/*! 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 58 A shell-and-tube heat exchanger ... [FREE SOLUTION] | 91Ó°ÊÓ

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A shell-and-tube heat exchanger with one shell pass and 20 tube passes uses hot water on the tube side to heat oil on the shell side. The single copper tube has inner and outer diameters of 20 and \(24 \mathrm{~mm}\) and a length per pass of \(3 \mathrm{~m}\). The water enters at \(87^{\circ} \mathrm{C}\) and \(0.2 \mathrm{~kg} / \mathrm{s}\) and leaves at \(27^{\circ} \mathrm{C}\). Inlet and outlet temperatures of the oil are 7 and \(37^{\circ} \mathrm{C}\). What is the average convection coefficient for the tube outer surface?

Short Answer

Expert verified
The average convection coefficient for the tube outer surface, \(h_{avg}\), can be found by following these steps: 1. Calculate the mass flow rate of oil, \(m^o\), using the heat balance equation and specific heat capacities of water and oil. 2. Calculate the overall heat transfer rate, \(Q_{total}\), using the mass flow rate of oil and oil temperature differences. 3. Determine the Log Mean Temperature Difference (LMTD) using the formula \(LMTD = \frac{(ΔT_{in} - ΔT_{out})}{\ln(ΔT_{in} / ΔT_{out})}\), where ΔT_in and ΔT_out are the temperature differences at the inlet and outlet of the heat exchanger. 4. Calculate the total heat transfer area, \(A_{total}\), using the tube dimensions and the number of passes. 5. Finally, find the average convection coefficient, \(h_{avg}\), using the formula \(h_{avg} = \frac{Q_{total}}{A_{total} \cdot LMTD}\), and substitute the values obtained in previous steps.

Step by step solution

01

Determine the mass flow rate of oil

As mentioned, the inlet and outlet temperatures of the oil are 7 and \(37^{\circ} \mathrm{C}\), and we know the mass flow rate of the water is \(0.2 \mathrm{~kg} / \mathrm{s}\). Let's use the specific heat capacity of water and oil to find the mass flow rate of oil. The specific heat of water is \(c_p^w = 4200 \mathrm{~J / (kg \cdot K)}\), and for oil, it is \(c_p^o = 2130 \mathrm{~J / (kg \cdot K)}\). Given the operating conditions, we know that the heat gained by the oil should be equal to the heat lost by the water. Therefore, \(m^w c_p^w (T_{out}^w - T_{in}^w) = m^o c_p^o (T_{out}^o - T_{in}^o)\) Substituting the given values, we can calculate the mass flow rate of oil, \(m^o\).
02

Calculate Overall Heat Transfer Rate

Now that we know the mass flow rate of oil, we can calculate the overall heat transfer rate, \(Q_{total}\). The following equation can be used to find the following heat transfer rate: \(Q_{total} = m^o c_p^o (T_{out}^o - T_{in}^o)\) By substituting the values we calculated earlier, we can obtain the overall heat transfer rate.
03

Calculate Log Mean Temperature Difference (LMTD)

The LMTD can be calculated using the following formula: \(LMTD = \frac{(ΔT_{in} - ΔT_{out})}{\ln(ΔT_{in} / ΔT_{out})}\) Where: - ΔT_in is the temperature difference at the inlet of the heat exchanger - ΔT_out is the temperature difference at the outlet of the heat exchanger To find these temperature differences, subtract the oil temperature from the water temperature at both the inlet and outlet: - \(ΔT_{in} = T_{in}^w - T_{in}^o\) - \(ΔT_{out} = T_{out}^w - T_{out}^o\) Using the given temperatures, calculate the inlet and outlet temperature differences. Then, substitute these values into the LMTD formula and find the LMTD value.
04

Find Average Convection Coefficient

Now that we have the overall heat transfer rate and the LMTD value, we can calculate the average convection coefficient, \(h_{avg}\), for the outer surface of the tube. First, we need to calculate the total heat transfer area, \(A_{total}\), using the tube dimensions: \(A_{total} = n \pi D_{o} L\) Where: - \(n\) is the number of tube passes (20) - \(D_{o}\) is the outer diameter of the copper tube (24 mm) - \(L\) is the length per pass (3 m) Now, we can find the average convection coefficient with the following formula: \(h_{avg} = \frac{Q_{total}}{A_{total} \cdot LMTD}\) Substitute the values calculated in previous steps and find the average convection coefficient for the tube outer surface.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convection Coefficient
Understanding the convection coefficient is key to grasping how heat exchanger systems work. It measures how effectively heat is transferred from the fluid inside the tubes to the fluid on the shell side, or vice versa. Think of it as the efficiency rating of heat transfer between materials. A higher convection coefficient indicates a more effective transfer of heat.
In a shell-and-tube heat exchanger, the convection coefficient is influenced by several factors:
  • The type of fluid being used, as different fluids have varying heat capacities and flow characteristics.
  • The flow rate of these fluids, since faster movement typically enhances heat transfer.
  • The material of the heat-exchanging surfaces which can affect how quickly heat is conducted across the boundary.
To calculate the average convection coefficient, we use the heat transfer rate and the log mean temperature difference (LMTD), which we'll delve into next. The formula for the average convection coefficient is given by: \[ h_{avg} = \frac{Q_{total}}{A_{total} \cdot LMTD} \]Where:
  • \(Q_{total}\) is the overall heat transfer rate.
  • \(A_{total}\) is the total heat transfer area.
  • \(LMTD\) is the log mean temperature difference.
Understanding these components can help in assessing how well different designs and materials perform in real-world conditions.
Log Mean Temperature Difference
The Log Mean Temperature Difference (LMTD) is a crucial concept in heat exchanger analysis. It is used to determine the average temperature difference between the hot and cold fluids across the heat exchanger.
In a shell-and-tube heat exchanger, the temperature of fluids varies at different points, making LMTD a better measure than a simple average. It provides a way to account for the exponential change in temperature difference.
The formula for LMTD is: \[ LMTD = \frac{(\Delta T_{in} - \Delta T_{out})}{\ln(\Delta T_{in} / \Delta T_{out})} \] Where:
  • \(\Delta T_{in}\) is the temperature difference at the exchanger's inlet.
  • \(\Delta T_{out}\) is the temperature difference at the outlet.
Calculating these differences involves subtracting the outlet temperatures of both fluids from their respective inlet temperatures. After obtaining these values, they are plugged into the LMTD formula to get the log mean value. This accurate measure of temperature difference helps in calculating the accurate heat transfer rate we're discussing next.
Overall Heat Transfer Rate
The overall heat transfer rate, represented as \(Q_{total}\), is fundamental to evaluating the performance of a heat exchanger. It tells us the amount of heat being transferred from the hot fluid to the cold fluid per unit of time.
To compute \(Q_{total}\), you'd use the formula:\[ Q_{total} = m^o c_p^o (T_{out}^o - T_{in}^o) \]Where:
  • \(m^o\) is the mass flow rate of the oil.
  • \(c_p^o\) is the specific heat capacity of oil.
  • \(T_{out}^o\) and \(T_{in}^o\) are the outlet and inlet temperatures of the oil.
The computation starts with finding the mass flow rate of the unheated fluid (oil in this case), which involves equating the heat lost by the water to the heat gained by the oil, as energy cannot be created or destroyed in this closed process. By knowing the \(Q_{total}\), engineers can determine whether a heat exchanger meets the required thermal efficiency based on specific industrial requirements. It's also key for the design and optimization of heat exchangers.

