A 100 kW belt-driven shunt generator running at 300 rpm on 220 V bus bars continues to run as a motor when the belt breaks then taking 10 kW from the mains. What will be its speed? Armature resistance is 0.025 ohm, field circuit resistance is 60 ohm contact drop per brush is 1 V and ignore

Question

A 100 kW belt-driven shunt generator running at 300 rpm on 220 V bus bars continues to run as a motor when the belt breaks then taking 10 kW from the mains. What will be its speed? Armature resistance is 0.025 ohm, field circuit resistance is 60 ohm contact drop per brush is 1 V and ignore armature reaction.

Accepted Answer

Shunt field current, [latex] I_{s h}=\frac{220}{60}=3.67 A[/latex]

(i) When the machine runs as a generator (see Fig. 5.33)

[latex] I_{L I} =\frac{100 imes 1000}{220}=454.54 A [/latex]

[latex] I_{a I} =I_{L I}+I_{s h}=454.54+3.67=458.21 A[/latex]

[latex] E_{g} =V+I_{a I} R_{a}+2 v_{b} [/latex]

[latex] =220+458.21 imes 0.025+2 imes 1=233.455 V[/latex]

(ii) When the machine runs as a motor (see Fig. 5.59.)

[latex] I_{L 2}=\frac{10 imes 1000}{220}=45.45 A ; I_{a 2}=I_{L 2}-I_{s h}=45.45-3.67=41.78 A[/latex]

[latex] E_{b}=V-I_{a 2} R_{a}-2 v_{b}=220-41.78 imes 0.025-2 imes 1=216.955 V[/latex]

In shunt machines, flux is constant;

[latex] herefore[/latex]

[latex] N_{1} \propto E_{g} ext { and } N_{2} \propto E_{b}[/latex]

or

[latex] \frac{N_{2}}{N_{1}} =\frac{E_{b}}{E_{g}} [/latex]

Or [latex] N_{2} =\frac{E_{b}}{E_{g}} imes N_{1}=\frac{216.955}{233.455} imes 300=278.8 \operatorname{rpm}[/latex]

Question 5.21: A 100 kW belt-driven shunt generator running at 300 rpm on 2...

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