Appendices 10.A & 10.B: An Educational Presentation Presented By: Joseph Ash Jordan Baldwin Justin Hirt Andrea Lance

History of Heat Conduction Jean Baptiste Biot ( )  French Physicist  Worked on analysis of heat conduction  Unsuccessful at dealing with the problem of incorporating external convection effects in heat conduction analysis

History of Heat Conduction Jean Baptiste Joseph Fourier (1768 – 1830) Read Biot’s work 1807 determined how to solve the problem Fourier’s Law  Time rate of heat flow (Q) through a slab is proportional to the gradient of temperature difference

History of Heat Conduction Ernst Schmidt German scientist Pioneer in Engineering Thermodynamics Published paper “Graphical Difference Method for Unsteady Heat Conduction” First to measure velocity and temperature field in free convection boundary layer and large heat transfer coefficients Schmidt Number  Analogy between heat and mass transfer that causes a dimensionless quantity

Derivation of the Heat Conduction Equation A first approximation of the equations that govern the conduction of heat in a solid rod.

Consider the following: A uniform rod is insulated on both lateral ends.  Heat can now only flow in the axial direction. It is proven that heat per unit time will pass from the warmer section to the cooler one. The amount of heat is proportional to the area, A, and to the temperature difference T 2 -T 1, and is inversely proportional to the separation distance, d.

The final consideration can be expressed as the following: is a proportionality factor called the thermal conductivity and is determined by material properties

Assumptions The bar has a length L so x=0 and x=L Perfectly insulated Temperature, u, depends only on position, x, and time, t  Usually valid when the lateral dimensions are small compared to the total length.

The differential equation governing the temperature of the bar is a physical balance between two rates:  Flux/Flow term  Absorption term

Flux The instantaneous rate of heat transfer from left to right across the cross sections x=x 0 where x 0 is arbitrary can be defined as: The negative is needed in order to show a positive rate from left to right (hot to cold)

Flux Similarly, the instantaneous rate of heat transfer from right to left across the cross section x=x 0 +Δx where Δx is small can be defined as:

Flux The amount of heat entering the bar in a time span of Δt is found by subtracting the previous two equations and then multiplying the result by Δt:

Heat Absorption The average change in temperature, Δu, can be written in terms of the heat introduced, Q Δt and the mass Δm of the element as: where s = specific heat of the material ρ = density

Heat Absorption The actual temperature change of the bar is simply the actual change in temperature at some intermediate point, so the above equation can also be written as: This is the heat absorption equation.

Heat Equation Equating the QΔt in the flux and absorption terms, we find the heat absorption equation to be:

If we divide the above equation by ΔxΔt and allow both Δx and Δt to both go to 0, we will obtain the heat conduction or diffusion equation: where and has the dimensions of length^2/time and called the thermal diffusivity

Boundary Conditions Certain boundary conditions may apply to the specific heat conduction problem, for example:  If one end is maintained at some constant temperature value, then the boundary condition for that end is u = T.  If one end is perfectly insulated, then the boundary condition stipulates u x = 0.

Generalized Boundary Conditions Consider the end where x=0 and the rate of flow of heat is proportional to the temperature at the end of the bar.  Recall that the rate of flow will be given, from left to right, as  With this said, the rate of heat flow out of the bar from right to left will be  Therefore, the boundary condition at x=0 is where h 1 is a proportionality constant if h 1 =0, then it corresponds to an insulated end if h 1 goes to infinity, then the end is held at 0 temp.

Generalized Boundary Conditions Similarly, if heat flow occurs at the end x = L, then the boundary condition is as follows: where, again, h 2 is a nonzero proportionality factor

Initial Boundary Condition Finally, the temperature distribution at one fixed instant – usually taken at t = 0, takes the form: occurring throughout the bar

Generalizations Sometimes, the thermal conductivity, density, specific heat, or area may change as the axial position changes. The rate of heat transfer under such conditions at x=x 0 is now: The heat equation then becomes a partial differential equation in the form: or

Generalizations Other ways for heat to enter or leave a bar must also be taken into consideration. Assume G(x,t,u) is a rate per unit per time.  Source G(x,t,u) is added to the bar G(x,t,u) is positive, non-zero, linear, and u does not depend on t G(x,t,u) must be added to the left side of the heat equation yielding the following differential equation

Generalizations Similarly,  Sink G(x,t,u) is subtracted from the bar G(x,t,u) is positive, non-zero, linear, and u does not depend on t G(x,t,u) then under this sink condition takes the form:

Generalizations Putting the source and sink equations together in the heat equation yields which is commonly called the generalized heat conduction equation

Multi-dimensional space Now consider a bar in which the temperature is a function of more than just the axial x- direction. Then the heat conduction equation can then be written:  2-D:  3-D:

Example 1: Section 10.6, Problem 9 Let an aluminum rod of length 20 cm be initially at the uniform temperature 25  C. Suppose that at time t=0, the end x=0 is cooled to 0  C while the end x=20 is heated to 60  C, and both are thereafter maintained at those temperatures. Find the temperature distribution in the rod at any time t

Example 1: Section 10.6, Problem 9 Find the temperature distribution, u(x,t)  2 u xx =u t, 0<x<20, t<0 u(0,t)=0 u(20,t)=60, t<0 u(x,0)=25, 0<x<20 From the initial equation we find that: L=20, T 1 =0, T 2 =60, f(x)=25 We look up the Thermal Diffusivity of aluminum→  2 =0.86

Example 1: Section 10.6, Problem 9 Using Equations 16 and 17 found on page 614, we find that where

Example 1: Section 10.6, Problem 9 Evaluating c n, we find that

Example 1: Section 10.6, Problem 9 Now we can solve for u(x,t)

Example 1: Section 10.6, Problem 9

Derivation of the Wave Equation Applicable for: One space dimension, transverse vibrations on elastic string Endpoints at x = 0 and x = L along the x-axis Set in motion at t = 0 and then left undisturbed

Schematic of String in Tension

Equation Derivation Since there is no acceleration in the horizontal direction However the vertical components must satisfy where is the coordinate to the center of mass and the weight is neglected Replacing T with V the and rearranging the equation becomes

Derivation continued Letting, the equation becomes To express this in terms of only terms of u we note that The resulting equation in terms of u is and since H(t) is not dependant on x the resulting equation is

Derivation Continued For small motions of the string, it is approximated that using the substitution that the wave equation takes its customary form of

Wave Equation Generalizations The telegraph equation where c and k are nonnegative constants cu t arises from a viscous damping force ku arises from an elastic restoring force F(x,t) arises from an external force The differences between this telegraph equation and the customary wave equation are due to the consideration of internal elastic forces. This equation also governs flow of voltage or current in a transmission line, where the coefficients are related to the electrical parameters in the line.

Wave Equations in Additional Dimensions For a vibrating system with more than on significant space coordinate it may be necessary to consider the wave equation in more than one dimension. For two dimensions the wave equation becomes For three dimensions the wave equation becomes

Example 2: Section 10.7, Problem 6 Consider an elastic string of length L whose ends are held fixed. The string is set in motion from its equilibrium position with an initial velocity g(x). Let L=10 and a=1. Find the string displacement for any time t.

Example 2: Section 10.7, Problem 6 From equations 35 and 36 on page 631, we find that where

Example 2: Section 10.7, Problem 6 Solving for k n, we find:

Example 2: Section 10.7, Problem 6 Now we can solve for u(x,t)

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