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Consider a tube for transporting steam that is not centered properly in a cylindrical insulation material \((k=\) \(0.73 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The tube diameter is \(D_{1}=20 \mathrm{~cm}\) and the insulation diameter is \(D_{2}=40 \mathrm{~cm}\). The distance between the center of the tube and the center of the insulation is \(z=5 \mathrm{~mm}\). If the surface of the tube maintains a temperature of \(100^{\circ} \mathrm{C}\) and the outer surface temperature of the insulation is constant at \(30^{\circ} \mathrm{C}\), determine the rate of heat transfer per unit length of the tube through the insulation.

Short Answer

Expert verified
Answer: The rate of heat transfer per unit length of the tube through the insulation is approximately \(466.67 \mathrm{~W} / \mathrm{m}\).

Step by step solution

01

Calculate the temperature drop through the insulation

The temperature difference between the inner surface of the tube and the outer surface of the insulation is the driving force for heat transfer. We can calculate it as follows: \(\Delta T = T_{1} - T_{2} = 100^{\circ}\mathrm{C} - 30^{\circ}\mathrm{C} = 70^{\circ}\mathrm{C}\)
02

Calculate the equivalent thermal resistance of the insulation

The insulation is not centered on the pipe; thus, the heat flow in the insulation is not radial. However, since its diameter (\(D_{2}\)) is large compared to the pipe diameter (\(D_{1}\)), it is reasonable to approximate the flow as radial. We can calculate the equivalent thermal resistance (\(R_{\text{th}}\)) of the insulation using the equation: \(R_{\text{th}} = \frac{\ln(\frac{D_{2}}{D_{1}})}{2\pi k}\) Using the given values, we get: \(R_{\text{th}} = \frac{\ln(\frac{0.4 \ \mathrm{m}}{0.2 \ \mathrm{m}})}{2\pi(0.73 \mathrm{~W} \ / \mathrm{m} \cdot \mathrm{K})} = \frac{\ln(2)}{4.581} = 0.150 \ \mathrm{m} \cdot \mathrm{K} / \mathrm{W}\)
03

Calculate the rate of heat transfer per unit length

Now that we have the equivalent thermal resistance and the temperature drop, we can calculate the rate of heat transfer per unit length (\(q'\)) using Fourier's Law: \(q' = \frac{\Delta T}{R_{\text{th}}}\) Substituting the values, we get: \(q' = \frac{70^{\circ}\mathrm{C}}{0.150 \ \mathrm{m} \cdot \mathrm{K} / \mathrm{W}} = 466.67 \mathrm{~W} / \mathrm{m}\) Thus, the rate of heat transfer per unit length of the tube through the insulation is approximately \(466.67 \mathrm{~W} / \mathrm{m}\).

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Key Concepts

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

Fourier's Law
Fourier's Law is the fundamental principle that governs heat conduction. It states that the rate at which heat energy is transferred through a material is directly proportional to the negative of the temperature gradient and the cross-sectional area through which heat is flowing, and is inversely proportional to the material's thickness. In simpler terms, heat flows from hot regions to cold ones, and the flow rate increases with larger temperature differences and cross-sectional areas, but decreases with greater distances.

Mathematically, Fourier's Law for one-dimensional heat flow can be expressed as: \[ q = -kA\frac{dT}{dx} \] where \( q \) is the heat transfer rate, \( k \) is the thermal conductivity of the material, \( A \) is the cross-sectional area perpendicular to the heat flow, and \( \frac{dT}{dx} \) is the temperature gradient along the direction of heat flow. In the context of the textbook exercise, this law helps us calculate the heat transfer rate per unit length once we know the thermal resistance and temperature difference.
Thermal Resistance
Thermal resistance, analogous to electrical resistance, is a measure of a material's ability to resist the flow of heat. When considering heat transfer through a material, thermal resistance describes how well the material insulates against the movement of heat. The higher the thermal resistance, the slower the heat flows through the material, and vice versa.

