Convective Heat Transfer

Transcription

1 Convective Heat Tranfer Example 1. Melt Spinning of Polymer fiber 2. Heat tranfer in a Condener 3. Temperature control of a Re-entry vehicle Fiber pinning The fiber pinning proce preent a unique engineering problem, primarily due to the effect of hape variation, heat and poibly the vicoelatic behavior of the material (polymer for example) typically ued. Thi become evident when the deign of the pinneret geometry i needed to produce a pecified fiber ize and hape. Determining the proper die geometry given the deired final fiber hape i further complicated by the heat and vicoelatic effect. In addition, ince the fiber i pulled from the pinneret, the final dimenion of the fiber are difficult to determine. The effect of vicou heating and air cooling mut be monitored to enure that the material doe not degrade becaue of extreme local temperature, often difficult to meaure becaue of the mall ize. The tree and deformation of the material mut alo be predicted to avoid the fiber from breaking. All thee effect complicate the deign of the fiber pinning proce. ChE 333 1

2 Definition of a Heat Tranfer Coefficient For heat tranfer in conducting ytem, we have een that we can expre the heat flux acro a urface S a = k T n We have ued a imilar repreentation to develop detailed decription of ma and tranfer in a number of ituation where the phyic or chemitry i well-undertood, e.g., permeation through a membrane, heat tranfer to a phere. We howed that we could get a decription of the macrocopic tranfer acro an interface by the ue of a Heat Tranfer Coefficient. = h T T b We do nor alway have uch a good model or undertanding. The are other equivalent phyical ituation, e.g., turbulent flow in a pipe. There we ue a meaure of the frictional lo in the pipe a a momentum tranfer coefficient. The dimenionle form wa the Friction Factor. The dimenionle ma tranfer coefficient i the Nuelt Number. Nu = hl k ChE 333 2

3 Method of Analyi 1. Detailed Solution of the Conervation Law 2. Approximate Analyi 3. Dimenional Analyi 4. Empirical Correlation of Data We have een everal example of Detailed Solution and we have done ome Approximate Analyi, e.g., ma tranfer to or from a flowing film, heat tranfer from a olid phere. What we did wa to tranform the exact problem into a impler more olvable one uing mathematical analyi. What we do in the next few lecture i examine in greater detail the lat three method a tool to analyze heat tranfer and to deign procee. Approximate Analyi and Film Theory Film Theory i the implet and oldet approach in the ue of ma tranfer coefficient and in their prediction. The Theory i attributed to Nernt. Examine the neighborhood of the phae boundary. We aume that the flow field conit of two region, a uniform region in the bulk of the fluid far from the urface and a region in the vicinity of the boundary where vicoity dominate (ince there i no lip at the boundary). ChE 333 3

4 Film Model The film model preume that the velocity field i linearized in ome ene near the boundary o that τ = µ v x y S µ U δ = F A Thi mean that the film ha a thickne δ µ U τ But recall the definition of the friction factor f τ 1 2 ρu 2 If we introduce that notion into our analyi δ µ U = 2µ 1 fρu 2 fρu2 Not urpriingly we can relate the fractional layer thickne to the Reynold number δ L = 2µ fρul = 2 f Re f A Re 1 4 ChE 333 4

5 Dimenionle Heat Tranfer Coefficient We defined the Heat Tranfer Coefficient, h, by = k T n = h T T b o that the Nuelt number i given a Nu L = k T T b = hl k Oberve that Nu L = hl k = L δ * = L δ o that δ δ * = 1 2 f Re δ δ * Nu L = 1 2 f Re δ δ * δ δ * = g(pr) Pr 1/3 Then we oberve a relation rather like the one we calculated in our more detailed model. Nu L = 1 2 f ReSc1/3 ChE 333 5

6 The Chilton-Colburn Analogy In the 1930, baed on the Nernt Film Theory, two dupont reearcher propoed an analogy between heat tranfer (and we have een, ma tranfer) and momentum tranfer. They defined a dimenionle number termed a j-factor, j H. H j = For ma tranfer, the relation wa j D Simple film theory, then, predict that Nu L RePr 1/ 3 = f 2 Sh L ReSc 1/3 = f 2 Nu L = 1 2 f RePr1/3 = Re 0.25 RePr 1 /3 or implified Nu L = 0.04 Re 0.75 Pr 1/3 ChE 333 6