Oil is being cooled from \(180^{\circ} \mathrm{F}\) to \(120^{\circ} \mathrm{F}\) in a 1 -shell and 2-tube heat exchanger with an overall heat transfer coefficient of \(40 \mathrm{Btu} / \mathrm{h} \mathrm{ft} 2{ }^{\circ} \mathrm{F}\). Water \(\left(c_{p c}=1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) enters at \(80^{\circ} \mathrm{F}\) and exits at \(100^{\circ} \mathrm{F}\) with a mass flow rate of \(20,000 \mathrm{lbm} / \mathrm{h}\), determine (a) the log mean temperature difference and \((b)\) the surface area of the heat exchanger.

Short Answer

Expert verified
Answer: The log mean temperature difference (LMTD) for this system is approximately \(48.27^{\circ}\mathrm{F}\), and the surface area of the heat exchanger is approximately \(207.6\ \mathrm{ft}^2\).

Step by step solution

01

Calculate the shell-side mass flow rate

Using the energy balance equation for water, we can calculate the shell-side mass flow rate: \(Q = m_c \cdot c_{pc} \cdot (T_{c,out} - T_{c,in})\), where \(Q\) is the heat transfer rate, \(m_c\) is the mass flow rate of water (cold fluid), \(c_{pc}\) is the specific heat capacity of water, and \(T_{c,out}\) and \(T_{c,in}\) represent the outlet and inlet water temperatures, respectively. Plug in the given values: \[ Q = (20000\ \mathrm{lbm}/\mathrm{h}) \cdot (1.0\ \mathrm{Btu}/\mathrm{lbm}{\cdot}^{\circ}\mathrm{F}) \cdot (100^{\circ}\mathrm{F} - 80^{\circ}\mathrm{F}) \] \[ Q = 400\,000\ \mathrm{Btu/h} \]
02

Calculate temperature differences at both ends of the heat exchanger

Calculate the temperature differences for hot and cold fluids at both ends of the heat exchanger: 1. Temperature difference at the inlet, \(\Delta T_1 = T_{h,in} - T_{c,in} = 180^{\circ}\mathrm{F} - 80^{\circ}\mathrm{F} = 100^{\circ}\mathrm{F}\). 2. Temperature difference at the outlet, \(\Delta T_2 = T_{h,out} - T_{c,out} = 120^{\circ}\mathrm{F} - 100^{\circ}\mathrm{F} = 20^{\circ}\mathrm{F}\).
03

Calculate the LMTD

Calculate the LMTD using the temperature differences obtained in step 2: \[ \text{LMTD} = \frac{(\Delta T_1 - \Delta T_2)}{\ln{(\Delta T_1/\Delta T_2)}} \] \[ \text{LMTD} = \frac{(100 - 20)}{\ln{(100/20)}} \] \[ \text{LMTD} \approx 48.27^{\circ}\mathrm{F} \]
04

Calculate the heat transfer rate using given parameters

According to LMTD, heat transfer rate can be expressed as follows: \[ Q = U \cdot A \cdot \text{LMTD} \], where \(U\) is the overall heat transfer coefficient, and \(A\) is the surface area of the heat exchanger. We have the values for \(Q\) and LMTD from previous steps, and the value of overall heat transfer coefficient is given as \(40\ \mathrm{Btu}/\mathrm{h}\mathrm{ft}^2{ }^{\circ}\mathrm{F}\).
05

Calculate the surface area of the heat exchanger

Rearranging the equation from step 4, we can solve for surface area \(A\): \[ A = \frac{Q}{U \cdot \text{LMTD}} \] \[ A = \frac{400\,000\ \mathrm{Btu/h}}{(40\ \mathrm{Btu}/\mathrm{h}\mathrm{ft}^2{ }^{\circ}\mathrm{F}) \cdot 48.27^{\circ}\mathrm{F}} \] \[ A \approx 207.6\ \mathrm{ft}^2 \] So, (a) the log mean temperature difference is approximately \(48.27^{\circ}\mathrm{F}\) and (b) the surface area of the heat exchanger is approximately \(207.6\ \mathrm{ft}^2\).

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Key Concepts

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

Overall Heat Transfer Coefficient
The overall heat transfer coefficient, denoted as U, is a measure of a heat exchanger's ability to transfer heat between two fluids that are at different temperatures. It encapsulates how well heat is being transferred through the walls of the heat exchanger and accounts for the conductive and convective heat transfers on both sides (hot and cold) of the heat exchanger.

For example, a high value of U indicates that the heat exchanger is efficient at transferring heat. When calculating the heat transfer rate, the coefficient is crucial as it represents the area-averaged heat transfer rate per unit area per unit temperature difference. The formula to calculate the heat transfer rate using the coefficient is: \[ Q = U \cdot A \cdot \text{LMTD}, \]
where Q is the heat transfer rate, A is the surface area, and LMTD stands for Log Mean Temperature Difference. The units of U are typically Btu/(h ft² °F) in the U.S. customary system or W/(m² K) in the Metric system.
Log Mean Temperature Difference (LMTD)
Log Mean Temperature Difference, or LMTD, is a logarithmic average of the temperature difference between the hot and cold fluids at each end of the heat exchanger. It provides a consistent temperature driving force for heat transfer when the temperature differences at each end vary.

