|1 Crore+ students have signed up on EduRev. Have you?|
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 = qi ei
Then for each of the components of q, we have the relation:
Thermal Properties of Matter
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
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 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
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, T1 = T10 = 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 qx 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-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
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)
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 qx , 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 = R1 , T = T1 ;
at r = R2 , T = T2
at r = R3 , T = T3
qr2 = qr2
qr3 = h(T3 - T0 )
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 R3?
Note that Q = f(R3 ).
To find an extremum,
It offers a critical radius for R3 = k2 /h beyond which the heat loss increases.