Sub: DC Machines and Transformer (2130904) Active Learning Assignment Topic: O.C & S.C Test, Sumpner or back to back Test, Condition for maximum efficiency, All day Efficiency

1. Sub: DC Machines and Transformer (2130904)
Active Learning Assignment
Topic: O.C & S.C Test, Sumpner or back to back Test,
Condition for maximum efficiency, All day Efficiency
Guided By: Prof. Hitesh Manani
Branch: Electrical Engineering
Div: B
Prepared By: (1) Abhishek Choksi 140120109005
(2) Himal Desai 140120109008
(3) Harsh Dedakiya 140120109012

2. Contents
• Open Circuit or no load test
• Short Circuit or Impedance Test
• Sumpner or Back to Back Test
• Condition for Maximum Efficiency
• All day Efficiency

3. Open-circuit Test
• A voltmeter, wattmeter, and an ammeter are connected in
LV side of the transformer as shown in the figure below.

4. • The voltage at rated frequency is applied to that LV side
with the help of a variac of variable ratio auto transformer.
• The HV side of the transformer is kept open. Now with help
of variac applied voltage is slowly increase until the
voltmeter gives reading equal to the rated voltage of the LV
side.
• After reaching at rated LV side voltage, all three instruments
reading (Voltmeter, Ammeter and Wattmeter readings) are
recorded.
• The ammeter reading gives the no load current I0 .
• As no load current I0 is quite small compared to rated
current of the transformer, the voltage drops due to this
electric current then can be taken as negligible.

5. • Since, voltmeter reading V can be considered equal to
secondary induced voltage of the transformer. The input
power during test is indicated by watt-meter reading.
• As the transformer is open circuited, there is no output
hence the input power here consists of core losses in
transformer and copper loss in transformer during no load
condition.
• The no load current in the transformer is quite small
compared to full load current so copper loss due to the
small no load current can be neglected.
• Hence the wattmeter reading can be taken as equal to core
losses in transformer.
• Therefore it is seen that the open circuit test on transformer
is used to determine core losses in transformer and
parameters of shunt branch of the equivalent circuit of
transformer.

6. Calculation
• As we know that,
W0 = V0 I0 cosφ
Cosφ0 =
𝑊0
𝑉0 𝐼0
= No load power factor
• Once Cosφ0 is known we can obtain,
IC = I0 Cosφ0 Im = I0 Sinφ0
• Once IC and Im are known we can determine exciting
circuit parameter as,
R0 =
𝑉0
𝐼 𝐶
X0 =
𝑉0
𝐼 𝑚

7. Short-circuit Test
• A voltmeter, wattmeter, and an ammeter are connected in
HV side of the transformer as shown in figure.

8. • The voltage at rated frequency is applied to that HV side
with the help of a variac of variable ratio auto transformer
• The LV side of the transformer is short circuited . Now with
help of variac applied voltage is slowly increase until the
ammeter gives reading equal to the rated current of the HV
side
• After reaching at rated current of HV side, all three
instruments reading (Voltmeter, Ammeter and Watt-meter
readings) are recorded
• The ammeter reading gives the primary equivalent of full
load current IL .
• As the voltage, applied for full load current in short circuit
test on transformer, is quite small compared to rated primary
voltage of the transformer, the core losses in transformer
can be taken as negligible here.

9. • Let’s, voltmeter reading is VSC . The input power during test
is indicated by watt-meter reading.
• As the transformer is short circuited, there is no output
hence the input power here consists of copper losses in
transformer
• Since, the applied voltage VSC is short circuit voltage in the
transformer and hence it is quite small compared to rated
voltage so core loss due to the small applied voltage can be
neglected.
• Hence the wattmeter reading can be taken as equal to copper
losses in transformer.
• Therefore it is seen that the Short Circuit test on transformer
is used to determine copper loss in transformer at full load
and parameters of approximate equivalent circuit of
transformer.

