A D C generator is connected to a 220 V D C mains. The current delivered by the generator to the mains is 100 A . The armature resistance is 0.1 ohm . The generator is driven at a speed of 500 rpm Calculate (i) the induced emf (ii) the electromagnetic torque (iii) the mechanical power input to the

Question

A D C generator is connected to a 220 V D C mains. The current delivered by the generator to the mains is 100 A . The armature resistance is 0.1 ohm . The generator is driven at a speed of 500 rpm Calculate (i) the induced emf (ii) the electromagnetic torque (iii) the mechanical power input to the armature neglecting iron, windage and friction losses, (iv) Electrical power output from the armature, (v) armature copper loss.

Accepted Answer

[latex] ext { (i) The induced emf, } \quad E_{g}=V+I_{a} R_{a}[/latex]

[latex]=220+0.1 imes 100=230 V[/latex]

[latex] ext { (ii) Using the relation, } \quad \omega T=E_{g} I_{a}[/latex]

[latex] ext { Electromagnetic torque, } \quad T=\frac{E_{g} I_{a}}{\omega}=\frac{E_{g} I_{a}}{2 \pi N} imes 60 \quad\left[\because \omega=\frac{2 \pi N}{60} ight][/latex]

[latex]=\frac{230 imes 100 imes 60}{2 \pi imes 500}=439 \cdot 27 Nm [/latex]

(iii) Neglecting iron, windage and friction losses,

[latex] ext { Input to armature } =\omega T ext { or } E_{g} I_{a}[/latex]

[latex]=\frac{2 \pi N T}{60}=\frac{2 \pi imes 500 imes 439 \cdot 27}{60}= 2 3 0 0 0 W[/latex]

(iv) [latex] ext { Electrical power output }=V I_{a}=220 imes 100=22000 W[/latex]

[latex] ext { Armature copper losses }=I_{a}^{2} R_{a}=(100)^{2} imes 0.1=1000 W [/latex]

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