A belt-driven DC shunt generator running at 300 rpm delivers 100 kW to a 200 V DC grid. It continues to run as a motor when the belt breaks, taking 10 kW from the DC grid. The armature resistance is 0.025 Ω, field resistance is 50 Ω and brush drop is 2 V. Ignoring armature reaction, the speed of the motor is _____ rpm. (Round off to 2 decimal places.)

The correct answer is 275.19 (or within the range of 275.00 to 275.25).

Here is the step-by-step derivation:

1. Identify the given parameters:
Terminal voltage, $V=200\text{V}$
Armature resistance, $R_a=0.025\text{Ω}$
Field resistance, $R_f=50\text{Ω}$
Total brush voltage drop, $V_{\text{brush}}=2\text{V}$
Generator speed, $N_g=300\text{rpm}$

2. Generator Operation:
The generator delivers an output power of $P_{\text{out}}=100\text{kW}=\mathrm{100,000}\text{W}$ to the grid.
The generator line current is:
$I_{\text{Lg}}=\frac{P_{\text{out}}}{V}=\frac{\mathrm{100,000}}{200}=500\text{A}$

The shunt field current (which remains constant in both generator and motor modes as it is connected directly across the constant grid voltage) is:
$I_f=\frac{V}{R_f}$

The generator armature current is:
$I_{\text{ag}}=I_{\text{Lg}}+I_f=500+4=504\text{A}$

The induced electromotive force (EMF) in the generator is:
$E_g=V+I_{\text{ag}}{R}_a+V_{\text{brush}}$
$E_g=200+(504\times 0.025)+2=200+12.6+2=214.6\text{V}$

3. Motor Operation:
When the belt breaks, the machine runs as a motor, drawing an input power of $P_{\text{in}}=10\text{kW}=\mathrm{10,000}\text{W}$ from the grid.
The motor line current is:
$I_{\text{Lm}}=\frac{P_{\text{in}}}{V}=\frac{\mathrm{10,000}}{200}=50\text{A}$

The motor armature current is:
$I_{\text{am}}=I_{\text{Lm}}-I_f=50-4=46\text{A}$

The back EMF of the motor is:
$E_m=V-I_{\text{am}}{R}_a-V_{\text{brush}}$
$E_m=200-(46\times 0.025)-2=200-1.15-2=196.85\text{V}$

4. Calculate the speed of the motor:
Since the armature reaction is ignored and the field current is constant, the magnetic flux ($\Phi$) is constant. Thus, the induced EMF is directly proportional to the rotational speed ($N$):
$E\propto N\Rightarrow \frac{E_m}{E_g}=\frac{N_m}{N_g}$

Substituting the known values:
$N_m=N_g\times \frac{E_m}{E_g}=300\times \frac{196.85}{214.6}\approx 275.186\text{rpm}$

Rounding off to 2 decimal places, the speed of the motor is 275.19 rpm.