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Chapter 8: Problem 68

Heat is to be removed from a reaction vessel operating at \(75^{\circ} \mathrm{C}\) by supplying water at \(27^{\circ} \mathrm{C}\) and \(0.12 \mathrm{~kg} / \mathrm{s}\) through a thin-walled tube of \(15-\mathrm{mm}\) diameter. The convection coefficient between the tube outer surface and the fluid in the vessel is \(3000 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). (a) If the outlet water temperature cannot exceed \(47^{\circ} \mathrm{C}\), what is the maximum rate of heat transfer from the vessel? (b) What tube length is required to accomplish the heat transfer rate of part (a)?

Short Answer

Expert verified
The maximum rate of heat transfer from the vessel is \(10,032\ W\). To accomplish this heat transfer rate, a tube length of approximately \(1.903\ m\) is required.

Step by step solution

01

a) Maximum rate of heat transfer from the vessel

To find the maximum heat transfer rate from the vessel, first, we need to find the maximum amount of heat the water can take. We can use the heat capacity formula for this calculation: \(Q = mC_p螖T\), where Q is the heat transfer rate, m is the mass flow rate, \(C_p\) is the specific heat of water, and 螖T is the temperature difference. Given, mass flow rate m = 0.12 kg/s, temperature of inlet water, \(T_{inlet}=27^{\circ}C\), temperature of outlet water, \(T_{outlet}=47^{\circ}C\) , and specific heat of water, \(C_p=4.18 kJ/kg^{\circ}\) 螖T = \(T_{outlet} - T_{inlet}\) 螖T = \(47^{\circ}C - 27^{\circ}C = 20^{\circ}C\) Now, calculate the heat transfer rate as follows: \(Q = mC_p螖T\) \(Q = (0.12\ kg/s)(4.18\ kJ/kg^{\circ}C)(20^{\circ}C)\) \(Q = 10.032\ kJ/s\) Converting to Watts: \(Q = 10.032 \times 1000 = 10,032\ W\) Therefore, the maximum rate of heat transfer from the vessel is \(10,032\ W\).
02

b) Required tube length for maximum heat transfer

Given that the convection coefficient between the tube outer surface and the fluid in the vessel is \(h_{o}=3000\ W/m^{2}K\). We will also need to find the surface area of the tube. First, we need to calculate the heat transfer rate per unit length of the tube, \(q_L\): \(q_L = h_o (T_{vessel} - T_{water})\) where \(T_{vessel}=75^{\circ}C\) and \(T_{water}\) is the average temperature of water, \(T_{water} = \frac{T_{inlet}+T_{outlet}}{2}\): \(T_{water} = \frac{27^{\circ}C + 47^{\circ}C}{2} = 37^{\circ}C\) Now calculate \(q_L\): \(q_L = h_o (T_{vessel} - T_{water})\) \(q_L = 3000\ W/m^{2}K(75^{\circ}C - 37^{\circ}C)\) \(q_L = 3000\ W/m^{2}K(38^{\circ}C)\) \(q_L = 114,000\ W/m^2\) Next, we need to find the tube surface area. The diameter of the tube is given as \(D=15\ mm (0.015\ m)\). The surface area of a cylinder per unit length is given by \(A_s = 蟺D\): \(A_s = 蟺(0.015\ m)\) \(A_s = 0.0471\ m^2/m\) Now, we will find the length of the tube using the calculated values of heat transfer rate (\(Q\)), heat transfer rate per unit length (\(q_L\)), and surface area of the tube per unit length (\(A_s\)). We can use the following formula: \(Q = A_s L q_L\) We need to find \(L\): \(L = \frac{Q}{A_s q_L}\) \(L = \frac{10,032\ W}{(0.0471\ m^2/m)(114,000\ W/m^2)}\) \(L \approx 1.903\ m\) Therefore, a tube length of approximately \(1.903\ m\) is required to accomplish the heat transfer rate determined in part (a).

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Key Concepts

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

Convection Coefficient
The convection coefficient is a measure of how effectively heat is transferred between a solid surface and a fluid moving over it. It is a key factor in calculating the rate of heat transfer in a reaction vessel like the one described in our exercise. This coefficient depends on the type of fluid, its velocity, temperature, and the properties of the surface it's flowing over.

