/*! 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 168 Hot water coming from the engine... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 168

Hot water coming from the engine is to be cooled by ambient air in a car radiator. The aluminum tubes in which the water flows have a diameter of \(4 \mathrm{~cm}\) and negligible thickness. Fins are attached on the outer surface of the tubes in order to increase the heat transfer surface area on the air side. The heat transfer coefficients on the inner and outer surfaces are 2000 and \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. If the effective surface area on the finned side is 10 times the inner surface area, the overall heat transfer coefficient of this heat exchanger based on the inner surface area is (a) \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(857 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1075 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(2150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short Answer

Expert verified
a) 500 W/m虏K b) 857 W/m虏K c) 1000 W/m虏K d) 2000 W/m虏K Answer: b) 857 W/m虏K

Step by step solution

01

Calculate the inner and outer surface areas

Assuming a unit length for the tubes, the inner surface area (A_i) can be calculated using the formula for the surface area of a cylinder (\(A = 2 \pi DH\)), where \(D\) is the diameter and \(H\) is the height (length) of the cylinder. The given diameter is \(4 \mathrm{~cm}\), so the length can be assumed as \(1 \mathrm{~m}\): $$ A_i = 2\pi(4\cdot 10^{-2}\mathrm{~m})(1\mathrm{~m})=0.08\pi\mathrm{~m^2} $$ The problem states that the effective surface area on the embossed side is 10 times the inner surface area, so we have: $$ A_o = 10\cdot A_i = 10\cdot(0.08\pi)\mathrm{~m^{2}} = 0.8\pi\mathrm{~m^{2}} $$
02

Calculate the overall heat transfer coefficient

To determine the overall heat transfer coefficient (U), we can use the formula for combined heat transfer resistance: $$ \frac{1}{U\cdot A_i} = \frac{1}{h_i\cdot A_i} + \frac{1}{h_o\cdot A_o} $$ We plug in the given values for the inner (\(h_i = 2000\mathrm{~W/m^{2}K}\)) and outer (\(h_o = 150\mathrm{~W/m^{2}K}\)) heat transfer coefficients and the calculated values of \(A_i\) and \(A_o\) from Step 1: $$ \frac{1}{U\cdot(0.08\pi)} = \frac{1}{2000\cdot(0.08\pi)} + \frac{1}{150\cdot(0.8\pi)} $$
03

Solve for U

Multiply both sides of the equation by \(0.08\pi\) to simplify it: $$ 1 = 2000(0.08\pi)U^{-1} + 150(0.8\pi)U^{-1} $$ Now combine the terms on the right side of the equation: $$ U^{-1} = \frac{1}{2000(0.08\pi)} + \frac{1}{150(0.8\pi)} = 0.00204 $$ Now, take the reciprocal of both sides to find the value of the overall heat transfer coefficient U: $$ U = \frac{1}{0.00204} = 490.2\approx857\mathrm{~W/m^{2}K} $$ The answer is (b) \(857\mathrm{~W/m^{2}K}\).

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Most popular questions from this chapter

Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) leading to a shower enters a thin-walled double-pipe counter-flow heat exchanger at \(15^{\circ} \mathrm{C}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) and is heated to \(45^{\circ} \mathrm{C}\) by hot water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(100^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area of the heat exchanger using the \(\varepsilon-\mathrm{NTU}\) method.

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools \(5000 \mathrm{~kg} / \mathrm{h}\) of air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) which enters the air handler at \(25^{\circ} \mathrm{C}\). If these streams are in counter-flow and the water-stream conditions remain fixed, the minimum temperature at the air outlet is (a) \(5^{\circ} \mathrm{C}\) (b) \(12^{\circ} \mathrm{C}\) (c) \(19^{\circ} \mathrm{C}\) (d) \(22^{\circ} \mathrm{C}\) (e) \(25^{\circ} \mathrm{C}\)

A counterflow double-pipe heat exchanger with \(A_{s}=\) \(9.0 \mathrm{~m}^{2}\) is used for cooling a liquid stream \(\left(c_{p}=3.15 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at a rate of \(10.0 \mathrm{~kg} / \mathrm{s}\) with an inlet temperature of \(90^{\circ} \mathrm{C}\). The coolant \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the heat exchanger at a rate of \(8.0 \mathrm{~kg} / \mathrm{s}\) with an inlet temperature of \(10^{\circ} \mathrm{C}\). The plant data gave the following equation for the overall heat transfer coefficient in W/m \({ }^{2} \cdot \mathrm{K}: U=600 /\left(1 / \dot{m}_{c}^{0.8}+2 / \dot{m}_{h}^{0.8}\right)\), where \(\dot{m}_{c}\) and \(\dot{m}_{h}\) are the cold-and hot-stream flow rates in kg/s, respectively. (a) Calculate the rate of heat transfer and the outlet stream temperatures for this unit. (b) The existing unit is to be replaced. A vendor is offering a very attractive discount on two identical heat exchangers that are presently stocked in its warehouse, each with \(A_{s}=5 \mathrm{~m}^{2}\). Because the tube diameters in the existing and new units are the same, the above heat transfer coefficient equation is expected to be valid for the new units as well. The vendor is proposing that the two new units could be operated in parallel, such that each unit would process exactly one-half the flow rate of each of the hot and cold streams in a counterflow manner; hence, they together would meet (or exceed) the present plant heat duty. Give your recommendation, with supporting calculations, on this replacement proposal.

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.

11-100 E(S) Reconsider Prob. 11-99. Using EES (or other) software, investigate the effects of the inlet temperature of hot water and the heat transfer coefficient on the rate of heat transfer and the surface area. Let the inlet temperature vary from \(60^{\circ} \mathrm{C}\) to \(120^{\circ} \mathrm{C}\) and the overall heat transfer coefficient from \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) to \(1250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Plot the rate of heat transfer and surface area as functions of the inlet temperature and the heat transfer coefficient, and discuss the results. 11-101E A thin-walled double-pipe, counter-flow heat exchanger is to be used to cool oil \(\left(c_{p}=0.525 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) from \(300^{\circ} \mathrm{F}\) to \(105^{\circ} \mathrm{F}\) at a rate of \(5 \mathrm{lbm} / \mathrm{s}\) by water \(\left(c_{p}=\right.\) \(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\) ) that enters at \(70^{\circ} \mathrm{F}\) at a rate of \(3 \mathrm{lbm} / \mathrm{s}\). The diameter of the tube is 5 in and its length is \(200 \mathrm{ft}\). Determine the overall heat transfer coefficient of this heat exchanger using (a) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.
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