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Chapter 11: Problem 84

Ethylene glycol and water, at 60 and \(10^{\circ} \mathrm{C}\), respectively, enter a shell-and-tube heat exchanger for which the total heat transfer area is \(15 \mathrm{~m}^{2}\). With ethylene glycol and water flow rates of 2 and \(5 \mathrm{~kg} / \mathrm{s}\), respectively, the overall heat transfer coefficient is \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the rate of heat transfer and the fluid outlet temperatures. (b) Assuming all other conditions to remain the same, plot the effectiveness and fluid outlet temperatures as a function of the flow rate of ethylene glycol for \(0.5 \leq \dot{m}_{h} \leq 5 \mathrm{~kg} / \mathrm{s}\).

Short Answer

Expert verified
In summary, to find the heat transfer rate and fluid outlet temperatures in the shell-and-tube heat exchanger, first calculate the log mean temperature difference, then use energy balance equations to relate the heat transfer rate to the flow rates and specific heat capacities of the fluids. Finally, for part (b), repeat the steps for each value of ethylene glycol flow rate and plot effectiveness and outlet temperatures as a function of ethylene glycol flow rate.

Step by step solution

01

Determine the heat transfer rate

To find the heat transfer rate, we will use the formula: $ Q = UA \Delta T_m $ where \(Q\) is the heat transfer rate, \(U\) is the overall heat transfer coefficient, \(A\) is the heat transfer area, and \(\Delta T_m\) is the log mean temperature difference. First, we must find the log mean temperature difference, given by: $ \Delta T_m = \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 at the inlet and outlet. Temperature differences at the inlet: $ \Delta T_1 = T_{h1} - T_{c1} = 60 - 10 = 50^{\circ}\mathrm{C} $ Temperature differences at the outlet: $ \Delta T_2 = T_{h2} - T_{c2} $ Since we don't know the outlet temperatures yet, we will reserve \(\Delta T_2\) for now. Next, we can plug the values we know into the formula for \(\Delta T_m\), as follows: $ \Delta T_m = \frac{(50 - \Delta T_2)}{\ln(50/\Delta T_2)} $ Now, we can find the heat transfer rate: $ Q = UA \Delta T_m = 800 \cdot 15 \cdot \frac{(50 - \Delta T_2)}{\ln(50/\Delta T_2)} = 12,000 \cdot \frac{(50 - \Delta T_2)}{\ln(50/\Delta T_2)} $
02

Determine the fluid outlet temperatures

Using energy balance equations, we can relate the heat transfer rate to the flow rates and specific heat capacities of the fluids. For the ethylene glycol: $ Q = \dot{m}_hC_{ph}(T_{h1} - T_{h2}) $ For the water: $ Q = \dot{m}_cC_{pc}(T_{c2} - T_{c1}) $ Using the relationship between heat transfer rate, flow rates, and specific heat capacities, we can rewrite the equations as: $ T_{h2} = T_{h1} - \frac{Q}{\dot{m}_hC_{ph}} $ $ T_{c2} = T_{c1} + \frac{Q}{\dot{m}_cC_{pc}} $ Now we can update the value of \(\Delta T_2\): $ \Delta T_2 = \left(T_{h1} - \frac{Q}{\dot{m}_hC_{ph}}\right) - \left(T_{c1} + \frac{Q}{\dot{m}_cC_{pc}}\right) $ Plug this value back into the equation for the heat transfer rate and solve for \(Q\). Once we have the value for \(Q\), we can find the outlet temperatures using the equations above.
03

Plot the effectiveness and outlet temperatures as a function of ethylene glycol flow rate

For part (b), we need to plot the following: 1. Effectiveness, given by: $ \varepsilon = \frac{Q}{Q_{\max}} = \frac{Q}{\dot{m}_hC_{ph}(T_{h1} - T_{c1})} $ 2. Fluid outlet temperatures, \(T_{h2}\) and \(T_{c2}\), as a function of ethylene glycol flow rate. We need to repeat steps 1 and 2 for each value of ethylene glycol flow rate in the interval \(0.5 \leq \dot{m}_{h} \leq 5 \mathrm{~kg} / \mathrm{s}\), and then plot effectiveness and outlet temperatures accordingly.

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Key Concepts

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

Understanding Shell-and-Tube Heat Exchangers
Shell-and-tube heat exchangers are widely used in various industries for heating or cooling fluids. They consist of a bundle of tubes enclosed by a shell. One fluid flows through the tubes, referred to as the tube-side fluid, while another fluid flows outside the tubes but within the shell, known as the shell-side fluid.

These heat exchangers are designed to maximize heat transfer between the two fluids. The key variables in calculating the performance of a shell-and-tube heat exchanger include the flow rates of the fluids, the specific heat capacities of the fluids, the temperature differences between the fluids, and the heat transfer area of the tubes.

Ease of Maintenance and Versatility

One of the reasons shell-and-tube heat exchangers are so prevalent is their ease of maintenance. The tubes can often be cleaned, replaced, or repaired without needing to dismantle the entire unit. Moreover, they are versatile, capable of handling high pressures and a wide range of temperatures.

