A 4 pole 240V, D.C series motor has a wave wound ammature with 180 conductors. the ammature resistance and field winding resistance are 0.1022 and 0.2022 respectively. If the mo tor is drawing 40A current at 0.015 webers flux I per pole, then the speed under this condition is___rpm. 1)2533.33 2) 2756.9 3) 2644.4 4) 2987.6?

Question

Answer

Solution:

Given data:

Number of poles, P = 4
Voltage, V = 240 V
Armature conductors, Z = 180
Armature resistance, Ra = 0.1022 Ω
Field resistance, Rf = 0.2022 Ω
Armature current, Ia = 40 A
Flux per pole, Φ = 0.015 Wb

1. Calculation of back e.m.f.:

Back e.m.f. is given by the expression,

Eb = PΦZN / 60A

where N is the speed of the motor in rpm.

Substituting the given values, we get

Eb = 4 × 0.015 × 180 / (60 × 40) = 0.027 V

2. Calculation of total voltage drop:

The total voltage drop in the motor is given by the expression,

Vt = Eb + IaRa + IaRf

Substituting the given values, we get

Vt = 0.027 + 40 × 0.1022 + 40 × 0.2022 = 12.97 V

3. Calculation of speed:

The speed of the motor is given by the expression,

N = (V - Vt) / (PΦZN / 60)

Substituting the given values, we get

N = (240 - 12.97) / (4 × 0.015 × 180 / 60) = 2533.33 rpm

Therefore, the speed of the motor under the given condition is 2533.33 rpm.

Answer: 1) 2533.33

Answer

2533.33

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