Thermal resistance can be calculated using the formula: \[ R_{\text{th}} = \frac{\Delta x}{kA} \] where \( R_{\text{th}} \) is the thermal resistance, \( \Delta x \) is the thickness of the material, \( k \) is the thermal conductivity, and \( A \) is the cross-sectional area. In practice, when dealing with non-uniform geometries like the insulation in the exercise, we make reasonable approximations to simplify calculations. Here, a radial heat flow assumption allows us to represent the thermal resistance of cylindrical insulation as an equivalent resistance, which is a crucial step in determining the heat transfer rate.
Radial Heat Flow
Radial heat flow refers to the movement of heat in a radial direction outward or inward through a cylindrical or spherical object. This type of heat flow is prevalent in systems where there is a non-uniform geometry such as pipes or tubes.

In the case of radial heat flow through a cylindrical object, the thermal resistance is different from that in one-dimensional heat flow situations, due to the varying cross-sectional area as heat moves radially. The formula to calculate thermal resistance for radial heat flow in a cylinder is: \[ R_{\text{th}} = \frac{\ln(\frac{r_2}{r_1})}{2\pi{kL}} \] where \( r_1 \) and \( r_2 \) are the inner and outer radii of the cylindrical layer through which heat is flowing, \( k \) is the thermal conductivity, and \( L \) is the length of the cylinder. This concept is crucial in the step-by-step solution provided, where approximating the insulation as a radially conducting layer allows us to calculate the heat transfer rate using the modified equation for thermal resistance in cylindrical coordinates.

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Most popular questions from this chapter

Steam in a heating system flows through tubes whose outer diameter is \(5 \mathrm{~cm}\) and whose walls are maintained at a temperature of \(180^{\circ} \mathrm{C}\). Circular aluminum alloy 2024-T6 fins \((k=186 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of outer diameter \(6 \mathrm{~cm}\) and constant thickness \(1 \mathrm{~mm}\) are attached to the tube. The space between the fins is \(3 \mathrm{~mm}\), and thus there are 250 fins per meter length of the tube. Heat is transferred to the surrounding air at \(T_{\infty}=25^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the increase in heat transfer from the tube per meter of its length as a result of adding fins.

The unit thermal resistances ( \(R\)-values) of both 40-mm and 90-mm vertical air spaces are given in Table 3-9 to be \(0.22 \mathrm{~m}^{2} \cdot \mathrm{C} / \mathrm{W}\), which implies that more than doubling the thickness of air space in a wall has no effect on heat transfer through the wall. Do you think this is a typing error? Explain.

Consider a flat ceiling that is built around \(38-\mathrm{mm} \times\) \(90-\mathrm{mm}\) wood studs with a center-to-center distance of \(400 \mathrm{~mm}\). The lower part of the ceiling is finished with 13-mm gypsum wallboard, while the upper part consists of a wood subfloor \(\left(R=0.166 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), a \(13-\mathrm{mm}\) plywood, a layer of felt \(\left(R=0.011 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\), and linoleum \(\left(R=0.009 \mathrm{~m}^{2} \cdot{ }^{\circ} \mathrm{C} / \mathrm{W}\right)\). Both sides of the ceiling are exposed to still air. The air space constitutes 82 percent of the heat transmission area, while the studs and headers constitute 18 percent. Determine the winter \(R\)-value and the \(U\)-factor of the ceiling assuming the 90 -mm-wide air space between the studs ( \(a\) ) does not have any reflective surface, (b) has a reflective surface with \(\varepsilon=0.05\) on one side, and ( ) has reflective surfaces with \(\varepsilon=0.05\) on both sides. Assume a mean temperature of \(10^{\circ} \mathrm{C}\) and a temperature difference of \(5.6^{\circ} \mathrm{C}\) for the air space.

Obtain a relation for the fin efficiency for a fin of constant cross-sectional area \(A_{c}\), perimeter \(p\), length \(L\), and thermal conductivity \(k\) exposed to convection to a medium at \(T_{\infty}\) with a heat transfer coefficient \(h\). Assume the fins are sufficiently long so that the temperature of the fin at the tip is nearly \(T_{\infty}\). Take the temperature of the fin at the base to be \(T_{b}\) and neglect heat transfer from the fin tips. Simplify the relation for \((a)\) a circular fin of diameter \(D\) and \((b)\) rectangular fins of thickness \(t\).

Two finned surfaces are identical, except that the convection heat transfer coefficient of one of them is twice that of the other. For which finned surface is the \((a)\) fin effectiveness and \((b)\) fin efficiency higher? Explain.

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