/*! 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

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

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.

A steam pipe of \(0.12-\mathrm{m}\) outside diameter is insulated with a layer of calcium silicate. (a) If the insulation is \(20 \mathrm{~mm}\) thick and its inner and outer surfaces are maintained at \(T_{s, 1}=800 \mathrm{~K}\) and \(T_{s, 2}=490 \mathrm{~K}\), respectively, what is the heat loss per unit length \(\left(q^{\prime}\right)\) of the pipe? (b) We wish to explore the effect of insulation thickness on the heat loss \(q^{\prime}\) and outer surface temperature \(T_{s, 2}\), with the inner surface temperature fixed at \(T_{s, 1}=\) \(800 \mathrm{~K}\). The outer surface is exposed to an airflow \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) that maintains a convection coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and to large surroundings for which \(T_{\text {sur }}=T_{\infty}=25^{\circ} \mathrm{C}\). The surface emissivity of calcium silicate is approximately \(0.8\). Compute and plot the temperature distribution in the insulation as a function of the dimensionless radial coordinate, \(\left(r-r_{1}\right) /\left(r_{2}-r_{1}\right)\), where \(r_{1}=0.06 \mathrm{~m}\) and \(r_{2}\) is a variable \(\left(0.06

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?

In a test to determine the friction coefficient \(\mu\) associated with a disk brake, one disk and its shaft are rotated at a constant angular velocity \(\omega\), while an equivalent disk/shaft assembly is stationary. Each disk has an outer radius of \(r_{2}=180 \mathrm{~mm}\), a shaft radius of \(r_{1}=20 \mathrm{~mm}\), a thickness of \(t=12 \mathrm{~mm}\), and a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). A known force \(F\) is applied to the system, and the corresponding torque \(\tau\) required to maintain rotation is measured. The disk contact pressure may be assumed to be uniform (i.e., independent of location on the interface), and the disks may be assumed to be well insulated from the surroundings. (a) Obtain an expression that may be used to evaluate \(\mu\) from known quantities. (b) For the region \(r_{1} \leq r \leq r_{2}\), determine the radial temperature distribution \(T(r)\) in the disk, where \(T\left(r_{1}\right)=T_{1}\) is presumed to be known. (c) Consider test conditions for which \(F=200 \mathrm{~N}\), \(\omega=40 \mathrm{rad} / \mathrm{s}, \tau=8 \mathrm{~N} \cdot \mathrm{m}\), and \(T_{1}=80^{\circ} \mathrm{C}\). Evaluate the friction coefficient and the maximum disk temperature.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.