Question

Armature of a 6-pole, 6-circuit D.C. shunt motor takes 400 A at a speed of 350 r.p.m. The flux per pole is 80 milli-webers, the number of armature turns is 600, and 3% of the

torque is lost in windage, friction and iron-loss. Calculate the brake-horse-power.

Answer 1

Solution:-

Number of armature conducter=1200

Electromagnetic torque developed is

=T Nw-m

Armature power =Tw

$$P=T\times2\pi\times\frac{350}{60}=36.67Twatt$$

To calculate armature power in term of electrical parameters E must be known

$E=\frac{\phi zN}{60}\times\frac{P}{A}$

$$E=80\times10^{-3}\times1200\times\frac{350}{60}\times\frac{6}{6}=560V$$

With the current of 400 A armature power =500 $\times$ 400 watt

$$T=560\times\frac{400}{36.67}=6108..5Nw-m$$

Since 3% of this torque is required for coming different loss term

Net torque $=0.97\times6180=5925Nw-m$

1hp=0.746kW

Where hp= horse power

Net output power in (kW) =

$$5925\times36.67\times10^{-3}=217.27KW$$

Converting in(brake horse power) BHP output =291.25HP