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Chapter 5: Problem 41

A heat pump maintains a dwelling at \(20^{\circ} \mathrm{C}\) when the outside temperature is \(0^{\circ} \mathrm{C}\). The heat transfer rate through the walls and roof is \(3000 \mathrm{~kJ} / \mathrm{h}\) per degree temperature difference between the inside and outside. Determine the minimum theoretical power required to drive the heat pump, in \(\mathrm{kW}\).

Short Answer

Expert verified
1.14 kW

Step by step solution

01

- Understand the problem

Determine the rate of heat transfer required to maintain the temperature difference between inside and outside the dwelling.
02

- Calculate the heat transfer rate, Q

The temperature difference between inside and outside is \ \ \text{\( \Delta T = 20^{\circ} \mathrm{C} - 0^{\circ} \mathrm{C} = 20^{\circ} \mathrm{C} \ \)}. Given that the heat transfer rate through the walls and roof is \ \(3000 \, \mathrm{kJ/h}^{-1} /^{\circ} \mathrm{C}\), the heat transfer rate, \ \(Q\), can be calculated as follows: \ \ \(Q = 3000 \, \mathrm{kJ/h} \times 20^{\circ} \, \mathrm{C} = 60000 \, \mathrm{kJ/h}\).
03

- Convert heat transfer rate Q to kJ/s

To find the rate in kJ/s, divide by the number of seconds in an hour: \ \ \( \frac{60000 \mathrm{kJ}}{3600 \, \mathrm{s}} = 16.67 \, \mathrm{kJ/s} \).
04

- Introduce the Coefficient of Performance (COP) for heating

The \ \text{Coefficient of Performance (COP)} for a heat pump is given by: \ \ \( \text{COP}_{\text{heating}} = \frac{T_H}{T_H - T_C} \), where \(T_H\) is the inside temperature and \(T_C\) is the outside temperature in Kelvin.
05

- Convert temperatures to Kelvin

Convert the temperatures to Kelvin: \ \(T_H = 20^{\circ}C + 273.15 = 293.15 \mathrm{K} \) \ and \ \(T_C = 0^{\circ}C + 273.15 = 273.15 \mathrm{K} \).
06

- Calculate the COP for heating

Using the formula for \ \text{COP}_{\text{heating}}: \ \ \( \text{COP}_{\text{heating}} = \frac{293.15}{293.15 - 273.15} = \frac{293.15}{20} = 14.66 \).
07

- Calculate the minimum theoretical power required

The minimum power required (W) is obtained by dividing the heat transfer rate by the COP: \ \ \(W = \frac{Q}{\text{COP}_{\text{heating}}} = \frac{16.67 \mathrm{kJ/s}}{14.66} = 1.14 \, \mathrm{kW} \).

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Key Concepts

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

Coefficient of Performance
The Coefficient of Performance (COP) is a measure of efficiency for heat pumps, air conditioners, and refrigerators. It indicates how effectively a system transfers heat compared to the work it requires. For heat pumps in heating mode, the COP is defined as the ratio of heat delivered to the work input required to deliver that heat. Mathematically, it's represented as: \ \( \text{COP}_{\text{heating}} = \frac{T_H}{T_H - T_C} \) where \( T_H \) is the high-temperature reservoir (inside temperature) and \( T_C \) is the low-temperature reservoir (outside temperature).
A higher COP value means the heat pump is more efficient. In the context of our problem, using the given temperatures, the high efficiency is indicated by the high COP value of 14.66.
Thermodynamic Temperature Scales
Understanding thermodynamic temperature scales is crucial for calculating the COP and other thermodynamic properties. The Kelvin scale is used in these calculations because it starts at absolute zero, ensuring that temperature values are always positive.
To convert from Celsius to Kelvin, the formula is simple: \ \( T(K) = T(^\circ C) + 273.15 \).
For our problem, the inside temperature of \(20^{\circ}C\) converts to \(293.15\ K\), and the outside temperature of \(0^{\circ}C\) converts to \(273.15\ K\). These conversions are then used in the COP calculation.
Heat Transfer Rate
The heat transfer rate indicates how much heat is being moved per unit of time. In our problem, this rate is given as \(3000\ \mathrm{kJ/h}\) for each degree Celsius difference between inside and outside temperatures.
Firstly, we calculate the total heat transfer rate based on the temperature difference: \ \( \Delta T = 20^{\circ}C - 0^{\circ}C = 20^{\circ}C \)
The heat transfer rate, \(Q\), becomes: \ \( Q = 3000 \ \mathrm{kJ/h} \times 20^{\circ}C = 60000\ \mathrm{kJ/h} \)
To align with the international unit system and make power calculations easier, we then convert this rate to joules per second (watts): \ \( \frac{60000\ \mathrm{kJ}}{3600 \ \mathrm{s}} = 16.67\ \mathrm{kJ/s} \).
Power Requirement Calculation
Calculating the power requirement involves using the heat transfer rate and the COP. The formula to determine the power, \(W\), is:
\ \( W = \frac{Q}{\text{COP}_{\text{heating}}} \)
Using our values: \ \( W = \frac{16.67 \ \mathrm{kJ/s}}{14.66} = 1.14 \ \mathrm{kW} \)
This power requirement is the minimum theoretical power needed to operate the heat pump efficiently under the given conditions. By focusing on the COP and precise heat transfer rates, we can understand and optimize how much energy will be used.

