Question

An electric motor operating at steady state draws a current of 10 amp with a voltage of \(220 \mathrm{~V}\). The output shaft rotates at 1000 RPM with a torque of \(16 \mathrm{~N} \cdot \mathrm{m}\) applied to an external load. The rate of heat transfer from the motor to its surroundings is related to the surface temperature \(T_{\mathrm{b}}\) and the ambient temperature \(T_{0}\) by \(\mathrm{hA}\left(T_{\mathrm{b}}-T_{0}\right)\), where \(\mathrm{h}=100 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}, \mathrm{A}=0.195 \mathrm{~m}^{2}\), and \(T_{0}=293 \mathrm{~K}\). Energy transfers are considered positive in the directions indicated by the arrows on Fig. P6.51. (a) Determine the temperature \(T_{\mathrm{b}}\), in \(\mathrm{K}\). (b) For the motor as the system, determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\). (c) If the system boundary is located to take in enough of the nearby surroundings for heat transfer to take place at temperature \(T_{0}\), determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for the enlarged system.

Short answer

1(a) 319.9 K, 1(b) 0.00164 kW/K, 1(c) 0.00179 kW/K

Step-by-step solution

Calculate the Input Power

The input power to the motor can be calculated using the formula: \[ P_{\text{input}} = V \times I \] where \( V = 220 \text{ V} \) and \( I = 10 \text{ A} \). Substituting the values, we get: \[ P_{\text{input}} = 220 \times 10 = 2200 \text{ W} \] Therefore, the input power is 2200 W.

Calculate the Output Power

The output power (mechanical power) can be calculated using the formula: \[ P_{\text{output}} = \tau \times \frac{2\text{π}N}{60} \] where \( \tau = 16 \text{ N} \text{ m} \) and \( N = 1000 \text{ RPM} \). Substituting the values, we get: \[ P_{\text{output}} = 16 \times \frac{2\text{π} \times 1000}{60} = 1675.52 \text{ W} \] Therefore, the output power is 1675.52 W.

Calculate the Rate of Heat Transfer

The rate of heat transfer (\( Q_{\text{out}} \)) can be calculated from the power balance: \[ Q_{\text{out}} = P_{\text{input}} - P_{\text{output}} \] Substituting the values, we get: \[ Q_{\text{out}} = 2200 - 1675.52 = 524.48 \text{ W} \] Therefore, the rate of heat transfer is 524.48 W.

Calculate the Surface Temperature Tb

The surface temperature \( T_{\text{b}} \) can be found using the heat transfer relation: \[ Q_{\text{out}} = hA(T_{\text{b}} - T_{0}) \] Rearranging the formula to solve for \( T_{\text{b}} \), we get: \[ T_{\text{b}} = \frac{Q_{\text{out}}}{hA} + T_{0} \] Substituting the given values: \[ T_{\text{b}} = \frac{524.48}{100 \times 0.195} + 293 = \frac{524.48}{19.5} + 293 = 26.9 + 293 = 319.9 \text{ K} \] Therefore, the surface temperature \( T_{\text{b}} \) is 319.9 K.

Calculate the Rate of Entropy Production for the Motor

The rate of entropy production (\( \frac{dS_{\text{gen}}}{dt} \)) for the motor can be calculated using: \[ \frac{dS_{\text{gen}}}{dt} = \frac{Q_{\text{out}}}{T_{\text{b}}} \] Substituting the values: \[ \frac{dS_{\text{gen}}}{dt} = \frac{524.48}{319.9} = 1.64 \text{ W/K} \] Therefore, the rate of entropy production for the motor is 1.64 W/K (or 0.00164 kW/K).

Calculate the Rate of Entropy Production for the Enlarged System

For the enlarged system, the boundary is set to take in enough surroundings so that heat transfer takes place at \( T_{0} \). The rate of entropy production can be calculated as: \[ \frac{dS_{\text{gen}}}{dt} = \frac{Q_{\text{out}}}{T_{0}} \] Substituting the values: \[ \frac{dS_{\text{gen}}}{dt} = \frac{524.48}{293} = 1.79 \text{ W/K} \] Therefore, the rate of entropy production for the enlarged system is 1.79 W/K (or 0.00179 kW/K).

Key concepts

Electric Motor Power Calculation

Calculating the power of an electric motor involves understanding both the electrical input power and the mechanical output power.
First, the input power is determined by using the formula \( P_{\text{input}} = V \times I \), where V is the voltage and I is the current. For our example, with a voltage of 220 V and a current of 10 A, the input power is \(2200 \text{ W} \).
Next, we calculate the output power—that is, the mechanical power that the motor generates. The formula used here is \( P_{\text{output}} = \tau \times \frac{2\pi N}{60} \), where \( \tau = 16 \text{ N} \cdot \text{m} \) represents the torque and \( N = 1000 \text{ RPM} \) is the rotational speed. This calculation then yields an output power of approximately \(1675.52 \text{ W} \).
Finally, the difference between the input and output power gives us the rate of heat transfer, calculated as \( Q_{\text{out}} = P_{\text{input}} - P_{\text{output}} \), resulting in \(524.48 \text{ W} \). This confirms that while the motor operates, it loses some energy as heat.

Heat Transfer

In thermodynamics, heat transfer is a crucial concept, especially when dealing with systems like electric motors.
In our motor example, the heat transfer from the motor is represented by the formula \( Q_{\text{out}} = hA(T_{\text{b}} - T_{0}) \). Here, 'h' is the heat transfer coefficient, 'A' is the surface area, \( T_{\text{b}} \) is the body (surface) temperature, and \( T_{0} \) is the ambient temperature.
In this exercise, we use \( h = 100 \text{ W}\text{ m}^{-2}\text{ K}^{-1} \), \( A = 0.195 \text{ m}^{2} \), and \( T_{0} = 293 \text{ K} \). To find the surface temperature \( T_{\text{b}} \), we rearrange the formula to solve for \( T_{\text{b}} \), yielding \( T_{\text{b}} = \frac{Q_{\text{out}}}{hA} + T_{0} = \frac{524.48}{19.5} + 293 = 319.9 \text{ K} \).
This step-by-step process highlights the importance of balancing energy in and out of the system to determine the surface temperature accurately.

Entropy Production

Entropy production is a measure of irreversible processes within a system, and it's central to understanding thermodynamic efficiency.
For the motor in our example, the rate of entropy production can be calculated in two ways—one for the motor itself, and another for an enlarged system boundary.
First, within the motor, the rate of entropy production is given by \( \frac{dS_{\text{gen}}}{dt} = \frac{Q_{\text{out}}}{T_{\text{b}}} = \frac{524.48}{319.9} = 1.64 \text{ W/K} \). This tells us how much disorder or inefficiency there is due to heat loss within the motor.
When we expand the system boundary to include enough surroundings so the heat transfer occurs at the ambient temperature \( T_{0} \), the formula changes slightly to \( \frac{dS_{\text{gen}}}{dt} = \frac{Q_{\text{out}}}{T_{0}} = \frac{524.48}{293} = 1.79 \text{ W/K} \). The increased entropy production indicates further inefficiencies introduced by including the surroundings.
Understanding these principles helps to optimize systems by minimizing entropy production, thus improving efficiency.