Torque in DC Motor - Armature Torque and Shaft Torque

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The torque is defined as the turning moment of a force about an axis. It is measure by the product of the force (F) and perpendicular distance (r) of the line of action of force from the axis of rotation, i.e.,

$$\mathrm{Torque,\: \tau \: = \: F \: \times \: r \:\: \dotso \: (1)}$$

The torque is measured in Newton-meters (Nm).

Armature Torque of DC motor

In a DC motor, a circumferential force (F) at a distance r which is the radius of the armature is acted on each conductor, tending to rotate the armature. The sum of the torques due to all the armature conductors is known as armature torque (τ_a).

Let,

P = Number of Poles

r = Radius of The Armature

l = Effective Length of Each Conductor

Z = Total Number of Armature Conductors

A = Number of Parallel Paths

i = Current in Each Conductor

B = magnetic Flux Density

ϕ = Flux Per Pole

Therefore,

$$\mathrm{\text{Force on each conductor, } \: F \: = \: B \: i \: l}$$

$$\mathrm{\text{Torque developed by one condcutor } \: = \: F \: \times \: r}$$

$$\mathrm{\therefore \: \text{ Total armature torque, } \: \tau_{a} \: = \: Z \: \times \: (F \: \times \: r) \: = \: Z B i l r}$$

Since,

$$\mathrm{\text{Current in each conductor, } \: i \: = \: \frac{I_{a}}{A}}$$

$$\mathrm{\text{magnetic flux density, } \: B \: = \: \frac{\varphi}{a}}$$

Where,

$$\mathrm{a \: = \: cross \: - \: \text{ sectional area of flux path at radius } \: r \: = \: \frac{2 \: \pi \: rl}{P}}$$

$$\mathrm{\therefore \: \tau_{a} \: = \: Z \: \times \: \frac{\varphi}{a} \: \times \: \frac{I_{a}}{A} \: \times \: l \: \times \: r \: = \: Z \: \times \: \frac{\varphi}{2\pi \: rl \: / \:P} \: \times \: \frac{I_{a}}{A} \: \times \: l \: \times \: r}$$

$$\mathrm{\Rightarrow \: \tau_{a} \: = \: \frac{PZ}{2 \:\pi \: A} \:\varphi \: I_{a} \: Nm \:\: \dotso \: (2)}$$

The expression in the equation (2) is known as the armature torque of DC motor.

For a given DC motor, the (PZ/2πA) is a constant. Hence,

$$\mathrm{\tau_{a} \: \varpropto \: \varphi \: I_{a} \:\: \dotso \: (3)}$$

Therefore, the armature torque developed in a DC motor is directly proportional to the flux per pole and the armature current.

Also, the back emf of a DC motor is given by,

$$\mathrm{E_{b} \: = \: \frac{NP \: \varphi \: Z}{60A}}$$

$$\mathrm{\Rightarrow \: \frac{P \: \varphi \: Z}{A} \: = \: \frac{60 \: \times \: E_{b}}{N} \:\: \dotso \: (4)}$$

From eqns. (2) & (4), we get,

$$\mathrm{\tau_{a} \: = \: \frac{1}{2\pi} \: \times \: \left(\frac{60 \: \times E_{b}}{N}\right) \: \times \: I_{a}}$$

$$\mathrm{\Rightarrow \: \tau_{a} \: = \: 9.55 \: \times \: \frac{E_bI_a}{N}Nm \:\: \dotso \: (5)}$$

Shaft Torque of DC motor

The torque in a DC motor which is available at the shaft of the machine for doing useful work is called as shaft torque (&tau_sh).

In a DC motor, the total torque developed in the armature of the motor is not transferred to the shaft of the motor as a part of it is lost in overcoming the mechanical losses in the machine. Hence, the shaft torque is somewhat less than the armature torque.

If the speed of the motor is N rpm, then the output shaft power (in watts) of the DC motor is given by,

$$\mathrm{P_{sh} \: = \: \frac{2 \:\pi \: N \:\tau_{sh}}{60}}$$

$$\mathrm{\Rightarrow \: \tau_{sh} \: = \: 9.55 \: \times \: \frac{P_{sh}}{N} \: Nm \:\: \dotso \: (6)}$$

Also, the difference of the armature torque and the shaft torque is known as lost torque i.e.

$$\mathrm{\text{Lost Torque } \: = \: \tau_{a} \: - \: \tau_{sh} \: = \: 9.55 \: \times \: \frac{\text{mechanical Losses}}{N} \:\: \dotso \: (7)}$$

Numerical Example

A 240 V DC shunt motor takes a total current of 30 A and runs at 1500 RPm. If the armature and shunt field resistances are 0.30 Ω and 240 Ω respectively. Find the torque developed by the armature. If the total shaft power output is 10 h.p. Also, determine the shaft torque and lost torque of the motor.

Solution

Here,

$$\mathrm{\text{Shunt field curren, } \: I_{sh} \: = \: \frac{240}{240} \: = \: 1 \:A}$$

$$\mathrm{\text{Armature current, } \: I_{a} \: = \: 30 \: - \: 1 \: = \: 29 \:A}$$

$$\mathrm{\text{Back EmF, } \: E_{b} \: = \: V \: - \: I_{a}R_{a} \: = \: 240 \: - \: (24 \: \times \: 0.30) \: = \: 232.8 \: V}$$

Therefore, the armature torque is,

$$\mathrm{\tau_{a} \: = \: 9.55 \: \times \: \frac{E_bI_a}{N} \: = \: 9.55 \: \times \: \left(\frac{232.8 \: \times \: 29}{1500 }\right) \: = \: 42.98 \: Nm}$$

The shaft torque is,

$$\mathrm{\tau_{sh} \: = \: 9.55 \: \times \: \frac{P_{sh}}{N} \: = \: 9.55 \: \times \: \left(\frac{746 \: \times \: 7}{1500}\right) \: = \: 33.25 \: Nm}$$

The lost torque is,

$$\mathrm{\text{Lost torque }\: = \: \tau_{a} \: - \: \tau_{sh} \: = \: 42.98 \: - \: 33.25 \: = \: 9.73 \:Nm}$$