/*! 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 50 A stainless steel (AISI 304) tub... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 50

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of \(36 \mathrm{~mm}\) and a wall thickness of \(2 \mathrm{~mm}\). The pharmaceutical and ambient air are at temperatures of \(6^{\circ} \mathrm{C}\) and \(23^{\circ} \mathrm{C}\), respectively, while the corresponding inner and outer convection coefficients are \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat gain per unit tube length? (b) What is the heat gain per unit length if a \(10-\mathrm{mm}\) thick layer of calcium silicate insulation \(\left(k_{\text {ins }}=\right.\) \(0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tube?

Short Answer

Expert verified
The heat gain per unit tube length without insulation is 80647 W/m, while the heat gain per unit tube length with insulation is 8854 W/m.

Step by step solution

01

Calculate the tube dimensions and areas

Firstly, let's calculate the tube's inner area, outer area, and the area of insulation. Inner radius (r1) = inner diameter / 2 = \( 36mm / 2 = 18mm = 0.018 m \) Outer radius (r2) = r1 + wall thickness = \( 0.018m + 0.002m = 0.02 m \) Insulation radius (r3) = r2 + insulation thickness = \( 0.02m + 0.01m = 0.03 m \) Inner area (A1) = \( 2\pi r1 = 2\pi(0.018m) = 0.113 m^2 \) Outer area (A2) = \( 2\pi r2 = 2\pi(0.02m) = 0.126 m^2 \) Insulation area (A3) = \( 2\pi r3 = 2\pi(0.03m) = 0.189 m^2 \) Now we have all the necessary dimensions and areas.
02

Calculate the thermal resistances without insulation

First, we need to compute the thermal resistance for conduction through the tube wall, and convection at the inner and outer surfaces. \( R_{cond} = \frac{\Delta x}{kA} \) \( R_{conv} = \frac{1}{hA} \) Let's calculate the thermal resistance for conduction through the stainless steel wall (R_wall) and convection at the inner (R_in) and outer surfaces (R_out) of the tube: Materials properties: Stainless steel thermal conductivity (k) = 16.3 \( \frac{W}{m\cdot K}\) Inner convection coefficient (h1)= 400 \( \frac{W}{m^2\cdot K} \) Outer convection coefficient (h2) = 6 \( \frac{W}{m^2\cdot K} \) \( R_{wall} = \frac{r2 - r1}{2\pi k(r2 + r1)} = \frac{0.02 - 0.018}{2\pi(16.3)(0.02 + 0.018)} = 4.71 \times 10^{-5} \frac{掳C}{W} \) \( R_{in} = \frac{1}{h1A1} = \frac{1}{400\times0.113} = 2.22 \times 10^{-5} \frac{掳C}{W} \) \( R_{out} = \frac{1}{h2A2} = \frac{1}{6\times0.126} = 1.32 \times 10^{-4} \frac{掳C}{W} \) Now we have the thermal resistances without insulation.
03

Calculate the heat gain without insulation

To calculate the heat gain (Q) per unit length without insulation, we use the formula: \( Q = \frac{螖T}{R_{total}} \) Where 螖T is the temperature difference between the pharmaceutical and the ambient air, and R_total is the sum of all thermal resistances in the system. \( R_{total} = R_{wall} + R_{in} + R_{out} = 4.71 \times 10^{-5} + 2.22 \times 10^{-5} + 1.32 \times 10^{-4} = 2.11 \times 10^{-4} \frac{掳C}{W} \) \( 螖T = 23 - 6 = 17掳C \) \( Q = \frac{17}{2.11 \times 10^{-4}} = 80647 W/m \) Thus, the heat gain per unit tube length without insulation is 80647 W/m.
04

Calculate the thermal resistance of insulation

Now, we will compute the additional thermal resistance of the insulation. Insulation thermal conductivity (k_ins) = 0.050 \( \frac{W}{m \cdot K} \) \( R_{insulation} = \frac{r3 - r2}{2\pi k_{ins}(r3 + r2)} = \frac{0.03 - 0.02}{2\pi(0.050)(0.03 + 0.02)} = 1.71 \times 10^{-3} \frac{掳C}{W} \) Now we have the insulation's thermal resistance.
05

