Unsteady State Heat Conduction in a Bounded Solid Consider a sphere of radius R. Initially the sphere is at a uniform temperature T. It is cooled by convection to an air stream at temperature T a. What do we do to describe the conduction and the temperature distribution within the sphere? You will recall that in the last lecture the uniform temperature approximation was quite unrealistic when it came to the time scale for cooling. There are a number of methods that we can use to attack this problem. The first is to do a shell balance on a differential shell within the sphere. That exercise leads to 4 r q r r 4 r q r r+ r = t ρu4r r If we divide both sides by 4r r, we obtain r q r r r q r r+ r r = r t ρu Recognizing that for a solid, C p C v and if the density is constant, we observe that r r q r = t ρu = ρ U T = ρc T T p t p t so that r r q r = ρc p T t At this point we should introduce Fourier s law of heat conduction ChE 333 - Lecture 7` Spring
q r = k T s The time-dependent equation for heat conduction becomes ρc T p = r t r k T s We can define the Fourier Diffusivity or thermal diffusivity as α = k s ρc p The conduction equation becomes mathematically equivalent to the diffusion equation T = α r t r T Boundary and Initial Conditions The problem becomes defined only when the initial conditions and the boundary conditions are specified. For t <, we have T = T in r ε (,R) For r =, the temperature field is bounded for all t > or T = (symmetry) For r = R, we have Newton s Law of Cooling, that is q r = k s T = h T T a ChE 333 - Lecture 7` Spring
The Dimensionless Description My passion for a dimensionless problem exists because I m fundamentally lazy and I don t want to solve a problem each time I formulate it. Let s put the equations in dimensionless form θ = T T a T T a ; ξ = r R X Fo = αt ; Bi = hr R k s Applying these definitions to the dimensionless conduction equation and its boundary and initial conditions, we obtain θ = ξ θ x Fo ξ ξ ξ θ = at x Fo < ; θ ξ = at ξ = θ ξ = Bi θ at ξ = It follows that Θ = f(ξ, x F, Bi). The solution procedure then uses a simple transformation z = ξ θ to obtain the form of the problem that we have already seen and solved in gory detail. The most common problem is when the Bi ->, then Θ = f(ξ, x F ) since the boundary condition 3 becomes Θ = ChE 333 - Lecture 7` 3 Spring
The Adimensional Problem Let's try to solve the problem by first making the equation adimensional. Define the following: z = ξ θ ; ξ = r R ; X Fo = α t R The differential equation and the initial and boundary conditions become: z z = X Fo ξ z = ξ at x Fo < ; z = at ξ = z ξ = Bi + z at ξ = The equation and its conditions are dimensionless and parameter free. Next Question How do we solve the equation? Suppose z has the form z = Y(X Fo )G(ξ) z z X Fo ξ = YG X Fo YG ξ = G dy dx Fo Y d G dξ = ChE 333 - Lecture 7` 4 Spring
The equation is separable in the form Y dy = d G = λ dx Fo G dξ dy = λ Y ; d G dx Fo dξ = λ G Integrating each of the equations we obtain Y(X Fo ) = Ke λ X Fo and G(ξ) = Asin(λξ) +Bcos(λξ) The solution for y(θ,η) has the form z(ξ, X Fo ) = Asin(λξ) +Bcos(λξ) e λ X Fo We can construct the exact solution using the boundary conditions z(,x Fo ) = Asin() +Bcos() e λ X Fo = It follows that B must be if the condition is true for all θ > Now the other boundary condition z(,x Fo ) = Asin(λ) e λ X Fo = Now this is true for all X Fo > if and only if sin(λ) = but sin(λ) = only where λ = n where n =,,,... ChE 333 - Lecture 7` 5 Spring
This means there are a countable infinity of solutions so that n= z(ξ, X Fo ) = e λ X Fo A n sin(nξ) To obtain the coefficients A n, we need to use the initial condition. z = ξ at x Fo < n= z(ξ, ) = A n sin(nξ) = ξ To determine the coefficients, we can use the orthogonality properties of the sine and cosine functions. (See Appendix) sin(nξ)sin(mξ)dξ = for m n for m = n ChE 333 - Lecture 7` 6 Spring
We integrate z(ξ,)sin(mξ)dξ = ξsin(mξ)dξ n= A n sin(nξ)sin(mξ)dξ = ξsin(mξ)dξ You might remember that the first sine integral is non-zero if and only if n = m. Now the equation for A n is A n = ξsin(nξ)dξ = n sin (nξ)dξ n n xsin(x)dx sin (x)dx The result for the definite integrals follow from what I give above. It follows that A n = 4 so that the solution can be described as : m z(ξ, X Fo ) = 4 n= n n e n X Fo sin(nξ) ChE 333 - Lecture 7` 7 Spring
A LITTLE EXERCISE Prove that the following is equivalent z(ξ, X Fo ) = 4 n= n + e n + X Fosin( n + ξ) A Return to the Original Problem How long does it take to get a fully developed temperatrure field in the sphere? Recall the initial definitions θ = T T a T T a ; ξ = r R X Fo = αt ; Bi = hr R k s When θ = everywhere, then the temperature profile is developed. The solution we saw was θ(ξ,x Fo ) = z(ξ,x Fo) ξ θ(ξ,x Fo ) = 4 ξ n= n + e n + X Fosin( n + ξ) ChE 333 - Lecture 7` 8 Spring
Now regardless of location in the sphere, θ is small when ξ is large. Now how large? The first term in the series is when n =, the second is when n = 3, etc. The first term is much greater than the subsequent one so that so as θ y(η, θ) 4 e θ sin(η) then X Fo z(ξ,x Fo) z(ξ,) e X Fo These solutions can be compared graphically. Temperature Response T e m p e r a t u r e.9.8.7.6.5.4.3...5.5 Dimensionless time It is apparent that by the time X Fo =.5, θ =., that is % of the original value. I ll presume this is long enough. ChE 333 - Lecture 7` 9 Spring
You should notice though that on a semi-log plot the data are almost a straight line. What does this suggest? Relative Temperature Plot......4.6.8 Time ChE 333 - Lecture 7` Spring
Appendix Problem --- evaluate A n The integrals in the definition of A n can be proven simply by noting that e ix = cos(x) + i sin(x) and using it to prove a number of identities involving sines and cosines. The result is a simple relation for the Fourier coefficients. The results are cos(nx)sin(mx)dx = for all m, n cos(nx)cos(mx)dx = for m n for m=n > for m=n = sin(nx)sin(mx)dx = for m=n > for m n xsin(mx)dx = m m It shows that sin(x) and cos(x) are orthogonal functions. Some algebra will get you the result for A n. A n = 4 m ChE 333 - Lecture 7` Spring