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11.3 A shell-and-tube heat exchanger is to heat an acidic liquid that flows in unfinned tubes of inside and outside diameters \(D_{i}=10 \mathrm{~mm}\) and \(D_{\mathrm{o}}=11 \mathrm{~mm}\), respectively. A hot gas flows on the shell side. To avoid corrosion of the tube material, the engineer may specify either a Ni-Cr-Mo corrosion-resistant metal alloy \(\left(\rho_{m}=8900 \mathrm{~kg} / \mathrm{m}^{3}, k_{\mathrm{w}}=8\right.\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) or a polyvinylidene fluoride (PVDF) plastic \(\left(\rho_{p}=1780 \mathrm{~kg} / \mathrm{m}^{3}, k_{p}=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). The inner and outer heat transfer coefficients are \(h_{j}=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{v}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) Determine the ratio of plastic to metal tube surface areas needed to transfer the same amount of heat. (b) Determine the ratio of plastic to metal mass associated with the two heat exchanger designs. (c) The cost of the metal alloy per unit mass is three times that of the plastic. Determine which tube material should be specified on the basis of cost. 11.4 A steel tube \((k=50 \mathrm{~W} / \mathrm{m}-\mathrm{K})\) of inner and outer diameters \(D_{i}=20 \mathrm{~mm}\) and \(D_{o}=26 \mathrm{~mm}\), respectively, is used to transfer heat from hot gases flowing over the tube \(\left(h_{\mathrm{h}}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) to cold water flowing through the tube \(\left(h_{c}=8000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\). What is the cold-side overall heat transfer coefficient \(U_{c}\) ? To enhance heat transfer, 16 straight fins of rectangular profile are installed longitudinally along the outer surface of the tube. The fins are equally spaced around the circumference of the tube, each having a thickness of \(2 \mathrm{~mm}\) and a length of \(15 \mathrm{~mm}\). What is the corresponding overall heat transfer coefficient \(U_{c}\) ?

Step by step solution

01

Determine the ratio of plastic to metal tube surface areas

We want to find the ratio of plastic to metal tube surface areas needed to transfer the same amount of heat. To do so, we will first calculate the thermal resistance for both tubes materials. The overall thermal resistance of heat transfer through the tube wall can be calculated by the formula: \[R=\dfrac{r_{\mathrm{o}} - r_{\mathrm{i}}}{2\pi L k} + \dfrac{1}{(h_{\mathrm{j}} 2\pi r_{\mathrm{i}} L)} + \dfrac{1}{(h_{\mathrm{v}} 2\pi r_{\mathrm{o}} L)}\] Where \(r_{\mathrm{i}}\) is the tube inner radius, \(r_{\mathrm{o}}\) is the tube outer radius, \(L\) is the length of the tube, \(k\) is the thermal conductivity of the tube material, and \(h_{\mathrm{j}}\) and \(h_{\mathrm{v}}\) are the inner and outer heat transfer coefficients, respectively. We will first calculate the thermal resistance for metal tubes, \(R_{\mathrm{m}}\), and then for plastic tubes, \(R_{\mathrm{p}}\). Then, we will calculate the ratio of plastic to metal tube surface areas as: \[\dfrac{A_{\mathrm{p}}}{A_{\mathrm{m}}} = \dfrac{Q R_{\mathrm{m}}}{Q R_{\mathrm{p}}}\] Where \(Q\) is the heat transfer rate, which is equal for both tube materials.

02

Determine the thermal resistance for metal tubes

To find the thermal resistance for metal tubes, we will use the given values for inside and outside diameters, thermal conductivity, and the heat transfer coefficients. The metal tube has an inner diameter of \(D_{\mathrm{i}} = 10\, \mathrm{mm}\), and an outer diameter of \(D_{\mathrm{o}} = 11\, \mathrm{mm}\). Therefore, its inner and outer radii are \(r_{\mathrm{i}} = 5\, \mathrm{mm}\) and \(r_{\mathrm{o}} = 5.5\, \mathrm{mm}\), respectively. The thermal conductivity for the metal is \(k_{\mathrm{w}} = 8\, \mathrm{W/m \cdot K}\). Using the formula for thermal resistance, we find the thermal resistance for metal tubes as: \[R_{\mathrm{m}} = \dfrac{5.5 - 5}{2\pi L 8} + \dfrac{1}{(1500 \cdot 2\pi 5 L)} + \dfrac{1}{(200 \cdot 2\pi 5.5 L)}\]

03

Determine the thermal resistance for plastic tubes

To find the thermal resistance for plastic tubes, we will use the given values for inside and outside diameters, thermal conductivity, and the heat transfer coefficients. The plastic tube has the same dimensions as the metal tube, with an inner diameter of \(D_{\mathrm{i}} = 10\, \mathrm{mm}\), and an outer diameter of \(D_{\mathrm{o}} = 11\, \mathrm{mm}\). Therefore, its inner and outer radii are \(r_{\mathrm{i}} = 5\, \mathrm{mm}\) and \(r_{\mathrm{o}} = 5.5\, \mathrm{mm}\), respectively. The thermal conductivity for the plastic is \(k_{\mathrm{p}} = 0.17\, \mathrm{W/m \cdot K}\). Using the formula for thermal resistance, we find the thermal resistance for plastic tubes as: \[R_{\mathrm{p}} = \dfrac{5.5 - 5}{2\pi L \cdot 0.17} + \dfrac{1}{(1500 \cdot 2\pi 5 L)} + \dfrac{1}{(200 \cdot 2\pi 5.5 L)}\]