Calculate the heat gain with insulation

To calculate the heat gain (Q_insulated) per unit length with insulation, we use the same formula as before with the total thermal resistance with insulation. \( R_{total\_insulated} = R_{wall} + R_{in} + R_{out} + R_{insulation} = 2.11 \times 10^{-4} + 1.71 \times 10^{-3} = 1.92 \times 10^{-3} \frac{掳C}{W} \) \( Q_{insulated} = \frac{17}{1.92 \times 10^{-3}} = 8854 W/m \) Thus, the heat gain per unit tube length with insulation is 8854 W/m. To summarize the results: - The heat gain per unit tube length without insulation is 80647 W/m. - The heat gain per unit tube length with insulation is 8854 W/m.

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Key Concepts

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

Thermal Resistance
In the context of heat transfer, thermal resistance is a measure of a material's ability to resist the flow of heat. Just like electrical resistance impedes the flow of electricity, thermal resistance hinders the passage of heat through materials. It depends on the material's thickness, thermal conductivity, and surface area. The concept is critical in calculating heat gain or loss in materials, as we've seen in the stainless steel tube exercise.

The equation used to determine thermal resistance (\(R_{th}\)) for conduction through a solid material is expressed as \(R_{th} = \frac{\Delta x}{kA}\), where \(\Delta x\) is the thickness of the material, \(k\) is the thermal conductivity, and \(A\) is the cross-sectional area perpendicular to the heat path. Meanwhile, for convection, thermal resistance is calculated using the formula \(R_{th} = \frac{1}{hA}\), where \(h\) is the convective heat transfer coefficient. In complex systems with multiple layers or interfaces, the total thermal resistance is the sum of the individual resistances.
Conduction
The process by which heat is directly transmitted through a substance when there is a difference of temperature between adjoining regions, without movement of the material, is known as conduction. In our exercise, conduction plays a key role in heat transfer through the wall of the stainless steel tube.

In solids, conduction is due to the vibrations of atoms and the movement of electrons. The thermal conductivity (\(k\)) is a property that indicates how well a material can conduct heat. Materials with high \(k\) values are good conductors of heat and have lower thermal resistances, whereas materials with low \(k\) values are poor conductors and therefore good insulators.

In the given problem, we calculate the conduction resistance through the tube wall considering the thermal conductivity of stainless steel. If the tube had been made of another material with a different \(k\) value, the thermal resistance, and thus the heat gain per unit tube length, would have been different. This illustrates the importance of material properties and design considerations in thermal systems.
Convection
On the other hand, convection is the heat transfer due to the bulk movement of molecules within fluids (gas or liquid), including the circulation currents caused by temperature variations. In the example of the pharmaceutical tube, convection applies to the interface between the tube surface and the air or the chilled pharmaceutical inside the tube.

The convection coefficient (\(h\)) quantifies the convective heat transfer between a surface and a fluid in contact with it. It's influenced by factors such as the nature of the fluid, its velocity, and its temperature. In our exercise, different coefficients for the inside (\(h_1\)) and outside (\(h_2\)) reflect the differing heat transfer rates, with the pharmaceutical inside the tube likely more efficient at transferring heat due to higher velocity or better mixing compared to the relatively static outdoor air.

The exercise demonstrates how we would account for both the resistance to heat flow of the steel tube's material by conduction and the resistance at the interface between tube surfaces and the surrounding media by convection. These principles are crucial for understanding and managing thermal management systems in various applications.

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Most popular questions from this chapter

Measurements show that steady-state conduction through a plane wall without heat generation produced a convex temperature distribution such that the midpoint temperature was \(\Delta T_{o}\) higher than expected for a linear temperature distribution. Assuming that the thermal conductivity has a linear dependence on temperature, \(k=k_{o}(1+\alpha T)\), where \(\alpha\) is a constant, develop a relationship to evaluate \(\alpha\) in terms of \(\Delta T_{o}, T_{1}\), and \(T_{2}\).

