/*! 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 31 A furnace wall is to operate wit... [FREE SOLUTION] | 91Ó°ÊÓ

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A furnace wall is to operate with inner and outer surface temperatures of \(1500 \mathrm{~K}\) and \(320 \mathrm{~K}\), respectively. Insulating bricks measuring \(20 \mathrm{~cm} \times 10 \mathrm{~cm} \times 8 \mathrm{~cm}\) are available in two kinds at the same price. Type A has a thermal conductivity of \(2.0\) \(\mathrm{W} / \mathrm{m} \mathrm{K}\) and a maximum allowable temperature of \(1600 \mathrm{~K}\). Type B has a thermal conductivity of \(1.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and a maximum allowable temperature of \(1000 \mathrm{~K}\). Determine how the bricks should be arranged so as not to exceed a heat flow per unit area of \(1000 \mathrm{~W} / \mathrm{m}^{2}\), and minimize the cost of the walls.

Short Answer

Expert verified
Use two layers of Type B bricks to minimize heat flow within limits.

Step by step solution

01

Understanding the Problem

We need to arrange bricks (Type A and Type B) such that the heat flow does not exceed 1000 W/m² and the cost is minimized. Both types are the same price, but have different thermal conductivities and maximum allowable temperatures.
02

Thermal Resistance Calculation

For materials used in series, the total thermal resistance is the sum of individual resistances. Calculate the thermal resistance for Type A: \( R_A = \frac{L}{k_A} \), where \( L \) is the thickness (0.10 m) and \( k_A = 2.0 \) W/mK. \( R_A = \frac{0.10}{2.0} = 0.05 \; \text{m}^2 \text{K/W} \).
03

Thermal Resistance for Type B

Similarly for Type B, \( R_B = \frac{L}{k_B} \), with \( L = 0.10 \; \text{m} \) and \( k_B = 1.0 \; \text{W/mK} \). Therefore, \( R_B = \frac{0.10}{1.0} = 0.10 \; \text{m}^2 \text{K/W} \).
04

Series Combination of Bricks

Consider a combination of both types in series. The total resistance \( R_{total} = R_A + R_B = 0.05 + 0.10 = 0.15 \; \text{m}^2 \text{K/W} \).
05

Calculate Heat Flow

Using the formula \( q = \frac{T_1 - T_2}{R_{total}} \) where \(T_1 = 1500 \; \text{K} \) and \( T_2 = 320 \; \text{K} \), calculate the heat flow: \( q = \frac{1500 - 320}{0.15} = 7866.67 \; \text{W/m}^2 \).
06

Evaluate Temperature Limits

Check if any part exceeds the maximum allowable temperature. Place Type A next to the hotter side (1500 K) due to its higher allowable temperature. Calculate intermediate temperature after Type A: \( T' = 1500 - q \times R_A = 1500 - 7866.67 \times 0.05 = 1116.67 \; \text{K} \).
07

Optimize Arrangement

Since both types cost the same, and the calculated setup already exceeds the heat flow limit, we need to adjust. Instead use only Type B, allowing twice the thickness \( R_B = \frac{0.20}{1.0} = 0.20 \; \text{m}^2 \text{K/W} \). Recalculate heat flow \( q = \frac{1500 - 320}{0.20} = 5900 \; \text{W/m}^2 \), appropriate but rearrange for best performance.
08

Final Setup

Place two layers of Type B bricks (0.10 m each) and avoid using Type A bricks to minimize heat flow and stay under the required threshold, as this material provides better resistance without exceeding thermal limits.

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Key Concepts

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

Heat Transfer
Heat transfer is a key concept in physics and engineering that describes how heat energy moves from one place to another. There are three primary modes of heat transfer: conduction, convection, and radiation. In this exercise, we focus on conduction, which is the transfer of heat through a solid medium.

When a temperature difference exists within an object or between objects that are in contact, heat will flow from the region of higher temperature to the region of lower temperature. This is governed by Fourier’s Law of Heat Conduction, which states that the rate of heat transfer (\( q \) ) through a material is proportional to the negative gradient of temperatures and to the area through which the heat flows.

To calculate the heat flow per unit area, we use the formula:
  • \( q = \frac{T_1 - T_2}{R_{total}} \)
Here, \( T_1 \) and \( T_2 \) are the temperatures on either side of the material, and \( R_{total} \) is the total thermal resistance. This exercise challenges students to apply this formula to maintain an optimal heat flow while considering material limitations.
Thermal Conductivity
Thermal conductivity is a material property that indicates how well a material can conduct heat. It is denoted by \( k \) and expressed in units of watts per meter-kelvin (W/m·K). The higher a material's thermal conductivity, the better it is at conducting heat. Conversely, a lower thermal conductivity means the material is a better insulator.

