/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 154 An air heater consists of a stee... [FREE SOLUTION] | 91Ó°ÊÓ

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An air heater consists of a steel tube \((k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), with inner and outer radii of \(r_{1}=13 \mathrm{~mm}\) and \(r_{2}=16\) \(\mathrm{mm}\), respectively, and eight integrally machined longitudinal fins, each of thickness \(t=3 \mathrm{~mm}\). The fins extend to a concentric tube, which is of radius \(r_{3}=\) \(40 \mathrm{~mm}\) and insulated on its outer surface. Water at a temperature \(T_{\infty, i}=90^{\circ} \mathrm{C}\) flows through the inner tube, while air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) flows through the annular region formed by the larger concentric tube. (a) Sketch the equivalent thermal circuit of the heater and relate each thermal resistance to appropriate system parameters. (b) If \(h_{i}=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the heat rate per unit length? (c) Assess the effect of increasing the number of fins \(N\) and/or the fin thickness \(t\) on the heat rate, subject to the constraint that \(N t<50 \mathrm{~mm}\).

Short Answer

Expert verified
In summary, to solve this problem, we need to determine the thermal resistances in the system, calculate the heat rate per unit length, and analyze the effect of changing the number of fins and fin thickness. The equivalent thermal circuit consists of the convection resistance at the inner tube surface, conduction resistance through the tube, and convective resistance at the outer tube surface. The heat rate per unit length can be calculated using \(q = \frac{T_{\infty,i} - T_{\infty,o}}{R_{total}}\). Lastly, increasing the number of fins and fin thickness within the constraint \(Nt < 50\text{ mm}\) can enhance the heat transfer rate, but an optimal balance must be found between these factors to achieve the best performance.

Step by step solution

01

Part (a) - Equivalent Thermal Circuit

To sketch the equivalent thermal circuit, we need to determine the thermal resistances in the system. There are three thermal resistances present in this air heater scenario: 1. Heat transfer from the water to the inner steel tube wall (\(R_{conv,i}\)). 2. Conduction through the steel tube (\(R_{cond}\)). 3. Heat transfer from the outer steel tube wall to the air (\(R_{conv,o}\)). Once we have the thermal resistances, we can draw the equivalent thermal circuit diagram.
02

Part (b) - Calculate the Heat Rate per Unit Length

We are given the convective heat transfer coefficients \(h_i\) and \(h_o\). The first step is to find all the thermal resistances. For the convection resistances, we have: \(R_{conv,i} = \frac{1}{h_i \cdot A_i}\) \(R_{conv,o} = \frac{1}{h_o \cdot A_o}\) where \(A_i = 2\pi r_1 L\) and \(A_o = 2\pi r_2 L\) are the surface areas of the inner and outer surfaces of the tube, respectively (and \(L\) is the length). For conduction resistance, we have: \(R_{cond} = \frac{\ln{(r_2 / r_1)}}{2\pi k L}\) Now, the overall resistance can be found using the relation: \(R_{total} = R_{conv,i} + R_{cond} + R_{conv,o}\) The heat rate per unit length can be found using the relation: \(q = \frac{T_{\infty,i} - T_{\infty,o}}{R_{total}}\)
03

Part (c) - Assess the Effect of Increasing the Number of Fins and Fin Thickness

It's important to analyze how increasing the number of fins and fin thickness will affect the heat rate while adhering to the \(Nt < 50\text{ mm}\) constraint. Increasing the number of fins will decrease the resistance due to convection and conduction at the outer surface by providing more surface area. However, it increases the overall constraint \(Nt < 50\text{ mm}\). Therefore, increasing the fin thickness will decrease the allowable number of fins according to the constraint. If the number of fins is increased without an increase in fin thickness, the available surface area for heat transfer increases, and thus the rate of heat transfer also increases. Increasing the fin thickness tends to decrease the conduction resistance locally at each added fin; however, the addition of more fins also restricts the available air flow space, which may increase the convective resistance to the flow. Considering the constraint \(Nt < 50\text{ mm}\), a balance must be struck between the number of fins and fin thickness to achieve optimal performance.

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Key Concepts

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

Thermal Resistance
Thermal resistance is a concept in thermal engineering that quantifies the temperature difference across a material when heat is transferred through it. Similar to electrical resistance, but for heat flow, it is defined as the ratio of the temperature difference across the material to the heat transfer rate (or heat current). In the context of fins, we deal with both conductive and convective thermal resistances.

For conductive resistance through the steel tube, the formula is \(R_{cond} = \frac{\ln{(r_2 / r_1)}}{2\pi k L}\).Conductive resistance represents the opposition to heat flow through a solid material, and it is directly related to the material's thermal conductivity (k), as well as the geometry of the material itself—this includes the lengths and radii in the case of the steel tube.

