Question

In a dc motor running at 1800 rpm, the hysteresis and eddy current losses are 400 W and 180 W respectively. If the flux remain constant, the speed at which the total iron losses are halved is _____rpm.

• A. 1038

Answer 1

The correct option is A 1038
For d.c.machines;

Hysteresis losses, $P_h=K_h{B}_m^{1.6}f$ .....(i)

Eddy current losses, $P_e=K_e{B}_m^2{f}^2$ ....(ii)

As we know that in the dc machine the induced emf and induced current in armature winding is alternating with frequency:
$f=\frac{PN}{120}$

or, $f\propto N$ ....(iii)
From equation (i),(ii) and (iii), we can get,

$P_h=K_{1}N$

$P_e=K_2{N}^2$

At, $N=1800rpm;$

$P_h=400W$

Now, $400=K_1\times 1800$

$K_1=\frac{2}{9}W/rpm$

Similarly, $N=1800rpm;$

$Pe=180W$

$180=K_2\times (1800{)}^2$
$K_2=\frac{180}{(1800{)}^2}=5.55\times {10}^{-5}W/(rpm{)}^2$

Total iron loss, $P_i=400+180=580W$

At the speed $N_2$;

The iron losses, $P_i^{\prime }=290W$

$P_i^{\prime }=K_1{N}_2+K_2{N}_2^2$

$290=\frac{2}{9}{N}^2+5.55\times {10}^{-5}{N}_2^2$

or, $5.55\times {10}^{-5}{N}_2^2+0.222N_2-290=0$

Solving, we get

$N_2=1037.30or-5037.30rpm$

only valid speed is;

$N_2=1037.30\approx 1038rpm$