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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 :

What we mean by the notation is the following:

q = q_{i} e_{i}

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

**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 CO_{2} to over 600 Watts/m-°K for Ag metal. The change may be as much as 5 orders of magnitude

**Thermal Conductivity of Gases**

**Thermal Conductivity of Solids**

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

**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 ft^{3} , the area of the window is 2 ft x 3 ft or 6 ft^{2} .

Recall that q_{x} is the heat flux and that k is the thermal conductivity

The energy at steady state yielded

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

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 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

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

At the outset, T_{1} = T_{10} = 60 °F The solution of the differential equation representing the energy balance is

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

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 q_{x} is the heat flux and that k is the thermal conductivity

The energy at steady state yielded

The heat flow through the glass is given by

Then we can rewrite the equations in this form

If we add the three equations, we obtain

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

The heat flow is then of the following form :

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-ft^{2} -°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

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. q_{x} = – h (T_{r} – T_{i})

Conduction through the glass is described by

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

Solving for the temperatures we get

Solving for q_{x} , the relation becomes

Which modified shows a correction to the heat transfer coefficient modulated by the conduction problem

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.

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

The total heat flow through any circular surface is constant

Rearranging we obtain a relation for the temperature gradient

which upon separation of variables is

An indefinite integration yields the temperature profile.

The boundary conditions are

at r = R_{1 }, T = T_{1} ;

at r = R_{2} , T = T_{2}

at r = R_{3} , T = T_{3}

q_{r2 }= q_{r2}

qr_{3 }= h(T_{3} - T_{0} )

so that

It follows that

This can be expressed as

**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 R_{3}?

Note that Q = f(R_{3 }).

To find an extremum,

It offers a critical radius for R_{3 }= k_{2} /h beyond which the heat loss increases.

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