a DC shunt machine ( 24kW, 240V, 1000 RPM) has Ra= .12 ohms, Nf= 600 turns/pol.

Question

a DC shunt machine ( 24kW, 240V, 1000 RPM) has Ra= .12 ohms, Nf= 600 turns/pol. The machine is operated as a seperately excited dc generator and is driven at 1000 RPM. When If = 1.8 amps the no load terminal voltage is 240V. When the generator delivers full load current the terminal voltage drops to 225V. Determine the generated voltage and developed torque at full load, the voltage drop due to armature reaction.

Explanation / Answer

We know the generated voltage is given by Eg=Vt+IaRa Eg=240 at no load as no current flows Ia=0 Eg=225+Ia(0.12) at full load 15/0.12=Ia =>Ia=125amps at full load Eg=240volts torque equation is given by T=0.159Ia(øPZ/A) but we know Eg=(øZN/60)(P/A) =>øZP/A=60xEg/N=60x240/1000=14.4 substituting this value in torque equation gives T=0.159x125x(14.4)=286.2N-m the voltage drop due to armature reaction is given by IaRa=125(0.12)=15volts

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