In this exercise, we compare two types of bricks:
  • Type A with a thermal conductivity of 2.0 W/m·K,
  • Type B with a thermal conductivity of 1.0 W/m·K.
Type B is a better insulator because it conducts heat less efficiently than Type A. Therefore, for minimizing heat transfer, materials with lower thermal conductivity are preferable.

To determine how the bricks should be arranged, we calculate each type's thermal resistance using the formula:
  • \( R = \frac{L}{k} \)
Where \( L \) is the thickness of the material. By comparing the thermal resistances, we can decide how to optimize the wall construction to limit heat flow effectively.
Insulating Materials
Insulating materials are used to reduce the rate of heat transfer. An effective insulating material has a low thermal conductivity, which means less heat is lost through it. In this exercise, Type B bricks are preferred as insulating materials due to their lower thermal conductivity compared to Type A.

Insulating materials are particularly important in scenarios where it is crucial to maintain specific temperature conditions, such as in the walls of a furnace or building. The goal is not only to minimize heat loss but also to maintain safety by preventing overheating or excessively high temperatures.

Here are some characteristics of good insulating materials:
  • Low thermal conductivity,
  • High temperature resistance,
  • Cost-effectiveness.
In this context, the choice of Type B bricks provides better thermal resistance without risking surpassing its maximum allowable temperature. This ensures the outer temperature doesn't exceed safe operational levels while maintaining an efficient heat transfer rate.

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Most popular questions from this chapter

The thermal resistance per unit area of clothing is often expressed in the unit clo, where \(1 \mathrm{clo}=0.88 \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{hr} / \mathrm{B}\) tu. (i) What is 1 clo in SI units? (ii) If a wool sweater is \(2 \mathrm{~mm}\) thick and has a thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m}\) \(\mathrm{K}\), what is its thermal resistance in clos? (iii) A cotton shirt has a thermal resistance of \(0.5 \mathrm{clo}\). If the inner and outer surfaces are at \(31^{\circ} \mathrm{C}\) and \(28^{\circ} \mathrm{C}\), respectively, what is the rate of heat loss per unit area?

A chemical reactor has a \(5 \mathrm{~mm}\)-thick mild steel wall and is lined inside with a 2 \(\mathrm{mm}\)-thick layer of polyvinylchloride. The contents are at \(80^{\circ} \mathrm{C}\), and the ambient air is at \(20^{\circ} \mathrm{C}\). The inside thermal resistance is negligible ( \(h_{c, i}\) very large), and the outside heat transfer coefficient for combined convection and radiation is \(7 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). (i) Draw the thermal circuit. (ii) Plot a graph of the temperature profile through the wall. (iii) Calculate the rate of heat loss for a surface area of \(10 \mathrm{~m}^{2} .\)

In a materials-processing experiment on a space station, a \(1 \mathrm{~cm}-\) diameter sphere of alloy is to be cooled from \(600 \mathrm{~K}\) to \(400 \mathrm{~K}\). The sphere is suspended in a test chamber by three jets of nitrogen at \(300 \mathrm{~K}\). The convective heat transfer coefficient between the jets and the sphere is estimated to be \(180 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Calculate the time required for the cooling process and the minimum quenching rate. Take the alloy density to be \(\rho=14,000 \mathrm{~kg} / \mathrm{m}^{3}\), specific heat \(c=140 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), and thermal conductivity \(k=240 \mathrm{~W} / \mathrm{m} \mathrm{K}\). Since the emittance of the alloy is very small, the radiation contribution to heat loss can be ignored.

A system consists of a body in which heat is continuously generated at a rate \(\underline{Q}_{v}\), while heat is lost from the body to its surroundings by convection. Using the lumped thermal capacity model, derive the differential equation governing the temperature response of the body. If the body is at temperature \(T_{o}\) when time \(t=0\), solve the differential equation to obtain \(T(t) .\) Also determine the steady-state temperature.

Estimate the heating load for a building in a cold climate when the outside temperature is \(-10^{\circ} \mathrm{C}\) and the air inside is maintained at \(20^{\circ} \mathrm{C}\). The \(350 \mathrm{~m}^{2}\) of walls and ceiling are a composite of \(1 \mathrm{~cm}\)-thick wallboard \((k=0.2 \mathrm{~W} / \mathrm{m} \mathrm{K}), 10\) \(\mathrm{cm}\) of vermiculite insulation \((k=0.06 \mathrm{~W} / \mathrm{m} \mathrm{K})\), and \(3 \mathrm{~cm}\) of wood \((k=0.15 \mathrm{~W} / \mathrm{m}\) K). Take the inside and outside heat transfer coefficients as 7 and \(35 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively.

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