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Chapter 11: Problem 115

A cross-flow heat exchanger consists of 80 thinwalled tubes of \(3-\mathrm{cm}\) diameter located in a duct of \(1 \mathrm{~m} \times 1 \mathrm{~m}\) cross section. There are no fins attached to the tubes. Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(18^{\circ} \mathrm{C}\) with an average velocity of \(3 \mathrm{~m} / \mathrm{s}\), while hot air \(\left(c_{p}=1010 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the channel at \(130^{\circ} \mathrm{C}\) and \(105 \mathrm{kPa}\) at an average velocity of \(12 \mathrm{~m} / \mathrm{s}\). If the overall heat transfer coefficient is \(130 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the outlet temperatures of both fluids and the rate of heat transfer.

Short Answer

Expert verified
Determine the outlet temperatures of both fluids (cold water and hot air) and the rate of heat transfer in a cross-flow heat exchanger given the fluid properties, mass flow rates, and heat transfer coefficients. Follow the steps below to solve the problem: Step 1: Calculate the mass flow rates of cold water and hot air using the formula, \(\dot{m} = \rho AV\), and the given fluid properties. Step 2: Apply the energy conservation equation for the heat exchanger, which states that the rate of heat transfer from hot air to cold water equals the difference in energy content between the inlet and outlet. Step 3: Apply the heat transfer formula to find the outlet temperatures, which states that the rate of heat transfer equals the product of the overall heat transfer coefficient \(U\), heat transfer surface area \(A_H\), and the logarithmic mean temperature difference \(\Delta T_{LM}\). Step 4: Calculate the rate of heat transfer using the energy conservation equation from Step 2 with the outlet temperatures of both fluids.

Step by step solution

01

Calculate mass flow rates of fluids

We need to calculate the mass flow rates of both cold water and hot air. We will use the formula: \(\dot{m} = \rho AV\), where \(\rho\) is the fluid density, \(A\) is the cross-sectional area, and \(V\) is the fluid velocity. Although we do not have the density of the fluids directly, we do know that for an ideal gas, \(PV = mRT\), so \(\rho = \frac{m}{V} = \frac{P}{RT}\). We'll assume water is incompressible, so its density remains constant at 1000 kg/m³. For cold water: \(\dot{m} = \rho AV\) For hot air: \(\rho = \frac{P}{RT}\) \(\dot{m} = \rho AV\) Calculating the mass flow rate for each fluid is the first step.
02

Apply the energy conservation equation

We need to apply the energy conservation equation for the heat exchanger, which states that the rate of heat transfer from hot air to cold water equals the difference in energy content between the inlet and outlet. \(q = \dot{m}_c c_{p_c}(T_{out_c} - T_{in_c}) = \dot{m}_h c_{p_h}(T_{in_h} - T_{out_h})\) Where \(q\) is the rate of heat transfer, \(\dot{m}_c\) and \(\dot{m}_h\) are the mass flow rates of cold water and hot air respectively, \(c_{p_c}\) and \(c_{p_h}\) are the specific heat capacities of cold water and hot air, and \(T_{in_c}\), \(T_{in_h}\), \(T_{out_c}\), and \(T_{out_h}\) are the inlet and outlet temperatures of cold water and hot air.
03

Apply the heat transfer formula

We will now apply the heat transfer formula to find the outlet temperatures, which states that the rate of heat transfer equals the product of the overall heat transfer coefficient \(U\), heat transfer surface area \(A_H\) and the logarithmic mean temperature difference \(\Delta T_{LM}\): \(q = UA_H \Delta T_{LM}\) Where, \(\Delta T_{LM} = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\Delta T_1 / \Delta T_2)}\) \(\Delta T_1 = T_{h1} - T_{c1}\) \(\Delta T_2 = T_{h2} - T_{c2}\) Applying the formula will help us find the outlet temperatures of both fluids.
04

Calculate the rate of heat transfer

With the outlet temperatures of both fluids, we can calculate the rate of heat transfer using the energy conservation equation from Step 2: \(q = \dot{m}_c c_{p_c}(T_{out_c} - T_{in_c}) = \dot{m}_h c_{p_h}(T_{in_h} - T_{out_h})\) #get_template-Solution#

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Most popular questions from this chapter

Consider a water-to-water counter-flow heat exchanger with these specifications. Hot water enters at \(95^{\circ} \mathrm{C}\) while cold water enters at \(20^{\circ} \mathrm{C}\). The exit temperature of hot water is \(15^{\circ} \mathrm{C}\) greater than that of cold water, and the mass flow rate of hot water is 50 percent greater than that of cold water. The product of heat transfer surface area and the overall heat transfer coefficient is \(1400 \mathrm{~W} / \mathrm{K}\). Taking the specific heat of both cold and hot water to be \(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the outlet temperature of the cold water, \((b)\) the effectiveness of the heat exchanger, \((c)\) the mass flow rate of the cold water, and \((d)\) the heat transfer rate.

A single-pass cross-flow heat exchanger is used to cool jacket water \(\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) of a diesel engine from \(190^{\circ} \mathrm{F}\) to \(140^{\circ} \mathrm{F}\), using air \(\left(c_{p}=0.245 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) with inlet temperature of \(90^{\circ} \mathrm{F}\). Both air flow and water flow are unmixed. If the water and air mass flow rates are \(92,000 \mathrm{lbm} / \mathrm{h}\) and \(400,000 \mathrm{lbm} / \mathrm{h}\), respectively, determine the log mean temperature difference for this heat exchanger.

A performance test is being conducted on a double pipe counter flow heat exchanger that carries engine oil and water at a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) and \(1.75 \mathrm{~kg} / \mathrm{s}\), respectively. Since the heat exchanger has been in service over a long period of time it is suspected that the fouling might have developed inside the heat exchanger that might have affected the overall heat transfer coefficient. The test to be carried out is such that, for a designed value of the overall heat transfer coefficient of \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a surface area of \(7.5 \mathrm{~m}^{2}\), the oil must be heated from \(25^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) by passing hot water at \(100^{\circ} \mathrm{C}\left(c_{p}=4206 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at the flow rates mentioned above. Determine if the fouling has affected the overall heat transfer coefficient. If yes, then what is the magnitude of the fouling resistance?

Reconsider Prob. 11-131. Using EES (or other) software, plot the number of tube passes as a function of water velocity as it varies from \(1 \mathrm{~m} / \mathrm{s}\) to \(8 \mathrm{~m} / \mathrm{s}\), and discuss the results.

Consider a heat exchanger that has an NTU of 4 . Someone proposes to double the size of the heat exchanger and thus double the NTU to 8 in order to increase the effectiveness of the heat exchanger and thus save energy. Would you support this proposal?
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