Open circuit and Short circuit Test Presentaion

Open and Short Circuit Test on Transformer
By:- Lalit Goyal
Gaurav Kumar
Ganesh Kumar
Manthan

1. Open circuit & Short circuit test BY Lalit Goyal

2. INTRODUCTION Necessit y of testing O pen circuit test Short circuit test Determination of parameters Advantages & disadvantages Applications

3. NECESSSITY OF TESTING IN TRANSFORMER AND OTHER EQUIPMENT HAVE INPUT AND OUTPUT . HENCE HOW WE HAVE TO FIND THE PERFOMACE OF TRANSFORMER AND OTHER EQUIPMENT . WE HAVE A SOLUTION EVEN WE WILL FIND THE LOSS OF TRANSFORMER. SO WE CAN EASILY FIND THE ALL PARAMETERS OF TRANSFORMER LIKE IRON LOSSES, Cu LOSSES, IMPEDENCE LOSSES, NO LOAD CURRENT ,SHORT CIRCUIT CURRENT .

4. To understand the basic principle of transformer To obtain the equivalent circuit parameters To estimate efficiency & voltage regulation at various load WHY WE ARE PERFORMING ?

5. CONCEPT AND THEORY THESETWO TRANSFORMETRS TESTS ARE PERFORMED TO FIND THE PARAMETERS OF TRANSFORMERS. OPEN CIRCUIT & SHORT CIRCUIT TEST ARE VERY CONVENIENT BECAUSE THEY RE PETRFORMED WITHOUT ACTUALLY LOADING OF TRANSFORMERS.

6. OPEN CIRCUIT :- TEST ON A TRANSFORMER IS PERFORMED TO ETERMINE NO LOAD LOSSES (CORE LOSS) & NO LOAD CURRENT FROM THIS WE CAN ALSO FIND SHUNT PARAMETERS OF EQUIVALENT CIRCUIT SHORT CIRCUIT :- IS TO CONUCT OR DETERMINING FULL LOAD Cu LOSSES EQUIVALENT RESISTANCE OF TRANSFORMER AS REFERRED TO METERING SIDE

7. OPEN CIRCUIT TEST

8. HOW TO PERFORM ? TO CONNECT WATTMETER IN PRIMARY SIDE TO CONNECT AMMETER IN SERIES OF PHASE TO CONNNECT VOLTMETER IN PARALLEL OF PRIMARY SIDE WE WILL OPEN THE SECONDARY SIDE THE PUPRPOSE OF THIS TEST IS TO FIND THE NO LOAD NO-LOAD LOSS OR CORE LOSS A WATTMETER ,VOLTMETER ,AMMETER ARE CONNECTED IN LOW VOLTAGE SIDE THIS TEST IS PERFORMED ON L.V. WINDING KEEPING H.V. WINING OPEN CIRCUITED

9. WHY WE PERORMING THIS?

10. IN OC TEST THE CURRENT MEASURED BY THIS AMMETER WILL BE VERY LESS SO IF KVA RATING OF THIS TRANSORMER IS SAME FROM PRIMARY AS WELL AS SECONDARY SIDE SO VERY LESS CURRENT YOU ARE GETTING. IN L.V. SIDE WE GIVE THE RATED VOLTAGE ,NORMAL FLUX WILL BE SET UP IN THE CORE ,HENCE NORMAL IRON LOSSES WILL OCCUR , WHICH WILL OCCUR WHICH ARE RECORDED BY THE WATTMETER AS THE PRIMARY NO-LOAD CURRENT I0 IS MEASURED BY AMMETER ND ITS VALUE IS SMALL Cu LOSS IS NEGLIGIBLY SMALL IN PRIMARY AN NIL IN SECONDARY (IT BEINGG OPEN). HENCE THE WATTMETER READING REPRESENTS PRACTICALLY THE CORE LOSS UNDER NO LOAD CONDITION.

11. SHORT CIRCUIT TEST

12. HOW TO MEASURE CONNECT WATTMETER PRIMARY SIDE (POWER) CONNECT THE AMMAETER IN SERIES OF PRIMRY SIDE ( LINE CURRENT). CONNECT THE VOLTMETER IN PARALLEL OF PRIMARY SIDE (VOLTGE) SHORT THE SECONDARY SIDE USING THICK CONDUCTOR (L.V. SHORTED). NOW GIVE THE SMALL VOLTAGE SUPPLY FOR THE MEASUREMENT OF Cu LOSSES

13. IMPORTANT TERMS A WATTMETER , VOLTMETER ,AN AMMETER ARE CONNECTED IN HIGH VOLTAGE SIDE . THE LOW VOLTAGE SIDE OF TRANSFORMER IS SHORT CIRCUITED. THE APPLIED VOLTAGE IS A SMALL PERCENTAGE NORMALLY 5-7 % OF THE RATED VOLTAGE . HENCE THE CORE LOSSES ARE VERY SMALL WITH THE RESULT OF THE WATTMETER READING REPRESENTS THE FULL LOAD Cu LOSS OR sq(I)*R LOSS FOR THE WHOLE TRNSFORMER. RATED CURRENT IS PROVIDED.

