Efficiency of a Transformer. (a) copper losses: These occur in the transformer primary and secondary windings. They may be calculated from R 2

Transcription

1 Efficiency of a Transfmer The losses that take place in a transfmer on load can be divided into two groups (a) copper losses: These occur in the transfmer primary and secondary windings. They may be calculated from p copper =p primary +p secondary =I 1 R1 +I R (b) Iron ce losses: These are constant losses due to hysteresis and eddy currents. P iron =constant the total power loss in a transfmer is p total =p iron +p copper =p iron +I 1 R1 +I R Since the efficiency of a transfmer is the ration of the output power to the input power, it may be expressed as output power input power = output power output power + losses But p=vi cosθ where cosθ = power fact Then output power =V I cosθ where: V =voltage across the load ; I =current drawn by the load. V I cosθ V I cosθ+p iron +I 1 R 1 +I R 1

2 Efficiency of a Transfmer on a Fraction Of The load If current I passes through a resistance R, then the power dissipated in R is p loss (full load )=I R If I is reduced to half, then power dissipated in the same R is p loss (half load )= I R = 1 I R= 1 p loss (full load ) In general if nth amount of the current passes through the same R, then and p loss (nth load )= 1 n p loss (full load ) p output (nth load )=n p output (full load ) Therefe the efficiency is nv I cosθ nv I cosθ+p iron +n I 1 R 1 +n I R Example The primary and secondary windings of a 500 kva transfmer have resistances of R 1 =0.4 Ω and R = Ω. The primary and secondary voltages are V 1 =6,600 V and V =400 V. The iron losses are p iron =.9 kw. Calculate the efficiency of the transfmer when it supplies (a) full load; (b) half load. Assume the load power fact to be 0.8 lagging. Solution (a)

3 I 1 (full load )= kva V 1 = 500x =75.8 A I (full load )= kva V = 500x =1,50 A p copper (full load )=I 1 (fulll load )R 1 +I (full load )R p copper =(75.8 ) x0.4 +(1,50 ) x p copper =,415 +1,70 =4.135 kw Therefe the transfmer total losses on full-load are p total =p iron +p copper = =7.035 kw output power(full load )=V I cosθ = 500x0.8 =400 kw V I cosθ V I cosθ+p iron +I R 1 1 +I R = Solution (b) 400, ,000 +7,035 =98.7% When the transfmer operates on half-load, the copper losses vary as follows p copper (half load )=( 0.5) p copper (full load ) The total power loss on half load is = ( 0.5 ) =1.034 kw p total (half load )=p iron +p copper (half load ) 3

4 = =3.934 kw The output power is now half of when the transfmer was on full-load, i.e. output power(half load )=V I cosθ = 50x0.8 =00 kw 0.5V I cosθ 0.5V I cosθ+p iron = =98.07% ( ) I 1 R 1 + ( 0.5) I R Condition of Maximum Efficiency Let the turn ratio of the transfmer be Then σ= N N 1 I 1 =σi and the efficiency is expressed as V I xpf V I xpf + p iron +σ I R 1 +I R V xpf V xpf + P iron +σ I R 1 +I R I F the efficiency to be maximum, the denominat must be minimum. In der to determine the load condition f which the denominat is minimum, we differentiate with respect to I and equate to zero to obtain 4

5 d V di xpf + p iron +σ I R 1 +I R = 0 I p iron I +σ R 1 +R =0 p iron =σ I R1 +I R =I 1 R1 +I R =p copper the condition f maximum efficiency is p copper =p iron Example Find the output at which the efficiency of the transfmer of the previous example is maximum and calculate its value. Solution Let n be the fraction of the load at which the efficiency is maximum, then p copper (nth load )=n p copper (full load ) = n x4.135 kw But at maximum efficiency p copper =p iron =.9 kw.9 =n x4.135 n= Therefe the output at maximum efficiency is p output (maxξ)=nxkva =0.837 x500 =418.5 kva 0.837x500x x500x0.8+x.9 = =98.3% 5

