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Chapter 11: Problem 56

Saturated process steam at 1 atm is condensed in a shell-and-tube heat exchanger (one shell, two tube passes). Cooling water enters the tubes at \(15^{\circ} \mathrm{C}\) with an average velocity of \(3.5 \mathrm{~m} / \mathrm{s}\). The tubes are thin walled and made of copper with a diameter of \(14 \mathrm{~mm}\) and length of \(0.5 \mathrm{~m}\). The convective heat transfer coefficient for condensation on the outer surface of the tubes is \(21,800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the number of tubes/pass required to condense \(2.3 \mathrm{~kg} / \mathrm{s}\) of steam. (b) Find the outlet water temperature. (c) Find the maximum possible condensation rate that could be achieved with this heat exchanger using the same water flow rate and inlet temperature. (d) Using the heat transfer surface area found in part (a), plot the water outlet temperature and steam condensation rate for water mean velocities in the range from 1 to \(5 \mathrm{~m} / \mathrm{s}\). Assume that the shell-side convection coefficient remains unchanged.

Short Answer

Expert verified
(a) To find the number of tubes/pass required to condense 2.3 kg/s of steam, we first find the total heat transfer rate: \(Q = 5191.1 \times 10^3 \, \mathrm{W}\). Next, we calculate the Reynolds number, tube-side heat transfer coefficient, and overall heat transfer coefficient. Using the overall heat transfer coefficient, we determine the heat transfer surface area needed for this process. We can find the number of tubes per pass: \[ n = \frac{A}{\pi D L} \] (b) To find the outlet water temperature, we use the energy balance for the water side: \[ m c_p \Delta T = Q \]. Solving for the outlet temperature, we find it using the given information. (c) To find the maximum possible condensation rate, we calculate the maximum heat transfer rate allowed by the water flow rate and temperature difference: \[Q_{max} = m c_p (\Delta T)_{max} \]. We find the maximum condensation rate by dividing the maximum heat transfer rate by the enthalpy of vaporization: \[ m_{max} = \frac{Q_{max}}{h_{fg}} \] (d) To plot the water outlet temperature and steam condensation rate as a function of water flow velocities, we create a function for each parameter with respect to water velocity and calculate them within the specified range of 1-5 m/s. Then, we create a plot of outlet temperature vs. flow velocity and steam condensation rate vs. flow velocity.

Step by step solution

01

(Step 1: Find the total heat transfer rate)

(We know the mass flow rate of steam, and we need to find the total heat transfer rate. Since it is saturated steam, we can use the enthalpy of vaporization to find the heat transfer rate: \[Q = m \cdot h_{fg}\] Knowing that the enthalpy of vaporization of water at 1 atm is approximately 2257 kJ/kg, and the mass flow rate of steam is 2.3 kg/s, so we can calculate the total heat transfer rate: \[Q = 2.3 \, \mathrm{kg/s} \times 2257 \, \mathrm{kJ/kg}\] \[Q = 5191.1 \, \mathrm{kJ/s} = 5191.1 \times 10^3 \, \mathrm{W}\])
02

(Step 2: Calculate the tube-side heat transfer coefficient)

(We need to find the tube-side heat transfer coefficient. For turbulent flows in tubes, the Dittus-Boelter equation can be used: \[h = 0.023 \frac{k^3}{D} Re^{0.8} Pr^{0.4}\] where \(k\) is the thermal conductivity of water (assume 0.6 W/m路K), \(D\) is the tube diameter, \(Re\) is the Reynolds number based on average velocity and tube diameter, and \(Pr\) is the Prandtl number (assume 6). We also need to calculate the Reynolds number: \[Re = \frac{4 \, m}{\pi D \mu}\] where \(m\) is the mass flow rate of the water and \(\mu\) is the dynamic viscosity of the water (assume 1.0016 脳 10\(^{-3}\) kg/m路s). Given the average velocity and diameter of the tubes, the mass flow rate can be determined as: \[m = \rho V \frac{\pi D^2}{4}\] where \(\rho\) is the density of water (assume 999 kg/m\(^{3}\)) and \(V\) is the average velocity of water. Using the given parameters, we can find the mass flow rate, Reynolds number, and tube-side heat transfer coefficient.)
03

