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Chapter 5: Problem 52

By supplying energy at an average rate of \(21,100 \mathrm{~kJ} / \mathrm{h}\), a heat pump maintains the temperature of a dwelling at \(21^{\circ} \mathrm{C}\). If electricity costs 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), determine the minimum theoretical operating cost for each day of operation if the heat pump receives energy by heat transfer from (a) the outdoor air at \(-5^{\circ} \mathrm{C}\). (b) well water at \(8^{\circ} \mathrm{C}\)

Short Answer

Expert verified
a) For outdoor air: \(3.57 per day, b) For well water: \)1.79 per day.

Step by step solution

01

Determine the heat pump鈥檚 Coefficient of Performance (COP) for each scenario

The Coefficient of Performance (COP) for a heat pump is given by \[ \text{COP} = \frac{T_{\text{inside}}}{T_{\text{inside}} - T_{\text{outside}}} \]where temperatures are in Kelvin. For the two cases, convert temperatures to Kelvin and then compute COP.
02

Convert temperatures from Celsius to Kelvin

Temperature indoors, \[ T_{\text{inside}} = 21 + 273.15 = 294.15 \text{ K} \](a) For the outdoor air, \[ T_{\text{outside}} = -5 + 273.15 = 268.15 \text{ K} \](b) For the well water, \[ T_{\text{outside}} = 8 + 273.15 = 281.15 \text{ K} \].
03

Calculate COP for each case

Plug the values into the COP formula for each case:a) For outdoor air,\[ \text{COP}_{\text{air}} = \frac{294.15}{294.15 - 268.15} = \frac{294.15}{26} = 11.32 \]b) For well water,\[ \text{COP}_{\text{water}} = \frac{294.15}{294.15 - 281.15} = \frac{294.15}{13} = 22.63 \].
04

Determine the electrical energy required per hour

Given the heat pump supplies energy at a rate of 21,100 kJ/h, convert this to kW:\[ 21,100 \text{ kJ/h} = 21.1 \text{ kW} \]The electrical energy (work) per hour is\[ W = \frac{Q}{\text{COP}} \]a) For outdoor air,\[ W_{\text{air}} = \frac{21.1 \text{ kW}}{11.32} = 1.86 \text{ kW} \]b) For well water,\[ W_{\text{water}} = \frac{21.1 \text{ kW}}{22.63} = 0.93 \text{ kW} \].
05

Convert hourly energy usage to daily energy usage

Multiply the hourly electrical energy required by 24 to get daily energy usage:a) For outdoor air,\[ \text{Energy}_{\text{air}} = 1.86 \text{ kW} \times 24 = 44.64 \text{ kW} \times \text{h} \]b) For well water,\[ \text{Energy}_{\text{water}} = 0.93 \text{ kW} \times 24 = 22.32 \text{ kW} \times \text{h} \].
06

Determine the cost of electricity

The cost per kilowatt-hour is given as 8 cents. Convert the daily energy usage to cost:a) For outdoor air,\[ \text{Cost}_{\text{air}} = 44.64 \text{ kW} \times \text{h} \times 0.08 = 3.57 \text{ dollars/day} \]b) For well water,\[ \text{Cost}_{\text{water}} = 22.32 \text{ kW} \times \text{h} \times 0.08 = 1.79 \text{ dollars/day} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Performance (COP)
The Coefficient of Performance (COP) is a crucial concept when evaluating heat pumps. It measures the efficiency of the heat pump by comparing the amount of heat delivered to the energy input required to operate the pump. Mathematically, you can express it as: \[ \text{COP} = \frac{T_{\text{inside}}}{T_{\text{inside}} - T_{\text{outside}}} \]In this formula, temperatures need to be in Kelvin (K). To convert from Celsius (掳C) to Kelvin, add 273.15 to the Celsius value. A higher COP means the heat pump is more efficient. For instance, the COP for outdoor air differs significantly from that of well water due to the varying temperature differences.
Temperature Conversion
Temperature conversion is essential in calculating the COP. The formula to convert Celsius to Kelvin is straightforward: \[ T(K) = T(掳C) + 273.15 \]For example, if the indoor temperature is 21掳C, convert it to Kelvin:\[ 21 + 273.15 = 294.15 \text{ K} \]Similarly, if the outdoor temperature is -5掳C:\[ -5 + 273.15 = 268.15 \text{ K} \]Converting these temperatures ensures the calculations for COP are accurate. This step eliminates the risk of errors that could arise from using Celsius values directly in the COP formula.
Energy Cost Calculation
To determine the operating cost of a heat pump, you start by identifying the electrical energy required. Given the energy supplied by the heat pump and its COP, the electrical energy per hour is:\[ W = \frac{Q}{\text{COP}} \]Where Q is the heat pump's energy supply. Convert this value from kJ/h to kW: \[ 21,100 \text{ kJ/h} = 21.1 \text{ kW} \]For the outdoor air scenario, if COP is 11.32: \[ W_{\text{air}} = \frac{21.1 \text{ kW}}{11.32} = 1.86 \text{ kW} \]Multiply by 24 hours to get daily usage: \[ \text{Energy}_{\text{air}} = 1.86 \text{ kW} \times 24 = 44.64 \text{ kW} \times \text{h} \]If electricity costs 8 cents/kWh, the daily cost is: \[ \text{Cost}_{\text{air}} = 44.64 \text{ kW} \times \text{h} \times 0.08 = 3.57 \text{ dollars/day} \]These steps can compare costs for different scenarios, such as using well water.

