Chapter 2.8, Problem 47P

A university campus has 200 classrooms and 400 faculty offices. The classrooms are equipped with 12 fluorescent tubes, each consuming 110 W, including the electricity used by the ballasts. The faculty offices, on average, have half as many tubes. The campus is open 240 days a year. The classrooms and faculty offices are not occupied an average of 4 h a day, but the lights are kept on. If the unit cost of electricity is $0.11/kWh, determine how much the campus will save a year if the lights in the classrooms and faculty offices are turned off during unoccupied periods.

To determine

The amount of electrical energy consumed per year during unoccupied work hours per year.

Answer to Problem 47P

The amount of electrical energy consumed per year during unoccupied work hours per year is $\underset{\_}{\text{\$55,757/yr}}$.

Explanation of Solution

Write the equation of electric power consumed by the lights in the classrooms.

$\begin{array}{c}{\dot{E}}_{\text{lighting,}\text{classroom}}=\left(P_{\text{classroom}}\right)\times n\\ =\left(\text{no}\text{.}\text{of}\text{classrooms}\times \text{no}\text{.}\text{of}\text{tubes}\right)\times n\end{array}$ (I)

Here, power consumed per lamp in the classroom is $P_{\text{classroom}}$ and number of lamps is n.

Write the equation of electric power consumed by the lights in the faculty offices.

$\begin{array}{c}{\dot{E}}_{\text{lighting,}\text{offices}}=\left(P_{\text{offices}}\right)n\\ =\left(\text{no}\text{.}\text{of}\text{offices}\times \text{no}\text{.}\text{of}\text{tubes}\text{in faculty}\text{offices}\right)\times n\end{array}$ (II)

Here, power consumed per lamp in the offices is $P_{\text{offices}}$.

Calculate the equation of total electric power consumed by the lights in the classrooms and faculty offices.

${\dot{E}}_{\text{lighting,}\text{total}}={\dot{E}}_{\text{lighting,}\text{classroom}}+{\dot{E}}_{\text{lighting,}\text{offices}}$ (III)

Since the campus is open for 24 hours a year, calculate the total number of unoccupied work hours per year.

$\text{Unoccupied}\text{hours}\text{per}\text{year}=\left(\text{no}\text{.}\text{of}\text{hours}\text{per}\text{day}\right)\left(\text{no}\text{.}\text{of}\text{days}\text{per}\text{year}\right)$ (IV)

Calculate the amount of electrical energy consumed per year during unoccupied work period and its cost.

$\text{Energy}\text{Savings}=\left({\dot{E}}_{\text{lighting,total}}\right)\left(\text{Unoccupied}\text{hours}\right)$ (V)

$\text{Cost}\text{Savings}=\left(\text{Energy}\text{savings}\right)\left(\text{Unit}\text{cost}\text{of}\text{energy}\right)$ (VI)

Conclusion:

Substitute 12 for $\text{no}\text{.}\text{of}\text{tubes}$, 200 classrooms for no. of classrooms, and 110 W for n in Equation (I).

$\begin{array}{c}{\dot{E}}_{\text{lighting,}\text{classroom}}=\left(\text{200}\times \text{12}\right)\times 110\text{W}\\ =264,000\text{W}\times \frac{\text{1}\text{kW}}{\text{1000}\text{W}}\\ =264\text{kW}\end{array}$

Substitute 6 for $\text{no}\text{.}\text{of}\text{tubes}$ in the faculty offices, 400 for no. of faculty offices, and 110 W for n in Equation (II).

$\begin{array}{c}{\dot{E}}_{\text{lighting,}\text{offices}}=\left(\text{400}\times 6\right)\times 110\text{W}\\ =264,000\text{W}\times \frac{\text{1}\text{kW}}{\text{1000}\text{W}}\\ =264\text{kW}\end{array}$

Substitute 264 kW for ${\dot{E}}_{\text{lighting,}\text{classroom}}$ and 264 kW for ${\dot{E}}_{\text{lighting,}\text{offices}}$ in Equation (III).

$\begin{array}{c}{\dot{E}}_{\text{lighting,}\text{total}}=264\text{kW}+264\text{kW}\\ =528\text{kW}\end{array}$

Substitute 4 for $\text{no}\text{.}\text{of}\text{hours}\text{per}\text{day}$ and 240 for $\text{no}\text{.}\text{of}\text{days}\text{per}\text{year}$ in Equation (IV).

$\begin{array}{c}\text{Unoccupied}\text{hours}\text{per}\text{year}=\left(\text{4}\right)\left(\text{240}\right)\\ =960\text{h/yr}\end{array}$

Substitute 960 h/yr for $\text{Unoccupied}\text{hours}\text{per}\text{year}$ and 528 kW for ${\dot{E}}_{\text{lighting,}\text{total}}$ in Equation (V).

$\begin{array}{c}\text{Energy}\text{Savings}=\left(528\text{kW}\right)\left(\text{960}\text{h/yr}\right)\\ =506,880\text{kWh}\end{array}$

Substitute 506,880 kWh for Energy savings and $\$\text{0}\text{.11/kWh}$ for unit cost of energy in Equation (VI).

$\begin{array}{c}\text{Cost}\text{Savings}=\left(506,880\text{kWh/yr}\right)\left(\$\text{0}\text{.11/kWh}\right)\\ =\$\text{55,757/yr}\end{array}$

Thus, the amount of electrical energy consumed per year during unoccupied work hours per year is $\underset{\_}{\text{\$55,757/yr}}$.