14. HOW SC TEST MEASURE Cu LOSS SHORTING L.V. SIDE WHOLE CURRENT PASSES THROUGH THE SHORTING CONDUCTOR & WHOLE CORE IS NEGLECTED . NOW CURRENT IS PASSING THROUGH ONLY WINDING AND NOT THROUGH CORE SO IT DEPICTS ONLY COPPER LOSS

15. WHY SHORTING IS DONE ON L.V. SIDE & NOT IN H.V. SIDE?

16. THE CURRENT WILL BE VERY HIGH I TEST IS CONDUCTED ON L.V. SIDE. WE CAN’T PROVIDE THIS MUCH CURRENT EASILY. WE ALSO HAVE TO USE A HUGE VARIAC FOR THIS. Cu LOSS WILL BE ALSO VERY LARGE. EXAMPLE:- Consider a 1 MVA 415V/ 11,000 V transformer If you were to perform SC test on LV side the current would be  (1 x 10^6)/(1.732 x 415) =  1391.2 A Whereas if you were to perform the Test on HV side the Current Would be (1 x 10^6)/ (1.732 x 11000) = 52.48 A

17. How to find the parameters of the Transformer ? KVA Rating of transformer= KVA FOR OPEN CIRCUIT TEST, Measured Power per Phase= Wo Line Voltage =Vo Line Current =Io FOR SHORT CIRCUIT TEST, Measured Power per Phase= Wsc Line Voltage = Vsc Line Current = Isc

18. CALCULATION:- FOR OPEN CIRCUIT TEST: cos A = W o /I o *V o I w = I o * cos A I m = I o *sin A R o = V o / I w X o = V o / I m FOR SHORT CIRCUIT TEST: Z sc = V sc / I sc R sc = W sc /I sc ˄2 X sc ˄2= (Z sc ˄2 – R sc ˄2) R 1 =R 2 = R sc /2 X 1 =X 2 = X sc /2

19. APPROX. EQUIVALENT CIRCUIT

20. VOLTAGE REGULATION The secondary terminal voltage changes with the change in the load due to the change in the voltage droop across the winding resistance and leakage reactances . The voltage regulation of transformer is defined as the net change in the secondary terminal voltage from no load to full load expressed as a percentage of its rated volage , for the same primary voltage. Percentage voltage regulation is given by , % Voltage Regulation = ( V snl – V sfl )/ V sfl * 100 *Where V snl and V sfl are the effective values across secondary terminal under no load and full load condition respectively. For given OC & SC test the equetion of Voltage Regulation is given by, % Votlage Regulation = ( I sc * R sc * cos A + I sc * X sc *sin A)/V 2(rated) * 100

21. EFFICIENCY OF TRANSFORMER Efficiency of transformer is defined as the ratio of output of the transformer to its input. The efficiency of transformer is much higher than that of a rotating electrical machine, because there is no friction and windage losses. thus, efficiency of transformer, η = Output power/Input power input power = output power + total losses total losses = iron losses + copper losses = W i +(I p ^2* R p + I s ^2*R s ) Now, η = Vs*Is* cos A / (Vs*Is* cos A + W i +(I p ^2* R p + I s ^2*R s )) CONDITION OF MAXIMUM EFFICIENCY totale copper losses = I s ^2*R s Now, η = Vs*Is* cos A / (Vs*Is* cos A + W i + I s ^2*R s )

22. APPLICATION By using open circuit test we can measure the loss of no-load losses means that iron losses. By using short circuit test we can measure the loss of Full-load losses means that Cu-losses We can find the efficiency by using of iron losses and cu-losses

23. ADVANTAGES No require of extra load This test are very economical and convenient because they furnish the required information without actually loading the transformer No much calculation is needed Efficiency can be found for any desired load without connecting it to full load

24. Disadvantages It’s not economical to do this test on large rating machines as, large amount of energy is wasted You can’t get information about the share of different types of losses It’s difficult task to perform different kinds of load (R,L,C,)form no load to full load to study performance at different situations.

25. THANK YOU