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Most popular questions from this chapter

Two reversible heat pump cycles operate in series. The first cycle receives energy by heat transfer from a cold reservoir at \(260 \mathrm{~K}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T\) greater than 260 K. The second cycle receives energy by heat transfer from the reservoir at temperature \(T\) and rejects energy by heat transfer to a higher- temperature reservoir at \(1200 \mathrm{~K}\). If the heat pump cycles have the same coefficient of performance, determine (a) \(T\), in \(\mathrm{K}\), and (b) the value of each coefficient of performance.

Two reversible refrigeration cycles are arranged in series. The first cycle receives energy by heat transfer from a cold reservoir at temperature \(T_{C}\) and rejects energy by heat transfer to a reservoir at an intermediate temperature \(T\), greater than \(T_{C .}\) The second cycle receives energy by heat transfer from the reservoir at temperature \(T\) and rejects energy by heat transfer to a higher-temperature reservoir at \(T_{\mathrm{H}}\). Obtain an expression for the coefficient of performance of a single reversible refrigeration cycle operating directly between cold and hot reservoirs at \(T_{\mathrm{C}}\) and \(T_{\mathrm{H}}\), respectively, in terms of the coefficients of performance of the two cycles.

Experiments by Australian researchers suggest that the second law can be violated at the micron scale over time intervals of up to 2 seconds. If validated, some say these findings could place a limit on the engineering of nanomachines because such devices may not behave simply like miniaturized versions of their larger counterparts. Investigate the implications, if any, for nanotechnology of these experimental findings. Write a report including at least three references.

\(5.60 \mathrm{~A}\) reversible power cycle receives \(Q_{\mathrm{H}}\) from a hot reservoir at temperature \(T_{\mathrm{H}}\) and rejects energy by heat transfer to the surroundings at temperature \(T_{0}\). The work developed by the power cycle is used to drive a reversible refrigeration cycle that removes \(Q_{\mathrm{C}}\) from a cold reservoir at temperature \(T_{C}\) and discharges energy by heat transfer to the same surroundings at \(T_{0}\). Develop an expression for the ratio \(Q_{\mathrm{C}} / Q_{\mathrm{H}}\) in terms of the temperature ratios \(T_{\mathrm{H}} / T_{0}\) ) and \(T_{\mathrm{C}} T_{0 .}\)

A reversible heat engine operates between two reservoirs at temperatures of \(650^{\circ} \mathrm{C}\) and \(35^{\circ} \mathrm{C}\). A reversible refrigerator operates between reservoirs at temperatures of \(35^{\circ} \mathrm{C}\) and \(-10^{\circ} \mathrm{C}\). The refrigerator is driven by the heat engine. The heat transfer to the heat engine is \(1500 \mathrm{~kJ}\). A net work output of \(300 \mathrm{~kJ}\) is obtained from the combined enginerefrigerator plant. Determine the net heat transfer to the reservoir at \(35^{\circ} \mathrm{C}\) and the heat transfer to the refrigerant from cold reservoir.
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