In our exercise, the convection coefficient between the tube outer surface and the fluid in the vessel is given as 3000 W/m2K, which is relatively high. This suggests that the heat transfer between the tube and the reaction fluid is quite efficient. To improve understanding, one could experiment by varying the convection coefficient and observing the impact it has on the heat exchange rate.
Mass Flow Rate
The mass flow rate is the amount of mass passing through a given surface per unit time. In the context of our exercise, it refers to the quantity of water flowing through the tube over time, and is given as 0.12 kg/s. The mass flow rate plays a direct role in determining how much heat energy can be carried away by the fluid.

Understanding that the greater the mass flow rate, the more heat energy can be transported, is essential in grasping the concepts of heat exchange. Delving into the mass flow rate and its effects can further enhance comprehension of fluid dynamics and heat transfer in real-world applications.
Specific Heat Capacity
Specific heat capacity, often symbolized as Cp, is the amount of heat required to raise the temperature of one kilogram of a material by one degree Celsius. Water has a high specific heat capacity, which means it can absorb a lot of heat without a significant change in temperature. In our exercise, the specific heat capacity of water is 4.18 kJ/kg掳C.

Understanding the role of water's specific heat capacity is crucial for students, as it explains why water is often used as a coolant. A practical exercise using different fluids with varying specific heat capacities could provide clearer insight into how this property affects the rate of heat transfer.
Heat Transfer Rate
Heat transfer rate, denoted by Q, is the amount of heat transferred per unit time. In our example, we can calculate it by multiplying the mass flow rate of water, its specific heat capacity, and the change in water temperature. The heat transfer rate is important as it dictates how quickly the reaction vessel can be cooled.

Understanding heat transfer rate is quintessential for students studying thermodynamics. By grasping this concept, students can apply it to other areas, such as designing cooling systems for electronic components or optimizing industrial processes for thermal management.
Tube Surface Area
The tube surface area is a critical parameter in determining the overall heat transfer rate from the vessel to the fluid inside the tube. In thermal systems, heat is transferred across the surfaces of objects, and larger surface areas allow for more heat exchange. In the exercise, we use the diameter of the tube to calculate the surface area per unit length.

Exploring different scenarios where the tube surface area varies can greatly enhance the students' understanding of the importance of surface area in heat exchangers. This concept is widely applicable, from simplistic classroom experiments to the complexities of heat exchanger design in industrial applications.

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Most popular questions from this chapter

A novel scheme for dissipating heat from the chips of a multichip array involves machining coolant channels in the ceramic substrate to which the chips are attached. The square chips \(\left(L_{c}=5 \mathrm{~mm}\right)\) are aligned above each of the channels, with longitudinal and transverse pitches of \(S_{L}=S_{T}=20 \mathrm{~mm}\). Water flows through the square cross section \((W=5 \mathrm{~mm}\) ) of each channel with a mean velocity of \(u_{m}=1 \mathrm{~m} / \mathrm{s}\), and its properties may be approximated as \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=855 \times 10^{-6} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}, k=0.610\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), and \(\operatorname{Pr}=5.8\). Symmetry in the transverse direction dictates the existence of equivalent conditions for each substrate section of length \(L_{s}\) and width \(S_{T}\). (a) Consider a substrate whose length in the flow direction is \(L_{s}=200 \mathrm{~mm}\), thereby providing a total of \(N_{L}=10\) chips attached in-line above each flow channel. To a good approximation, all the heat dissipated by the chips above a channel may be assumed to be transferred to the water flowing through the channel. If each chip dissipates \(5 \mathrm{~W}\), what is the temperature rise of the water passing through the channel? (b) The chip-substrate contact resistance is \(R_{\mathrm{t}, c}^{\mathrm{r}}=\) \(0.5 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the three-dimensional conduction resistance for the \(L_{s} \times S_{T}\) substrate section is \(R_{\text {cond }}=0.120 \mathrm{~K} / \mathrm{W}\). If water enters the substrate at \(25^{\circ} \mathrm{C}\) and is in fully developed flow, estimate the temperature \(T_{c}\) of the chips and the temperature \(T_{s}\) of the substrate channel surface.

