The motor is running at D) 1594.8 rpm with the external resistance of 50 Ω.Therefore , D) 1594.8 rpm is correct .
Here's the solution:
Given: Power (P) = 20 kW = 20,000 W ,Voltage (V) = 300, VArmature resistance (Ra) = 0.02 Ω ,Field resistance (Rf) = 100 Ω ,Line current (I) = 100 A ,Speed (N) = 1600 rpm.
Step 1: Calculate armature current (Ia) Ia = I - If where If is the field current.
If = V / Rf = 300 V / 100 Ω = 3 A
Ia = 100 A - 3 A = 97 A
Step 2: Calculate back emf (Eb) Eb = V - IaRa
Eb = 300 V - 97 A * 0.02 Ω = 283.4 V
Step 3: Calculate torque (T) T = (Eb * Ia) / ϕ
where ϕ is the magnetic flux per pole.
Assuming a linear magnetization curve, the flux per pole is proportional to the field current. Since the field current is reduced by 50 Ω, the flux per pole is also reduced by 50%.
Therefore, the new flux per pole is ϕ = 0.5 * ϕ₀, where ϕ₀ is the original flux per pole.
T = (283.4 V * 97 A) / (0.5 * ϕ₀)
T = 54,608.8 ϕ₀ Nm / rad
Step 4: Calculate speed (N) with reduced field current
For a shunt motor, the speed is inversely proportional to the field current. Therefore, when the field current is reduced by 50%, the speed increases by 100%.
N = 1600 rpm * 2 = 3200 rpm
Step 5: Calculate new speed (N') with external resistance
The external resistance of 50 Ω adds to the field resistance, effectively reducing the field current even further.
To calculate the new field current, use the following equation:
If' = V / (Rf + 50 Ω)
If' = 300 V / (150 Ω) = 2
A The new back emf is:
Eb' = V - IaRa - If'Rf'
Eb' = 300 V - 97 A * 0.02 Ω - 2 A * 100 Ω = 186 V
The new torque is:
T' = (Eb' * Ia) / (0.5 * ϕ₀)
T' = (186 V * 97 A) / (0.5 * ϕ₀) = 35,292 ϕ₀ Nm / rad
The new speed is: N' = (T' * 60) / (2πϕ₀)
N' = (35,292 ϕ₀ Nm / rad * 60) / (2πϕ₀)
N' = 1594.8 rpm
Therefore, the motor is running at 1594.8 rpm with the external resistance of 50 Ω.
Question
A 20 kW, 300 V dc shunt motor is driving a constant torque load with line current of 100 A at a speed of 1600 rpm. Motor has armature and field winding resistance as 0.02 Ω and 100 Ω respectively. If 50 Ω external resistance is added in field circuit, then motor is running at (Assume linear magnetzation curve)
A) 1812.9 rpm
B) 2100 rpm
C) 2392.18 rpm
D) 1594.8 rpm