O.C & S.C Test, Sumpner or back to back Test, Condition for maximum efficiency, All day Efficiency

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 I . As no load current I 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, W = V I cosφ Cosφ = = No load power factor Once Cosφ is known we can obtain, I C = I Cosφ I m = I Sinφ Once I C and I m are known we can determine exciting circuit parameter as, R = X =  

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 I L . 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 V SC . 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 V SC 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 , W SC = V SC I SC cosφ SC Cosφ = = Short circuit power factor = Copper loss While, Thus we get the equivalent circuit parameter , and . 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 T 1  and T 2 . 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 S 2  is open and switch S 1  is closed, thus the circulating current in the transformer secondary circuit loop is zero (i.e. I 2 =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 2I . The wattmeter reading W 1  and the core loss of both the transformers are equal . I = No load current of every transformer W 1  = Core losses of both the transformers.

15. Now switch S 2  also closed and the voltage at the output of the regulating transformer is varied until the full load current I 2  drift in the secondary circuit loop. The secondary full load current will cause full load current I 1 in the primary circuit . The full load current I 1  will circulates in the primary winding alone. The two transformers full load copper losses is equal wattmeter W 2  reading since the full load current circulating through the primary winding and secondary windings . W 2  = Transformer full load copper losses W 2 +W 1  = 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 =  

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 I 2  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 = 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 = Or, Ƞ = 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 = or Iron loss = Hysteresis loss + Eddy current loss = Considering primary side, Primary output = Ƞ = Differentiating both the side with respect to , we get  

22. For to be maximum, = 0. Hence, the above equation becomes or or Cu loss = Iron loss The output current corresponding to maximum efficiency is . 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.  

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 * 100  

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

26. THANK YOU