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Chapter 19: Problem 39

A heat pump extracts energy from groundwater at \(10^{\circ} \mathrm{C}\) and transfers it to water at \(70^{\circ} \mathrm{C}\) to heat a building. Find (a) its COP and (b) its electric power consumption if it supplies heat at the rate of \(20 \mathrm{kW}\). (c) Compare the pump's hourly operating cost with that of an oil furnace if electricity costs \(15.5 \notin / \mathrm{kWh}\) and oil costs \(\$ 2.60 /\) gallon and releases about \(30 \mathrm{kWh} / \mathrm{gal}\) when burned.

Short Answer

Expert verified
The COP of the heat pump is 34.3. Its electric power consumption is 0.58 kW. The heat pump costs 8.99 cents per hour to operate, while the oil furnace costs 173 cents per hour to operate. Therefore, using the heat pump is much cheaper than using the oil furnace.

Step by step solution

01

Calculate the Coefficient of Performance (COP)

The COP of a heat pump is given by \(\frac{Q_h}{W}\), where \(Q_h\) is the thermal energy transferred to the hot reservoir and \(W\) is the work done. Here, since the heat pump moves heat from the cold reservoir (groundwater) to the hot reservoir (building), we use the formula: \(COP = \frac{T2}{T2 - T1}\), where \(T1\) is the temperature of the cold reservoir and \(T2\) is the temperature of the hot reservoir. But first, we need to convert the temperatures from Celsius to Kelvin as \(T_{Kelvin} = T_{Celsius} + 273\). So, \(T1 = 10 + 273 = 283 \, Kelvin\) and \(T2 = 70 + 273 = 343 \, Kelvin\). Now we can calculate the COP as \(\frac{343}{343 - 283} = 34.3\)
02

Calculate electric power consumption

With the COP and the rate of heat transfer \(Q_h\), we can calculate the work done \(W = \frac{Q_h}{COP}\). Here, \(Q_h = 20 \, kW\) and \(COP = 34.3\), so \(W = \frac{20}{34.3} = 0.58 \, kW\). This is the electric power consumption of the heat pump.
03

Compare operating costs of heat pump and oil furnace

The cost of operating the heat pump for an hour is the electric power consumption times the cost of electricity, which is \(0.58 \, kW * 15.5 \, cents/kWh = 8.99 \, cents\). A gallon of oil releases \(30 \, kWh\) of energy when burned, so the oil furnace supplies heat at the rate of \(30 \, kW\), and the cost of operating it for an hour is the oil consumption times the cost of oil, which is \(\frac{20}{30} * 2.60 = 1.73 \, dollars = 173 \, cents\). So, the heat pump is much cheaper to operate than the oil furnace.

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Most popular questions from this chapter

It costs \(\$ 180\) to heat a house with electricity in a typical winter month. (Electric heat simply converts all the incoming electrical energy to heat.) What would the monthly heating bill be after switching to an electrically powered heat-pump system with \(\mathrm{COP}=3.1 ?\)

Name some irreversible processes that occur in a real engine.

A Carnot engine absorbs \(900 \mathrm{J}\) of heat each cycle and provides \(350 \mathrm{J}\) of work. (a) What's its efficiency? (b) How much heat is rejected each cycle? (c) If the engine rejects heat at \(10^{\circ} \mathrm{C},\) what's its maximum temperature?

You're engineering an energy-efficient house that will require an average of \(4.6 \mathrm{kW}\) to heat on cold winter days. You've designed a photovoltaic system for electric power, which will supply on average \(2.0 \mathrm{kW}\). You propose to heat the house with an electrically operated groundwater-based heat pump. What should you specify as the minimum acceptable COP for the pump if the photovoltaic system supplies its energy?

Find an expression for the entropy gain when hot and cold water are irreversibly mixed. A corresponding reversible process you can use to calculate this change is to bring each water sample slowly to their common final temperature \(T_{\mathrm{f}}\) and then mix them. Express your answer in terms of the initial temperatures \(T_{\mathrm{h}}\) and \(T_{\mathrm{c}} .\) Assume equal masses of hot and cold water, with constant specific heat \(c .\) What's the sign of your answer?
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