Question

For full load operation of 60Hz, 380V, 4 pole 30kW induction motor,
the speed is 1746rpm, input current 56.2A power factor 90%.
In no load operation, input current 17.3A, power factor 13.5%.
In full load operation, assume that the mechanical loss is 25% of the total loss and no other losses.

a) Iron loss load efficiency
b) Find a rotor copper loss under full load
c) Iron loss
d) Find a fixed automatic hand under full load
e) Find the proportion of stator loss, iron loss, rotor copper loss and machine loss in total losses.

answer -
a) 90.1%
b) 953W
c) 714W
d) 801W
e) 24.3%

Answer 1

Given f = 6oita P=4 line input voltage = 380V NE 1746 R.P.m. input current 56.2 A - cosp=0.9 At no load I. - 17.3 A cosde13:51. - 0.135 mo-load loss Iron toss : a) 3x3 80x17-3 X0-135 1537,1770 - powes = input 6x 380 x 56-axo.9 33290-70934 W X 100 efficiency = 30000 1 33290-70934 b? Fatt - goolust b) full load Rotor Coppes loss Toral losses= 33290.70934- 30000 - 3290.70934 W mechanical loss - 0.25x 3290-70934 822.6773 W me chassical powes developed (Pm) - 30000 + 822-6723 om - 30829-6773 W Rotor Copper loss = sx = sx Pg pm Sz 120x60 * -1746 30.03 a20xGo

& Rolf coppes loss = 0.03 x 30822.6773 1-0.03 - 953,2786w () ison loss no-load losses me charsical loss = 1537.177 - 822.6723 : 714.49w - Fixed automatic hand - Total loss – 953.2786-1537.00 -- 3290.709 - 2490.44 800.26 W 1. e) n o copper loss = stator loss - iron loss -714-4 = 33290.70934 - (30822.6773) 1-0-03 800.26 W Proportion of statos col.loss in total losses - 800-26 x 100 = 24-37. 3290-7 Porop of tion of iron Coss in total loss : 714.4 voor 21:37. 3290.7 Proportion of rolor cu-loss in total loss - 953.278 vious 28,961. 329017 Propoitim of mechanical loss in total loss - 25%.