Chapter 6, Problem 9PEA

(a)

To determine

The rate at which all the three motors are using the electrical energy.

Answer to Problem 9PEA

The rate at which all the three motors are using the electrical energy is $3258\text{W}$.

Explanation of Solution

Write the expression for the total power.

    $P=P_1+P_2+P_3$                                                                                                 (I)

Here, $P$ is the rate at which all the three motors are using the electrical energy, $P_1$ is the power of the fan motor for blowing air over the inside cooling coils, $P_2$ is the power of the fan motor for blowing air over the outside condenser coils and $P_3$ is the power of the compressor motor.

Conclusion:

Determine the power of the fan motor for blowing air over the inside cooling coils in Watt.

    $\begin{array}{c}{P}_1=\frac{1}{3}\text{hp}\left(\frac{746\text{W}}{1\text{hp}}\right)\\ =248.67\text{W}\\ \approx \text{249}\text{W}\end{array}$

Determine the power of the fan motor for blowing air over the outside condenser coils in Watt.

    $\begin{array}{c}{P}_2=\frac{1}{3}\text{hp}\left(\frac{746\text{W}}{1\text{hp}}\right)\\ =248.67\text{W}\\ \approx \text{249}\text{W}\end{array}$

Determine the power of the compressor motor in Watt.

    $\begin{array}{c}{P}_3=3.70\text{hp}\left(\frac{746\text{W}}{1\text{hp}}\right)\\ =2760.2\text{W}\\ \approx \text{2760}\text{W}\end{array}$

Substitute $\text{249}\text{W}$ for $P_1$, $\text{249}\text{W}$ for $P_2$ and $\text{2760}\text{W}$ for $P_3$ in equation (I) to calculate the total power.

    $\begin{array}{c}P=\text{249}\text{W}+\text{249}\text{W}+\text{2760}\text{W}\\ =3258\text{W}\end{array}$

Therefore, the rate at which all the three motors are using the electrical energy is $3258\text{W}$.

(b)

To determine

The cost of running the unit per hour.

Answer to Problem 9PEA

The cost of running the unit per hour is $\$0.33$.

Explanation of Solution

Write the expression for the cost of running the unit per hour.

    $c=Pr$

Here, $c$ is the cost of running the unit per hour and $r$ is the rate per kilowatt hour.

Conclusion:

Substitute $3258\text{W}$ for $P$ and $\$0.10/\text{kWh}$ for $r$ in the above equation to find the cost per hour..

    $\begin{array}{c}c=\left(3258\text{W}\right)\left(\$0.10/\text{kWh}\right)\\ =\left(3258\text{W}\times \left(\frac{1\text{kW}}{{10}^3\text{W}}\right)\right)\left(\$0.10/\text{kWh}\right)\\ =\$0.3258/\text{h}\\ \approx \$0.33/\text{h}\end{array}$

Therefore, the cost of running the unit per hour is $\$0.33$.

(c)

To determine

The cost of running the unit for $12\text{hours}$ a day for a 30-day month.

Answer to Problem 9PEA

The cost of running the unit is $118.80.

Explanation of Solution

Write the expression for the cost of running the unit.

    C = ct

Here, C is the cost and t is the operation time.

Conclusion:

Calculate the total operation time.

$\begin{array}{ccc}t& =& (12h/day)(30day)\\ & =& 360\text{h}\end{array}$

Substitute $0.33/h for c and 360 h for t in the above equation to find the cost.

$\begin{array}{ccc}c& =& (\$0.33/\text{h})(360\text{h})\\ & =& \$118.80\end{array}$

Therefore, the cost of running the unit is $118.80.