Question

A 230 V, dc series has 500 lap connected armature conductor. It is supplying a shaft load of 1.5 kW. Armature and field winding resistances are 0.02 $\Omega$/phase and 0.04$\Omega$/phase respectively. Friction and windage losses are 100 W. At this load flux per pole is 0.035 Wb. Then speed of motor at this load is

• A. 787.13 rpm
• B. 813.22 rpm
• C. 790.1 rpm
• D. 900.32 rpm

Answer 1

The correct option is A 787.13 rpm


$\text{Armature power developed,}$

$\text{= shaft power + rotational losses}$

$\text{= 1.5 kW + 0.1 kW}$

$E_b{I}_a=1.6kW$

$\text{Back emf is given by,}$

$E_b=\frac{NP\varphi Z}{A\times 60}=\frac{N\times 0.035\times 500}{60}$

$\text{(For lap A = P)}$

$\text{and armature current,}{I}_a=\frac{230-E_b}{0.06}$

$E_b\left(\frac{230-E_b}{0.06}\right)=1600$

$230E_b-E_b^2-96=0$

$E_b=229.58V$

$\text{Speed, N =}\frac{60E_b}{\varphi Z}=\frac{60\times 229.58}{0.035\times 500}$

$N=787.13rpm$