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Chapter 11: Problem 156

The condenser of a room air conditioner is designed to reject heat at a rate of \(15,000 \mathrm{~kJ} / \mathrm{h}\) from refrigerant-134a as the refrigerant is condensed at a temperature of \(40^{\circ} \mathrm{C}\). Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows across the finned condenser coils, entering at \(25^{\circ} \mathrm{C}\) and leaving at \(35^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the refrigerant side is \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer area on the refrigerant side.

Short Answer

Expert verified
Answer: The heat transfer area on the refrigerant side is approximately 2.78 m².

Step by step solution

01

Convert heat rejection rate to Watts

We are given that the heat rejection rate is \(15,000 \mathrm{~kJ} / \mathrm{h}\). To convert this value to Watts, we will use the following conversion factors: 1 kJ = 1000 J 1 h = 3600 s So, the heat rejection rate in Watts can be calculated as: Heat rejection rate (W) = \(\dfrac{15,000 \times 1000}{3600} \mathrm{~W} = 4166.67 \mathrm{~W}\)
02

Calculate the temperature difference

We are given that the air enters the condenser at a temperature of \(25^{\circ} \mathrm{C}\) and leaves at \(35^{\circ} \mathrm{C}\). Since the refrigerant is at a constant temperature of \(40^{\circ} \mathrm{C}\), the average temperature difference between the refrigerant and the air is: Temperature difference (\(\Delta T\)) = \(\dfrac{(40 - 25) + (40 - 35)}{2} = \dfrac{15 + 5}{2} = 10^{\circ} \mathrm{C}\)
03

Apply the heat transfer equation

The rate of heat transfer (\(Q\)) can be written as: \(Q = U \cdot A \cdot \Delta T\) where: - \(Q\): heat transfer rate - \(U\): overall heat transfer coefficient - \(A\): heat transfer area - \(\Delta T\): temperature difference between the refrigerant and the air We need to find the heat transfer area \(A\). So, we will rearrange the equation above to get: \(A = \dfrac{Q}{U \cdot \Delta T}\) Now, we can substitute the values we have found in Step 1 and Step 2, and the given overall heat transfer coefficient, into the equation above to calculate the heat transfer area: \(A = \dfrac{4166.67}{150 \cdot 10} = \dfrac{4166.67}{1500} = 2.78 \mathrm{~m}^2\)
04

Final answer for heat transfer area

The heat transfer area on the refrigerant side is approximately 2.78 square meters.

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Most popular questions from this chapter

Saturated water vapor at \(100^{\circ} \mathrm{C}\) condenses in the shell side of a 1 -shell and 2-tube heat exchanger with a surface area of \(0.5 \mathrm{~m}^{2}\) and an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Cold water \(\left(c_{p c}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(0.5 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(15^{\circ} \mathrm{C}\), determine the outlet temperature of the cold water and the heat transfer rate for the heat exchanger.

In a one-shell and eight-tube pass heat exchanger, the temperature of water flowing at rate of \(50,000 \mathrm{lbm} / \mathrm{h}\) is raised from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\). Hot air \(\left(c_{p}=0.25 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) that flows on the tube side enters the heat exchanger at \(600^{\circ} \mathrm{F}\) and exits at \(300^{\circ} \mathrm{F}\). If the convection heat transfer coefficient on the outer surface of the tubes is \(30 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\), determine the surface area of the heat exchanger using both LMTD and \(\varepsilon-\mathrm{NTU}\) methods. Account for the possible fouling resistance of \(0.0015\) and \(0.001 \mathrm{~h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu on water and air side, respectively.

Air at \(18^{\circ} \mathrm{C}\left(c_{p}=1006 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be heated to \(58^{\circ} \mathrm{C}\) by hot oil at \(80^{\circ} \mathrm{C}\left(c_{p}=2150 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a cross-flow heat exchanger with air mixed and oil unmixed. The product of heat transfer surface area and the overall heat transfer coefficient is \(750 \mathrm{~W} / \mathrm{K}\) and the mass flow rate of air is twice that of oil. Determine \((a)\) the effectiveness of the heat exchanger, \((b)\) the mass flow rate of air, and \((c)\) the rate of heat transfer.

Explain how the maximum possible heat transfer rate \(\dot{Q}_{\max }\) in a heat exchanger can be determined when the mass flow rates, specific heats, and the inlet temperatures of the two fluids are specified. Does the value of \(\dot{Q}_{\max }\) depend on the type of the heat exchanger?

How do heavy clothing and extreme environmental conditions affect the cardiovascular counter-current exchanger?
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