GATE Electrical Engineering (EE) 2021 Solved Papers

Question : 49 of 65

Marks: +1, -0

A belt-driven DC shunt generator running at

300 {RPM}

delivers

100 {kW}

to a

200 {V} {DC}

grid. It continues to run as a motor when the belt breaks, taking

10 {kW}

from the DC grid. The armature resistance is

0.025 Ω

, field resistance is

50 Ω

, and brush drop is

2 {V}

. Ignoring armature reaction, the speed of the motor is ________________ RPM. (Round off to 2 decimal places.)

Solution:

{P}_{0}=100 × 10^{3}

I_{L}=\;{100 × 10^{3}}/{200}

{I}_{ {L}}=500 {A}

→ I_{s h}=\;{200}/{50}=4 {A}

{I}_{ {a}}=500+4=504 {A}

By KVL

{E}_{ {g}}- {I}_{ {a}} {R}_{ {a}}- {V}_{ {B}}= {V}_{ {Bus}}

{E}_{ {g}}- {V}_{ {BUS}}+ {I}_{ {a}} {R}_{ {a}}+ {B} . {D}

=200+504 × 0.025+2

{E}_{ {g}}=214.6 {V}

The machine starts consuming 10 kW Power due to absence of mechanical input.

{P}=10 × 10^{3}

I_{L}=\;{10 × 10^{3}}/{200}=50 {A}

→ I_{s h}=\;{200}/{50}=4 {A}

{I}_{ {a}}= {I}_{ {L}}- {I}_{ {sh}}=50-4=46 {A}

By KVL

{V}- {I}_{ {a}} {R}_{ {a}}- {B} ⋅ {D}= {E}_{ {b}}

200-46 × 0.025-2= {E}_{ {b}}

{E}_{ {m}}= {E}_{ {b}}=196.85

Volts

N ∝ \;{E}/{φ}

' φ

' is constant since current through shunt field winding is constant.

∴ \;{N_{m}}/{N_{g}}=\;{E_{m}}/{E_{g}}

N_{m}=\;{196.85}/{214.6} × 300

N_{m}=275.18 {rpm}

Go to Question:

Prev Question

Next Question