The heat transfer rate calculation in a shell-and-tube heat exchanger takes the specific characteristics of the exchange鈥攕uch as the thermal conductivities of the tube material鈥攊nto account, along with the aforementioned key variables.
Log Mean Temperature Difference (LMTD) Explained
The Log Mean Temperature Difference (LMTD) is a critical factor in the thermal performance calculation of a heat exchanger. It signifies the driving force behind the heat transfer process. In simple terms, LMTD represents an 'average' temperature difference between the hot and cold streams over the length of the exchanger.

To calculate LMTD, you need to know the inlet and outlet temperatures of both the hot and cold fluids. The formula for LMTD is as follows: \[\Delta T_m = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\Delta T_1/\Delta T_2)}\]where \(\Delta T_1\) and \(\Delta T_2\) are the temperature differences between the hot and cold fluids at the inlet and outlet, respectively.

Importance in Heat Exchanger Design

Understanding and accurately determining LMTD is vital. Engineers use it to size heat exchangers and to estimate how effective a heat exchanger will be under certain operating conditions. It's a standard way of representing temperature difference when the temperatures at either end of the heat exchanger vary.
Overall Heat Transfer Coefficient (U-Factor)
The overall heat transfer coefficient, commonly referred to as U-factor, is a measure of the total thermal resistance between the two fluids in a heat exchanger. It incorporates the conductive, convective, and sometimes radiative heat transfer mechanisms. The U-factor is critical for determining the rate of heat transfer through the unit's surface area.

Mathematically, the heat transfer rate \(Q\) can be expressed using U-factor as:\[Q = UA\Delta T_m\]where \(A\) is the heat transfer area and \(\Delta T_m\) is the log mean temperature difference. The higher the U-factor, the more efficient the heat exchanger, as it implies lower thermal resistance and thus a greater capability to transfer heat.

Factors Affecting U-Factor

The U-factor can be influenced by several parameters, including the types of fluids involved, the velocity of the fluids, the nature of the flow (laminar or turbulent), and the material properties of the tubes. It's also affected by the cleanliness of the heat transfer surfaces鈥攆ouling can significantly reduce the U-factor over time.
Energy Balance Equations in Heat Exchangers
In the context of heat exchangers, energy balance equations are fundamental. They state that the rate of energy loss by the hot fluid is equal to the rate of energy gain by the cold fluid. This principle adheres to the law of conservation of energy and is fundamental in solving for unknown variables in heat exchanger calculations.

The general form of the energy balance equation for a fluid in a heat exchanger is given by:\[Q = \dot{m}C_p(T_{in} - T_{out})\]where \(Q\) is the rate of heat transfer, \(\dot{m}\) is the mass flow rate, \(C_p\) is the specific heat at constant pressure, and \((T_{in} - T_{out})\) are the inlet and outlet temperatures of the fluid.

Practical Application

In practice, if you know the mass flow rates and specific heat capacities of the fluids, as well as either the inlet or outlet temperatures, you can calculate the heat transfer rate and the other unknown temperatures using these energy balance equations. They are pivotal for design, analysis, and troubleshooting of heat exchangers in any thermal system.

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Most popular questions from this chapter

Hot water for an industrial washing operation is produced by recovering heat from the flue gases of a furnace. A cross-flow heat exchanger is used, with the gases passing over the tubes and the water making a single pass through the tubes. The steel tubes \((k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) have inner and outer diameters of \(D_{i}=15 \mathrm{~mm}\) and \(D_{o}=20 \mathrm{~mm}\), while the staggered tube array has longitudinal and transverse pitches of \(S_{T}=S_{L}=40 \mathrm{~mm}\). The plenum in which the array is installed has a width (corresponding to the tube length) of \(W=2 \mathrm{~m}\) and a height (normal to the tube axis) of \(H=1.2 \mathrm{~m}\). The number of tubes in the transverse plane is therefore \(N_{T} \approx H / S_{T}=30\). The gas properties may be approximated as those of atmospheric air, and the convection coefficient associated with water flow in the tubes may be approximated as \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If \(50 \mathrm{~kg} / \mathrm{s}\) of water are to be heated from 290 to \(350 \mathrm{~K}\) by \(40 \mathrm{~kg} / \mathrm{s}\) of flue gases entering the exchanger at \(700 \mathrm{~K}\), what is the gas outlet temperature and how many tube rows \(N_{L}\) are required? (b) The water outlet temperature may be controlled by varying the gas flow rate and/or inlet temperature. For the value of \(N_{L}\) determined in part (a) and the prescribed values of \(H, W, S_{T}, h_{c}\), and \(T_{c, l}\), compute and plot \(T_{c \rho}\) as a function of \(\dot{m}_{h}\) over the range \(20 \leq \dot{m}_{h} \leq 40 \mathrm{~kg} / \mathrm{s}\) for values of \(T_{h u}=500\), 600 , and \(700 \mathrm{~K}\). Also plot the corresponding variations of \(T_{h \rho}\). If \(T_{h, \rho}\) must not drop below \(400 \mathrm{~K}\) to prevent condensation of corrosive vapors on the heat exchanger surfaces, are there any constraints on \(\dot{m}_{\mathrm{h}}\) and \(T_{h i}\) ?

