/*! 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 52 Two identical aluminum plates wi... [FREE SOLUTION] | 91Ó°ÊÓ

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Two identical aluminum plates with thickness of \(30 \mathrm{~cm}\) are pressed against each other at an average pressure of \(1 \mathrm{~atm}\). The interface, sandwiched between the two plates, is filled with glycerin. On the left outer surface, it is subjected to a uniform heat flux of \(7800 \mathrm{~W} / \mathrm{m}^{2}\) at a constant temperature of \(50^{\circ} \mathrm{C}\). On the right outer surface, the temperature is maintained constant at \(30^{\circ} \mathrm{C}\). Determine the thermal contact conductance of the glycerin at the interface, if the thermal conductivity of the aluminum plates is \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Discuss whether the value of the thermal contact conductance is reasonable or not.

Short Answer

Expert verified
Answer: The thermal contact conductance of the glycerin at the interface between two aluminum plates is 790.7 W/m²K.

Step by step solution

01

Calculate the temperature difference across the glycerin layer

Using the given temperature values on the outer surfaces, we can calculate the temperature difference across the glycerin layer as follows: \(\Delta T = T_{left} - T_{right} = 50^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C} = 20^{\circ} \mathrm{C}\)
02

Determine the heat transfer through one aluminum plate

According to Fourier's Law of heat conduction, the heat transfer through one aluminum plate can be calculated as follows: \(q = k \frac{A \Delta T}{d}\) Here, q is the heat flux, k is the thermal conductivity of the aluminum plates, A is the surface area, ΔT is the temperature difference, and d is the thickness of one aluminum plate. We are given the values: \(q = 7800 \mathrm{~W} / \mathrm{m}^{2}\) \(k = 237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) \(\Delta T = 20^{\circ} \mathrm{C} = 20 \mathrm{~K}\) (converting to Kelvin) \(d = 30 \mathrm{~cm} = 0.3 \mathrm{~m}\) (converting to meters) We can rearrange the formula to solve for A: \(A = \frac{q \cdot d}{k \cdot \Delta T} = \frac{7800 \mathrm{~W} / \mathrm{m}^{2} \cdot 0.3 \mathrm{~m}}{237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot 20 \mathrm{~K}}\) Calculating the surface area: \(A = \frac{2340 \mathrm{~W}}{4740 \mathrm{~W} / \mathrm{m}} = 0.493 \mathrm{~m}^{2} \)
03

Calculate the thermal contact conductance of the glycerin at the interface

Now, we can find the thermal contact conductance of the glycerin at the interface using the following formula: \(h_c = \frac{q}{A \cdot \Delta T} = \frac{7800 \mathrm{~W} / \mathrm{m}^{2}}{0.493 \mathrm{~m}^{2} \cdot 20 \mathrm{~K}}\) Calculating the thermal contact conductance: \(h_c = \frac{7800 \mathrm{~W} / \mathrm{m}^{2}}{9.86 \mathrm{~K} \cdot \mathrm{m}^{2}} = 790.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)
04

Discuss the reasonability of the value obtained

The calculated thermal contact conductance value is 790.7 W/m²K for the glycerin at the interface between the aluminum plates. Usually, this value depends on the surface roughness, contact pressure, temperature, and the materials in contact. Glycerin is known to have a good thermal conductivity, which should lead to a relatively high thermal contact conductance. Comparing this value, which is specific for this problem, with values found in literature or other similar setups can help in determining the reasonability of the calculated thermal contact conductance.

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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 fundamental concept in thermal sciences that involves the movement of thermal energy from one place to another due to temperature differences. It ensures that warmer areas transfer energy to cooler ones until thermal equilibrium is achieved. There are three primary modes of heat transfer: conduction, convection, and radiation.

In the context of the given exercise featuring aluminum plates and glycerin, conduction is the key mode of interest. Conduction occurs when the thermal energy is transferred through a material without the actual movement of the substance. This is what's happening when the heat is passed from the hotter side of an aluminum plate, through the glycerin, to the cooler side.

Understanding heat transfer is crucial when attempting to determine the efficiency and rate at which heat is conducted through different materials, as well as when selecting materials for engineering applications where temperature control is vital.
Fourier's Law of Heat Conduction
Fourier's Law of heat conduction is a principle that quantifies the heat flux passing through a material as a result of a temperature gradient. It is expressed mathematically by the equation:
\[q = -k \frac{\Delta T}{d}\]
where:
  • \(q\) is the heat flux (the amount of heat transferred per unit area per unit time),
  • \(k\) is the thermal conductivity of the material (a measure of its ability to conduct heat),
  • \(\Delta T\) is the temperature difference across the material, and
  • \(d\) is the thickness of the material.
Fourier's Law is the cornerstone of heat conduction analysis and provides a foundation for solving engineering problems involving thermal energy transfer. This relationship is applied in the textbook exercise to calculate the heat transfer through the aluminum plates.
Thermal Conductivity
Thermal conductivity, denoted as \(k\), is a material property that indicates how well a substance can conduct heat. It is defined as the amount of heat that passes through a unit area of the material with a unit temperature gradient per unit time. The SI unit for thermal conductivity is watts per meter per Kelvin (W/m·K).

Different materials have different thermal conductivities. For example, metals typically have high thermal conductivities, which means they are good conductors of heat, while materials like wood or plastic have low thermal conductivities, making them good insulators.

In our exercise, we are told that the thermal conductivity of the aluminum plates is \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), indicating aluminum's excellent ability to conduct heat, which is expected given its metallic nature.
Temperature Difference Calculation
Calculating the temperature difference is a straightforward yet essential process in thermal analysis. It is the driving force behind heat transfer by conduction and is represented by the symbol \(\Delta T\), indicating the difference between the temperatures of two points.

In the given exercise, the temperature difference calculation is the first step, being the difference between the uniform heat flux applied to one side of the aluminum plates and the maintained temperature on the other side. Precisely, the temperature difference across the glycerin layer is \(20^\circ \mathrm{C}\).

Knowing the temperature difference allows us to apply Fourier's Law of heat conduction effectively and to predict the behavior of heat flow through the materials involved, an important aspect when addressing real-world engineering challenges.

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Most popular questions from this chapter

Consider a surface of area \(A\) at which the convection and radiation heat transfer coefficients are \(h_{\text {conv }}\) and \(h_{\mathrm{rad}}\), respectively. Explain how you would determine \((a)\) the single equivalent heat transfer coefficient, and \((b)\) the equivalent thermal resistance. Assume the medium and the surrounding surfaces are at the same temperature.

Consider two metal plates pressed against each other. Other things being equal, which of the measures below will cause the thermal contact resistance to increase? (a) Cleaning the surfaces to make them shinier. (b) Pressing the plates against each other with a greater force. (c) Filling the gap with a conducting fluid. (d) Using softer metals. (e) Coating the contact surfaces with a thin layer of soft metal such as tin.

Chilled water enters a thin-shelled 5-cm-diameter, 150-mlong pipe at \(7^{\circ} \mathrm{C}\) at a rate of \(0.98 \mathrm{~kg} / \mathrm{s}\) and leaves at \(8^{\circ} \mathrm{C}\). The pipe is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipe is to be insulated with glass wool insulation \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in order to decrease the temperature rise of water to \(0.25^{\circ} \mathrm{C}\), determine the required thickness of the insulation.

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 a 3-m-high, 6-m-wide, and \(0.3\)-m-thick brick wall whose thermal conductivity is \(k=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be \(14^{\circ} \mathrm{C}\) and \(2^{\circ} \mathrm{C}\), respectively. Determine the rate of heat loss through the wall on that day.

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