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Most popular questions from this chapter

A refrigeration cycle having a coefficient of performance of 3 maintains a computer laboratory at \(18^{\circ} \mathrm{C}\) on a day when the outside temperature is \(30^{\circ} \mathrm{C}\). The thermal load at steady state consists of energy entering through the walls and windows at a rate of \(30,000 \mathrm{~kJ} / \mathrm{h}\) and from the occupants, computers, and lighting at a rate of \(6000 \mathrm{~kJ} / \mathrm{h}\). Determine the power required by this cycle and compare with the minimum theoretical power required for any refrigeration cycle operating under these conditions, each in \(\mathrm{kW}\).

A reversible power cycle receives energy \(Q_{\mathrm{H}}\) from a reservoir at temperature \(T_{\mathrm{H}}\) and rejects \(Q_{\mathrm{C}}\) to a reservoir at temperature \(T_{\mathrm{C}}\). The work developed by the power cycle is used to drive a reversible heat pump that removes energy \(Q_{\mathrm{C}}^{\prime}\) from a reservoir at temperature \(T_{\mathrm{C}}^{\prime}\) and rejects energy \(Q_{\mathrm{H}}^{\prime}\) to a reservoir at temperature \(T_{\mathrm{H}^{\prime}}^{\prime}\) (a) Develop an expression for the ratio \(Q_{H}^{\prime} / Q_{H}\) in terms of the temperatures of the four reservoirs. (b) What must be the relationship of the temperatures \(T_{\mathrm{H}}, T_{\mathrm{C}}\) \(T_{\mathrm{C}}^{\prime}\), and \(T_{\mathrm{H}}^{\prime}\) for \(Q_{\mathrm{H}}^{\prime} / Q_{\mathrm{H}}\) to exceed a value of unity? (c) Letting \(T_{\mathrm{H}}^{\prime}=T_{\mathrm{C}}=T_{0}\), plot \(Q_{\mathrm{H}}^{\prime} / Q_{\mathrm{H}}\) versus \(T_{\mathrm{H}} / T_{0}\) for \(T_{\mathrm{C}}^{\prime} / T_{0}=0.85,0.9\), and \(0.95\), and versus \(T_{\mathrm{C}}^{\prime} / T_{0}\) for \(T_{\mathrm{H}} / T_{0}\) \(=2,3\), and 4 .

A method for generating electricity using gravitational energy is described in U.S. Patent No. \(4,980,572\). The method employs massive spinning wheels located underground that serve as the prime mover of an alternator for generating electricity. Each wheel is kept in motion by torque pulses transmitted to it via a suitable mechanism from vehicles passing overhead. What practical difficulties might be encountered in implementing such a method for generating electricity? If the vehicles are trolleys on an existing urban transit system, might this be a cost-effective way to generate electricity? If the vehicle motion were sustained by the electricity generated, would this be an example of a perpetual motion machine? Discuss.

The preliminary design of a space station calls for a power cycle that at steady state receives energy by heat transfer at \(T_{\mathrm{H}}=600 \mathrm{~K}\) from a nuclear source and rejects energy to space by thermal radiation according to Eq. 2.33. For the radiative surface, the temperature is \(T_{\mathrm{C}}\), the emissivity is \(0.6\), and the surface receives no radiation from any source. The thermal efficiency of the power cycle is one- half that of a reversible power cycle operating between reservoirs at \(T_{\mathrm{H}}\) and \(T_{\mathrm{C}}\) - (a) For \(T_{\mathrm{C}}=400 \mathrm{~K}\), determine \(\dot{W}_{\text {cycle }} / \mathrm{A}\), the net power developed per unit of radiator surface area, in \(\mathrm{kW} / \mathrm{m}^{2}\), and the thermal efficiency. (b) Plot \(\dot{W}_{\text {cycle }} / \mathrm{A}\) and the thermal efficiency versus \(T_{\mathrm{C}}\), and determine the maximum value of \(\dot{W}_{\text {cycle }} / \mathrm{A}\). (c) Determine the range of temperatures \(T_{\mathrm{C}}\), in \(\mathrm{K}\), for which \(\dot{W}_{\text {cycle }} / \mathrm{A}\) is within 2 percent of the maximum value obtained in part (b). The Stefan-Boltzmann constant is \(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\).

Ocean temperature energy conversion (OTEC) power plants generate power by utilizing the naturally occurring decrease with depth of the temperature of ocean water. Near Florida, the ocean surface temperature is \(27^{\circ} \mathrm{C}\), while at a depth of \(700 \mathrm{~m}\) the temperature is \(7^{\circ} \mathrm{C}\). (a) Determine the maximum thermal efficiency for any power cycle operating between these temperatures. (b) The thermal efficiency of existing OTEC plants is approximately 2 percent. Compare this with the result of part (a) and comment.
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