A process fluid having a specific heat of \(3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and flowing at \(2 \mathrm{~kg} / \mathrm{s}\) is to be cooled from \(80^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) with chilled water, which is supplied at a temperature of \(15^{\circ} \mathrm{C}\) and a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\). Assuming an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), calculate the required heat transfer areas for the following exchanger configurations: (a) parallel flow, (b) counterflow, (c) shell-and-tube, one shell pass and two tube passes, and (d) cross-flow, single pass, both fluids unmixed. Compare the results of your analysis. Your work can be reduced by using IHT.

A cross-flow heat exchanger used in a cardiopulmonary bypass procedure cools blood flowing at \(5 \mathrm{~L} / \mathrm{min}\) from a body temperature of \(37^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\) in order to induce body hypothermia, which reduces metabolic and oxygen requirements. The coolant is ice water at \(0^{\circ} \mathrm{C}\), and its flow rate is adjusted to provide an outlet temperature of \(15^{\circ} \mathrm{C}\). The heat exchanger operates with both fluids unmixed, and the overall heat transfer coefficient is \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The density and specific heat of the blood are \(1050 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3740 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. a) Determine the heat transfer rate for the exchanger. b) Calculate the water flow rate. c) What is the surface area of the heat exchanger? d) Calculate and plot the blood and water outlet temperatures as a function of the water flow rate for the range 2 to \(4 \mathrm{~L} / \mathrm{min}\), assuming all other parameters remain unchanged. Comment on how the changes in the outlet temperatures are affected by changes in the water flow rate. Explain this behavior and why it is an advantage for this application.

A liquefied natural gas (LNG) regasification facility utilizes a vertical heat exchanger or vaporizer that consists of a shell with a single-pass tube bundle used to convert the fuel to its vapor form for subsequent delivery through a land-based pipeline. Pressurized LNG is off-loaded from an oceangoing tanker to the bottom of the vaporizer at \(T_{c, i}=-155^{\circ} \mathrm{C}\) and \(\dot{m}_{\mathrm{LNG}}=150 \mathrm{~kg} / \mathrm{s}\) and flows through the shell. The pressurized LNG has a vaporization temperature of \(T_{f}=-75^{\circ} \mathrm{C}\) and specific heat \(c_{p l}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The specific heat of the vaporized natural gas is \(c_{p, v}=2210 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) while the gas has a latent heat of vaporization of \(h_{f g}=575 \mathrm{~kJ} / \mathrm{kg}\). The LNG is heated with seawater flowing through the tubes, also introduced at the bottom of the vaporizer, that is available at \(T_{h, i}=20^{\circ} \mathrm{C}\) with a specific heat of \(c_{\mu \mathrm{Sw}}=3985 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). If the gas is to leave the vaporizer at \(T_{c o}=8^{\circ} \mathrm{C}\) and the seawater is to exit the device at \(T_{\text {hot }}=10^{\circ} \mathrm{C}\), determine the required vaporizer heat transfer area. Hint: Divide the vaporizer into three sections, as shown in the schematic, with \(U_{\mathrm{A}}=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), \(U_{\mathrm{B}}=260 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(U_{\mathrm{C}}=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

In open heart surgery under hypothermic conditions, the patient's blood is cooled before the surgery and rewarmed afterward. It is proposed that a concentric tube, counterflow heat exchanger of length \(0.5 \mathrm{~m}\) be used for this purpose, with the thin-walled inner tube having a diameter of \(55 \mathrm{~mm}\). The specific heat of the blood is \(3500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) If water at \(T_{h j}=60^{\circ} \mathrm{C}\) and \(\dot{m}_{h}=0.10 \mathrm{~kg} / \mathrm{s}\) is used to heat blood entering the exchanger at \(T_{c A}=18^{\circ} \mathrm{C}\) and \(\dot{m}_{c}=0.05 \mathrm{~kg} / \mathrm{s}\), what is the temperature of the blood leaving the exchanger? The overall heat transfer coefficient is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) The surgeon may wish to control the heat rate \(q\) and the outlet temperature \(T_{c, 0}\) of the blood by altering the flow rate and/or inlet temperature of the water during the rewarming process. To assist in the development of an appropriate controller for the prescribed values of \(\hat{m}_{c}\) and \(T_{c \jmath}\), compute and plot \(q\) and \(T_{c, \rho}\) as a function of \(\dot{m}_{h}\) for \(0.05 \leq \dot{m}_{\mathrm{h}} \leq 0.20 \mathrm{~kg} / \mathrm{s}\) and values of \(T_{h, l}=50,60\), and \(70^{\circ} \mathrm{C}\). Since the dominant influence on the overall heat transfer coefficient is associated with the blood flow conditions, the value of \(U\) may be assumed to remain at \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Should certain operating conditions be excluded?
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