Question

A 50 Hp, 250 V, 1200 rpm dc shunt motor without compensating windings has an armature resistance (including brushes and interpoles) of 0.06 $\Omega$. Its field circuit has a resistance of 50 $\Omega$, which produces a no load speed of 1200 rpm. There are 1200 turns per pole on the shunt field winding, and the armature reaction produces a demagnetizing magneto motive force of 840 AT at a load current of 200 A. The magnetizing curve of this machine is shown below,

The speed of this motor when its input current is 200 A is

• A. 1160 rpm
• B. 1412 rpm
• C. 1227 rpm
• D. 11200 rpm

Answer 1

The correct option is C 1227 rpm

$I_a=I_L-I_f=200-\frac{250}{50}=195A$

$E_a=V_t-I_a{R}_a=250-(195\times 0.06)=238.3V$

$\text{At}{I}_L=200A\text{, the demagnetizing mmf is 840 AT.}$
$\text{So, effective shunt field current of the motor is}$

$I_f^{\ast }=I_f-\frac{mmf}{N_f}=5-\frac{840}{1200}=4.3A$

$\text{From the given graph, if}{I}_f^{\ast }=4.3A$
$\text{then}{E}_{noload}\text{=233 V}$

$\frac{E_{1,nl}}{E_2}=\frac{N_{1,nl}}{N_2}$

$\frac{233}{238.3}=\frac{1200}{N_2}$

$\text{Speed of the motor, N = 1227.3 rpm}$