ω. The input voltage is 6600 V and the turn ratio is 10.6:1. The secondary feeds a load which draws 300 A at 0.8 power factor lagging. Find the secondary terminal voltage and the kW output. Neglect no-load current I0.
Solution. Impedance Z1 = 1.4 + j5.2 Ω
Impedance Z2 = 0.0117 + j0.0465 Ω
Input voltage V1 = 6600 V
Turn ratio, K = 10.6 :1
Secondary and load current, I2 = 300 A
Power factor, cos ɸ2 = 0.8
Secondary terminal voltage, V2 :
Example 16. The full load copper loss on h.v. side of 100 kVA, 11000/317 V 1-phase transformer is 0.62 kW and on the l.v. side is 0.48 kW.
(i) Calculate R1, R2 and R2’ in ohms:
(ii) The total reactance is 4 percent, find X1, X2 and X2’ in ohms if the reactance is divided in the same proportion as resistance.
Solution. Given : Rated kVA = 100
V1 = 11000 V
V2 = 317 V
loss on h.v. side = 0.62 kW
loss of l.v. side = 0.48 kW
total reactance = 4%
(i) R1, R2 and R2’ :
Hysteresis and Eddy Current Losses
Example 17. A 230 V, 2.5 kVA single-phase transformer has an iron loss of 100 W at 40 Hz and 70 W at 30 Hz. Find the hysteresis and eddy current losses at 50 Hz.
Solution. Iron loss at 40 Hz = 110 W
Iron loss at 30 Hz = 75 W
Hysteresis and eddy current loss at 50 Hz :
We know that hysteresis loss,
Example 18. When a transformer is supplied at 400 V, 50 Hz the hysteresis loss is found be 310 W and eddy current loss is found to be 260 W. Determine the hysteresis loss and eddy current loss when the transformer is supplied at 800 V, 100-Hz.
Solution. Hysteresis loss at 400 V, 50 Hz = 310 W
Eddy current loss at 400 V, 50-Hz = 260 W
We know that,
Transformer Tests (O.C. and S.C.) and Equivalent Circuit
Example 19. A 50 Hz, single phase transformer has a turn ratio of 5. The resistance are 0.8 Ω, 0.02 Ω and reactance are 4 Ω and 0.12 Ω for high-voltage and low-voltage windings respectively. Find:
(i) The voltage to be applied to the h.v. side to obtain full-load current of 180 A in the l.v., winding on short circuit.
(ii) The power factor on short circuit.
Draw the equivalent circuit and vector diagram.
Equivalent circuit and vector diagram as shown in Figs. 45, 46 and 47,
Example 20. A 12 kVA 4000/400 V transformer has primary and secondary winding resistance of 13 Ω and 0.15 Ωand leakage reactance of 20 Ωand 0.25 Ω respectively. The magnetising reactance is 6000 Ω and the resistance representing core loss is 12000 Ω. Determine :
(i) Equivalent resistance and reactance as referred to primary.
(ii) Input current with secondary terminals open circuited.
(iii) Input current when the secondary load current is 30 A at 0.8 power factor lagging.
Example 21. Obtain the approximate equivalent circuit of a given 200/2000 V single-phase 30 kVA transformer having the following test results:
O.C. test : 200 V, 6.2 A, 360 W on l.v. side
S.C.test : 75 V, 18 A, 600 W on h.v. side
Solution. O.C. test (l.v.side)
Example 22. The following readings were obtained on O.C. and S.C. tests on a 200/400 V, 50 Hz single-phase transformer.
O.C. test (l.v. side) : 200 V, 0.6 A, 60 W
S.C. test (h.v.side) : 15 V, 9 A, 80 W
Calculate the secondary voltage when delivering 4 kW at 0.8 power factor lagging, the primary voltage being 200 V.
Solution. O.C.test – l.v. side (Instruments in the primary side and secondary open)
Primary voltage, V1 = 200 V
No-load current, I0 = 0.6 A
No-load loss, P0 = 60 W
Now, P0 = V1I0 cos ɸ0
60 = 200 × 0.6 × cos ɸ0
Hence, secondary voltage = 380 V. (Ans.)
Approximate equivalent circuit is shown in Fig. 49.
