The lighting needs of a storage room are being met by six fluorescent light fixtures, each...

Question:

The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated at 60 W each. All the lamps are on during operating hours of the facility, which are 6 a.m. to 6 p.m. 365 days a year. The storage room is actually used for an average of 3 h a day. If the price of electricity is $0.11/kWh, determine the amount of energy and money that will be saved as a result of installing motion sensors. Also, determine the simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $40

Electric current

When there is the net flow of electric charge through the region then an electric current is said to exist. In electrical circuits, this electric charge is carried by the electrons that flow through the wire.

Answer and Explanation:


Given data:


  • Number of fixtures is, f = 6.
  • Number of lamps is, n = 4.
  • Power rating is, {eq}P = 60\;{\rm{W}} {/eq}
  • Operating period is, t = 365 days.
  • Average operating time in hours is, {eq}\Delta t = 3 {/eq}.
  • Price of electricity is, {eq}p = 0.11 \;${{\rm{\ }} {\left/ {\vphantom {{\rm{\ }} {{\rm{kWh}}}}} \right. } {{\rm{kWh}}}} {/eq}.
  • Purchase price of sensor is, {eq}{p_1} = 32 \;${\rm{\ }} {/eq}.
  • Installation cost for 1 hour is, {eq}{p_2} = 40 \;${\rm{\ }} {/eq}.


Formula for the energy savings is,

{eq}\begin{align*} {E_s} &= f \times n \times P \times t \times \Delta t\\ {E_s} &= 6 \times 4 \times 60 \times {10^{ - 3}} \times 365 \times 3\\ {E_s} &= 1576.8\;{{{\rm{kWh}}} {\left/ {\vphantom {{{\rm{kWh}}} {{\rm{Yr}}}}} \right. } {{\rm{yr}}}} \end{align*} {/eq}

Hence, the energy savings is {eq}1576.8\;{{{\rm{kWh}}} {\left/ {\vphantom {{{\rm{kWh}}} {{\rm{Yr}}}}} \right. } {{\rm{yr}}}} {/eq}.


The formula for money savings from energy saving and unit cost is,

{eq}\begin{align*} {M_s} &= {E_s}{M_{unit}}\\ {M_s} &= 4730.4 \times 0.11\\ {M_s} &= 520.344 \;${{\rm{\ }} {\left/ {\vphantom {{\rm{\ }} {{\rm{Yr}}}}} \right. } {{\rm{yr}}}} \end{align*} {/eq}

Hence, the money savings is {eq}520.344 \;${{\rm{\ }} {\left/ {\vphantom {{\rm{\ }} {{\rm{Yr}}}}} \right. } {{\rm{yr}}}} {/eq}.


The formula for payback period is,

{eq}{t_{PB}} = \dfrac{{{E_{ic}}}}{{{M_s}}} {/eq}

Here, {eq}{E_{ic}} {/eq} is the excess initial cost.

{eq}{E_{ic}} = {\rm{ }}32{\rm{ }} + {\rm{ }}40{\rm{ }} = {\rm{ }}72 \;${\rm{ \ }} {/eq}

Substituting the above value to obtain payback period as,

{eq}\begin{align*} {t_{PB}} &= \dfrac{{72}}{{520.344}}\\ {t_{PB}} &= 0.138\;{\rm{yr}} \end{align*} {/eq}

Hence, the payback period is 0.138 year.


Learn more about this topic:

Loading...
What is Electric Current? - Definition, Unit & Types

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 6 / Lesson 7
306K

Related to this Question

Explore our homework questions and answers library