(Step 3: Determine the number of tubes per pass)

(Now we can calculate the overall heat transfer coefficient: \[ U = \frac{1}{\frac{1}{h_t} + \frac{1}{h_o}} \] here, \(h_t\) is the tube-side heat transfer coefficient, and \(h_o\) is the outer-side heat transfer coefficient given as 21800 W/m虏路K. Using this overall heat transfer coefficient, we can find the heat transfer surface area needed for this process: \[A = \frac{Q}{U \Delta T_{lm}} \] where \(\Delta T_{lm}\) is the logarithmic mean temperature difference between the steam and water. To find this, we need to determine the outlet water temperature, inlet steam temperature (100 掳C at 1 atm), and inlet water temperature. After finding the total heat transfer area, we can find the number of tubes per pass by dividing the total area by the area of a single tube: \[ n = \frac{A}{\pi D L} \] where \(L\) is the length of the tube (0.5 m).)
04

(Step 4: Find the outlet water temperature)

(We can use the energy balance for the water side to find the outlet water temperature: \[ m c_p \Delta T = Q \] where \(c_p\) is the specific heat capacity of water (assume 4186 J/kg路K) and \(\Delta T\) is the temperature difference between inlet and outlet. Solving for the outlet temperature, we can find it using the given information.)
05

(Step 5: Find the maximum possible condensation rate)

(The maximum possible condensation rate can be found by calculating the maximum heat transfer rate allowed by the water flow rate and temperature difference: \[Q_{max} = m c_p (\Delta T)_{max} \] where \((\Delta T)_{max}\) is the maximum temperature difference between inlet and outlet water (100 掳C - 15 掳C). Then, we can find the maximum condensation rate by dividing the maximum heat transfer rate by the enthalpy of vaporization: \[ m_{max} = \frac{Q_{max}}{h_{fg}} \] Solving for the maximum condensation rate, we can find it using the given information.)
06

(Step 6: Plot the water outlet temperature and steam condensation rate)

(To plot the water outlet temperature and steam condensation rate as a function of water flow velocities, we can create a function for each parameter with respect to water velocity and calculate them within the specified range of 1-5 m/s. Then, we can create a plot of outlet temperature vs. flow velocity and steam condensation rate vs. flow velocity, assuming the constant shell-side convection coefficient.)

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Key Concepts

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

Convective Heat Transfer Coefficient
The convective heat transfer coefficient is a key parameter in the design and analysis of heat exchangers. It represents the heat transferred per unit area per unit temperature difference between the solid surface and the surrounding fluid. Higher coefficients imply more efficient heat transfer.

In our textbook example, the convective heat transfer coefficient on the shell side where condensation occurs is given as 21,800 W/m虏路K. This value is notably high due to the phase change from vapor to liquid, which usually results in enhanced heat transfer. When optimizing heat exchange systems, it's crucial to increase this coefficient, which can be done by increasing fluid velocity, using fins, or selecting materials with better thermal properties.

Improvement can be achieved by considering the fluid properties and flow regime when selecting and installing heat exchangers. Factors like fluid velocity, viscosity, and thermal conductivity play a substantial role in determining this coefficient.
Logarithmic Mean Temperature Difference (LMTD)
LMTD is a concept crucial in the heat exchanger design process. It represents an average temperature difference between the hot and cold streams across the heat exchanger. A higher LMTD allows for a smaller heat exchanger to achieve the same heat transfer.

In our scenario, the LMTD would be calculated based on the temperatures of the condensing steam and the water entering and exiting the tubes. To optimize a heat exchanger, aim for a higher LMTD by adjusting the flow arrangement or the temperatures of the fluids. However, one must consider the practical operating conditions and the limitations of the materials used in the heat exchanger.