Example 23. A 4 kVA 400/200 V, 50 Hz single phase transformer has the following test data :
O.C.test (l.v. side) 200 V, 1 A, 64 W
S.C. test (h.v. side) 15 V, 10 A, 80 W
(i) Equivalent circuit referred to l.v. side, and
(ii) Secondary load voltage on full loat at 0.8 power factor lagging
Solution (i) O.C. test – l.v. side :
Example 24. A single-phase step-down transformer has a turn ratio of 3. The resistance and reactance of the primary winding are 1.2Ω and 6 Ω and those of the secondary winding are 0.05 Ω and 0.03 Ω respectively. If the h.v. winding is supplied at 230 V, 50 Hz with l.v. winding short circuited, find :
(i) Current in the l.v. winding
(ii) Coppor loss in the transformer, and
(iii) Power factor.
Example 25. A single, phase, 3 kVA, 230/115 V, 50 Hz transformer has the following constants :
Resistance : Primary 0.3 Ω, secondary 0.09 Ω
Reactance : Primary 0.4 Ω, secondary 0.1 Ω
Resistance of equivalent exciting circuit referred to primary, R0 = 600 Ω
Reactance of equivalent exciting circuit referred to primary, X0 = 200 Ω
What would be the reading of the instruments when the transformer is connected for :
(i) O.C. test,
In both the tests supply is given to h.v. side.
Ammeter reading=13.04 A. (Ans.)
Wattmeter reading=112.2 W.(Ans.)
Regulation and Efficiency of a Transformer
Example 26. A 4kVA 220/440 V, 50 Hz, single-phase transformer gave the following test figures. No load test performed on 220 V side keeping 440 V side open : 220 V, 0.7 A, 60 W. Short circuit test performed short circuiting the 440 V side through an ammeter : 9 V, 6A, 21.6 W calculate.
(a) The magnetising current and the component corresponding to iron loss at normal voltage and frequency.
(b) The efficiency at full-load at unity power factor and the corresponding secondary terminal voltage.
(c) Draw the phasor diagram corresponding to the full load operation.
(PTU, May, 2000)
Solution. Open circuit test on 220 V (l.v.) side : V1 = 220 V
P0 = 60 W
I0 = 0.7 A
Short circuit test on 440 V (h.v.) side : VSC = 9 V
ISC = 6 A
PSC = 21.6 W
Example 27. The parameters of the equivalent circuit of a 100 kVA, 2000/200 volt single-phase transformer are as follows :
Primary resistance = 0.2 ohms ;
Secondary resistance = 2 milli ohms.
Primary leakage reactance = 0.45 ohms ;
Secondary leakage reactance = 4.5 milli ohms.
Core loss resistance = 10 kilo ohms ;
Magnetizing reactance = 1.55 kilo ohms.
Using the circuit referred to primary, determine the :
(i) Voltage regulation.
(ii) Efficiency of the transformer operating at rated load with 0.8 lagging power factor.
(PTU, June and Dec, 2000)
Solution. Given : Rating 100 kVA
R1 = 0.2 Ω
R2 = 0.002 Ω
X1 = 0.45 Ω
X2 = 0.0045 Ω
R0 = 10000 Ω
X0 = 1550 Ω
(i) Voltage regulation:
Example 28. In a 25 kVA, 2000/200 V transformer, the constant and variable losses are 350 W and 400 W respectively. Calculate the efficiency on u.p.f. at
(i) Full load, and
(ii) Half full load.
Solution. Rating of transformer = 25 kVA
Constant of iron losses, Pi = 350 W or 0.35 kW
Variable or copper losses, Pc = 400 W or 0.4 kW
(i) At full load:
Example 29. The following readings were obtained from O.C. and S.C. tests on 8 kVA 400/120 V, 50 Hz transformer,
O.C. test (l.v. side) : 120 V ; 4 A ; 75 W.
S.C.test (h.v. side) : 9.5 V ; 20 A ; 110 W.
(i) The equivalent circuit (approximate) constants,
(ii) Voltage regulation and efficiency for 0.8 lagging power factor load, and
(iii) The efficiency at half full-load and 0.8 power factor load.
