Question

220 v dc shunt motor has Ra= 0.25 ohms and Rf=120 ohms . At no-load the motor draws a current It=6A. And runs at 1500 rpm. At full-load the motor draws 62 A. the armature reaction at full load reduces the flux by 10%.

Determine the rotational losses , motor speed at full-load, the shaft torque and efficiency.

Answer 1

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page Solution Given : V = 220v, Ra=0.252, Rf=1202 - Load It = 6A , tt Speed (No) = 15000.p.m. - 6 * At Full load It = 62 and the flux reduces by 10%. That means ***if the flux at no load is to then flux at full load [Of = 0.96 To Find: Rotational losses,. - Full load speed and shaft torque and efficiency. If YIA ♡ If = 1.83A/ 120 = R18 At No load: I a=6-1.83 = 417 A I a=4.17 A V = E + I aRa (Eo = No load back emf 220=Eo +4.17X0.25 of motor) > E. = 218.95 Volt → *** Rotational losses will be equal to the power dissipated through Arma- - Armature.

page ② page : Rotational losses (Pr ) = Eola (At no load) = (2.18.95) x 4.17 walt I P = 913.02 Walt Ansi since Ex 0 N ; N= speed in o.p.m. At full load: v=EfRt Iafe Ra 220 = Efl + I afera lac - It fi- If = 62-1083 Iafl = 60.17 A/ Ec, = 220 - 60:17)(0.25) Efl = 204.95 V since ExNO SO NEL = full load 2 speed EFL - NALDEL I PFC = full load flux Eo No 0. 6. = No load 204,95 NEL X 0.9% flux No= No load 218.95 1500x 0. speed A : PEL=0.90) - ANS. - INFL= 1560 R.P.M.

page power output = Elace Po Py = Rotational losses ... Poutput = (204.95)*(60-17) - 913.02. = 11418.82 Watt Pout = 11.42 kW] since w - WEL Where Shaft Torque = Pout WL = 2 NELT 2XT X 1560 – 163.36 vad/sec 60 11.42x103 shaft Torque - 163636 Shaft torque = 69.9 N-m) . And. Efficiency = Efficiency - Pout X100% Pin : Pin= V It = 220x62=13640 watt 11.42 x 103 Efficiency - 13640 Efficiency = 83.72% | Ans.