/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 An electric motor draws a curren... [FREE SOLUTION] | 91Ó°ÊÓ

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An electric motor draws a current of 10 amperes (A) with a voltage of \(110 \mathrm{~V}\). The output shaft develops a torque of \(9.5 \mathrm{~N} \cdot \mathrm{m}\) and a rotational speed of \(1000 \mathrm{rpm}\). All operating data are constant with time. Determine (a) the electric power required by the motor and the power developed by the output shaft, each in kilowatts; (b) the net power input to the motor, in kilowatts; (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 \(\mathrm{kW} \cdot \mathrm{h}\) and Btu, during \(2 \mathrm{~h}\) of operation.

Short Answer

Expert verified
(a) The electric power required by the motor is 1.1 kW and the power developed by the output shaft is 0.995 kW. (b) The net power input to the motor is 0.105 kW. (c) The amount of energy transferred to the motor by electrical work is 2.2 kWh or 7513.2 Btu and the amount of energy transferred out of the motor by the shaft is 1.99 kWh or 6793.2 Btu.

Step by step solution

01

Compute Electric and Shaft Power

The electric power (\(P_{elec}\)) required by the motor is given by the formula \(P_{elec}=IV\), where \(I\) is the current and \(V\) is the voltage. Thus, \(P_{elec}=10A*110V = 1100 W = 1.1 kW\). Now, we compute the power developed by the output shaft, which is given by the formula \(P_{shaft}=T\omega\), where \(T\) is the torque and \(\omega\) is the angular speed. Remember to convert rotations per minute (rpm) into radians per second (rad/s) using \(\omega=2\pi N/60\), where \(N\) is the rpm. Thus, \(\omega = 2\pi*1000rpm/60 = 104.72 rad/s\). Hence, \(P_{shaft}= 9.5Nm*104.72 rad/s = 994.84 W = 0.995 kW\).
02

Compute Net Power Input

The net power input to the motor is the difference between the electrical power input and the power developed by the output shaft, which is \(P_{net}=P_{elec}-P_{shaft}=1.1 kW - 0.995 kW = 0.105 kW\).
03

Determine Energy Transferred

The energy transferred to and from the motor can be computed by multiplying the powers calculated in Steps 1 and 2 by the time (\(t = 2h = 2*60*60 = 7200 s\)). Thus, the energy transferred to the motor by electrical work (\(E_{elec}\)) is \(E_{elec}= P_{elec}*t = 1.1 kW * 2 h = 2.2 kWh\), and the energy transferred from the motor by the shaft (\(E_{shaft}\)) is \(E_{shaft}= P_{shaft}*t = 0.995 kW * 2 h = 1.99 kWh\). To convert these energies to British Thermal Units (Btu), use the conversion factor of 1 Btu = 0.00029307107 kWh. Thus, \(E_{elec}= 2.2 kWh / 0.00029307107 = 7513.2 Btu\) and \(E_{shaft}= 6793.2 Btu\).

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