Example 30. Consider a 20 kVA, 2200/220 V, 50 Hz transformer. The OC/SC test results are as follows :
O.C. Test : 2200 V, 4.5 A, 148 W (l.v. side)
S.C. Test : 86 V, 10.5 A, 360 W (h.v. side)
(i) Determine the regulations at 0.8 p.f. lagging at full load.
(ii) What is the p.f. on short-circuit?
(Nagpur University, 1998)
Example 31. High voltage side short circuit test data for 20 kVA, 2300/230 V transformer are :
Power = 250 watts
Current = 8.7 A
Voltage = 50 V
Calculate equivalent impedance, resistance, reactance referred to h.v. side. Find the transformer regulation at 0.7 lagging power factor.
Solution. Given: h.v. side :
Example 32. A 20 kVA 2300/230 V, 50 Hz, distribution transformer has h.v. winding resistance 3.96 Ω and leakage resistance of 15.8 Ω. The l.v. winding has corresponding value of 0.0396 Ω and 0.158 Ω respectively. The transformer has a core loss of 58 W under normal operation conditions. Find :
(i) Load terminal voltage when transformer delivers rated current at 0.8 p.f. lagging to a load on l.v. side, with h.v. side voltage held at rated value and compute efficiency at this load.
(ii) The h.v. side voltage necessary to maintain rated voltage at load terminals when the transformer is loaded as above. Is efficiency same as in (i)?
(AMIE Summer, 1990)
Solution. Given :
R1 = 3.96 Ω
X1 = 15.8 Ω
R2 = 0.0396 Ω
X2 = 0.158 Ω
Pi = 58 W
p.f. = 0.8 lagging
(i) Load terminal voltage :
Example 33. The following test results were obtained for a 1000/100 V, 100 kVA single-phase transformer :
O.C. test :
Primary volts = 1000, secondary volts = 100, watts in primary = 950
S.C. test :
Primary volts for full-load current = 20, watts in primary = 1000
Determine the regulation and efficiency of the transformer at full-load and at 0.8 power factor logging.
Example 34. The primary and secondary winding resistance of a 30 kVA, 6600/250 V single-phase transformer are 8 Ω and 0.015 Ω respectively. The equivalent leakage reactance as referred to the primary winding is 30 Ω. Find the full-load regulation for load power factors of:
(iii) 0.8 leading
Example 35. Short-circuit test is conducted on a 5 kVA, 400 V/100 V single phase transformer with 100 V winding shorted. The input voltage at full load current is 40 V. The wattmeter, on the input reads 250 W. Find the power factor for which regulation at full load is zero.
Solution. Given : Rating of transformer : 5 kVA, 400 V /100 V
Input power, on short-circuit, PSC = 250 W
The input voltage at full load current, VSC = 40 V
Power factor for which regulation at full load is zero :
Example 36. A 50 MVA, 76.2 V/33 kV, 1-phase, 50 Hz, two-winding transformer with tap changer has percentage impedance of 0.5+j7.0. What tapping must be used to maintain rated voltage at the secondary on
(i) full load at 0.8 lagging power factor, and
(ii) 40 MVA load at 0.6 lagging power factor.
Assume that the top changer is provided on the h.v. side.
Solution. Given : Rated MVA = 50
f = 50 Hz
E1 = 76.2 V
E2 = 33 kV
Percentage impedance = 0.5 + j7.0.
Note: If the tap changer is on the primary side, tap setting will be down and if it is to be provided on secondary side, tap setting will be up for raising the secondary voltage on load.
Example 37. The percentage resistance and reactance of a transformer are 2% and 4% respectively. Find the approximate regulation on full-load at:
(i) Unity power factor,
(ii) 0.8 power factor lagging, and
(iii) 0.8 power factor leading.
Solution. Percentage resistance = 2%
Percentage reactance = 4%
Approximate % regulation = % resistance × cos ɸ % reactance × sin ɸ
(i) Unity power factor :
Approximate % regulation = % resistance cos ɸ
=2 × 1 = 2%. (Ans.)
(ii) 0.8 power factor lagging :
Approximate % regulation = % resistance × cos ɸ + % reactance × sin ɸ
= 2 × 0.8 + 4 × 0.6 = 1.6 + 2.4 = 4%. (Ans.)