The LMTD is a crucial component in the design and analysis of heat exchangers, particularly when the process involves parallel or counter-flow arrangements. The expression to calculate LMTD is given by:\[ \text{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln{(\Delta T_1/\Delta T_2)}} \]
where \(\Delta T_1\) and \(\Delta T_2\) are the temperature differences between the fluids at the two ends of the heat exchanger. It is important to use the natural logarithm (ln) when calculating this value. In our exercise, the LMTD was found to be approximately 48.27°F, which represents the average temperature difference for the heat transfer process.
Heat Transfer Rate
The heat transfer rate, often denoted as Q, represents the amount of heat that is transferred per unit time in a heat exchange process. In a heat exchanger, it is essential to understand how quickly energy is moving from the hot fluid to the cold fluid. The rate can vary depending on the overall heat transfer coefficient, the surface area for heat exchange, and the temperature difference.

Following the formula mentioned earlier, the heat transfer rate can be quantified as:
\[ Q = U \cdot A \cdot \text{LMTD} \]
In the given example, we calculated the heat transfer rate to be 400,000 Btu/h. This value was then used to determine the necessary surface area of the heat exchanger to ensure efficient operation.
Specific Heat Capacity
Specific heat capacity, which has a symbol cp, is a property that represents the amount of heat required to raise the temperature of one unit of mass of a substance by one degree of temperature (typically in °F or °C). In the context of a heat exchanger, specific heat capacity is significant because it affects how much heat a fluid can transport.

In our exercise, the specific heat capacity of water was given as 1.0 Btu/(lbm·°F), meaning each pound mass of water needs 1.0 Btu of energy to increase its temperature by 1°F. This value, along with the mass flow rate and temperature change, provided us with the heat transfer rate for water, which is crucial for sizing the heat exchanger:
\[ Q = m_c \cdot c_{pc} \cdot (T_{c,out} - T_{c,in}) \]
Understanding specific heat capacity is necessary not only for determining heat transfer rates but also for thermal energy calculations in various engineering applications.

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Most popular questions from this chapter

Steam is to be condensed on the shell side of a 1 -shellpass and 8-tube-passes condenser, with 50 tubes in each pass, at \(30^{\circ} \mathrm{C}\left(h_{f g}=2431 \mathrm{~kJ} / \mathrm{kg}\right)\). Cooling water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the tubes at \(15^{\circ} \mathrm{C}\) at a rate of \(1800 \mathrm{~kg} / \mathrm{h}\). The tubes are thin-walled, and have a diameter of \(1.5 \mathrm{~cm}\) and length of \(2 \mathrm{~m}\) per pass. If the overall heat transfer coefficient is \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine \((a)\) the rate of heat transfer and \((b)\) the rate of condensation of steam.

In a parallel-flow, water-to-water heat exchanger, the hot water enters at \(75^{\circ} \mathrm{C}\) at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\) and cold water enters at \(20^{\circ} \mathrm{C}\) at a rate of \(09 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient and the surface area for this heat exchanger are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(6.4 \mathrm{~m}^{2}\), respectively. The specific heat for both the hot and cold fluid may be taken to be \(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). For the same overall heat transfer coefficient and the surface area, the increase in the effectiveness of this heat exchanger if counter-flow arrangement is used is (a) \(0.09\) (b) \(0.11\) (c) \(0.14\) (d) \(0.17\) (e) \(0.19\)

Write an essay on the static and dynamic types of regenerative heat exchangers and compile information about the manufacturers of such heat exchangers. Choose a few models by different manufacturers and compare their costs and performance.

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 10 times the inner surface area, the overall heat transfer coefficient of this heat exchanger based on the inner surface area is (a) \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(857 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1075 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(2150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

11-100 E(S) Reconsider Prob. 11-99. Using EES (or other) software, investigate the effects of the inlet temperature of hot water and the heat transfer coefficient on the rate of heat transfer and the surface area. Let the inlet temperature vary from \(60^{\circ} \mathrm{C}\) to \(120^{\circ} \mathrm{C}\) and the overall heat transfer coefficient from \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) to \(1250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Plot the rate of heat transfer and surface area as functions of the inlet temperature and the heat transfer coefficient, and discuss the results. 11-101E A thin-walled double-pipe, counter-flow heat exchanger is to be used to cool oil \(\left(c_{p}=0.525 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) from \(300^{\circ} \mathrm{F}\) to \(105^{\circ} \mathrm{F}\) at a rate of \(5 \mathrm{lbm} / \mathrm{s}\) by water \(\left(c_{p}=\right.\) \(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\) ) that enters at \(70^{\circ} \mathrm{F}\) at a rate of \(3 \mathrm{lbm} / \mathrm{s}\). The diameter of the tube is 5 in and its length is \(200 \mathrm{ft}\). Determine the overall heat transfer coefficient of this heat exchanger using (a) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

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