For enhanced understanding, remember that different flow configurations, such as parallel flow, counter flow, or cross flow, will affect the LMTD and, hence, the performance of the heat exchanger.
Reynolds Number
The Reynolds number is a dimensionless quantity used to predict flow patterns in different fluid flow situations. It's defined as the ratio of inertial forces to viscous forces and provides a measure for whether a flow will be laminar or turbulent.

In the given textbook problem, the Reynolds number helps us determine the fluid dynamics inside the tubes, which in turn influences the convective heat transfer coefficient of water. For turbulent flow, heat transfer is generally more efficient due to the mixing effect. Optimizing the Reynolds number is about finding the right balance to ensure turbulent flow without causing excessive pressure drop or potential damage to the heat exchanger.

To increase the Reynolds number, and hence the potential for heat transfer, operators can increase fluid velocity or decrease the fluid's viscosity by heating it. Practical applications might include pre-heating the fluid before it enters the heat exchanger or selecting tubes of an appropriate diameter.
Enthalpy of Vaporization
The enthalpy of vaporization is the energy required to change a substance from a liquid to a gas at a constant temperature and pressure. This property is central to the operation of condensers in heat exchangers where phase change occurs, like in our example.

It's crucial to know the enthalpy of vaporization to calculate the heat transfer for condensing steam. The higher the enthalpy of vaporization, the more heat is removed per kilogram of steam, which translates into a more efficient condensation process. In optimization efforts, understanding and utilizing the enthalpy of vaporization can inform decisions on operating pressures and temperatures to maximize energy transfer.

For educational purposes, linking this informative concept with the practical example of the heat exchanger helps illustrate the importance of thermodynamic properties in engineering applications. When a substance has a high enthalpy of vaporization, like water, heat exchangers must be designed to manage and utilize this energy removal effectively.

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Most popular questions from this chapter

A shell-and-tube exchanger (two shells, four tube passes) is used to heat \(10,000 \mathrm{~kg} / \mathrm{h}\) of pressurized water from 35 to \(120^{\circ} \mathrm{C}\) with \(5000 \mathrm{~kg} / \mathrm{h}\) pressurized water entering the exchanger at \(300^{\circ} \mathrm{C}\). If the overall heat transfer coefficient is \(1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the required heat exchanger area.

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?

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 concentric tube heat exchanger of length \(L=2 \mathrm{~m}\) is used to thermally process a pharmaceutical product flowing at a mean velocity of \(u_{\mathrm{mcc}}=0.1 \mathrm{~m} / \mathrm{s}\) with an inlet temperature of \(T_{c, i}=20^{\circ} \mathrm{C}\). The inner tube of diameter \(D_{i}=10 \mathrm{~mm}\) is thin walled, and the exterior of the outer tube \(\left(D_{o}=20 \mathrm{~mm}\right)\) is well insulated. Water flows in the annular region between the tubes at a mean velocity of \(u_{\mathrm{mhh}}=0.2 \mathrm{~m} / \mathrm{s}\) with an inlet temperature of \(T_{h, i}=60^{\circ} \mathrm{C}\). Properties of the pharmaceutical product are \(\nu=10 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \quad k=0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=2460 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Evaluate water properties at \(\bar{T}_{\mathrm{h}}=50^{\circ} \mathrm{C}\). (a) Determine the value of the overall heat transfer coefficient \(U\). (b) Determine the mean outlet temperature of the pharmaceutical product when the exchanger operates in the counterflow mode. (c) Determine the mean outlet temperature of the pharmaceutical product when the exchanger operates in the parallel-flow mode.

Hot exhaust gases are used in a shell-and-tube exchanger to heat \(2.5 \mathrm{~kg} / \mathrm{s}\) of water from 35 to \(85^{\circ} \mathrm{C}\). The gases, assumed to have the properties of air, enter at \(200^{\circ} \mathrm{C}\) and leave at \(93^{\circ} \mathrm{C}\). The overall heat transfer coefficient is \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the effectiveness-NTU method, calculate the area of the heat exchanger.
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