Water \(\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) is to be heated by solarheated hot air \(\left(c_{p}=0.24 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) in a double- pipe counterflow heat exchanger. Air enters the heat exchanger at \(190^{\circ} \mathrm{F}\) at a rate of \(0.7 \mathrm{lbm} / \mathrm{s}\) and leaves at \(135^{\circ} \mathrm{F}\). Water enters at \(70^{\circ} \mathrm{F}\) at a rate of \(0.35 \mathrm{lbm} / \mathrm{s}\). The overall heat transfer coefficient based on the inner side of the tube is given to be \(20 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine the length of the tube required for a tube internal diameter of \(0.5 \mathrm{in}\).

Short Answer

Expert verified
For a double-pipe counterflow heat exchanger with the given fluids, inlet and outlet temperatures, mass flow rates, and overall heat transfer coefficient, calculate the required length of the tube. Solution: The length of the tube required for the given conditions is approximately 730.24 feet.

Step by step solution

01

Determine the rate of heat transfer

To find the rate of heat transfer for both fluids, we can use the equation $$Q = mc_p\Delta T$$ Let's calculate the rate of heat transfer for air, $$Q_{air}$$: $$Q_{air} = m_{air}c_{p, air}\Delta T_{air}$$ $$Q_{air} = (0.7\,\mathrm{lbm/s})(0.24\,\mathrm{Btu/lbm\,^\circ F})(190\,^\circ F - 135\,^\circ F)$$ $$Q_{air} = 8.82\,\mathrm{Btu/s}$$ Now, calculate the rate of heat transfer for water, $$Q_{water}$$: $$Q_{water} = m_{water}c_{p, water}\Delta T_{water}$$ Since heat transfer is balanced, we can use $$Q_{air}$$ to determine the final temperature of the water: $$Q_{water} = 8.82\,\mathrm{Btu/s} = (0.35\,\mathrm{lbm/s})(1.0\,\mathrm{Btu/lbm\,^\circ F})(T_{water, out} - 70\,^\circ F)$$ $$T_{water, out} = 95\,^\circ F$$
02

Calculate the Log Mean Temperature Difference (LMTD)

LMTD is calculated using the inlet and outlet temperature of both fluids: $$\mathrm{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln \frac{\Delta T_1}{\Delta T_2}}$$ Where: $$\Delta T_1 = T_{air, in} - T_{water, in}$$ $$\Delta T_2 = T_{air, out} - T_{water, out}$$ Let's calculate LMTD: $$\mathrm{LMTD} = \frac{(190\,^\circ F - 70\,^\circ F) - (135\,^\circ F - 95\,^\circ F)}{\ln \frac{190\,^\circ F - 70\,^\circ F}{135\,^\circ F - 95\,^\circ F}}$$ $$\mathrm{LMTD} = 93.25\,^\circ F$$
03

Determine the length of the tube

To calculate the length of the tube, we will use the equation $$Q = U A LMTD$$ Where: $$A = \pi d_i L$$ (heat transfer area considering inner diameter) Rearranging for the length, L: $$L = \frac{Q}{U \pi d_i \mathrm{LMTD}}$$ Now, plug in the given values and solve for L: $$L = \frac{8.82\,\mathrm{Btu/s}}{(20\,\mathrm{Btu/h\,ft^2\,^\circ F}) (\pi)(0.5\,\mathrm{in})(\frac{1\,\mathrm{ft}}{12\,\mathrm{in})} (93.25\,^\circ F)}$$ $$L = 730.24\,\mathrm{ft}$$ The required length of the tube for the given conditions is approximately 730.24 feet.

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Key Concepts

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

Log Mean Temperature Difference
Understanding the Log Mean Temperature Difference (LMTD) is crucial for analyzing heat exchangers. It is a theoretical measure used to compare the temperature difference between the hot and cold fluids at each end of the heat exchanger.

The formula for LMTD is:\[\begin{equation}LMTD = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\end{equation}\]where \(\Delta T_{1}\) and \(\Delta T_{2}\) are the temperature differences at either end of the exchanger. If the temperatures of both fluids change linearly, which rarely happens, the LMTD would be a simple average of these differences.

However, since the temperature of fluids often varies, using LMTD provides an effective temperature difference for calculating the heat transfer rate over the length of the exchanger. In scenarios like a counterflow heat exchanger, where fluids flow in opposite directions, the LMTD method comprehensively accounts for the varying temperatures and enhances the accuracy of the design and analysis process.
Heat Transfer Rate Calculation
The heat transfer rate, often denoted by \(Q\), is a measure of the thermal energy transferred per unit time. It is a fundamental concept in designing and evaluating the performance of heat exchangers.