A thin-walled tube with a diameter of \(6 \mathrm{~mm}\) and length of \(20 \mathrm{~m}\) is used to carry exhaust gas from a smoke stack to the laboratory in a nearby building for analysis. The gas enters the tube at \(200^{\circ} \mathrm{C}\) and with a mass flow rate of \(0.003 \mathrm{~kg} / \mathrm{s}\). Autumn winds at a temperature of \(15^{\circ} \mathrm{C}\) blow directly across the tube at a velocity of \(5 \mathrm{~m} / \mathrm{s}\). Assume the thermophysical properties of the exhaust gas are those of air. (a) Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube. (b) Estimate the heat transfer coefficient for the air flowing across the outside of the tube. (c) Estimate the overall heat transfer coefficient \(U\) and the temperature of the exhaust gas when it reaches the laboratory.

Many of the solid surfaces for which values of the thermal and momentum accommodation coefficients have been measured are quite different from those used in micro- and nanodevices. Plot the Nusselt number \(N u_{D}\) associated with fully developed laminar flow with constant surface heat flux versus tube diameter for \(1 \mu \mathrm{m} \leq D \leq 1 \mathrm{~mm}\) and (i) \(\alpha_{s}=1, \alpha_{p}=1\), (ii) \(\alpha_{t}=0.1, \alpha_{p}=0.1\), (iii) \(\alpha_{t}=1, \alpha_{p}=0.1\), and (iv) \(\alpha_{t}=0.1, \alpha_{p}=1\). For tubes of what diameter do the accommodation coefficients begin to influence convection heat transfer? For which combination of \(\alpha_{t}\) and \(\alpha_{p}\) does the Nusselt number exhibit the least sensitivity to changes in the diameter of the tube? Which combination results in Nusselt numbers greater than the conventional fully developed laminar value for constant heat flux conditions, \(N u_{D}=4.36\) ? Which combination is associated with the smallest Nusselt numbers? What can you say about the ability to predict convection heat transfer coefficients in a small-scale device if the accommodation coefficients are not known for material from which the device is fabricated? Use properties of air at atmospheric pressure and \(T=300 \mathrm{~K}\).

Water at \(20^{\circ} \mathrm{C}\) and a flow rate of \(0.1 \mathrm{~kg} / \mathrm{s}\) enters a heated, thin-walled tube with a diameter of \(15 \mathrm{~mm}\) and length of \(2 \mathrm{~m}\). The wall heat flux provided by the heating elements depends on the wall temperature according to the relation $$ q_{s}^{\prime \prime}(x)=q_{s, o}^{\prime \prime}\left[1+\alpha\left(T_{s}-T_{\mathrm{ref}}\right)\right] $$ where \(q_{s, \rho}^{\prime \prime}=10^{4} \mathrm{~W} / \mathrm{m}^{2}, \alpha=0.2 \mathrm{~K}^{-1}, T_{\text {ref }}=20^{\circ} \mathrm{C}\), and \(T_{s}\) is the wall temperature in \({ }^{\circ} \mathrm{C}\). Assume fully developed flow and thermal conditions with a convection coefficient of \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Beginning with a properly defined differential control volume in the tube, derive expressions for the variation of the water, \(T_{m}(x)\), and the wall, \(T_{s}(x)\), temperatures as a function of distance from the tube inlet. (b) Using a numerical integration scheme, calculate and plot the temperature distributions, \(T_{m}(x)\) and \(T_{s}(x)\), on the same graph. Identify and comment on the main features of the distributions. Hint: The \(I H T\) integral function \(D E R\left(T_{m}, x\right)\) can be used to perform the integration along the length of the tube. (c) Calculate the total rate of heat transfer to the water.

A bayonet cooler is used to reduce the temperature of a pharmaceutical fluid. The pharmaceutical fluid flows through the cooler, which is fabricated of \(10-\mathrm{mm}-\) diameter, thin-walled tubing with two 250 -mm-long straight sections and a coil with six and a half turns and a coil diameter of \(75 \mathrm{~mm}\). A coolant flows outside the cooler, with a convection coefficient at the outside surface of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) and a coolant temperature of \(20^{\circ} \mathrm{C}\). Consider the situation where the pharmaceutical fluid enters at \(90^{\circ} \mathrm{C}\) with a mass flow rate of \(0.005 \mathrm{~kg} / \mathrm{s}\). The pharmaceutical has the following properties: \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, \quad \mu=4 \times 10^{-3} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\), \(c_{p}=2000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).
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