Long shunt compound generator speed, lap and wave

In a long shunt compound generator the shunt field is across the armature plus series field combination (i.e. across the load terminals), so the series field carries the armature current.

       Ia        Rse
  +--->---[~~~~]---+---------+----o +
  |                |         |
 (E,Ra)          [Rsh]     Load
  |                |         |
  +----------------+---------+----o -

Currents

$$I_L = \frac{10000}{250} = 40\ \text{A}, \qquad I_{sh} = \frac{250}{250} = 1\ \text{A}$$ $$I_a = I_{se} = I_L + I_{sh} = 40 + 1 = 41\ \text{A}$$

Generated emf

$$E_g = V + I_a(R_a + R_{se}) = 250 + 41(0.1 + 0.15) = 260.25\ \text{V}$$

Speed with lap winding

Slots $\times$ conductors per slot: $Z = 50\times 6 = 300$. For lap winding $A = P = 4$.

$$E_g = \frac{\Phi Z N P}{60A} \;\Rightarrow\; N = \frac{60AE_g}{\Phi Z P} = \frac{60\times 4\times 260.25}{0.05\times 300\times 4} = 1041.0\ \text{rpm}$$

Speed with wave winding

For wave winding $A = 2$; $E_g$, $\Phi$, $Z$ and $P$ are unchanged.

$$N = \frac{60\times 2\times 260.25}{0.05\times 300\times 4} = 520.5\ \text{rpm}$$

The wave-wound machine needs only half the speed because it has half the parallel paths (twice the conductors in series per path).

Answer: lap wound: $N = 1041.0\ \text{rpm}$; wave wound: $N = 520.5\ \text{rpm}$. (Brush drop is not given, so it is neglected.)