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Long shunt compound generator speed, 300 conductors

Short answer

Armature current = 41 A, generated emf = 260.25 V. With lap winding (4 poles, 300 conductors, A=4), speed N = 60×4×260.25/(0.05×300×4) = 1041.0 rpm.

A 4 pole, 250 V long shunt dc compound generator supplies a load of 10 kW at the rated voltage. The armature, series and shunt field resistances are 0.1 Ω0.1\ \Omega, 0.15 Ω0.15\ \Omega and 250 Ω250\ \Omega respectively. The armature is lap wound with 300 conductors. If the flux per pole is 50 mWb, calculate the speed of the generator.

In a long shunt machine the shunt field is across the load terminals, so the series field carries the full armature current.

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

Currents

IL=10000250=40 A,Ish=250250=1 AI_L = \frac{10000}{250} = 40\ \text{A}, \qquad I_{sh} = \frac{250}{250} = 1\ \text{A} Ia=Ise=IL+Ish=41 AI_a = I_{se} = I_L + I_{sh} = 41\ \text{A}

Generated emf

Eg=V+Ia(Ra+Rse)=250+41(0.1+0.15)=260.25 VE_g = V + I_a(R_a + R_{se}) = 250 + 41(0.1 + 0.15) = 260.25\ \text{V}

Speed

For lap winding A=P=4A = P = 4:

Eg=ΦZNP60A  ⇒  N=60AEgΦZP=60×4×260.250.05×300×4=1041.0 rpmE_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}

Answer: generator speed =1041.0 rpm= 1041.0\ \text{rpm} (brush drop not given, so neglected).

✓ Answer checked by Claude Opus 5.5 (high effort)

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