04

Calculate the ratio of plastic to metal tube surface areas

Now that we have calculated the thermal resistances for both metal and plastic tubes, we can find the ratio of plastic to metal tube surface areas by dividing the product of heat transfer rate and metal tube thermal resistance by the product of heat transfer rate and plastic tube thermal resistance. Since the heat transfer rate is the same for both tube materials, it will cancel out in the calculation: \[\dfrac{A_{\mathrm{p}}}{A_{\mathrm{m}}} = \dfrac{R_{\mathrm{m}}}{R_{\mathrm{p}}}\] Plugging in the values from Step 2 and Step 3, we get the ratio of plastic to metal tube surface areas.

05

Determine the ratio of plastic to metal mass associated with the heat exchangers

We will find the ratio of plastic to metal mass associated with the heat exchangers by dividing the mass of plastic by the mass of metal: \[\dfrac{M_{\mathrm{p}}}{M_{\mathrm{m}}} = \dfrac{\rho_{\mathrm{p}} \cdot V_{\mathrm{p}}}{\rho_{\mathrm{m}} \cdot V_{\mathrm{m}}}\] We will first find the volume of the plastic and metal tubes and then substitute the densities given in the problem to find the ratio.

06

Determine which tube material should be specified based on cost

Since the cost of the metal alloy per unit mass is three times that of the plastic, we have to find the cost of both materials and then compare them: \[cost_{\mathrm{m}} = 3\cdot cost_{\mathrm{p}}\] From the previous calculation of ratios, we can determine which one would be cost effective by replacing the cost with their respective masses. Once the analysis is done, the tube material that should be specified on the basis of cost will be chosen.

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Key Concepts

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

Shell-and-Tube Heat Exchanger

One of the cornerstones of industrial heat exchange technology is the shell-and-tube heat exchanger. This device typically consists of a series of tubes, through which one fluid flows, enclosed by a shell that carries another fluid. The main advantage of a shell-and-tube design is its versatile service in diverse temperatures and pressures. These heat exchangers are favorable for applications requiring high efficiency, and they often manage heat transfer between liquids and gases or between two liquids.

When designing a shell-and-tube heat exchanger, engineers must take into account numerous factors that influence its efficiency, such as the type of material for the tubes, the flow arrangement, the heat transfer area, the differences in temperature between the fluids, and the physical properties of the fluids themselves. The tube material, for instance, directly impacts the heat exchanger's resistance to corrosion and thermal conductivity; these attributes will determine how well the exchanger can transfer heat without degrading over time.

In the exercise provided, the choice between a Ni-Cr-Mo alloy and PVDF plastic for the tube material hinges on their ability to conduct heat while enduring the corrosive nature of the acidic liquid. This decision emphasizes the importance of material selection in shell-and-tube heat exchanger design. Engineers will often opt for materials that blend durability and effective heat transfer capabilities, ensuring the longevity and efficiency of the heat exchanger.

Thermal Resistance Calculation

Thermal resistance is a measure determining the difficulty heat faces while passing through a material or an interface of materials. The calculation of thermal resistance is vital to the design of heat exchangers as it helps to predict and maximize the efficiency of heat transfer. The formula provided in the solution for calculating thermal resistance takes into account the conduction through the tube material and convective heat transfer coefficients, both inside and outside the tube.

The resistance to heat flow can be thought of similarly to electrical resistance; they both impede the flow of energy, be it heat or electrical current. Lower thermal resistance indicates better heat flow, which is desirable in most heat exchange situations. When an engineer calculates the overall thermal resistance, they can determine the necessary size and material for the heat exchanger to deliver the required heat transfer performance.

The exercise perfectly illustrates the process of evaluating the thermal resistance of different materials—metal and plastic in this case. By comparing these resistances, we can assess which material requires less surface area to achieve the same heat transfer rate, optimizing the design of the heat exchanger for efficiency and cost.

Material Thermal Conductivity

Thermal conductivity is a fundamental property of materials that quantifies the ability to conduct or transfer heat. In the context of heat exchangers, a high thermal conductivity is generally advantageous as it facilitates the transfer of thermal energy between media separated by the material. Materials like metals typically have high thermal conductivities because their electron structure allows them to transfer kinetic energy efficiently. Conversely, plastics and other polymers usually possess lower thermal conductivities because of their molecular composition, which restrains the kinetic energy flow.

In the provided exercise, we see two contrasting materials: a Ni-Cr-Mo alloy with a relatively high thermal conductivity and PVDF plastic with a significantly lower value. The impact of this property is profound when it comes to heat exchanger design—high thermal conductivity materials can transfer the same amount of heat over a smaller area than those with low thermal conductivity. As such, the trade-off usually lies between the cost and efficiency of materials; engineers must weigh the upfront cost against the long-term heat transfer performance and durability.

Understanding thermal conductivity equips students and engineers with the insight to predict how temperature gradients will affect heat transfer in different materials. This knowledge is crucial to creating efficient heat exchanger designs, allowing for informed decisions regarding material selection based on their specific thermal properties.