Question

A 1500 kW, 600 V, 16 pole, dc generator runs at 200 rpm.There are 2500 lap connected conductors and full load copper losses are 25kW. Average air gap flux density is 0.85T. The area of the pole shoes is_____ $c{m}^2$ (Neglect field current)

• A. 861

Answer 1

The correct option is A 861
$\text{Full load armature current,}$
$I_a=\frac{1500\times 1000}{600}=2500A$

$\text{copper-loss}=I_a^2{R}_a=25\times 1000=25kW$

where $R_a$ is the effective resistance of the armature circuit,

$R_a=\frac{25\times 1000}{2500\times 2500}=4\times {10}^{-3}\Omega$

$\text{Induced emf,}{E}_a=V_t+I_a{R}_a$
$=600+2500\times 4\times {10}^{-3}=610V$

Also, $E_a=\left(\frac{\varphi NZ}{60}\right)\left(\frac{P}{A}\right)$

$\Rightarrow 610=\frac{\varphi \times 200\times 2500}{60}\times \left(\frac{16}{16}\right)\text{(for lap connected A = P)}$

$\Rightarrow \varphi =0.0732Wb$

$\text{Area of the pole shoe}=\frac{0.0732}{0.85}=861c{m}^2$