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Chapter 15: Q38P (page 412)

Question: (II) How much less per year would it cost a family to operate a heat pump that has a coefficient of performance of 2.9 than an electric heater that costs \(2000 to heat their home for a year? If the conversion to the heat pump costs \)15,000, how long would it take the family to break even on heating costs? How much would the family save in 20 years?

Short Answer

Expert verified

Operating a heat pump would cost $1310 less per year than running the electric heater to the family. It would take 11.45 years for the family to break even on heating costs. The family would save approximately $11,000 in 20 years.

Step by step solution

01

Understanding the coefficient of performance of heat pump

The ratio of heat delivered inside the room to the work done by the heat pump to deliver the heat is termed the coefficient of performance of the heat pump (COP).

The expression for COP is

\({\rm{COP}} = \frac{{{Q_{\rm{H}}}}}{W}\).

The work (or energy) required to deliver the required amount of heat by the heat pump can be calculated using this expression.

02

Given information

The coefficient of performance is\({\rm{COP}} = 2.9\).

The cost per year for the electric heater is $2000.

The conversion cost is $15,000.

03

Evaluation of money saved per year by the family

The cost of heating the home for a year by the electric heater is $2000. Suppose it delivers \({Q_{\rm{H}}}\) heat in one year.

Energy (W) required by the heat pump to deliver the same amount of heat (\({Q_{\rm{H}}}\)) can be calculated below:

\(\begin{aligned}{c}{\rm{COP}} &= \frac{{{Q_{\rm{H}}}}}{W}\\W &= \frac{{{Q_{\rm{H}}}}}{{COP}}\\ &= \frac{{{Q_{\rm{H}}}}}{{2.9}}\end{aligned}\)

This energy is required by the heat pump to deliver the same amount of heat as delivered by the electric heater in one year. If \({Q_{\rm{H}}}\) heat costs $2000, the energy required by the heat pump would cost

\(\frac{{\$ 2000}}{{2.9}} = \$ 689.7 \approx \$ 690\).

Thus, the cost of operating the heat pump for one year is approximately $690.

Operating a heat pump would cost $1310 (= $2000 - $690) less per year than running the electric heater to the family. So, the family would save $1310 per year.

04

Evaluation of time taken to break even on heating costs

The cost of conversion to the heat pump is $15,000, and the family saves $1310 per year, So the time taken by the family to cover this cost is

\(\begin{aligned}{c}T &= \frac{{\$ 15,000}}{{\$ 1310}}\\ &= 11.45\;{\rm{years}}{\rm{.}}\end{aligned}\)

Thus, the time taken by the family to break even on heating costs is 11.45 years.

05

Evaluation of money saved in 20 years

Since the family saves $1310 in one year, the money saved in 20 years would be

\(\begin{aligned}{c}{\rm{Amount}}\;{\rm{saved}} &= \left( {\$ 1310} \right) \times 20\\ &= \$ 26{\rm{,200}}{\rm{.}}\end{aligned}\)

It takes $15,000 for conversion to the heat pump. Thus, the net amount that the family saves in 20 years would be

\(\$ 26,200 - \$ 15,000 = \$ 11,200\).

Thus, the family would save approximately $11,000 in 20 years.

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Most popular questions from this chapter

Question: (III) The PV diagram in Fig. 15鈥�23 shows two possible states of a system containing 1.75 moles of a monatomic ideal gas. \(\left( {{P_1} = {P_2} = {\bf{425}}\;{{\bf{N}} \mathord{\left/{\vphantom {{\bf{N}} {{{\bf{m}}^{\bf{2}}}}}} \right.} {{{\bf{m}}^{\bf{2}}}}},\;{V_1} = {\bf{2}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}},\;{V_2} = {\bf{8}}{\bf{.00}}\;{{\bf{m}}^{\bf{3}}}.} \right)\) (a) Draw the process which depicts an isobaric expansion from state 1 to state 2, and label this process A. (b) Find the work done by the gas and the change in internal energy of the gas in process A. (c) Draw the two-step process which depicts an isothermal expansion from state 1 to the volume \({V_2}\), followed by an isovolumetric increase in temperature to state 2, and label this process B. (d) Find the change in internal energy of the gas for the two-step process B.

(II) An inventor claims to have built an engine that produces 2.00 MW of usable work while taking in 3.00 MW of thermal energy at 425 K, and rejecting 1.00 MW of thermal energy at 215 K. Is there anything fishy about his claim? Explain.

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