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Chapter 15: Q63GP (page 412)

Refrigeration units can be rated in 鈥渢ons.鈥� A 1-ton air conditioning system can remove sufficient energy to freeze 1 ton (2000 pounds = 909 kg) of 0掳C water into 0掳C ice in one 24-h day. If, on a 35掳C day, the interior of a house is maintained at 22掳C by the continuous operation of a 5-ton air conditioning system, how much does this cooling cost the homeowner per hour? Assume the work done by the refrigeration unit is powered by electricity that costs $0.10 per kWh and that the unit鈥檚 coefficient of performance is 18% that of an ideal refrigerator.\({\bf{1}}\;{\bf{kWh = 3}}{\bf{.60 \times 1}}{{\bf{0}}{\bf{6}}}\;{\bf{J}}\).

Short Answer

Expert verified

The cost of cooling per hour is \(\$ 0.43{\rm{/h}}\).

Step by step solution

01

Understanding the coefficient of performance

The ratio of the heat removed from a low temperature area\(\left( {{{\bf{Q}}_{\bf{L}}}} \right)\)to work done (W) to remove the heat.

The expression for the COP is given as:

\({\rm{COP}} = \frac{{{Q_{\rm{L}}}}}{W}\) 鈥� (i)

Here,\({Q_{\rm{L}}}\)is the heat removed and W is the work done.

02

Given Data

The time is, \(t = 24\;{\rm{h}}\).

The temperature outside the house is, \({T_{\rm{H}}} = 35\circ {\rm{C}}\).

The temperature of the interior of the house is, \({T_{\rm{L}}} = 22\circ {\rm{C}}\).

The cost of electricity per kWh is \(p = \$ 0.10\;{\rm{/kWh}}\).

The coefficient of performance is 18% of the ideal refrigerator.

The amount of heat is, \({Q_{\rm{L}}} = 5\;{\rm{tons}}\).

03

Determination of work done per unit time

From equation (i), the work done per unit time is,

\(\frac{W}{t} = \frac{{{Q_{\rm{L}}}}}{{{\rm{COP}} \times t}}\) 鈥� (ii)

The expression for the heat required to transfer water into ice is,

\({Q_{\rm{L}}} = m{L_{\rm{f}}}\) 鈥� (iii)

Here, m is the mass and\({L_{\rm{f}}}\)is the latent heat of fusion.

Again, the expression for COP for an ideal refrigerator is given as:

\({\rm{COP}} = \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}\) 鈥� (iv)

Rewrite expression (ii) for 18% COP and substitute values from expressions (iii) and (iv).

\(\begin{array}{c}\frac{W}{t} = \frac{{{Q_{\rm{L}}}{\rm{/}}t}}{{0.18{\rm{COP}}}}\\\frac{W}{t} = \frac{{5\;{\rm{ton}} \times \frac{m}{t}{L_{\rm{f}}}}}{{0.18\left( {\frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}} - {T_{\rm{L}}}}}} \right)}}\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{l}W = \frac{{5\;{\rm{tons}} \times \frac{{909\;{\rm{kg}}}}{{1\;{\rm{day}}}} \times \frac{{3.33 \times {{10}5}\;{\rm{J}}}}{{1\;{\rm{kg}}}}}}{{0.18\left( {\frac{{\left( {22 + 273} \right)\;{\rm{K}}}}{{\left( {35 + 273} \right)\;{\rm{K}} - \left( {22 + 273} \right)\;{\rm{K}}}}} \right)}}\\W = 3.705 \times {108}\;{\rm{J/day}}\end{array}\)

04

Determination of cooling cost per hour

The cooling cost per hour is calculated as:

\(\begin{array}{l}{\rm{cost/h}} = \left( {3.705 \times {{10}8}\;{\rm{J/day}}} \right)\left( {\frac{{1\,\;{\rm{day}}}}{{24\;{\rm{h}}}}} \right)\left( {\frac{{1\,\;{\rm{kWh}}}}{{3.6 \times {{10}6}\;{\rm{J}}}}} \right)\left( {\frac{{\$ 0.10}}{{1\;{\rm{kWh}}}}} \right)\\{\rm{cost/h}} = \$ 0.43{\rm{/h}}\end{array}\)

Thus, the cost of cooling per hour is\(\$ 0.43{\rm{/h}}\).

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Most popular questions from this chapter

(II) When\({\bf{5}}{\bf{.80 \times 1}}{{\bf{0}}{\bf{5}}}\;{\bf{J}}\)of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from\({\bf{1}}{\bf{.9}}\;{{\bf{m}}{\bf{3}}}\)to\({\bf{4}}{\bf{.1}}\;{{\bf{m}}{\bf{3}}}\). Calculate

(a) the work done by the gas, and

(b) the change in internal energy of the gas.

(c) Graph this process on a PV diagram.

Question:(II) Which will improve the efficiency of a Carnot engine more: a 10 掳C increase in the high-temperature reservoir, or a 10 掳C decrease in the low temperature reservoir? Give detailed results. Can you state a generalization?

Question: An ideal heat pump is used to maintain the inside temperature of a house at \({T_{{\rm{in}}}} = 22{\rm{^\circ C}}\) when the outside temperature is \({T_{{\rm{out}}}}\). Assume that when it is operating, the heat pump does work at a rate of 1500 W. Also assume that the house loses heat via conduction through its walls and other surfaces at a rate given by \(\left( {650\;{{\rm{W}} \mathord{\left/

{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)\). (a) For what outside temperature would the heat pump have to operate all the time in order to maintain the house at an inside temperature of 22掳C? (b) If the outside temperature is 8掳C, what percentage of the time does the heat pump have to operate in order to maintain the house at an inside temperature of 22掳C?

It has been suggested that a heat engine could be developed that made use of the temperature difference between water at the surface of the ocean and water several hundred meters deep. In the tropics, the temperatures may be 27掳C and 4掳C, respectively.

(a) What is the maximum efficiency such an engine could have?

(b) Why might such an engine be feasible in spite of the low efficiency?

(c) Can you imagine any adverse environmental effects that might occur?

Question: An ideal monatomic gas is contained in a tall cylindrical jar of cross-sectional area \({\bf{0}}{\bf{.080}}\;{{\bf{m}}^{\bf{2}}}\) fitted with an airtight frictionless 0.15-kg movable piston. When the gas is heated (at constant pressure) from 25掳C to 55掳C, the piston rises 1.0 cm. How much heat was required for this process? Assume atmospheric pressure outside. (Hint: See Section 14鈥�2.)
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