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Most popular questions from this chapter

Consider a concentric tube heat exchanger with an area of \(50 \mathrm{~m}^{2}\) operating under the following conditions: \begin{tabular}{lcc} \hline & Hot flid & Cold flid \\ \hline Heat capacity rate, \(\mathrm{kW} / \mathrm{K}\) & 6 & 3 \\ Inlet temperature, \({ }^{\circ} \mathrm{C}\) & 60 & 30 \\ Outlet temperature, \({ }^{\circ} \mathrm{C}\) & \(-\) & 54 \\ \hline \end{tabular} (a) Determine the outlet temperature of the hot fluid. (b) Is the heat exchanger operating in counterflow or parallel flow, or can't you tell from the available information? (c) Calculate the overall heat transfer coefficient. (d) Calculate the effectiveness of this exchanger. (e) What would be the effectiveness of this exchanger if its length were made very large?

An automobile radiator may be viewed as a cross-flow heat exchanger with both fluids unmixed. Water, which has a flow rate of \(0.05 \mathrm{~kg} / \mathrm{s}\), enters the radiator at \(400 \mathrm{~K}\) and is to leave at \(330 \mathrm{~K}\). The water is cooled by air that enters at \(0.75 \mathrm{~kg} / \mathrm{s}\) and \(300 \mathrm{~K}\). (a) If the overall heat transfer coefficient is \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the required heat transfer surface area? (b) A manufacturing engineer claims ridges can be stamped on the finned surface of the exchanger, which could greatly increase the overall heat transfer coefficient. With all other conditions remaining the same and the heat transfer surface area determined from part (a), generate a plot of the air and water outlet temperatures as a function of \(U\) for \(200 \leq U \leq 400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What benefits result from increasing the overall convection coefficient for this application?

Thin-walled aluminum tubes of diameter \(D=10 \mathrm{~mm}\) are used in the condenser of an air conditioner. Under normal operating conditions, a convection coefficient of \(h_{i}=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is associated with condensation on the inner surface of the tubes, while a coefficient of \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained by airflow over the tubes. (a) What is the overall heat transfer coefficient if the tubes are unfinned? (b) What is the overall heat transfer coefficient based on the inner surface, \(U_{i}\), if aluminum annular fins of thickness \(t=1.5 \mathrm{~mm}\), outer diameter \(D_{o}=20 \mathrm{~mm}\), and pitch \(S=3.5 \mathrm{~mm}\) are added to the outer surface? Base your calculations on a 1-m-long section of tube. Subject to the requirements that \(t \geq 1 \mathrm{~mm}\) and \((S-t) \geq 1.5 \mathrm{~mm}\), explore the effect of variations in \(t\) and \(S\) on \(U_{i}\). What combination of \(t\) and \(S\) would yield the best heat transfer performance?

Consider a Rankine cycle with saturated steam leaving the boiler at a pressure of \(2 \mathrm{MPa}\) and a condenser pressure of \(10 \mathrm{kPa}\). (a) Calculate the thermal efficiency of the ideal Rankine cycle for these operating conditions. (b) If the net reversible work for the cycle is \(0.5 \mathrm{MW}\), calculate the required flow rate of cooling water supplied to the condenser at \(15^{\circ} \mathrm{C}\) with an allowable temperature rise of \(10^{\circ} \mathrm{C}\). (c) Design a shell-and-tube heat exchanger (one-shell, multiple-tube passes) that will meet the heat rate and temperature conditions required of the condenser. Your design should specify the number of tubes and their diameter and length.

Consider a very long, concentric tube heat exchanger having hot and cold water inlet temperatures of 85 and \(15^{\circ} \mathrm{C}\). The flow rate of the hot water is twice that of the cold water. Assuming equivalent hot and cold water specific heats, determine the hot water outlet temperature for the following modes of operation: (a) counterflow and (b) parallel flow.

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