Convective thermal resistance, on the other hand, accounts for the resistance to heat transfer between a surface and a fluid moving past it. This is computed using the inverse of the product of the convective heat transfer coefficient (\(h\)) and the surface area (\(A\)) through which convection is occurring, as seen in the formula \(R_{conv} = \frac{1}{h \cdot A}\).It is crucial to minimize thermal resistance in systems designed for efficient heat transfer. In the exercise, one aim is to modify the fin parameters to adjust this resistance for optimal heat transfer.
Convective Heat Transfer Coefficient
The convective heat transfer coefficient, represented as \(h\), is a crucial parameter in the calculation of convective thermal resistance and ultimately in determining how effectively a heat transfer device operates. It measures the efficiency of heat transfer from a solid surface to a fluid or from a fluid to a surrounding fluid. High values of \(h\) indicate efficient heat transfer, while low values suggest poor heat transfer.

In our exercise, two different convective heat transfer coefficients are given: one for the inner fluid \(h_i\), which is relatively high, suggesting efficient heat transfer, and another for the outer fluid \(h_o\), which is lower, indicating less effective heat transfer. The effectiveness of fins in improving the heat transfer is partially contingent upon these coefficients because they influence the convective resistance across the surface of the fins—both at the inner and outer sides of the tube. The overall efficiency of the finned surface critically depends on the balance between these coefficients and the physical design of the fins.
Heat Rate Per Unit Length
Heat rate per unit length is a term used to express how much heat energy is transferred per unit length of a thermal system in a given time. It's an important measure for devices such as heat exchangers and radiators, including the finned air heater in our exercise. The formula to calculate the heat rate per unit length \(q\) is given by \(q = \frac{T_{\text{infinity,i}} - T_{\text{infinity,o}}}{R_{\text{total}}}\).This calculation encapsulates the effects of thermal resistances and the temperature difference between the hot and cold fluids. In the context of the exercise, optimizing the heat rate per unit length could involve adjusting the fin design—altering parameters like the number and thickness of fins to maximize the rate of heat transfer within certain physical constraints (e.g., \(Nt < 50\text{ mm}\)). Understanding how each factor, such as thermal resistance and convective heat transfer coefficients, influences this heat rate is key to designing an efficient thermal system.

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Most popular questions from this chapter

A nuclear fuel element of thickness \(2 L\) is covered with a steel cladding of thickness \(b\). Heat generated within the nuclear fuel at a rate \(\dot{q}\) is removed by a fluid at \(T_{\infty}\), which adjoins one surface and is characterized by a convection coefficient \(h\). The other surface is well insulated, and the fuel and steel have thermal conductivities of \(k_{f}\) and \(k_{s}\), respectively. (a) Obtain an equation for the temperature distribution \(T(x)\) in the nuclear fuel. Express your results in terms of \(\dot{q}, k_{f}, L, b, k_{s}, h\), and \(T_{\infty}\). (b) Sketch the temperature distribution \(T(x)\) for the entire system.

In Problem 3.48, the electrical power required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C}\) depends on the thermal conductivity of the wall material \(k\), the thermal contact resistance \(R_{t, c}^{\prime}\) and the convection coefficient \(h\). Compute and plot the separate effect of changes in \(k\) \((1 \leq k \leq 200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), \quad R_{t, c}^{\prime} \quad\left(0 \leq R_{t, c}^{\prime} \leq 0.1 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\right)\), and \(h\left(10 \leq h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) on the total heater power requirement, as well as the rate of heat transfer to the inner surface of the tube and to the fluid.

A brass rod \(100 \mathrm{~mm}\) long and \(5 \mathrm{~mm}\) in diameter extends horizontally from a casting at \(200^{\circ} \mathrm{C}\). The rod is in an air environment with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=30\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the temperature of the rod 25,50 , and \(100 \mathrm{~mm}\) from the casting?

An uninsulated, thin-walled pipe of \(100-\mathrm{mm}\) diameter is used to transport water to equipment that operates outdoors and uses the water as a coolant. During particularly harsh winter conditions, the pipe wall achieves a temperature of \(-15^{\circ} \mathrm{C}\) and a cylindrical layer of ice forms on the inner surface of the wall. If the mean water temperature is \(3^{\circ} \mathrm{C}\) and a convection coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained at the inner surface of the ice, which is at \(0^{\circ} \mathrm{C}\), what is the thickness of the ice layer?

An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of \(25 \mathrm{~mm}\) and a surface temperature of \(250^{\circ} \mathrm{C}\). The fin is \(1 \mathrm{~mm}\) thick and \(10 \mathrm{~mm}\) long, and the temperature and the convection coefficient associated with the adjoining fluid are \(25^{\circ} \mathrm{C}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at \(5-\mathrm{mm}\) increments along the tube length, what is the heat loss per meter of tube length?

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