A DC motor takes an armature current of $I_a = 110$ A at a voltage of $V = 480$ V. The armature circuit resistance is $R_a = 0.2 \Omega$. The motor has $P = 6$ poles, and the armature is lap connected with $Z = 864$ conductors. The flux per pole is $\phi = 0.05$ Wb. We need to calculate (i) the speed $N$ and (ii) the gross torque $T_a$ developed by the armature.

1. Problem Description

A DC motor takes an armature current of $I_a = 110$ A at a voltage of $V = 480$ V. The armature circuit resistance is $R_a = 0.2 \Omega$. The motor has $P = 6$ poles, and the armature is lap connected with $Z = 864$ conductors. The flux per pole is $\phi = 0.05$ Wb. We need to calculate (i) the speed $N$ and (ii) the gross torque $T_a$ developed by the armature.

2. Solution Steps

(i) Calculation of Speed $N$:

First, we need to find the back EMF $E_b$. We know that

$V = E_b + I_a R_a$

So, $E_b = V - I_a R_a$

$E_b = 480 - (110 \times 0.2) = 480 - 22 = 458$ V

For a DC machine, the generated EMF is given by

$E_b = \frac{P \phi N Z}{60 A}$

Since the armature is lap connected, the number of parallel paths $A = P$. Therefore, $A=6$.

$E_b = \frac{P \phi N Z}{60 P} = \frac{\phi N Z}{60}$

Now, we can solve for the speed $N$:

$N = \frac{E_b \times 60}{\phi Z}$

$N = \frac{458 \times 60}{0.05 \times 864} = \frac{27480}{43.2} = 636.0 \text{ rpm}$

(ii) Calculation of Gross Torque $T_a$:

The gross torque developed by the armature is given by

$T_a = \frac{E_b I_a}{2 \pi N / 60}$

$T_a = \frac{E_b I_a \times 60}{2 \pi N}$

Substituting the known values:

$T_a = \frac{458 \times 110 \times 60}{2 \pi \times 636.0} = \frac{3022800}{4000.0} = 755.7 Nm$

Alternatively,

$T_a = \frac{P Z \phi I_a}{2 \pi A}$

Since it is lap wound, $A=P$,

$T_a = \frac{Z \phi I_a}{2 \pi}$

$T_a = \frac{864 \times 0.05 \times 110}{2 \pi} = \frac{4752}{2 \pi} = 756.3 Nm$

3. Final Answer

(i) Speed, $N = 636.0$ rpm

(ii) Gross torque, $T_a = 756.3$ Nm