Chapter 11: Problem 39
Explain how the LMTD method can be used to determine the heat transfer surface area of a multipass shelland-tube heat exchanger when all the necessary information, including the outlet temperatures, is given.
Chapter 11: Problem 39
Explain how the LMTD method can be used to determine the heat transfer surface area of a multipass shelland-tube heat exchanger when all the necessary information, including the outlet temperatures, is given.
All the tools & learning materials you need for study success - in one app.
Get started for freeA shell-and-tube heat exchanger with two shell passes and eight tube passes is used to heat ethyl alcohol $\left(c_{p}=2670 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( in the tubes from \)25^{\circ} \mathrm{C}\( to \)70^{\circ} \mathrm{C}\( at a rate of \)2.1 \mathrm{~kg} / \mathrm{s}$. The heating is to be done by water $\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( that enters the shell at \)95^{\circ} \mathrm{C}$ and leaves at \(60^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is $800 \mathrm{~W} / \mathrm{m}^{2} . \mathrm{K}$, determine the heat transfer surface area of the heat exchanger using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method. Answer: \((a) 11.4 \mathrm{~m}^{2}\)
A shell-and-tube heat exchanger with one 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 a length of \(3.2 \mathrm{~m}\) per pass. Water \(\left(c_{p}=4180\right.\) \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) ) enters the tubes at $18^{\circ} \mathrm{C}\( at a rate of \)6.2 \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.
The cardiovascular countercurrent heat exchanger has an overall heat transfer coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Arterial blood enters at \(37^{\circ} \mathrm{C}\) and exits at \(27^{\circ} \mathrm{C}\). Venous blood enters at \(25^{\circ} \mathrm{C}\) and exits at $34^{\circ} \mathrm{C}$. Determine the mass flow rates of the arterial blood and venous blood in \(\mathrm{g} / \mathrm{s}\) if the specific heat of both arterial and venous blood is constant and equal to $3475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, and the surface area of the heat transfer to occur is \)0.15 \mathrm{~cm}^{2}$.
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 12 times the inner surface area, the overall heat transfer coefficient of this heat exchanger based on the inner surface area is (a) \(760 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(832 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(947 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(1075 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(1210 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)
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 inlet temperatures of the two fluids are specified. Does the value of \(\dot{Q}_{\max }\) depend on the type of the heat exchanger?
What do you think about this solution?
We value your feedback to improve our textbook solutions.