Heat Transfer in a Slab Consider a large planar solid whose thickness (y-direction) is L. What is the temperature history of the slab if it is suddenly brought into contact with a fluid at temperature T? The transient conduction equation is T t = α 2 T y 2 at t =,T = T at y =, T = T 1 for t > at y = L, T = T 1 Let s make the problem dimensionless. θ = T T 1 T T 1 ; η = y L ; τ = αt L 2 The temperature can be expressed as so that the problem reposed is t = α 2 θ y 2 θ = 1 at τ = θ = at η = θ = as η = 1 ChE 333 1
How do we solve the equation? τ 2 θ η 2 = YG τ 2 YG η 2 = G dy dτ Yd2 G dη 2 = Suppose z has the form Θ = Y(τ)G(η) 1 dy Y dτ = 1 d 2 G = λ 2 G dη 2 dy dτ = λ2 Y ; d2 G dη 2 The equation is separable in the form = λ 2 G Integrating each of the equations we obtain Y(τ) = Ke λ2τ and G(η) = Asin(λη) +Bcos(λη) The solution for y(θ,η) has the form θ(η, τ) = Asin(λη) +Bcos(λη) e λ2 τ We can construct the exact solution using the boundary conditions θ(, τ) = Asin() +Bcos() e λ2τ = It follows that B must be if the condition is true for all θ > ChE 333 2
Now the other boundary condition θ(1, τ) = Asin(λ) e λ2τ = Now this is true for all τ > if and only if sin(λ) = but sin(λ) = only where λ = nπ where n =, 1, 2,... This means there are a countable infinity of solutions so that θ(η, τ) = Σ n= 1 e λ2τ A n sin(nπη) To obtain the coefficients A n, we need to use the initial condition. θ = 1 at τ < Σ n= 1 z(η, ) = A n sin(nπη) = 1 To determine the coefficients, we can use the orthogonality properties of the sine and cosine functions. (See Appendix) 1 1 sin(nπξ)sin(mπξ)dξ = for m n π for m = n ChE 333 3
We integrate 1 θ(η, )sin(mπη)dξ 1 = sin(mπη)dη You might remember that the first sine integral is non-zero if and only if n = m. Now the equation for A n is Σ n= 1 A n 1 sin(nπη)sin(mπη)dη 1 = sin(mπη)dη A n = 1 sin(nπη)dη = 1 sin 2 (nπη)dη nπ nπ sin(x)dx sin 2 (x)dx The result for the definite integrals follow from what I gave above. It follows that A n = 4 π 1 n n + 1 2 We saw earlier that the solution can be described as: ChE 333 4
θ(η, τ) = 4 π Σ n = 1 2n +1 e 2n + 1 2 π 2 τ sin( 2n + 1 πη) We have already examined how the sum converges. For τ >.2, only one term suffices to describe the solution. We can look at many different classes of problems. The general problem for transient heat transfer in a slab is one posed as T t = α 2 T y 2 at t =, T = T at y =, k T y = h T T 1 for t > at y = L 2, T y = The dimensionless form is: The solution is of the form = α t 2 θ y 2 θ = 1 at τ= = Bi θ at η = η η = as η = 1 θ(η, τ) = C n e ζ n 2 τ sin(ζ 2 n η) Σn = C n = 4 sin ζ n 2ζ n + sin 2 ζ n and ζ n tan ζ n = Bi ChE 333 5
Again the approximate solution is the one-term solution θ(η, τ) C 1 e ζ 1 2 τ sin(ζ 2 1 η) This argument is the same for any transient 1-dimensional heat transfer problems involving cylinders, planes or spheres. Examples Infinite Cylinder = 1 τ η η η η θ = 1 at τ= in η, 1 η = Bi θ at η = 1 η = at η = The solution is θ(η, τ) = C n e ζ n 2 τ J (ζ n η) Σn = C n = 2 ζ n J 1 ζ n 2 J 2 ζ n + J 1 2 ζ n and ζ n J 1 ζ n J ζ n = Bi An approximate one-term solution is θ(η, τ) = C 1 e ζ 1 2 τ J (ζ 1 η) Note that along the center line θ (, τ) = C 1 e ζ 1 2 τ ChE 333 6
So that the simpler representation is θ(η, τ) = θ J (ζ 1 η) ChE 333 7
Sphere τ = η 1 2 η η 2 η θ = 1 at τ= in η, 1 η = Bi θ at η = 1 η = at η = The solution is θ(η, τ) = Σn = C n e ζ n τsin(ζ 2 n η) ζ n η C n = 4 sin ζ n ζ n cos ζ n 2 ζ n + sin 2ζ n and 1 ζ n cot ζ n = Bi The Approximate Solution The center temperature is θ(η, τ) = C 1 e ζ 1 2 τsin(ζ 1 η) ζ 1 η θ (, τ) = C 1 e ζ 1 2 τ So that the temperature can be expressed as sin(ζ θ(η, τ) = θ 1 η) ζ 1 η ChE 333 8
