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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.

Short Answer

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**Question:** Explain how to use the Log Mean Temperature Difference (LMTD) method to determine the heat transfer surface area of a multi-pass shell and tube heat exchanger given all necessary information, including outlet temperatures. **Answer:** To determine the heat transfer surface area using the LMTD method, follow these steps: 1. Understand the LMTD method: The LMTD method is used for determining the heat transfer rate and required surface area of a heat exchanger, considering the temperature gradients between the hot and cold fluids. 2. Calculate the LMTD: Use the inlet and outlet temperatures of the hot and cold fluids to determine the LMTD using the formula: LMTD = \[\frac{(ΔT_1 - ΔT_2)}{ln(\frac{ΔT_1}{ΔT_2})}\], where ΔT_1 and ΔT_2 are the temperature differences at the inlet and outlet, respectively. 3. Determine the heat transfer rate (Q): Calculate Q using the formula Q = m × Cp × ∆T, where m is the mass flow rate, Cp is the specific heat capacity, and ∆T is the temperature difference. 4. Calculate the overall heat transfer coefficient (U): The overall heat transfer coefficient combines the fluid's conductivity and the material's resistance and is provided in the exercise or obtained from a reference source or experimental data. 5. Determine the required heat transfer surface area (A): Calculate the surface area using the formula A = \[\frac{Q}{U × LMTD}\], where Q is the heat transfer rate, U is the overall heat transfer coefficient, and LMTD is the Log Mean Temperature Difference.

Step by step solution

01

Understand the LMTD method

The Log Mean Temperature Difference (LMTD) method is used to determine the heat transfer rate and the required surface area of a heat exchanger. The LMTD is the average driving force for heat transfer across the heat exchanger, considering the temperature gradients between the hot and cold fluids.
02

Calculate the LMTD

Use the given inlet and outlet temperatures of the hot and cold fluids in the heat exchanger to determine the LMTD using the formula: LMTD = \[\frac{(ΔT_1 - ΔT_2)}{ln(\frac{ΔT_1}{ΔT_2})}\] Where: - ΔT_1 is the temperature difference between the hot and cold fluids at the heat exchanger's inlet - ΔT_2 is the temperature difference between the hot and cold fluids at the heat exchanger's outlet
03

Determine the heat transfer rate

To find the heat transfer rate, Q, you need to know the mass flow rate (m) and the specific heat capacity (Cp) of the hot and cold fluids. With this information, use the formula: Q = m × Cp × ∆T Where: - m is the mass flow rate of either the hot or cold fluid (whichever has known values) - Cp is the specific heat capacity of the fluid - ∆T is the temperature difference between inlet and outlet for the fluid
04

Calculate the overall heat transfer coefficient

The overall heat transfer coefficient (U) represents the effectiveness of heat transfer in the heat exchanger and combines the fluid's conductivity and the material's resistance. This coefficient should be provided in the exercise or obtained from a reference source or experimental data.
05

Determine the required heat transfer surface area

To calculate the heat transfer surface area (A) of the heat exchanger, use the formula: A = \[\frac{Q}{U × LMTD}\] Where: - Q is the heat transfer rate, calculated in Step 3 - U is the overall heat transfer coefficient, calculated in Step 4 - LMTD is the Log Mean Temperature Difference, calculated in Step 2

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Most popular questions from this chapter

A 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?

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