Calculating \(Q\) involves the specific heat capacities of the fluids (\(c_p\)), their mass flow rates (\(m\)), and the change in temperature (\(\Delta T\)) as the fluids pass through the exchanger:\[\begin{equation}Q = m c_p \Delta T\end{equation}\]In our example, the heat transfer rate calculation for both air and water was crucial to determining the change in temperature of the water, which could then be used to further compute other parameters such as the required tube length. It's essential to consider that in an ideal heat exchange process without energy losses, the heat lost by the hot fluid is equal to the heat gained by the cold fluid.
Counterflow Heat Exchanger
Counterflow heat exchangers are designed in a way that the two fluids involved move in opposite directions. This setup increases the efficiency of heat transfer compared to parallel flow exchangers.

In a counterflow heat exchanger, the hot fluid exits the exchanger at a lower temperature, while the cold fluid exits at a higher temperature, thus transferring heat more effectively. This occurs because the temperature gradient between the fluids, which drives the heat transfer, remains relatively high along the entire length of the heat exchanger.

When considering the design of a counterflow exchanger, it becomes evident that not only does it often provide better performance, but it may also require less space, as illustrated by the compact design in our exercise example. The efficiency of a counterflow design is particularly evident in the improved LMTD, which can result in a more compact heat exchanger for the same heat transfer rate.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, often symbolized by \(U\), is a comprehensive measure of the heat transfer capability of a heat exchanger. It takes into account the heat transfer by conduction through the materials involved, as well as the convective heat transfer from the fluids. The unit of \(U\) is typically expressed in \(\mathrm{Btu/h\cdot ft^2 \cdot{ }^\circ F}\).

The value of \(U\) depends on the properties of the fluids, the materials used in the construction of the heat exchanger, the condition of the heating surfaces, and the type of flow. A higher \(U\) represents a more effective heat exchanger, able to transfer more heat between the fluids with a smaller surface area.

In the exercise, the given value of \(U\) based on the inner side of the tube is a determined factor in calculating the necessary length of the tube for the desired rate of heat transfer. By combining the overall heat transfer coefficient with the LMTD and the heat transfer area, we can accurately determine the tube length required to achieve the heating or cooling needed in a specific application.

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Most popular questions from this chapter

What does the effectiveness of a heat exchanger represent? Can effectiveness be greater than one? On what factors does the effectiveness of a heat exchanger depend?

A test is conducted to determine the overall heat transfer coefficient in a shell-and-tube oil-to-water heat exchanger that has 24 tubes of internal diameter \(1.2 \mathrm{~cm}\) and length \(2 \mathrm{~m}\) in a single shell. Cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(20^{\circ} \mathrm{C}\) at a rate of \(3 \mathrm{~kg} / \mathrm{s}\) and leaves at \(55^{\circ} \mathrm{C}\). Oil \(\left(c_{p}=2150 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows through the shell and is cooled from \(120^{\circ} \mathrm{C}\) to \(45^{\circ} \mathrm{C}\). Determine the overall heat transfer coefficient \(U_{i}\) of this heat exchanger based on the inner surface area of the tubes.

Oil in an engine is being cooled by air in a cross-flow heat exchanger, where both fluids are unmixed. Oil \(\left(c_{p h}=\right.\) \(2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) flowing with a flow rate of \(0.026 \mathrm{~kg} / \mathrm{s}\) enters the heat exchanger at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters at \(30^{\circ} \mathrm{C}\) with a flow rate of \(0.21 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient of the heat exchanger is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total surface area is \(1 \mathrm{~m}^{2}\). Determine \((a)\) the heat transfer effectiveness and \((b)\) the outlet temperature of the oil.

Consider the flow of saturated steam at \(270.1 \mathrm{kPa}\) that flows through the shell side of a shell-and-tube heat exchanger while the water flows through 4 tubes of diameter \(1.25 \mathrm{~cm}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) through each tube. The water enters the tubes of heat exchanger at \(20^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Due to the heat exchange with the cold fluid, steam is condensed on the tubes external surface. The convection heat transfer coefficient on the steam side is \(1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the fouling resistance for the steam and water may be taken as \(0.00015\) and \(0.0001 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. Using the NTU method, determine \((a)\) effectiveness of the heat exchanger, \((b)\) length of the tube, and \((c)\) rate of steam condensation.

A 1-shell-pass and 8-tube-passes heat exchanger is used to heat glycerin \(\left(c_{p}=0.60 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) from \(65^{\circ} \mathrm{F}\) to \(140^{\circ} \mathrm{F}\) by hot water \(\left(c_{p}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) that enters the thinwalled \(0.5\)-in-diameter tubes at \(175^{\circ} \mathrm{F}\) and leaves at \(120^{\circ} \mathrm{F}\). The total length of the tubes in the heat exchanger is \(500 \mathrm{ft}\). The convection heat transfer coefficient is \(4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the glycerin (shell) side and \(50 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\) on the water (tube) side. Determine the rate of heat transfer in the heat exchanger \((a)\) before any fouling occurs and \((b)\) after fouling with a fouling factor of \(0.002 \mathrm{~h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu on the outer surfaces of the tubes.

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