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Chapter 11: Problem 160

A cross-flow heat exchanger with both fluids unmixed has an overall heat transfer coefficient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a heat transfer surface area of \(400 \mathrm{~m}^{2}\). The hot fluid has a heat capacity of \(40,000 \mathrm{~W} / \mathrm{K}\), while the cold fluid has a heat capacity of \(80,000 \mathrm{~W} / \mathrm{K}\). If the inlet temperatures of both hot and cold fluids are \(80^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine (a) the exit temperature of the hot fluid and \((b)\) the rate of heat transfer in the heat exchanger.

Short Answer

Expert verified
Answer: The exit temperature of the hot fluid is approximately \(27^{\circ} \mathrm{C}\), and the rate of heat transfer in the heat exchanger is approximately \(2,280,800 \mathrm{~W}\).

Step by step solution

01

Identify the given parameters

We are given the following parameters: - Overall heat transfer coefficient (U): \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Heat transfer surface area (A): \(400 \mathrm{~m}^{2}\) - Heat capacity of hot fluid (Cp_h): \(40,000 \mathrm{~W} / \mathrm{K}\) - Heat capacity of cold fluid (Cp_c): \(80,000 \mathrm{~W} / \mathrm{K}\) - Inlet temperature of hot fluid (T1_h): \(80^{\circ} \mathrm{C}\) - Inlet temperature of the cold fluid (T1_c): \(20^{\circ} \mathrm{C}\)
02

Determine the minimum heat capacity rate and the heat capacity rate ratio

First, we need to find the minimum heat capacity rate, which is the smaller of Cp_h and Cp_c: \(C_{min} = min(Cp_h, Cp_c) = min(40,000, 80,000) = 40,000 \mathrm{~W} / \mathrm{K}\) Next, calculate the heat capacity rate ratio (C_r): \(C_r = \dfrac{C_{min}}{C_{max}} = \dfrac{40,000}{80,000} = 0.5 \)
03

Calculate the heat transfer effectiveness of the heat exchanger (ε)

The heat transfer effectiveness (ε) of a cross-flow heat exchanger with both fluids unmixed can be found using the following formula: \(\varepsilon = 1 - \mathrm{exp} \left(- Ntu \cdot (1 + C_r) \right)\) Where Ntu is the number of transfer units and is calculated as: \(Ntu = \dfrac{UA}{C_{min}}\) Calculate Ntu: \(Ntu = \dfrac{200 \cdot 400}{40,000} = 2\) Now, substitute the Ntu and C_r values back into the formula for ε: \(\varepsilon = 1 - exp(-2 \cdot (1 + 0.5)) = 1 - exp(-3) \approx 0.9502\)
04

Calculate the actual heat transfer rate (Q)

Now that we have the heat transfer effectiveness (ε), we can calculate the actual heat transfer rate (Q). The formula for Q is: \(Q = \varepsilon \cdot C_{min} \cdot (T1_h - T1_c)\) Substitute the given values and the calculated ε into the formula: \(Q = 0.9502 \cdot 40,000 \cdot (80 - 20) \approx 2,280,800 \mathrm{~W}\)
05

Determine the exit temperature of the hot fluid (T2_h)

Finally, we can find the exit temperature of the hot fluid (T2_h) using the equation: \(T2_h = T1_h - \dfrac{Q}{Cp_h}\) Substitute the given values and the calculated Q into the equation: \(T2_h = 80 - \dfrac{2,280,800}{40,000} \approx 27^{\circ} \mathrm{C}\) Thus, (a) the exit temperature of the hot fluid is approximately \(27^{\circ} \mathrm{C}\) (b) the rate of heat transfer in the heat exchanger is approximately \(2,280,800 \mathrm{~W}\).

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Most popular questions from this chapter

Consider a heat exchanger in which both fluids have the same specific heats but different mass flow rates. Which fluid will experience a larger temperature change: the one with the lower or higher mass flow rate?

A shell-and-tube heat exchanger with 1-shell pass and 14-tube passes is used to heat water in the tubes with geothermal steam condensing at \(120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)\) on the shell side. The tubes are thin-walled and have a diameter of \(2.4 \mathrm{~cm}\) and length of \(3.2 \mathrm{~m}\) per pass. Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(3.9 \mathrm{~kg} / \mathrm{s}\). If the temperature difference between the two fluids at the exit is \(46^{\circ} \mathrm{C}\), determine (a) the rate of heat transfer, \((b)\) the rate of condensation of steam, and \((c)\) the overall heat transfer coefficient.

Consider a shell-and-tube water-to-water heat exchanger with identical mass flow rates for both the hotand cold-water streams. Now the mass flow rate of the cold water is reduced by half. Will the effectiveness of this heat exchanger increase, decrease, or remain the same as a result of this modification? Explain. Assume the overall heat transfer coefficient and the inlet temperatures remain the same.

Steam is to be condensed on the shell side of a 1 -shellpass and 8-tube-passes condenser, with 50 tubes in each pass, at \(30^{\circ} \mathrm{C}\left(h_{f g}=2431 \mathrm{~kJ} / \mathrm{kg}\right)\). Cooling water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(15^{\circ} \mathrm{C}\) at a rate of \(1800 \mathrm{~kg} / \mathrm{h}\). The tubes are thin-walled, and have a diameter of \(1.5 \mathrm{~cm}\) and length of \(2 \mathrm{~m}\) per pass. If the overall heat transfer coefficient is \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine \((a)\) the rate of heat transfer and \((b)\) the rate of condensation of steam.

Hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be cooled by water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a 2 -shell-passes and 12 -tube-passes heat exchanger. The tubes are thin-walled and are made of copper with a diameter of \(1.8 \mathrm{~cm}\). The length of each tube pass in the heat exchanger is \(3 \mathrm{~m}\), and the overall heat transfer coefficient is \(340 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Water flows through the tubes at a total rate of \(0.1 \mathrm{~kg} / \mathrm{s}\), and the oil through the shell at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The water and the oil enter at temperatures \(18^{\circ} \mathrm{C}\) and \(160^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.
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