Unsteady State Heat Conduction in a Bounded Solid How Does a Solid Sphere Cool?

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Louise Hunt
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1 Unstead State Heat Conduction in a Bounded Solid How Does a Solid Sphere Cool? We examined the cooling a sphere of radius R. Initiall the sphere is at a uniform temperature T 0. It is cooled b convection to an air stream at temperature T a. How long does it take to cool to T a? The answer is and was simple... an infinitel long time. In a sense because it is the answer equivalent to the solution of the Archimedean Paradox. If I walk half the distance to a wall, how man steps will I have to take to reach the wall? The answer is an infinite number! However, if I get within a millimeter of the wall, for intents and purposes, I am there. So if the temperature is within 1 % of the final temperature, it will have reached the final temperature. What temperature am I taking about...the center line temperature, the surface temperature, the average temperature??? We will choose the average temperature, θ. The first model we looked at should be valid for small Biot numbers θ = T T a T 0 T a = e ha ρc p V t = e 3Bix Fo To get this form we had to recognize that for spheres, A/V = 3/R. If θ = 0.01, then 3 Bi x Fo = ln (0.01), so that x Fo = 1.535/Bi At large Biot numbers, the suitable model was θ sph = e 9.87x Fo ChE 333 1

2 Now for this case the value of x Fo it takes to reach Θ = 0.01 is x Fo = A plot of the response is shown below 100 Time for a temperature drop of 99% 10 XFo Bi Some Dimensional Arguments At large Biot numbers, the dimensionless time is constant, that is, x Fo = 0.415, but x Fo = αt R 2 so that for two spheres one of size R 1 and another R 2, the ratio of the cooling times is t 1 t 2 = R 2 R 1 2 ChE 333 2

3 Heat Transfer in a Semi-Infinite region The questions we have posed thus far and the solutions have been reall onl applicable for Long times. That is, when the temperature field in the sphere, for example, has developed to the center of the sphere. But what happens at Short times? Consider a large planar solid whose extent (-direction) is ver large. What is the temperature histor of the slab if it is suddenl brought into contact with a fluid at temperature T a? The transient conduction equation is T t = α 2 T at t = 0, T = T 0 at = 0, T = T a as, T = T 0 Let s make the problem dimensionless. The temperature can be expressed as θ = T T a T 0 T a so that the problem reposed is θ t = α 2 θ θ = 1 at t = 0 θ = 0 at = 0 θ = 1 as ChE 333 3

4 Solution Let Θ = f(η(,t)) where η = c m t n, then we can introduce that into the differential equation. θ = dθ η t dη t = dη dθ η t = dθ dη cnm t n 1 θ = dθ η dη = dη dθ η = dη dθ cmm 1 t n 2 θ = dθ dη 2 η + d 2 θ dη 2 η 2 2 θ = dθ dη cm m 1 m 2 t n + d2 θ dη 2c2 m 2 2m 2 t 2n putting these derivatives into an equation we get dθ cn dη m t n 1 m 2 2m 2 t 2n We can divide b c m t n so that we obtain dθ nt 1 dη = α dθ dη cmm 1 t n + α dθ dη cm m 1 m 2 t n + d2 θ dη 2c2 = α dθ dη m 1 + α dθ dη m m d2 θ dη 2cm 2 2 Grouping we obtain a more compact form dθ nt 1 αm dη 1 α m m 1 2 = d2 θ dη 2 2cm2 Note that things simplif if we pick m = 1. A bit of exercise will show that the appropriate choice for η is η = 4αt ChE 333 4

5 Solution of the Differential Equation The equation becomes and ordinar differential equation The boundar conditions are 2η dθ dη + d2 θ dη 2 = 0 θ = 1 for η θ = 0 for η = 0 The solution we have seen is related to an error function defined as erf (x) = 1 π e t 2 dt 0 This solution affords us the opportunit to talk of an effective penetration depth, δ T, that is, the distance at which the dimensionless temperature goes from 0 to The solution for Θ is Θ = erf(η), so that for the penetration depth η θ = 0.99 = erf δ T 4αt It follows that δ T 4αt = 2 This means that if δ T is less than the thickness of the slab, it behaves as a semi-infinite region. ChE 333 5

6 Heat Conduction with a Convective Boundar Condition The boundar condition at the cooling surface can have a major effect on the process. The problem in this instance is posed as T t = α 2 T at t = 0, T = T 0 at = 0, T = h T T a as, T = T 0 The problem can again be solved using combination of variables and the same transformation as above to ield θ = erf 4t + exp + t erfc 4t + t where = h and t = h 2 αt ChE 333 6

7 Surface temperature of a Cooling Sheet Polethlene is extruded and coated onto an insulated substrate, moving at 20 cm/sec. The molten polmer is coated at a uniform temperature T 0 of 400 F. Cooling is achieved b blowing air at a temperature T a of 80 F. Earlier heat transfer studies determined that the heat transfer coefficient, h, is 0.08 cal/cm-sec- F. The coating thickness B is 0.1 cm. At what point downstream does the surface temperature, T(0) fall to 144 F? Data T 0 = 400 F h = 3.35 kw/m 2 - K B = 0.1 cm. T a = 80 F = 0.33 W/m- K α = m 2 /sec The Biot number can be estimated as: Bi = hb = = The dimensionless surface temperature ratiois θ s = T 0 T a T 0 T a = = 0.2 The Gurne-Lurie Chart 11.4c ields for Bi 10, the ratio of the surface temperature to the mid-plane temperature However, since θ θ = we can calculate the mid-plane temperature 1, 0 from the relation for θ s which is θ s = θ θ 1 0 = 0.2 θ 1 This gives a midplane-temperature of θ 0 1 = 0.2/0.15 > 1...Nonsense What s wrong??? We did a lot of things wrong. ChE 333 7

8 First of all the solution we used involved onl 1 term of an infinite series... θ 1 = A 1 e β 1 2 x Fo sin β 1 ξ = θ 1 0 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 4t + exp + t erfc 4t + t where = hb B and t = hb 2 αt B 2 Now for this case, = 0 and we can use figure We can determine that the value of t at which Θ = 0.2. We observe that t 1/2 = 2.65 and consequentl t = 7.65 Recall that ρc p = k/α = 2.5 MJ/m 2 - K. This leads to t = hb 2 αt B 2 = 7.65 = h 2 αt We calculate that the time passed is t = 0.52 seconds and since d = Vt, the distance is d = (20 cm/s) 0.52 sec = 10.4 cm. ChE 333 8

9 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 ln If we calculate the first three terms of the Fourier expansion, we obtain θ(1) = 0.178e 2.04x Fo e 18.5x Fo For Θ = 0.2, b trial and error, we obtain x Fo = If we calculate the time, we get 0.52 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 9

Heat Transfer in a Slab

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