/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 An air conditioner in a very hot... [FREE SOLUTION] | 91Ó°ÊÓ

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An air conditioner in a very hot region uses a power input of \(2.5 \mathrm{~kW}\) to \(\operatorname{cool}\) a \(5^{\circ} \mathrm{C}\) space with the high temperature in the cycle at \(40^{\circ} \mathrm{C}\). The \(Q_{H}\) is pushed to the ambient air at \(30^{\circ} \mathrm{C}\) in a heat exchanger where the transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Find the required minimum heat transfer area.

Short Answer

Expert verified
The required minimum heat transfer area is approximately 8.14 m².

Step by step solution

01

Understand the Problem

To find the required minimum heat transfer area, we need to use the thermodynamic principles of heat transfer. The air conditioner extracts heat ( Q_c ) from a space maintained at  T_c = 5^{ { ext{}} } C  and rejects heat ( Q_H ) to the ambient air at  T_H = 30^{ { ext{}} } C . The power input is  2.5  {kW} , and we need to calculate the minimum area  A  required by using the heat transfer coefficient  50  {W/m^2K} .
02

Apply Thermodynamic Formulas

Use the formula for the Coefficient of Performance (COP) for a refrigerator:  ext{COP} =  Q_c/W  where  W = 2.5  {kW} . The ideal Carnot COP is given by  ext{COP}_{ ext{Carnot}} =  T_c/(T_H - T_c)  where temperatures are in Kelvin. Convert temperatures:  T_c = 5 + 273 = 278  K ,  T_H = 30 + 273 = 303  K .
03

Calculate Q_c and Q_H

First, calculate  ext{COP}_{ ext{Carnot}}  =  278/(303 - 278) = 11.12 .  ext{COP}_{ ext{Carnot}}  of 11.12 implies that  ext{COP}_{ ext{actual}}  is slightly lower assuming no losses in the heat exchanger. Thus  Q_c = ext{COP}_{ ext{Carnot}}   imes   2.5  {kW}  = 27.8  {kW} . Then,  Q_H = W + Q_c = 2.5 + 27.8 = 30.3  {kW} .
04

Use Heat Transfer Equation

Use the heat transfer equation:  Q_H = U imes A imes   heta_{m}  where  U = 50  {W/m^2K} . Calculate  heta_{m} , the logarithmic mean temperature difference, using:  heta_{m} =  [(30 - 5) - (40 - 5)]/ ext{ln}[(30 - 5)/(40 - 5)] = 25/ ext{ln}(25/35).
05

Calculate Minimum Heat Transfer Area

Substitute into the heat transfer equation and solve:  30.3  {kW} = 50  {W/m^2K}   imes   A   imes   ([25/ ext{ln}(25/35)]). Solve for  A  to find: \[ A = rac{30.3 imes 10^3}{50 imes (25/ ext{ln}(25/35))} \]
06

Solve for A

Calculate  ext{ln}(25/35) , which gives approximately  ext{ln}(0.714)  = -0.3365. Thus, \[ A = rac{30.3 imes 10^3} {50 imes (25/-0.3365)}\] This results in  A = rac{30.3 imes 10^3}{50 imes 74.3} = 8.1371  {m^2} .