Circular copper rods of diameter \(D=1 \mathrm{~mm}\) and length \(L=25 \mathrm{~mm}\) are used to enhance heat transfer from a surface that is maintained at \(T_{s, 1}=100^{\circ} \mathrm{C}\). One end of the rod is attached to this surface (at \(x=0\) ), while the other end \((x=25 \mathrm{~mm})\) is joined to a second surface, which is maintained at \(T_{s, 2}=0^{\circ} \mathrm{C}\). Air flowing between the surfaces (and over the rods) is also at a temperature of \(T_{\infty}=0^{\circ} \mathrm{C}\), and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. (a) What is the rate of heat transfer by convection from a single copper rod to the air? (b) What is the total rate of heat transfer from a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) section of the surface at \(100^{\circ} \mathrm{C}\), if a bundle of the rods is installed on 4 -mm centers?

A wire of diameter \(D=2 \mathrm{~mm}\) and uniform temperature \(T\) has an electrical resistance of \(0.01 \Omega / \mathrm{m}\) and a current flow of \(20 \mathrm{~A}\). (a) What is the rate at which heat is dissipated per unit length of wire? What is the heat dissipation per unit volume within the wire? (b) If the wire is not insulated and is in ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=20^{\circ} \mathrm{C}\), what is the temperature \(T\) of the wire? The wire has an emissivity of \(0.3\), and the coefficient associated with heat transfer by natural convection may be approximated by an expression of the form, \(h=C\left[\left(T-T_{\infty}\right) / D\right]^{1 / 4}, \quad\) where \(C=1.25\) \(\mathrm{W} / \mathrm{m}^{7 / 4} \cdot \mathrm{K}^{5 / 4}\). (c) If the wire is coated with plastic insulation of 2-mm thickness and a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what are the inner and outer surface temperatures of the insulation? The insulation has an emissivity of \(0.9\), and the convection coefficient is given by the expression of part (b). Explore the effect of the insulation thickness on the surface temperatures.

The cross section of a long cylindrical fuel element in a nuclear reactor is shown. Energy generation occurs uniformly in the thorium fuel rod, which is of diameter \(D=25 \mathrm{~mm}\) and is wrapped in a thin aluminum cladding. (a) It is proposed that, under steady-state conditions, the system operates with a generation rate of \(\dot{q}=\) \(7 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\) and cooling system characteristics of \(T_{\infty}=95^{\circ} \mathrm{C}\) and \(h=7000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Is this proposal satisfactory? (b) Explore the effect of variations in \(\dot{q}\) and \(h\) by plotting temperature distributions \(T(r)\) for a range of parameter values. Suggest an envelope of acceptable operating conditions.

An air heater may be fabricated by coiling Nichrome wire and passing air in cross flow over the wire. Consider a heater fabricated from wire of diameter \(D=\) \(1 \mathrm{~mm}\), electrical resistivity \(\rho_{e}=10^{-6} \Omega \cdot \mathrm{m}\), thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and emissivity \(\varepsilon=0.20\). The heater is designed to deliver air at a temperature of \(T_{\infty}=50^{\circ} \mathrm{C}\) under flow conditions that provide a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for the wire. The temperature of the housing that encloses the wire and through which the air flows is \(T_{\text {sur }}=50^{\circ} \mathrm{C}\). If the maximum allowable temperature of the wire is \(T_{\max }=1200^{\circ} \mathrm{C}\), what is the maximum allowable electric current \(I\) ? If the maximum available voltage is \(\Delta E=110 \mathrm{~V}\), what is the corresponding length \(L\) of wire that may be used in the heater and the power rating of the heater? Hint: In your solution, assume negligible temperature variations within the wire, but after obtaining the desired results, assess the validity of this assumption.
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