10. Calculation
• From S.C. test we can write ,
WSC = VSC ISC cosφSC
Cosφ0 =
𝑊 𝑆𝐶
𝑉 𝑆𝐶 𝐼 𝑆𝐶
= Short circuit power factor
𝑊𝑆𝐶 = 𝐼2
𝑆𝐶 𝑅1𝑒 = Copper loss
𝑅1𝑒 =
𝑊 𝑆𝐶
𝐼2
𝑆𝐶
While,
𝑍1𝑒 =
𝑉 𝑆𝐶
𝐼 𝑆𝐶
= 𝑅2
1𝑒 + 𝑋2
1𝑒 and 𝑋1𝑒 = 𝑍2
1𝑒 − 𝑅2
1𝑒
• Thus we get the equivalent circuit parameter 𝑅1𝑒, 𝑋1𝑒 and
𝑍1𝑒. Knowing the transformation ratio K, the equivalent
circuit parameter refer to secondary also can be obtained.

11. Draw back of Open Circuit
and Short Circuit
• In O.C. test, there is no load on the transformer while in
S.C. circuit test only fractional load gets applied. In all O.C.
and S.C. tests, the loading conditions are absent. Hence the
results are inaccurate.
• In open and short circuit test iron losses and copper losses
are determined separately but in actual use both losses
occurs simultaneously.
• The temperature rise in the transformer is due to total loss
that occurs simultaneously during actual use, it can’t be
determined by O.C and S.C tests.

12. Sumpner or Back-to-Back Test
• Sumpner's test or back to back test on transformer is
another method for determining transformer efficiency,
voltage regulation and heating under loaded conditions.

13. • The figure above shows that the connection diagram of
back to back test on two similar transformers named T1 and
T2.
• Both transformers are connected to supply such that one
transformer is loaded on another.
• Primaries of the two identical transformers are connected
in parallel across a supply.
• Secondary are connected in series such that emf's of them
are opposite to each other.
• Another low voltage supply is connected in series with
secondary to get the readings, as shown in the circuit
diagram.

14. Working
• The transformer secondary sides are in phase opposition.
• Then switch S2 is open and switch S1 is closed, thus the
circulating current in the transformer secondary circuit loop
is zero (i.e. I2=0).
• This is due to EMF induced in the secondary’s are in equal
and opposition.
• This circumstance is just like an open circuit test. Hence the
current drawn from the source is 2I0.
• The wattmeter reading W1 and the core loss of both the
transformers are equal.
• I0= No load current of every transformer
• W1 = Core losses of both the transformers.

15. • Now switch S2 also closed and the voltage at the output of
the regulating transformer is varied until the full load
current I2 drift in the secondary circuit loop.
• The secondary full load current will cause full load current
I1in the primary circuit.
• The full load current I1 will circulates in the primary
winding alone.
• The two transformers full load copper losses is equal
wattmeter W2 reading since the full load current circulating
through the primary winding and secondary windings.
• W2 = Transformer full load copper losses
• W2+W1 = Total losses of the two transformers at full load.

16. • From above test results, the full load efficiency of each
transformer can be given as –
% full load efficiency of each transformer
=
𝒐𝒖𝒕𝒑𝒖𝒕
𝒐𝒖𝒕𝒑𝒖𝒕+
𝑾 𝟏
𝟐
+
𝑾 𝟐
𝟐
∗ 𝟏𝟎𝟎

17. Advantages of Sumpner’s Test
• Little much of power is required to conduct this test
• Under full load conditions transformers can be test using
this test.
• Simultaneously full load copper losses and iron losses are
measured
• The secondary current I2 can be varied at any value of the
current. Hence we can determine the copper losses at full
load condition or at any load.
• The transformer temperature increase can be noted.