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Key Concepts

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

Heat Transfer
Heat transfer is an essential concept in thermodynamics that explains how heat energy moves from one body or system to another. In the context of air conditioners, heat is extracted from a cooler space and expelled into a warmer environment. This process is governed by the principles of thermodynamics and can occur in three primary ways: conduction, convection, and radiation.
Heat exchangers, like the ones in air conditioning systems, often rely on convection to transfer heat between fluids.
To determine the minimum heat transfer area, we utilize the heat transfer coefficient, which is a measure of a material's ability to conduct heat. The formula incorporates this coefficient along with temperature differences to calculate the necessary surface area for effective heat dissipation.
Coefficient of Performance (COP)
The Coefficient of Performance (COP) is a crucial metric in evaluating the efficiency of devices like refrigerators and air conditioners. It represents the ratio of useful heating or cooling provided to the work energy input. A higher COP indicates a more efficient system.
In refrigerators, the COP is calculated using:
  • COP = \(\frac{Q_c}{W}\)
where \(Q_c\) is the heat absorbed from the cold space and \(W\) is the work input.
For the exercise given, COP helps us understand how effectively the air conditioner cools down the environment by using the power input of \(2.5\, \mathrm{kW}\). Calculating the actual COP involves comparing it to the ideal Carnot cycle, giving insight into possible inefficiencies.
Carnot Cycle
The Carnot cycle is a theoretical thermodynamic cycle that serves as a standard of performance for all reversible cycles. It demonstrates the maximum possible efficiency that any engine operating between two temperature reservoirs can attain.
In the context of the exercise, the Carnot cycle helps to determine the ideal COP by setting the upper limit for the possible efficiency of the air conditioning system.
Calculations for any cycle's efficiency using the Carnot principle involve the temperatures of the hot and cold reservoirs:
  • \(\text{COP}_{\text{Carnot}} = \frac{T_c}{T_H - T_c}\)
where \(T_c\) and \(T_H\) are the absolute temperatures of the cold and hot reservoirs respectively.
Understanding the Carnot cycle allows us to recognize the limitations imposed by thermodynamic laws and helps optimize practical systems within these confines.
Logarithmic Mean Temperature Difference
The logarithmic mean temperature difference (LMTD) is a crucial concept in sizing heat exchangers and determining the heat transfer rate. It effectively accounts for the variation in temperature differences that occur throughout the heat exchange process.
Unlike an average, the LMTD provides an accurate mean temperature difference, considering the exponential nature of thermal gradients across the exchanger surface.
The formula is given by:
  • \(\Delta T_m = \frac{\Delta T_1 - \Delta T_2}{\ln(\frac{\Delta T_1}{\Delta T_2})}\)
where \(\Delta T_1\) and \(\Delta T_2\) are temperature differences at each end of the heat exchanger.
In the original problem, the LMTD is used to calculate the required heat transfer area, ensuring sufficient cooling provided by the air conditioner system. By understanding LMTD, engineers can design systems that maximize heat transfer efficiency.

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Most popular questions from this chapter

A water cooler for drinking water should cool 10 gal/h water from \(65 \mathrm{~F}\) to \(50 \mathrm{~F}\) using a small refrigeration unit with a COP of 2.5 . Find the rate of cooling required and the power input to the unit.

A car engine burns 5 kg fuel (equivalent to addition of \(Q_{H}\) ) at \(1500 \mathrm{~K}\) and rejects energy to the radiator and the exhaust at an average temperature of \(750 \mathrm{~K}\). If the fuel provides \(40000 \mathrm{~kJ} / \mathrm{kg}\), what is the maximum amount of work the engine can provide?

A house should be heated by a heat pump, \(\beta^{\prime}=2.2\), and maintained at \(20^{\circ} \mathrm{C}\) at all times. It is estimated that it loses \(0.8 \mathrm{~kW}\) for each degree that the ambient is lower than the inside. Assume an outside temperature of \(-10^{\circ} \mathrm{C}\) and find the needed power to drive the heat pump.

A small house kept at \(77 \mathrm{~F}\) inside loses \(12 \mathrm{Btu} / \mathrm{s}\) to the outside ambient at \(32 \mathrm{~F}\). A heat pump is used to help heat the house together with possible electric heat. The heat pump is driven by a \(2.5-\mathrm{kW}\) motor, and it has a COP that is one-fourth that of a Carnot heat pump unit. Find the actual COP for the heat pump and the amount of electric heat that must be used (if any) to maintain the house temperature.

A refrigerator maintaining a \(5^{\circ} \mathrm{C}\) inside temperature is located in a \(30^{\circ} \mathrm{C}\) room. It must have a high temperature \(\Delta T\) above room temperature and a low temperature \(\Delta T\) below the refrigerated space in the cycle to actually transfer the heat. For a \(\Delta T\) of \(0^{\circ}\) \(5^{\circ},\) and \(10^{\circ} \mathrm{C},\) respectively, calculate the \(\mathrm{COP},\) assuming a Carnot cycle.

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