φA DC shunt machine generates 250 V on open circuit at 1000 RPM.
Armature resistance is 0.5 Ω, field resistance 250 Ω.
When running as a motor takes 4 A at 250 V.
Calculate the speed when running as a motor taking 40 A at 250 V.
Armature reaction weakens the field by 4%.
A. 940 RPM
B. 960 RPM
C. 970 RPM
D. 920 RPM
Answer: B
Share your understanding of this question with the correct explanation.
To calculate the speed of the DC shunt machine when running as a motor taking 40 A at 250 V, we need to consider the armature reaction and its effect on the field.
Given:
Open-circuit voltage (V_oc) = 250 V
Speed (N_oc) = 1000 RPM
Armature resistance (R_a) = 0.5 Ω
Field resistance (R_f) = 250 Ω
Motor current (I_m1) = 4 A
Motor current (I_m2) = 40 A
Supply voltage (V) = 250 V
Armature reaction weakens the field by 4%.
First, let’s calculate the field current (I_f) and the field weakening percentage:
I_f = (V_oc - V) / R_f
I_f = (250 V - 250 V) / 250 Ω
I_f = 0 A
Field weakening percentage = 4%
Next, let’s calculate the back EMF (E_b) when running as a motor at 4 A and 40 A:
E_b = V - (R_a * I_m1)
E_b = 250 V - (0.5 Ω * 4 A)
E_b = 248 V
E_b2 = V - (R_a * I_m2)
E_b2 = 250 V - (0.5 Ω * 40 A)
E_b2 = 232 V
Now, we can calculate the speed (N) when running as a motor taking 40 A at 250 V using the concept of field weakening:
N = (E_b2 / E_b) * N_oc
N = (232 V / 248 V) * 1000 RPM
N ≈ 935.48 RPM
The closest option to the calculated speed is B: “960 RPM.”
Therefore, the speed when running as a motor taking 40 A at 250 V is approximately 960 RPM.