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Chapter 3: Problem 197

Heat is lost at a rate of \(275 \mathrm{~W}\) per \(\mathrm{m}^{2}\) area of a \(15-\mathrm{cm}-\) thick wall with a thermal conductivity of \(k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature drop across the wall is (a) \(37.5^{\circ} \mathrm{C}\) (b) \(27.5^{\circ} \mathrm{C}\) (c) \(16.0^{\circ} \mathrm{C}\) (d) \(8.0^{\circ} \mathrm{C}\) (e) \(4.0^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: The temperature drop across the wall is approximately \(37.5^{\circ} C\).

Step by step solution

01

Write down the given parameters.

We are given the following information: - Heat loss rate (Q): \(275\, W/m^2\) - Wall thickness (d): \(15\, cm = 0.15\, m\) (convert to meters) - Thermal conductivity (k): \(1.1\, W/m\cdot K\)
02

Write the formula for Fourier's law of heat conduction.

Fourier's law of heat conduction can be stated as: \(Q = -k\cdot A \cdot \frac{\Delta T}{d}\) Where Q is the rate of heat loss, k is the thermal conductivity, A is the area of the wall, 螖T is the temperature drop across the wall, and d is the wall thickness. We want to find the temperature drop (螖T), so we'll rearrange the formula for 螖T: \(\Delta T = -\frac{Q\cdot d}{k\cdot A}\) However, since Q is given as \(275\, W/m^2\), we don't have to worry about the area, and the formula becomes: \(\Delta T = -\frac{Q\cdot d}{k}\)
03

Calculate the temperature drop (螖T).

Now that we have the formula, we can plug in the given values to find the temperature drop: \(\Delta T = -\frac{275\, W/m^2\cdot 0.15\, m}{1.1\, W/m\cdot K}\) \(\Delta T = -\frac{41.25\, W/m^2}{1.1\, W/m\cdot K}\) \(\Delta T \approx 37.5\, ^\circ C\)
04

Choose the correct option.

The calculated temperature drop is \(\approx 37.5\, ^\circ C\). Thus, the correct option is: (a) \(37.5^{\circ} \mathrm{C}\)

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a material property that measures a substance's ability to conduct heat. It is typically denoted by the symbol \( k \) and is defined as the amount of heat (in watts) transferred through a unit area of the substance (in square meters) for a temperature difference (in Kelvin) per unit thickness (in meters). Materials with high thermal conductivity, like metals, are excellent heat conductors, whereas materials like wood or fiberglass with low thermal conductivity are considered insulators.

Understanding thermal conductivity is crucial when designing thermal insulation systems or analyzing heat transfer in materials. Higher values of \( k \) indicate that the material can transfer heat more efficiently, which is often desirable in heat sinks but undesirable in thermal insulation scenarios.
Temperature Gradient
A temperature gradient is the rate at which temperature changes from one point to another in space. Mathematically, it is expressed as \( \frac{\Delta T}{d} \), where \( \Delta T \) is the temperature difference between two points and \( d \) is the distance between those points. The temperature gradient is the driving force behind heat transfer by conduction, and the direction of heat flow is from the region of higher temperature to the region of lower temperature.

This concept is essential when calculating the rate of heat loss or gain in materials, such as in our exercise, where you need to find the temperature drop across a wall. A steeper temperature gradient indicates a more significant temperature change over a short distance, which results in a higher rate of heat transfer.
Heat Transfer
Heat transfer is a fundamental concept in thermodynamics that involves the movement of thermal energy from one place to another. There are three main mechanisms of heat transfer: conduction, convection, and radiation. In the context of Fourier's law, we're concerned with conduction, which is the transfer of heat through a material without any actual movement of the material itself. It occurs at the molecular level as atoms and molecules vibrate and transfer energy to neighboring atoms and molecules.

The rate of heat transfer (\( Q \)) for conduction can be calculated using Fourier's law, which relates the heat transfer to the temperature gradient, the thermal conductivity, and the material's cross-sectional area. This allows us to deduce the amount of heat that flows through a specific material over time, given the temperature difference across it, as exemplified by the exercise provided.

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Most popular questions from this chapter

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 \(t\).

Consider an insulated pipe exposed to the atmosphere. Will the critical radius of insulation be greater on calm days or on windy days? Why?

A mixture of chemicals is flowing in a pipe \(\left(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\right.\), and \(L=10 \mathrm{~m}\) ). During the transport, the mixture undergoes an exothermic reaction having an average temperature of \(135^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent any incident of thermal burn, the pipe needs to be insulated. However, due to the vicinity of the pipe, there is only enough room to fit a \(2.5\)-cm-thick layer of insulation over the pipe. The pipe is situated in a plant, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the insulation for the pipe such that the thermal conductivity of the insulation is sufficient to maintain the outside surface temperature at \(45^{\circ} \mathrm{C}\) or lower.

Steam at \(235^{\circ} \mathrm{C}\) is flowing inside a steel pipe \((k=\) \(61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose inner and outer diameters are \(10 \mathrm{~cm}\) and \(12 \mathrm{~cm}\), respectively, in an environment at \(20^{\circ} \mathrm{C}\). The heat transfer coefficients inside and outside the pipe are \(105 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine ( \(a\) ) the thickness of the insulation \((k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) needed to reduce the heat loss by 95 percent and \((b)\) the thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to \(40^{\circ} \mathrm{C}\) for safety reasons.

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum ( \(k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a 10 -cm-thick layer of insulation \((k=\) \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the exterior surface of the insulation is (a) \(13^{\circ} \mathrm{C}\) (b) \(9^{\circ} \mathrm{C}\) (c) \(2^{\circ} \mathrm{C}\) (d) \(-3^{\circ} \mathrm{C}\) (e) \(-12^{\circ} \mathrm{C}\)
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