A motor running at its rated power provides 2.6 lb-ft torque at its rated speed of 2020 rpm. The motor draws 5.6-A from a 230-V supply at a phase angle of 37 degrees. Determine the following: The horsepower delivered by the motor, The efficiency of the motor.
Get the motor electrical input power Pin (in Watts)from the product of current, voltage and the cosine of the phase angle.
Get the delivered power Pout (in Watts) from the product of torque and angular velocity. Make sure the torque is in Newton meters and the angular velocity is in radians/sec
The motor efficiency is Pout/Pin
To determine the horsepower delivered by the motor, we need to calculate the power output. The power output can be obtained using the formula:
Power (in horsepower) = (Torque * Speed) / 5250
Given that the motor provides 2.6 lb-ft torque at a speed of 2020 rpm, we can calculate the power output as follows:
Power = (2.6 * 2020) / 5250
Now, we can calculate the power drawn by the motor from the supply using the formula:
Power input (in watts) = Voltage * Current * Power factor
The power factor (cos φ) can be obtained from the given phase angle (37 degrees) using trigonometry:
Power factor (cos φ) = cos(37 degrees)
Next, we calculate the power input:
Power input = 230 * 5.6 * cos(37 degrees)
To convert the power input from watts to horsepower, we divide the value by 746 (1 horsepower = 746 watts):
Power input (in horsepower) = Power input / 746
Finally, we can calculate the efficiency of the motor using the formula:
Efficiency = (Power output / Power input) * 100
Now we can plug in the values to get the answers:
1. Calculate the horsepower delivered by the motor:
Power = (2.6 * 2020) / 5250
2. Calculate the power input:
Power input = 230 * 5.6 * cos(37 degrees)
3. Convert the power input from watts to horsepower:
Power input (in horsepower) = Power input / 746
4. Calculate the efficiency of the motor:
Efficiency = (Power output / Power input) * 100