An electric motor drawing 10 amps at 110 V in steady state produces shaft power at 9.7 Nm and 1000 RPM. For a first Law analysis and conside

Question

An electric motor drawing 10 amps at 110 V in steady state produces shaft power at 9.7 Nm and 1000 RPM. For a first Law analysis and considering the motor as the control volume How much heat will be produced from the motor (in Watts)

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Thiên Thanh 3 years 2021-08-17T06:45:36+00:00 1 Answers 10 views 0

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  1. kimcuc
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    2021-08-17T06:47:35+00:00
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    Answer:

    The heat rate produced from the motor is 84.216 watts.

    Explanation:

    The electric motor receives power from electric current and releases power in the form of mechanical energy (torque) and waste heat and can be considered an stable-state system. The model based on the First Law of Thermodynamics for the electric motor is:

    \dot W_{e} - \dot W_{T} -\dot Q = 0

    Where:

    \dot Q – Heat transfer from the electric motor, measured in watts.

    \dot W_{e} – Electric power, measured in watts.

    \dot W_{T} – Mechanical power, measured in watts.

    The heat transfer rate can be calculated in terms of electric and mechanic powers, that is:

    \dot Q = \dot W_{e} - \dot W_{T}

    The electric and mechanic powers are represented by the following expressions:

    \dot W_{e} = i \cdot V

    \dot W_{T} = T \cdot \omega

    Where:

    i – Current, measured in amperes.

    V – Steady-state voltage, measured in volts.

    T – Torque, measured in newton-meters.

    \omega – Angular speed, measured in radians per second.

    Now, the previous expression for heat transfer rate is expanded:

    \dot Q = i \cdot V - T \cdot \omega

    The angular speed, measured in radians per second, can be obtained by using the following expression:

    \omega = \frac{\pi}{30}\cdot \dot n

    Where:

    \dot n – Rotational rate of change, measured in revolutions per minute.

    If \dot n = 1000\,rpm, then:

    \omega = \left(\frac{\pi}{30} \right)\cdot (1000\,rpm)

    \omega \approx 104.720\,\frac{rad}{s}

    Given that i = 10\,A, V = 110\,V, T = 9.7\,N\cdot m and \omega \approx 104.720\,\frac{rad}{s}, the heat transfer rate from the electric motor is:

    \dot Q = (10\,A)\cdot (110\,V) -(9.7\,N\cdot m)\cdot \left(104.720\,\frac{rad}{s} \right)

    \dot Q = 84.216\,W

    The heat rate produced from the motor is 84.216 watts.

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