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)

Answer 1

The heat rate produced from the motor is 84.216 watts.

Step-by-step 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 = (\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((\pi)/(30) \right)\cdot (1000\,rpm)`

`\omega \approx 104.720\,(rad)/(s)`

Given that
`i = 10\,A`,
`V = 110\,V`,
`T = 9.7\,N\cdot m` and
`\omega \approx 104.720\,(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\,(rad)/(s) \right)`

`\dot Q = 84.216\,W`

The heat rate produced from the motor is 84.216 watts.