Question

The short-circuit test readings for a 50-kVA 2400:240-V transformer are 48 V, 20.8 A, and 617 W, with the instruments located on the high-voltage side and the low-voltage side short- circuited. An open-circuit test with the low-voltage side energized gives instrument readings on that side of 240 V, 5.41 A, and 186 W. 1. Calculate the equivalent parameters of the transformer as referred to the high-voltage side. 2. Draw the equivalent approximate circuit of that transformer referred to the high voltage side. (Label the impedances numerically). 3. If the transformer is to operate at maximum efficiency with 0.8 power factor lagging load, find the load in kVA that should be connected to achieve maximum efficiency if rated voltage is applied at the primary (H.V) side and then find the value of that efficiency.

Answer 1

48V, 20 glade 611 W across Hence, the equivalent resistance/ Reactance are Shord Circuit rest> Voltage ei LV Shorted, restered to HV. 2.3081 22 (48) (208) RE 1.42but 316 617 (20.882 I X z²_R² Х 1815U Rse - 1.4260/ xml |Xse = tj 1.8154 mili ya Reffered to ld V. Open Circuit desh → HV opened , Vollage across LV The equivalent resistance I reaclince die reffered to LV. They has to be reffered to HtV. 240V : 8.41A and 1860 Ro (240) ? 309.7 P 186 IRO 240V so.775 A 309.74 Ixo (5.412-6.7752 : S.35 A X • Bảo : 44.86 u 5.35 RO 309.74 Referred tolu Xo j 44.8611 CS Scanned with CamScanner

we get Reflessing to HV side , we Re: 364,7 (84 12400 20019 نم Ro 3.lou lilly Xo : j0.45 u - Reflessing to hu. b) → Xse : j1.8154 mm V2 V #zu > Rse 1.4264 Xo: 10.45 Ro-3.lou The equivalent diageam referred to HV side > PF:08 Al maximum efficiency Chloss: Fe loss (sonio? Cn loss = 1 x Rse = 2:39 x 1.426 2400 = 619 w Fe loss = 619 w. load at max eld full load, Iron loss Chloss (full load) • Sorios x 619 619 Lood at max eff Is achieved So kVA 50x103 z n = Output Output & Feloss + In loss Sox 103 + 2x619 1 = 97.6% CS Scanned with CamScanner