/*! 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 26 A composite wall separates combu... [FREE SOLUTION] | 91Ó°ÊÓ

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A composite wall separates combustion gases at \(2600^{\circ} \mathrm{C}\) from a liquid coolant at \(100^{\circ} \mathrm{C}\), with gas- and liquid-side convection coefficients of 50 and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall is composed of a \(10-\mathrm{mm}\)-thick layer of beryllium oxide on the gas side and a 20 -mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is \(0.05 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.

Short Answer

Expert verified
The heat loss per unit surface area of the composite is \(34474.1\, W/m^2\). The temperature distribution can be sketched by finding the temperature drop across each resistance. The temperature drops are \(\Delta T_{gconv} = 689.5^{\circ}C\), \(\Delta T_{BeO} = 11.4^{\circ}C\), \(\Delta T_{c} = 1723.7^{\circ}C\), \(\Delta T_{AISI} = 45.9^{\circ}C\), and \(\Delta T_{lconv} = 34.5^{\circ}C\). Plot these values on the y-axis and the corresponding layer on the x-axis.

Step by step solution

01

Identify Relevant Parameters

For this exercise, the relevant parameters given are: 1. Temperatures: \(T_g = 2600^{\circ}C\) (Combustion gas) and \(T_l = 100^{\circ}C\) (Liquid coolant) 2. Convection coefficients: \(h_g = 50 W/m^2\cdot K\) (Gas side) and \(h_l = 1000 W/m^2\cdot K\) (Liquid side) 3. Wall thicknesses: \(L_1 = 10 mm\) (Beryllium oxide) and \(L_2 = 20 mm\) (Stainless steel) 4. Contact resistance: \(R_{c} = 0.05 m^2\cdot K/W\) (Between oxide and steel)
02

Calculate Thermal Resistances

Next, we need to calculate the thermal resistance for each section of the system. These include convection resistances at the gas and liquid side, and the conduction resistances for the beryllium oxide, stainless steel, and the contact resistance between them. For convection resistances: \[R_{gconv} = \frac{1}{h_g A} ,\, R_{lconv} = \frac{1}{h_l A}\] For conduction resistances: \[R_{BeO} = \frac{L_1}{k_1 A} ,\, R_{AISI} = \frac{L_2}{k_2 A}\] The material properties of the beryllium oxide and AISI 304 stainless steel are needed to determine the conduction resistances. The corresponding thermal conductivities are \(k_1 = 30 \, W/m\cdot K\) for the beryllium oxide and \(k_2 = 15 \, W/m\cdot K\) for the stainless steel. Calculating the resistances: - \(R_{gconv} = \frac{1}{(50)(1)} = 0.02\, m^2\cdot K/W\) - \(R_{lconv} = \frac{1}{(1000)(1)} = 0.001\, m^2\cdot K/W\) - \(R_{BeO} = \frac{0.01}{(30)(1)} = 0.00033\, m^2\cdot K/W\) - \(R_{AISI} = \frac{0.02}{(15)(1)} = 0.00133\, m^2\cdot K/W\)
03

Calculate Overall Resistance and Heat Transfer Rate

Find the overall resistance by summing all individual resistances: \[R_{total} = R_{gconv} + R_{BeO} + R_{c} + R_{AISI} + R_{lconv}\] \[R_{total} = 0.02 + 0.00033 + 0.05 + 0.00133 + 0.001 = 0.07266\, m^2\cdot K/W\] Now that the total resistance has been determined, we can calculate the heat transfer rate per unit surface area using the heat transfer equation: \[q = \frac{T_g - T_l}{R_{total}}\] \[q = \frac{2600 - 100}{0.07266} = 34474.1\, W/m^2\]
04

Sketch Temperature Distribution

To sketch the temperature distribution graphically, we need to find the temperature drop across each resistance. This can be done using the equation: \[\Delta T_i = q \cdot R_i\] - \(\Delta T_{gconv} = 34474.1 \cdot 0.02 = 689.5^{\circ}C\) - \(\Delta T_{BeO} = 34474.1 \cdot 0.00033 = 11.4^{\circ}C\) - \(\Delta T_{c} = 34474.1 \cdot 0.05 = 1723.7^{\circ}C\) - \(\Delta T_{AISI} = 34474.1 \cdot 0.00133 = 45.9^{\circ}C\) - \(\Delta T_{lconv} = 34474.1 \cdot 0.001 = 34.5^{\circ}C\) Plot the temperature distribution by marking the temperature drop across each section on the y-axis and layer (conduction or convection part) on the x-axis.

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Key Concepts

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

Thermal Resistance Calculation
Understanding thermal resistance is crucial when it comes to heat transfer in composite walls. Think of thermal resistance as an insulator's effectiveness to combat heat flow; the higher the resistance, the slower the heat transfer. The total thermal resistance, often denoted as 'R', is the sum of individual resistances in series.

In practice, for a wall with multiple layers, we calculate resistance for each layer individually. Convection resistance, associated with the fluid motion on either side of the wall, is given by \( R_{conv} = \frac{1}{hA} \), where 'h' is the convection heat transfer coefficient and 'A' is the area through which heat is being transferred. Conduction resistance, on the other hand, is derived from \( R_{cond} = \frac{L}{kA} \), with 'L' being the thickness of the material, and 'k' its thermal conductivity.

When walls are in contact, an additional thermal resistance known as contact resistance may be present. It accounts for the imperfect contact between materials.

