Question

QUESTION 1 [25 MARKS] The data in the Table 1 were obtained when open and short circuit test were performed on a single phase 50 kVA, 2400/240 V, 60 Hz transformer. Table 1 Voltage Current Power (V) (A) (Watt) 240 4.85 173 52 20.83 650 Test A Test B (a) Determine tests A, B and the sides the tests were done. Calculate the non-ideal transformer parameters referred to the low voltage side. Draw the equivalent circuit by labeling the parameters and their values. (b) (c) [15 MARKS] If the secondary side is connected to a half rated inductive load at 0.8 PF, calculate Web [5 MARKS] A second load is to be connected in parallel to the first load and then connected to the secondary side of the transformer. Determine is it possible to connect the second load if the second load is consuming half of the rated current at 0.85 power factor leading. (5 MARKS]

Answer 1

Que 1-0 , 50kVA 2400 / 240, 60Hz ohen araut test 240 V 4,85 A 173W so Zo = > (coste Р VI T 240 17 3 Cort (! La 4.85 240X485 Za - 49.484/81.452 referred to Secondary and short circut test and 52 V, 20.83 , 650 watt 552 650 Zse (cost / 52 x 20.82 6-82) 20.82 - 2.497/ 53.10 10499 +11.99 ZSL (1.5 +1 2.0) Zo referred to primary = zo' 2 4948.45 81-452 V

a hom (1:5+12,0) T les 1 v2 load Zo = 49.484/81-45°, to ad side. b a load of o-spf. is connected to coor ji wo 0.75 j K 0.75 R I a vi' 240 vola Zo =49.484/81.45 50 XOSX 100 2-costo.8 guo Ig 104.167/-3 1-36.87 = 10-41672 -36.87 X10 up 240 + (10.75 +1)x10467/-36.8687 = a Vp Oº v ó 370.20832 = a so slo c 370.2083 v a

c two wads are Connents so Ig 50X0.5 X 1000 50X6'5X100 (Costo 8 t (lost 0.85 २५० Ia = 172.04 ) -2.54* Amy so Up Vat (x + R₂ ) to = a = 240 + ( 3170.75) 172.04/-2.54 Je II 411.56/23-81 a So Vq 411:56 Volt a and o 1650.8 + 50/-cos?0.85 ( 56 s] to X0.5 Th = ol 82.58/ 2.59 = 41.2972-54 kVA is S 82.58 kVA which Q S = 41. 29 KVA < 50 KVA I ruled to Corment So It is possible