(iii) 0.8 power factor leading :
Approximate % regulation = % resistance × cos ɸ – % reactance × sin ɸ
= 2 × 0.8 – 4 × 0.6 = 1.6 – 2.4 = -0.8%. (Ans.)
Example 38. A single phase 80 kVA, 2000/200 V, 50 Hz transformer has impedance drop of 8% and resistance drop of 4% :
(i) Find the regulation at full-load 0.8 power factor lagging.
(ii) At what power factor is the regulation zero.
Solution. Impedance drop = 8%
Resistance drop = 4%
Example 39. The high voltage of a single-phase 200 kVA 4400/220 V transformer takes a current of 35 A and power of 1250 W at 80 V when the low voltage winding is short-circuited. Determine:
(i) The voltage to be applied to the high voltage winding on full-load at 0.8 power factor lagging if the full load secondary terminal voltage is to be kept at 230 V.
(ii) % regulation.
Example 40. The high voltage and low voltage windings of a 50 kVA, 4400/220 V, 50 Hz transformer have resistances of 2.2 Ω and 0.005 Ω respectively. The full-load current is obtained when 160 V at 50 Hz is applied to h.v. winding with l.v. winding short circuited. Find :
(i) The equivalent resistance and reactance of the transformer to h.v. side and
(ii) Reactance of each winding.
Assume that the ratio of resistance to reactance is the same for each winding and the full-load efficiency of transformer is 0.98.
Example 41. The following test results were obtained in a 250/500 V transformer :
O.C. test (l.v.side) : 250 V, 1 A, 80 W
S.C. test (l.v. winding short-circuited) : 20 V, 12 A, 100 W
(i) The circuit constants.
(ii) The applied voltage and efficiency when the output is 10 A at 500 V and 0.8 power factor lagging.
Example 42. A 25-kVA, 2200/220 V, 50-Hz distribution transformer is tested for efficiency and regulations as follows:
O.C. test (l.v. side) : 22 V, 4 A, 150 W.
S.C. test (h.v. side) : 90 V, 10 A, 350 W.
(i) Core loss,
(ii) Equivalent resistance referred to primary,
(iii) Equivalent reactance referred to secondary
(iv) Equivalent reactance referred to primary
(v) Equivalent reactance referred to secondary
(vi) Regulation of transformer at 0.8 power factor lagging current, and
Efficiency at full-load and half-load at 0.8 power factor lagging current.
Example 43. Two similar 100 kVA, single-phase transformers gave the following test readings when tested by Sumpner’s test.
Supply power = 2.4 kW
Power supplied to secondary circuit in passing full-load current through it = 3.2 kW
Find the efficiency and regulation of each transformer at unity power factor.
Solution. Core loss (or iron loss) of the transformers
= 2.4 kW
Iron loss of each transformer,
Example 44. A 10 kVA, 2500/250 V, single phase transformer gave the following test results :
Open circuit test : 250 V, 0.8 A, 50 W
Short circuit test : 60 V, 3 A, 45 W
(i) Calculate the efficiency of half full-load at 0.8 p.f.
(ii) Calculate the load kVA at which maximum efficiency occurs and also the maximum efficiency at 0.8 p.f.
(iii) Compute the voltage regulation at 0.8 p.f. leading.
(AMIE Winter, 1998)
Example 45. A 50 kVA single-phase transformer has a full-load primary current of 250 A and total resistance referred to primary is 0.006 ohm. If the iron loss amounts to 200 W, find the efficiency on full-load and half-load at :
(i) Unity power factor, and
(ii) 0.8 power factor.
Solution. Full load primary current,
I1 = 250 A
Example 46. A 40 kVA, single-phase transformer has an iron loss of 300 W and full-load copper loss of 600 W,
(i) Find the load at which maximum efficiency occurs and the value of maximum efficiency at unity power factor.
(ii) If the maximum efficiency occurs at 90% of full-load, fund the new core loss and full-load copper loss assuming that total full-load loss is a constant.
Solution. Rating of transformer = 40 kVA
Example 47. The primary and secondary resistance of a 1100/220 V transformer are 0.3 Ω and 0.02 respectively. If iron loss amounts to 260 W determine the secondary current at which maximum efficiency occurs and find the maximum efficiency at 0.8 power factor.