Question

An electric motor draws a current of 10 amp with a voltage of 110 V. The output shaft develops a torque of 10.2 N-m and a rotational speed of 1000 rpm. For steady-state operation, determine:________.a) the electric power required by the motor and the power developed by the output shaft, each in kW. b) the net power input to the motor, in kW. c) the amount of energy transferred to the motor by electrical work and the amount of energy transferred out of the motor by the shaft.

Answer 1

The electric power required by the motor is 1.1 kW, and the power developed by the output shaft is 1.07 kW. The net power input to the motor is 0.03 kW. The amount of energy transferred to the motor by electrical work is 1.1 kWh, and the amount of energy transferred out of the motor by the shaft is 1.07 kWh.

Step-by-step explanation:

Given information:

• Current drawn by the electric motor (I) = 10 A
• The voltage across the motor (V) = 110 V
• Torque developed by the output shaft (T) = 10.2 N-m
• The rotational speed of the output shaft (ω) = 1000 rpm

(a) To calculate the electric power required by the motor, we use the formula: Power (P) = Voltage (V) x Current (I). Substituting the given values, we get P = 110 V x 10 A = 1100 W = 1.1 kW. Therefore, the electric power required by the motor is 1.1 kW. To calculate the power developed by the output shaft, we use the formula: Power (P) = Torque (T) x Angular speed (ω). Substituting the given values, we get P = 10.2 N-m x (1000 rpm x 2π / 60) = 1072.3 W = 1.07 kW. Therefore, the power developed by the output shaft is 1.07 kW.

(b) The net power input to the motor can be calculated by subtracting the power developed by the output shaft from the electric power required by the motor. Net power input = Electric power required - Power developed = 1.1 kW - 1.07 kW = 0.03 kW. Therefore, the net power input to the motor is 0.03 kW.

(c) The amount of energy transferred to the motor by electrical work can be calculated using the formula: Electrical work = Power x Time. Substituting the given values, we get Electrical work = 1.1 kW x 1 hour = 1.1 kWh. The amount of energy transferred out of the motor by the shaft can be calculated using the formula: Mechanical work = Power x Time. Substituting the given values, we get Mechanical work = 1.07 kW x 1 hour = 1.07 kWh. Therefore, the amount of energy transferred to the motor by electrical work is 1.1 kWh and the amount of energy transferred out of the motor by the shaft is 1.07 kWh.

Answer 2

Answer 1:

The input power into the mother is 1.1 kW

The output power at the shaft is = 1.068 kW

Step-by-step explanation:

Electrical power = input = Volt*current = 10*110 = 1100W = 1.1kW

Motor shaft power = output = Torque*angular speed

Angular speed = 2¶N/60 = (2*3.142*1000)/60 = 104.4 rad/sec

Ouput = 10.2*104.4 =1068W = 1.068kW

Answer 2: net power input = 0.032kW

Explanation: net power input = power input-power output.

Answer 3: energy transfered into the motor = 1100J in one sec

Energy transfer to shaft is 1068J in one sec

Explanation: power is energy expended per unit sec.