Question

An electric motor draws a current of 10 amp with a voltage of \(110 \mathrm{~V}\). The output shaft develops a torque of \(10.2 \mathrm{~N} \cdot \mathrm{m}\) and a rotational speed of 1000 RPM. For operation at steady state, determine (a) the electric power required by the motor and the power developed by the output shaft, each in \(\mathrm{kW}\). (b) the net power input to the motor, in \(\mathrm{kW}\). (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, in \(\mathrm{kW} \cdot \mathrm{h}\) during \(2 \mathrm{~h}\) of operation.

Short answer

(a) 1.1 kW, 1.067 kW (b) 0.033 kW (c) 2.2 kW h, 2.134 kW h

Step-by-step solution

- Electric Power Required by the Motor

To find the electric power required by the motor, use the formula: \( P_{\text{electric}} = V \times I \)where \( V \) is the voltage and \( I \) is the current. Substituting the given values:\( P_{\text{electric}} = 110 \text{ V} \times 10 \text{ A} = 1100 \text{ W} = 1.1 \text{ kW} \)

- Power Developed by the Output Shaft

To find the power developed by the output shaft, use the formula: \( P_{\text{shaft}} = \frac{T \times \text{RPM} \times 2\text{π}}{60} \)where \( T \) is the torque and RPM is the rotational speed. Substituting the given values:\( P_{\text{shaft}} = \frac{10.2 \text{ Nm} \times 1000 \times 2\text{π}}{60} \) Calculating this gives: \( P_{\text{shaft}} = 1.067 \text{ kW} \)

- Net Power Input to the Motor

The net power input to the motor is the difference between the electric power required and the power developed by the output shaft:\( P_{\text{net}} = P_{\text{electric}} - P_{\text{shaft}} \)Substituting the values from previous steps:\( P_{\text{net}} = 1.1 \text{ kW} - 1.067 \text{ kW} = 0.033 \text{ kW} \)

- Energy Transferred to the Motor

To find the energy transferred to the motor by electrical work during 2 hours of operation, use the formula: \( E_{\text{in}} = P_{\text{electric}} \times t \)where \( t \) is time in hours. Substituting the given values:\( E_{\text{in}} = 1.1 \text{ kW} \times 2 \text{ h} = 2.2 \text{ kW} \times \text{h} \)

- Energy Transferred Out by the Shaft

To find the energy transferred out of the motor by the shaft during 2 hours of operation, use the formula: \( E_{\text{out}} = P_{\text{shaft}} \times t \)Substituting the given values:\( E_{\text{out}} = 1.067 \text{ kW} \times 2 \text{ h} = 2.134 \text{ kW} \times \text{h} \)

Key concepts

Electric Power

Electric power is the rate at which electrical energy is converted into another form of energy, such as mechanical energy in an electric motor. The unit of electric power is the watt (W), and it is typically measured in kilowatts (kW) for larger applications. To determine the electric power required by a motor, you use the formula: \( P_{\text{electric}} = V \times I \)where \( V \) is the voltage and \( I \) is the current. For the given problem, with a voltage of 110 V and a current of 10 A, the electric power is calculated as:\( P_{\text{electric}} = 110 \times 10 = 1100\;\text{W} = 1.1\;\text{kW} \).It's straightforward: just multiply the voltage by the current to get the power.

Torque and Rotational Speed

Torque and rotational speed are crucial for understanding the mechanical power developed by an electric motor. Torque (\(T\)) is a measure of how much rotational force is applied, measured in newton-meters (Nm). Rotational speed, on the other hand, is how fast the motor shaft is spinning, usually measured in revolutions per minute (RPM). Together, they help determine the power output via the formula:\( P_{\text{shaft}} = \frac{T \times \text{RPM} \times 2\pi}{60} \)In our specific example, the motor develops a torque of 10.2 Nm and a rotational speed of 1000 RPM. Plugging these values into the formula gives us:\( P_{\text{shaft}} = \frac{10.2 \times 1000 \times 2\pi}{60} \approx 1.067\;\text{kW} \).So, 1.067 kW of mechanical power is developed by the motor's output shaft.

Energy Transfer

Energy transfer measures how much energy moves from one form to another over a certain period. For electric motors, this often means converting electrical energy into mechanical energy. This is particularly important for efficiency calculations.
Electric energy transferred to the motor can be calculated using the formula:\( E_{\text{in}} = P_{\text{electric}} \times t \) where \(t\) is time in hours. For 2 hours of operation, with an electric power input of 1.1 kW, the energy transferred to the motor is:\( E_{\text{in}} = 1.1 \times 2 = 2.2\;\text{kW}\cdot \text{h} \).
Mechanical energy transferred out via the motor's shaft is calculated similarly:
\( E_{\text{out}} = P_{\text{shaft}} \times t \). With a shaft power of 1.067 kW over 2 hours, the energy transferred out is:\( E_{\text{out}} = 1.067 \times 2 = 2.134\;\text{kW}\cdot \text{h} \).The difference between energy input and energy output highlights losses like heat due to inefficiencies.