18. • Only limitation is that two identical transformers are
required. In practice exact identical transformers cannot be
obtained and as two transformers are required, the test is
not economical.
Drawback of Sumpner’s Test

19. Efficiency of a transformer
• As in the case of electrical machines, the efficiency of a
transformer at particular load and power factor is defined as
the output divided by the input the two being measured in
the same units i.e in watts or kilowatts.
Efficiency =
𝑜𝑢𝑡𝑝𝑢𝑡
𝐼𝑛𝑝𝑢𝑡
• But the transformer being highly piece of equipment, has
very small loss, hence it is impractical to try to measure
transformer, efficiency by measuring input and output.
These quantities are nearly of the same size.
• A better method is to determine the losses and then to
calculate the efficiency from,

20. Efficiency =
𝑜𝑢𝑡𝑝𝑢𝑡
𝐼𝑛𝑝𝑢𝑡+𝑙𝑜𝑠𝑠𝑒𝑠
=
𝑜𝑢𝑡𝑝𝑢𝑡
𝑜𝑢𝑡𝑝𝑢𝑡+𝐶𝑢 𝑙𝑜𝑠𝑠𝑒𝑠+𝑖𝑟𝑜𝑛 𝑙𝑜𝑠𝑠𝑒𝑠
Or,
Ƞ =
𝑖𝑛𝑝𝑢𝑡−𝑙𝑜𝑠𝑠𝑒𝑠
𝐼𝑛𝑝𝑢𝑡
= 1 −
𝑙𝑜𝑠𝑠𝑒𝑠
𝑖𝑛𝑝𝑢𝑡
• It may be noted here that efficiency is based on power
output in watts and not in volt-amperes, although losses are
proportional to VA.
• Hence at any volt-ampere load, the efficiency depend on
power factor, being maximum at a power factor of unity.

21. Condition for Maximum Efficiency
Cu loss = 𝐼1
2
𝑅01 or 𝐼2
2
𝑅02 = 𝑊𝐶𝑈
Iron loss = Hysteresis loss + Eddy current loss
= 𝑊ℎ + 𝑊𝑒 = 𝑊𝑖
• Considering primary side,
Primary output = 𝑉1 𝐼1 cos φ1
Ƞ =
𝑉1 𝐼1 cos ∅1−𝑙𝑜𝑠𝑠𝑒𝑠
𝑉1 𝐼1 cos ∅1
=
𝑉1 𝐼1 cos ∅1−𝐼2
1 𝑅01−𝑊 𝑖
𝑉1 𝐼1 cos ∅1
= 1 −
𝐼1 𝑅01
𝑉1 cos ∅1
−
𝑊 𝑖
𝑉1 𝐼1 cos ∅1
Differentiating both the side with respect to 𝐼1, we get

22. 𝑑Ƞ
𝑑𝐼1
= 0 −
𝑅01
𝑉1 cos ∅1
−
𝑊 𝑖
𝑉1 𝐼2
1 cos ∅1
• For Ƞ to be maximum,
𝑑Ƞ
𝑑𝐼1
= 0. Hence, the above equation
becomes
𝑅01
𝑉1 cos ∅1
=
𝑊 𝑖
𝑉1 𝐼2
1 cos ∅1
or 𝑊𝑖 = 𝐼2
1 𝑅01 or 𝐼2
2 𝑅02
Cu loss = Iron loss
• The output current corresponding to maximum efficiency is
𝐼2 =
𝑊1
𝑅02
.
• It is this value of the output current which will make the Cu
loss equal to iron loss.

23. All day efficiency
• The commercial Efficiency of a transformer is given by the ratio
of output power to input power
• The losses in the transformer can be classified into copper losses and
iron losses. The copper losses (hysteresis and Eddy Current Losses)
are independent of the load. The iron losses though are dependent on
the load.
•
In the case of the distribution transformers, the load is continually
varying. It is low in the day time and high in the evenings and night.
Therefore, efficiency measured at any one point of the day would not
be an accurate reflection of the transformer's capability.
Efficiency =
output power in watts
input power in watts

24. • Hence, We have the all day efficiency measurement of the
distribution transformers. The formula for the all day efficiency of the
distribution transformers is
• The All day Efficiency is always lesser than the commercial
efficiency of the transformer.
Efficiency =
Output in kWh for 24 hours
Output in kWh for 24 hours
* 100

25. Reference
• U.A. Bakshi
• B.L. Theraja
• http://www.mytech-info.com
• https://en.wikipedia.org