The actual torque available at the motor shaft for driving the mechanical equipment is known as **shaft torque(T _{sh})**. In fact, the same electromagnetic or gross torque(T

_{a}) developed by the DC motor can not be available at the shaft, however, it is somewhat less than the electromagnetic torque. The reason for this is that some part of it was lost in overcoming the iron and mechanical losses. The shaft torque is the net torque available for driving the mechanical equipment. We shall study about shaft torque of the DC motor in this article.

The below-given expression shows the **relation between power and torque.**

## Formula of shaft torque of DC Motor

Shaft torque is less than the armature torque in the case of DC motor.

The difference between the armature torque and the shaft torque shows the lost torque.

### Lost Torque Formula of DC Motor

**Lost Torque = Armature or Gross Torque – Shaft Torque**

The mechanical power available at the shaft is the Brake Horsepower of the motor.

## Brake Horse Power (B.H.P) of DC Motor

We can express the motor output power in BHP as follows.

### Solved problem on Shaft Torque of DC Motor

*Determine* *developed* *torque* *and shaft torque of 220-V, the 4-pole series motor with 900 conductors wave-connected supplying a load of 9.0 kW by taking 50 A from the mains. The flux per pole is 30 mWb and its armature circuit resistance is 0.5 Ω*

Gross torque = Armature torque

T_{a }= 0.159 X Φ ZI_{a }(P/A)

T_{a}= 0.159 X 30 x 10^{-3} x 900 x 50 (4/2) = 429.3 N-m

E_{b }= V – Ia Ra = 220 – 50 X 0.5 = 195 V

**E _{b} = 195 Volts**

Now,

E_{b }= Φ ZN (P/A)= 195 = 30 x 10^{-3} x 900 x N x (4/2)

N = 195 /(30 X10^{-3} X 900 x 2)

N = 3.611 r.p.s.

Also,

2πN T_{sh} = Power output

2π X 3.611 Tsh =9 x 10^{3
}T_{sh} =9 x 10^{3} / ( 2π X 3.611)

**T _{sh} = 396.87 N-m**

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