/*! 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] | 91Ó°ÊÓ

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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 \( \frac{W}{m^2\cdot K} \) \( R_{wall} = \frac{r2 - r1}{2\pi k(r2 + r1)} = \frac{0.02 - 0.018}{2\pi(16.3)(0.02 + 0.018)} = 4.71 \times 10^{-5} \frac{°C}{W} \) \( R_{in} = \frac{1}{h1A1} = \frac{1}{400\times0.113} = 2.22 \times 10^{-5} \frac{°C}{W} \) \( R_{out} = \frac{1}{h2A2} = \frac{1}{6\times0.126} = 1.32 \times 10^{-4} \frac{°C}{W} \) 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: \( Q = \frac{ΔT}{R_{total}} \) Where ΔT is the temperature difference between the pharmaceutical and the ambient air, and R_total is the sum of all thermal resistances in the system. \( R_{total} = R_{wall} + R_{in} + R_{out} = 4.71 \times 10^{-5} + 2.22 \times 10^{-5} + 1.32 \times 10^{-4} = 2.11 \times 10^{-4} \frac{°C}{W} \) \( ΔT = 23 - 6 = 17°C \) \( Q = \frac{17}{2.11 \times 10^{-4}} = 80647 W/m \) 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 \( \frac{W}{m \cdot K} \) \( R_{insulation} = \frac{r3 - r2}{2\pi k_{ins}(r3 + r2)} = \frac{0.03 - 0.02}{2\pi(0.050)(0.03 + 0.02)} = 1.71 \times 10^{-3} \frac{°C}{W} \) 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. \( R_{total\_insulated} = R_{wall} + R_{in} + R_{out} + R_{insulation} = 2.11 \times 10^{-4} + 1.71 \times 10^{-3} = 1.92 \times 10^{-3} \frac{°C}{W} \) \( Q_{insulated} = \frac{17}{1.92 \times 10^{-3}} = 8854 W/m \) 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

Consider a plane composite wall that is composed of two materials of thermal conductivities \(k_{\mathrm{A}}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{B}}=0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thicknesses \(L_{\mathrm{A}}=10 \mathrm{~mm}\) and \(L_{\mathrm{B}}=20 \mathrm{~mm}\). The contact resistance at the interface between the two materials is known to be \(0.30 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Material A adjoins a fluid at \(200^{\circ} \mathrm{C}\) for which \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and material \(\mathrm{B}\) adjoins a fluid at \(40^{\circ} \mathrm{C}\) for which \(h=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) What is the rate of heat transfer through a wall that is \(2 \mathrm{~m}\) high by \(2.5 \mathrm{~m}\) wide? (b) Sketch the temperature distribution.

The walls of a refrigerator are typically constructed by sandwiching a layer of insulation between sheet metal panels. Consider a wall made from fiberglass insulation of thermal conductivity \(k_{i}=0.046 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{i}=50 \mathrm{~mm}\) and steel panels, each of thermal conductivity \(k_{p}=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(L_{p}=3 \mathrm{~mm}\). If the wall separates refrigerated air at \(T_{\infty, i}=4^{\circ} \mathrm{C}\) from ambient air at \(T_{\infty, o}=25^{\circ} \mathrm{C}\), what is the heat gain per unit surface area? Coefficients associated with natural convection at the inner and outer surfaces may be approximated as \(h_{i}=h_{o}=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate \(\dot{q}\). As shown in the sketch for Case 1 , the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{o}\). For Case 2 , a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(\mathrm{A}\) and \(\mathrm{B}\). The thermal resistance of the strip is \(R_{t}^{\prime \prime}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=\) \(20 \mathrm{~mm}, \dot{q}=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{o}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for Case 1 on \(T-x\) coordinates. Describe the key features of this distribution. Identify the location of the maximum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Case 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2 ? (d) What is the location of the maximum temperature in the composite wall of Case 2 ? Calculate this temperature.

An electrical current of 700 A flows through a stainless steel cable having a diameter of \(5 \mathrm{~mm}\) and an electrical resistance of \(6 \times 10^{-4} \mathrm{\Omega} / \mathrm{m}\) (i.e., per meter of cable length). The cable is in an environment having a temperature of \(30^{\circ} \mathrm{C}\), and the total coefficient associated with convection and radiation between the cable and the environment is approximately \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the cable is bare, what is its surface temperature? (b) If a very thin coating of electrical insulation is applied to the cable, with a contact resistance of \(0.02 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), what are the insulation and cable surface temperatures? (c) There is some concern about the ability of the insulation to withstand elevated temperatures. What thickness of this insulation \((k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) will yield the lowest value of the maximum insulation temperature? What is the value of the maximum temperature when this thickness is used?

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?

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