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:
[tex]\dot W_{e} - \dot W_{T} -\dot Q = 0[/tex]
Where:
[tex]\dot Q[/tex] - Heat transfer from the electric motor, measured in watts.
[tex]\dot W_{e}[/tex] - Electric power, measured in watts.
[tex]\dot W_{T}[/tex] - Mechanical power, measured in watts.
The heat transfer rate can be calculated in terms of electric and mechanic powers, that is:
[tex]\dot Q = \dot W_{e} - \dot W_{T}[/tex]
The electric and mechanic powers are represented by the following expressions:
[tex]\dot W_{e} = i \cdot V[/tex]
[tex]\dot W_{T} = T \cdot \omega[/tex]
Where:
[tex]i[/tex] - Current, measured in amperes.
[tex]V[/tex] - Steady-state voltage, measured in volts.
[tex]T[/tex] - Torque, measured in newton-meters.
[tex]\omega[/tex] - Angular speed, measured in radians per second.
Now, the previous expression for heat transfer rate is expanded:
[tex]\dot Q = i \cdot V - T \cdot \omega[/tex]
The angular speed, measured in radians per second, can be obtained by using the following expression:
[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]
Where:
[tex]\dot n[/tex] - Rotational rate of change, measured in revolutions per minute.
If [tex]\dot n = 1000\,rpm[/tex], then:
[tex]\omega = \left(\frac{\pi}{30} \right)\cdot (1000\,rpm)[/tex]
[tex]\omega \approx 104.720\,\frac{rad}{s}[/tex]
Given that [tex]i = 10\,A[/tex], [tex]V = 110\,V[/tex], [tex]T = 9.7\,N\cdot m[/tex] and [tex]\omega \approx 104.720\,\frac{rad}{s}[/tex], the heat transfer rate from the electric motor is:
[tex]\dot Q = (10\,A)\cdot (110\,V) -(9.7\,N\cdot m)\cdot \left(104.720\,\frac{rad}{s} \right)[/tex]
[tex]\dot Q = 84.216\,W[/tex]
The heat rate produced from the motor is 84.216 watts.