To find the total thermal resistance, we simply add up all the resistances: \[ R_{total} = R_{conv1} + R_{cond1} + R_{contact} + R_{cond2} + R_{conv2} \]With the total resistance known, we can then determine the heat loss using the formula \( q = \frac{\Delta T}{R_{total}} \) where \( \Delta T \) is the temperature difference across the composite wall.
Conduction and Convection in Heat Transfer
When heat traverses a solid material, it does so primarily by conduction. The ease with which heat moves through a material is measured by its thermal conductivity, 'k'. Materials like copper have a high 'k', indicating efficient heat transfer. Conversely, insulators like beryllium oxide have a low 'k', reflecting their resistance to heat flow.

Conduction can be mathematically described by Fourier's law, and the rate at which heat transfers through a material depends on its thermal conductivity, the temperature gradient, and the cross-sectional area perpendicular to the heat flow.

Convection is the transfer of heat through a fluid (gas or liquid), which occurs when a fluid moves from a warm location to a cooler one. The convection heat transfer coefficient 'h' quantifies how well a fluid can carry away heat from a surface. High 'h' values, like those for liquids, typically signify efficient heat removal.

Convection can be natural, driven by buoyancy forces that arise from temperature differences, or forced, where a pump or a fan propels the fluid. In calculations, the distinction between conduction and convection is important as it affects the computation of thermal resistances and therefore the overall heat transfer rate.
Temperature Distribution Sketching
A temperature distribution sketch visually represents how temperature changes across different layers of a composite wall. It depicts the temperature drop in each section, essential for understanding the heat transfer process.

To create this sketch, you start by plotting temperature on the y-axis against the wall layers on the x-axis. Mark the initial temperature on the hot side of the wall and begin to subtract the temperature drops, computed by multiplying the heat transfer rate by the corresponding thermal resistance (\( \Delta T_i = q \cdot R_i \) ), from this value at each layer or interface.

For example, the largest temperature drop usually occurs where the thermal resistance is highest—typically at a contact resistance or on the side with the lowest convection coefficient. The slope of the line between points in the sketch indicates how quickly temperature changes: a steeper slope means a faster temperature change, linked to lower resistance. The sketch provides a clear visual cue for where the most significant insulation is needed or where heat is most readily lost, aiding in the design of efficient thermal systems.

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Most popular questions from this chapter

The outer surface of a hollow sphere of radius \(r_{2}\) is subjected to a uniform heat flux \(q_{2}^{\prime \prime}\). The inner surface at \(r_{1}\) is held at a constant temperature \(T_{s, 1}\). (a) Develop an expression for the temperature distribution \(T(r)\) in the sphere wall in terms of \(q_{2}^{\prime \prime}, T_{s, 1}, r_{1}, r_{2}\), and the thermal conductivity of the wall material \(k\). (b) If the inner and outer tube radii are \(r_{1}=50 \mathrm{~mm}\) and \(r_{2}=100 \mathrm{~mm}\), what heat flux \(q_{2}^{\prime \prime}\) is required to maintain the outer surface at \(T_{s, 2}=50^{\circ} \mathrm{C}\), while the inner surface is at \(T_{s, 1}=20^{\circ} \mathrm{C}\) ? The thermal conductivity of the wall material is \(k=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of \(5^{\circ} \mathrm{C}\). The tube wall has inner and outer radii of 25 and \(75 \mathrm{~mm}\), respectively, and a thermal conductivity of \(10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is \(R_{t, c}^{\prime}=\) \(0.01 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\). The outer surface of the heater is exposed to a fluid with \(T_{\infty}=-10^{\circ} \mathrm{C}\) and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heater power per unit length of tube required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C} .\)

A thin metallic wire of thermal conductivity \(k\), diameter \(D\), and length \(2 L\) is annealed by passing an electrical current through the wire to induce a uniform volumetric heat generation \(\dot{q}\). The ambient air around the wire is at a temperature \(T_{\infty}\), while the ends of the wire at \(x=\pm L\) are also maintained at \(T_{\infty}\). Heat transfer from the wire to the air is characterized by the convection coefficient \(h\). Obtain expressions for the following: (a) The steady-state temperature distribution \(T(x)\) along the wire, (b) The maximum wire temperature. (c) The average wire temperature.

The air inside a chamber at \(T_{\infty, i}=50^{\circ} \mathrm{C}\) is heated convectively with \(h_{i}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) by a 200 -mm-thick wall having a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a uniform heat generation of \(1000 \mathrm{~W} / \mathrm{m}^{3}\). To prevent any heat generated within the wall from being lost to the outside of the chamber at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) with \(h_{o}=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a very thin electrical strip heater is placed on the outer wall to provide a uniform heat flux, \(q_{\sigma^{\prime}}\) (a) Sketch the temperature distribution in the wall on \(T-x\) coordinates for the condition where no heat generated within the wall is lost to the outside of the chamber. (b) What are the temperatures at the wall boundaries, \(T(0)\) and \(T(L)\), for the conditions of part (a)? (c) Determine the value of \(q_{o}^{\prime \prime}\) that must be supplied by the strip heater so that all heat generated within the wall is transferred to the inside of the chamber. (d) If the heat generation in the wall were switched off while the heat flux to the strip heater remained constant, what would be the steady-state temperature, \(T(0)\), of the outer wall surface?

A high-temperature, gas-cooled nuclear reactor consists of a composite cylindrical wall for which a thorium fuel element \((k \approx 57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is encased in graphite \((k \approx 3\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) and gaseous helium flows through an annular coolant channel. Consider conditions for which the helium temperature is \(T_{\infty}=600 \mathrm{~K}\) and the convection coefficient at the outer surface of the graphite is \(h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If thermal energy is uniformly generated in the fuel element at a rate \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), what are the temperatures \(T_{1}\) and \(T_{2}\) at the inner and outer surfaces, respectively, of the fuel element? (b) Compute and plot the temperature distribution in the composite wall for selected values of \(\dot{q}\). What is the maximum allowable value of \(\dot{q}\) ?

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