Short Time Solutions Consider a large planar solid whose extent (y-direction) is very large. What is the temperature history of the slab if it is suddenly brought into contact with a fluid at temperature T a? The transient conduction equation is T t = α 2 T y 2 at t =, T = T at y =, k T y = h T T a T a as y, T = T Let s make the problem dimensionless. The temperature can be expressed as θ = T T a T T a so that the problem reposed is t = α 2 θ y 2 θ = 1 at t = y = Bi θ at y = θ = 1 as y ChE 333 9
Solutions We noted earlier that the equation can be solved by a combination of variables supposing that Τ = Τ(η,t) and we saw that the the appropriate choice for η is η = y 4αt The solutions for a number of different cases are as follows: Case 1 Constant Surface Temperature (T = T s ) T T s T i T s = erf y αt q " s t = k T s T i παt Case 2 Constant Surface Heat Flux (q s = q ) T T s = 2 q" k αt π e y 2 4αt q" y k erfc y 4αt Case 3 Surface Convection θ = erf y 4t + exp y + t erfc y 4t + t ChE 333 1
where y = hy k s and t = h k s 2 αt ChE 333 11
Surface temperature of a Cooling Sheet Polyethylene is extruded and coated onto an insulated substrate, moving at 2 cm/sec. The molten polymer is coated at a uniform temperature T of 4 F. Cooling is achieved by blowing air at a temperature T a of 8 F. Earlier heat transfer studies determined that the heat transfer coefficient, h, is.8 cal/cm-sec- F. The coating thickness B is.1 cm. At what point downstream does the surface temperature, T() fall to 144 F? Data T = 4 F h = 3.35 kw/m 2 - K B =.1 cm. T a = 8 F k s =.33 W/m- K α = 1.3 1-7 m 2 /sec The Biot number can be estimated as: Bi = hb k s = 335.1.33 = 1.15 The dimensionless surface temperature ratio is θ s = T T a T T a = 144 8 4 8 =.2 The Gurney-Lurie Chart 11.4c yields for Bi 1, the ratio of the surface temperature to the mid-plane temperature However, since θ θ =.15 we can calculate the mid-plane temperature 1, from the relation for θ s which is θ s = θ θ 1 =.2 θ 1 This gives a midplane-temperature of θ 1 =.2/.15 > 1...Nonsense What s wrong??? We did a lot of things wrong. ChE 333 12
First of all the solution we used involved only 1 term of an infinite series... θ 1 = A 1 e β 1 2 x Fo sin β 1 ξ = θ 1 sin β 1 ξ We also get into trouble if we use such an equation for a short time solution. Therefore avoid the charts for small x Fo and large Biot numbers. The short time solution we presented in the last lecture had the form. θ = erf y 4t + exp y + t erfc y 4t + t where y = hb k s y B and t = hb k s 2 αt B 2 Now for this case, y = and we can use figure 11.3.2. We can determine that the value of t at which Θ =.2. We observe that t 1/2 = 2.65 and consequently t = 7.65 Recall that ρc p = k/α = 2.5 MJ/m 2 - K. This leads to t = hb k s 2 αt B 2 = 7.65 = h k s 2 αt We calculate that the time passed is t =.52 seconds and since d = Vt, the distance is d = (2 cm/s).52 sec = 1.4 cm. ChE 333 13
An alternative method We can use the complete Fourier expansion, not just one term. θ = Σn =1 4 sin λ n cos λ 2λ n + sin 2λ n ξ e λ 2 nx Fo n λ n tan λ n = Bi The first set of eigenvalues are n λ n 1 1.429 2 4.36 3 7.228 If we calculate the first three terms of the Fourier expansion, we obtain θ(1) =.178e 2.4x Fo +.155e 18.5x Fo For Θ =.2, by trial and error, we obtain x Fo =.68. If we calculate the time, we get.52 sec. The same as the short time solution. This allows us a measure of short time. as for a slab 4 αt B 2 1 or x Fo 1 16 ChE 333 14