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Chapter 11: Problem 22

A counter-flow heat exchanger is stated to have an overall heat transfer coefficient, based on outside tube area of \(50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) when operating at design and clean conditions. After a period of use, scale built-up in the heat exchanger gives a fouling factor of \(0.002 \mathrm{~h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} /\) Btu. Determine \((a)\) the overall heat transfer coefficient of the heat exchanger and \((b)\) the percentage change in the overall heat transfer coefficient due to the scale built-up.

Short Answer

Expert verified
Answer: The overall heat transfer coefficient with the scale built-up (Uf) is approximately 45.45 Btu/h·ft²·°F, and the percentage change in the overall heat transfer coefficient due to the scale built-up is approximately 9.09%.

Step by step solution

01

Given Parameters

We are given the following information: - The clean overall heat transfer coefficient (U): \(50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) - The fouling factor (Rf): \(0.002 \mathrm{~h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} /\) Btu
02

Calculate the overall heat transfer coefficient with the scale built-up (Uf)

We can use the following formula to calculate the overall heat transfer coefficient with the fouling factor: \(1/Uf = 1/U + Rf\) Now, we can substitute the given values into this equation to find the overall heat transfer coefficient with scale built up (Uf): \(1/Uf = 1/50 + 0.002\) \(1/Uf = 0.022\) To find Uf, take the reciprocal of the value obtained: \(Uf = 1/(0.022)\) \(Uf = 45.45 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) So, the overall heat transfer coefficient with the scale built-up is approximately \(45.45 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).
03

Calculate the percentage change in the overall heat transfer coefficient due to the scale built-up

To calculate the percentage change, we can use the following formula: Percentage change= \(((U - Uf) / U) \cdot 100\) Now, we can substitute the given and calculated values to find the percentage change: Percentage change = \(((50 - 45.45) / 50) \cdot 100\) Percentage change = \((4.55/50) \cdot 100\) Percentage change ≈ \(9.09\%\) The percentage change in the overall heat transfer coefficient due to the scale built-up is approximately \(9.09\%\).

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Most popular questions from this chapter

A shell-and-tube heat exchanger with 2-shell passes and 8 -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 side at \(95^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area of the heat exchanger.

Hot water \(\left(c_{p h}=4188 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with mass flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) at \(100^{\circ} \mathrm{C}\) enters a thin-walled concentric tube counterflow heat exchanger with a surface area of \(23 \mathrm{~m}^{2}\) and an overall heat transfer coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Cold water \(\left(c_{p c}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with mass flow rate of \(5 \mathrm{~kg} / \mathrm{s}\) enters the heat exchanger at \(20^{\circ} \mathrm{C}\), determine \((a)\) the heat transfer rate for the heat exchanger and \((b)\) the outlet temperatures of the cold and hot fluids. After a period of operation, the overall heat transfer coefficient is reduced to \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine (c) the fouling factor that caused the reduction in the overall heat transfer coefficient.

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.

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.

Consider a water-to-water double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold water enters at \(20^{\circ} \mathrm{C}\) and leaves at \(50^{\circ} \mathrm{C}\), while the hot water enters at \(80^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). Do you think this is a parallel-flow or counterflow heat exchanger? Explain.
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