6 Efficiency of a Transfmer From Open- and Sht-Circuit Tests The open circuit test and sht circuit tests enable the efficiency and voltage regulation to be determined without actually loading the transfmer. The power required to carry out these tests is small compared to the fullload output of the transfmer. The tests produce highly accurate results compared to load tests. Open-Circuit Test The transfmer is connected as shown in Figure (1). Open-circuit test on a transfmer V 1 =V rated name plate f= f rated name plate V 1 (reading) V (reading) = N 1 N =turns ratio A(reading)=I 1 =I o ; usually < 5% I 1 (full load ). p copper =I (no load )R< I (full load )R 400 p copper may be igned. W(reading ) p iron 6

7 Sht-Circuit Test The transfmer is connected as shown in Figure (). Figure (): Sht-circuit test on a transfmer V 1 =V sc enough to ciculate full current Usually V sc 1 5 V rated f= f rated name plate A 1 (reading) A (reading) = N N 1 A 1 (reading)=i 1 (rated ) A (reading )=I (rated ) W(reading )=p copper =I 1 R1 +I R V sc 1 5 V rated p iron <<p copper p iron can be igned. p iron are negligible because of the reduced applied voltage and therefe the flux. Calculation of Efficiency From Open-Circuit and Sht-Circuit Tests 7

8 Since p iron =p open cicuit =W open circuit p copper =p sht cicuit =W sht circuit the efficiency on full-load is calculated from full load VA x pf full load VA x pf + p open-circuit +p sht -circuit And f an nth-load n x full load VA x pf n x full load VA x pf + p open-circuit +n x p sht -circuit Calculation of Voltage Regulation From Sht-Circuit Tests Consider the simplified equivalent circuit of a transfmer shown in Figure (3). Figure (3): simplified equivalent circuit of a transfmer It was established that the voltage regulation of a transfmer can be expressed as V.R= I 1Z e cos(φ e Φ ) V 1 V 1 = primary rated voltage I 1 = primary current 8

9 Z e = equivalent impedance of the primary and secondary windings referred to the primary. Φ e = phase angle between I 1 and I 1 Z e. Φ = load phase angle. From figure 3, it is clear that p sc =I 1 V sc cosφ e and V sc =I 1 Z e ;since V =0. V.R= V sc cos( Φ e Φ ) V 1 Example The following results were obtained from tests on a 50 kva transfmer Open-circuit test Primary voltage = 3300 V. Secondary voltage = 400 V. Primary power = 430 W. Sht-circuit test Primary voltage = 14 V. Primary current = 15.3 A. Secondary current = full load value. Primary power = 55 W. Calculate (a) the efficiencies at full load and half load f 0.7 power fact. (b) the voltage regulation f 0.7 power fact, lagging and leading. (c) the secondary terminal voltage f 0.7 power fact, lagging and leading. 9

10 Solution (a) From the test results we have p iron =430 W p copper (full load )=55 W p total (full load )= =0.955 kw 50x0.7 ξ( full load )= 50x =97.34% 0.5x50x0.7 ξ( half load )= 0.5x50x x55 =96.9% Solution (b) p sc =I 1 V sc cosφ e cosφ e = Φ e =74 o p sc I 1 V sc = 55 14x15.3 =0.765 Also from cosφ =0.7 we find Φ =45.5 o V.R= V sc cos( Φ e Φ ) V 1 f 0.7 pf lagging VR= 14cos( ) =3.3% 3300 f 0.7 pf leading VR= 14cos( ) = 1.85%

11 Solution (c) Since the secondary voltage on open circuit (no-load) = rated voltage = 400 V, the secondary voltage on full-load is V (full load )=400(1 VR) f 0.7 power fact lagging V (full load )=400( )=386.8 V f 0.7 power fact leading V (full load )=400( )=407.4 V 11