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Energy Equation & Fourier's Law Notes | Study Heat Transfer - Mechanical Engineering

Document Description: Energy Equation & Fourier's Law for Mechanical Engineering 2022 is part of Introduction to Heat Transfer for Heat Transfer preparation. The notes and questions for Energy Equation & Fourier's Law have been prepared according to the Mechanical Engineering exam syllabus. Information about Energy Equation & Fourier's Law covers topics like and Energy Equation & Fourier's Law Example, for Mechanical Engineering 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Energy Equation & Fourier's Law.

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Fourier's Law 

The constitutive equation for conduction, we have see, is Fourier's Law. It says that the heat flux vector is a linear function of the temperature gradient, that is :

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

What we mean by the notation is the following:

q = qi ei     Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Then for each of the components of q, we have the relation:

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Thermal Properties of Matter

Thermal conductivity

The conductivity is a material property that is a very strong function of the state of the material. The range of values goes from less than 0.01 Watts/m-°K for gaseous CO2 to over 600 Watts/m-°K for Ag metal. The change may be as much as 5 orders of magnitude

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Thermal Conductivity of Gases

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Thermal Conductivity of Solids

The thermal conductivity of solids differ significantly as the next figure shows

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Conductive Loss through a Window Pane

Examine the simple one-dimensional conduction problem as heat flow through a windowpane. The window glass thickness, L, is 1/8 in. If this is the only window in a room 9x12x8 or 864 ft3 , the area of the window is 2 ft x 3 ft or 6 ft2 .

Recall that qx is the heat flux and that k is the thermal conductivity

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The energy at steady state yielded

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The room is well heated and the temperature is uniform, so the heat low through the windowpane is

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

If the room temperature is 60 °F and the exterior temperature is 20 °F, and k is 0.41 Btu/hr-ft2-°F then

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Energy balance on the Room

How long does it take for the room temperature to change from 60 °F to 45 °F?

To make this estimate, we need to solve an energy balance on the room. A simple analysis yields

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Recognizing that heat capacity density are essentially constant, the equation becomes

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

At the outset, T1 = T10 = 60 °F The solution of the differential equation representing the energy balance is

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

To solve for the time required to get to 46 °F, we need all the data in the table.

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

It follows that τ = 0.47, τ = 1.75 minutes

Heat Conduction in a Composite Solid

Examine the simple one-dimensional conduction problem as heat flow through a thermally insulated windowpane. Each layer of window glass thickness, L, is 1/16 in. The insulation layer of air between the two panes is also 1/16 in. Recall that qx is the heat flux and that k is the thermal conductivity

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The energy at steady state yielded

  Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The heat flow through the glass is given by

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Then we can rewrite the equations in this form

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

If we add the three equations, we obtain

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

We can consider the thickness/conductivity as a resistance so that

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The heat flow is then of the following form :

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

This is like a problem of current flow in a series circuit. In the single pane problem discussed in Lecture 1, we noted that the resistance, δ/k, was 1/(192(0.41) = 0.254 hr-ft2 -°F/Btu. Recall that for the problem of cooling the room, t was 1.75 minutes. The thermal conductivity of air is 0.014 Btu/hr-ft-°F. so that

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

as a consequence the reciprocal of the overall resistance is 0.744 + (0.0254) = 0.746.

Then we see that τ = (1 min) (0.746/0.0254) = 29.37 min

The Convective Boundary Condition 

Again consider a windowpane, but now there is a heat transfer limitation at one boundary described by a boundary condition. qx = – h (Tr – Ti)

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Conduction through the glass is described by

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The flux is constant at any cross-section so that we can write

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Solving for the temperatures we get

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Solving for qx , the relation becomes  Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Which modified shows a correction to the heat transfer coefficient modulated by the conduction problem  Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The dimensionless number in the denominator is the Biot number, a ratio of the convective heat transfer coefficient to the equivalent heat transfer coefficient due to conduction.

Heat Transfer across a Composite Cylindrical Solid. 

In the case of heat transfer in a cylinder, there is radial symmetry do that heat conduction is important only in the radial direction.

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The heat flux in the radial direction is given by Fourier’s law

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The total heat flow through any circular surface is constant

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Rearranging we obtain a relation for the temperature gradient

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

which upon separation of variables is

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

An indefinite integration yields the temperature profile.

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

The boundary conditions are

at r = R, T = T1 ;

at r = R2 , T = T2

at r = R3 , T = T3

qr2 = qr2

qr= h(T3 - T0 )

so that

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

It follows that

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

This can be expressed as

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Optimal Insulation on a Pipe

Is there an optimal thickness for the exterior insulation? In the context of the problem just formulated, is there a best value for R3?

Note that Q = f(R).

To find an extremum,

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

Energy Equation & Fourier`s Law Notes | Study Heat Transfer - Mechanical Engineering

It offers a critical radius for R= k2 /h beyond which the heat loss increases.

 

The document Energy Equation & Fourier's Law Notes | Study Heat Transfer - Mechanical Engineering is a part of the Mechanical Engineering Course Heat Transfer.
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