Solutions to DC Motor and Generator Problems
Problem 1
Calculate the torque in newton-meters developed by a 440-V DC motor having an armature
resistance of 0.25 Ω and running at 750 RPM when taking a current of 60 A.
Solution:
The torque equation for a DC motor is given by:
T = P_armature / ω
where:
- P_armature is the power developed in the armature (in watts).
- ω is the angular velocity (in rad/s).
Step 1: Calculate the armature voltage:
V_armature = V - I_a * R_a
V_armature = 440 - (60 * 0.25) = 425 V
Step 2: Calculate the armature power:
P_armature = V_armature * I_a = 425 * 60 = 25,500 W
Step 3: Calculate the angular velocity:
ω = (2 * π * N) / 60, where N = 750 RPM
ω = (2 * π * 750) / 60 = 78.54 rad/s
Step 4: Calculate the torque:
T = P_armature / ω = 25,500 / 78.54 = 324.6 Nm
Final Answer: The torque developed is 324.6 Nm.
Problem 2
A 4-pole, lap-connected DC motor has 576 conductors and draws an armature current of 10
A. If the flux per pole is 0.02 Wb, calculate the armature torque developed.
Solution:
The torque equation for a DC motor in terms of flux and current is given by:
T = (P * φ * I_a * Z) / (2 * π * A)
where:
- P = number of poles = 4
- φ = flux per pole (in Wb) = 0.02 Wb
- I_a = armature current (in A) = 10 A
- Z = total number of armature conductors = 576
- A = number of parallel paths = P (for lap connection)
Step 1: Substitute the values:
T = (4 * 0.02 * 10 * 576) / (2 * π * 4)
T = (460.8) / (25.1327)
T = 18.34 Nm
Final Answer: The armature torque developed is 18.34 Nm.
Problem 3
A DC shunt machine has armature and field resistances of 0.025 Ω and 80 Ω respectively.
When connected to constant 400-V bus-bars and driven as a generator at 450 RPM, it
delivers 120 kW. Calculate its speed when running as a motor and absorbing 120 kW from
the same bus-bars.
Solution:
Step 1: Calculate the generated EMF (E_g) when the machine is operating as a generator:
P = V * I, where P = 120 kW, V = 400 V
I = P / V = 120,000 / 400 = 300 A
Armature current: I_a = I - I_f
I_f = V / R_f = 400 / 80 = 5 A
I_a = 300 - 5 = 295 A
Generated EMF: E_g = V + I_a * R_a
E_g = 400 + (295 * 0.025) = 407.375 V
Step 2: Calculate the generated EMF (E_m) when operating as a motor:
P = V * I, where P = 120 kW, V = 400 V
I = P / V = 120,000 / 400 = 300 A
Armature current: I_a = I - I_f
I_a = 300 - 5 = 295 A
E_m = V - I_a * R_a
E_m = 400 - (295 * 0.025) = 392.625 V
Step 3: Use the speed relation to calculate the motor speed:
N_m / N_g = E_m / E_g
N_m = N_g * (E_m / E_g) = 450 * (392.625 / 407.375) = 433.61 RPM
Final Answer: The motor runs at approximately 433.61 RPM.
Problem 4
The armature current of a series motor is 60 A when on full-load. If the load is adjusted so
that this current decreases to 40 A, find the new torque expressed as a percentage of the
full-load torque. The flux for a current of 40 A is 70% of that when current is 60 A.
Solution:
Step 1: Torque in a DC series motor is proportional to the square of the current times the
flux (T ∝ I_a * φ).
Let T_f and T_n represent the full-load and new torques, respectively.
T_f ∝ I_f^2 * φ_f, where I_f = 60 A, φ_f = flux at 60 A
T_n ∝ I_n^2 * φ_n, where I_n = 40 A, φ_n = 0.7 * φ_f
Torque ratio: T_n / T_f = (I_n^2 * φ_n) / (I_f^2 * φ_f)
T_n / T_f = (40^2 * 0.7) / (60^2 * 1) = (1600 * 0.7) / 3600 = 1120 / 3600 = 0.3111
Step 2: Express the new torque as a percentage of the full-load torque:
T_n (%) = 0.3111 * 100 = 31.11%
Final Answer: The new torque is 31.11% of the full-load torque.
