sci.math.num-analysis SUNA40: Numerical Heat Exchange =============================== Wrap a control-volume around a couple of tubes. Set up the energy balances for this volume. Let the volume become "infinitesimally small", though remaining bigger than the distance between the tubes. Or throw away the integral signs after applying Gauss theorems. Or whatever. Then the following set of Partial Differential Equations may be inferred for Heat Exchange in a tube bundle: c.Gp.[ u.dTp/dr + v.dTp/dz ] + a.(Tp - Ts ) = 0 : primary sodium c.Gs.dTs/dz + a.(Ts - Tp ) = 0 : secondary sodium Where: c = heat capacity; G = mass-flow; T = temperature; (r,z) = coordinates; (u,v) = velocities; a = total heat transfer coefficient; p = primary, s = secondary sodium. Ideal internal Flow ------------------- Velocities (u,v) are calculated by assuming Ideal internal Flow. The flow field is invariant for scaling with a factor Gp (independent of the flow magnitude) and therefore will be normed in such a way that the absolute vaule | (u,v) | is unity (1) in the middle of the bundle. Now we have seen that the Neratoom heat exchanger is designed in such a way that the flow velocity in the middle of the bundle is the same as the flow velocity at the inlet perforation. So our Ideal Flow calculations, with u = -1 at the inlet opening, can immediately be used. BTW, boundary conditions and elements for problem "15" are defined in such a way that the number of unsquared equations _precisely_ equals the total number of unknowns. I forgot to mention this in a previous poster. There is an argument why ideal internal flow is litteraly IDEAL for calculating temperature fields in heat exchangers. The total power transferred is expressed by the volume integral at the left hand side, in the following nice application of Schwartz inequality: _ _ | (a.b)^2 <= /// | _ _ _ _ { ||| [ u.dTp/dr + v.dTp/dz ] dV }^2 <= | (a.a).(b.b) /// |__________________ /// /// { ||| [ u^2 + v^2 ] dV } . { ||| [ (dTp/dr)^2 + (dTp/dz)^2 ] dV } /// /// /// Or: ||| [ (dTp/dr)^2 + (dTp/dz)^2 ] dV = mean | temperature gradient |^2 >= /// /// >= [ transferred power ]^2 / ||| [ u^2 + v^2 ] dV /// Quoting without permission from Marvin Shinbrod "Lectures on fluid mechanics", Gordon and Breach 1973, chapter 3-5: " ... potential flow has the least kinetic energy of _any_ flow having the same normal velocity on the boundary and satis- fying the condition of conservation of mass". As a consequence, the denominator at the right hand side is minimal for Ideal Flow. If the power delivered by the apparatus is predescribed, then we may conclude: the mean temperature gradients are maximal for ideal internal flow. The calculations are thus _safe_, because they give rise to maximal temperature stresses with given operating conditions. Heat exchange constants ----------------------- The constants a in the above temperature equations are calculated as follows: a = N.a1 where N = number of tubes 1/a1 = 1/as + 1/aw + 1/ap where s,w,p = Secondary, Wall, Primary. Here the tube Wall constant is given by: aw = lamda.2.pi.1/log(du/di) ; lambda = heat transfer coefficient di, du = inner, outer tube wall diameter For thin walls this reduces to: aw = lamda * circumference / thickness as = 2.pi.di.alpha ; ap = 2.pi.du.alpha ; alpha = assumed constant So far so good for the bulk of the apparatus. Inlet conditions ---------------- The only proper Boundary Conditions for the temperature problem are given by: Tp = TpL at the primary inlet ; Ts = Ts0 at the secondary inlet Because the whole problem is linear in both Tp and Ts , the solutions simply _scale up_ with different such boundary conditions. To be precise, if Tp and Ts are solutions of the analytical model, then it is easily verified that also (a.Tp + b) and (a.Ts + b) are solutions of the model: by substitution. Now suppose that we have made a calculation for boundary conditions Tp1 , Ts1 and we want to know the outcome of a calculation for Tp2 , Ts2 then, assuming all other things constant: Tp2 = a.Tp1 + b ; Ts2 = a.Ts1 + b : two equations with two unknowns. Solving for a and b : a = (Tp2 - Ts2)/(Tp1 - Ts1) b = (Ts2.Tp1 - Ts1.Tp2)/(Tp1 - Ts1) In particular, if the heat exchanger problem is solved for _normed_ boundary conditions Tp1 = 1 and Ts1 = 0 , then a = (Tp2 - Ts2) and b = Ts2 . So: Tp'= (TpL - Ts0).Tp + Ts0 ; Ts'= (TpL - Ts0).Ts + Ts0 Where TpL and Ts0 are the true inlet temperatures, Tp and Ts are normed temperature fields, and Tp', Ts' are the true temperature fields. Conclusion: only the _normed_ temperature fields have to be calculated numerically. BTW. The difference TpL - Ts0 is called: temperature SPAN. It is the maximum possible temperature difference in a heat exchanger. Treatment of the other boundaries consists of special one-dimensional elements, or specializations of the 2-D case, based upon the following analysis: c.Gp.u.dTp/dr + a.(Tp - Ts ) = 0 : horizontal streamlines primary sodium c.Gs.dTs/dz + a.(Ts - Tp ) = 0 upper and lower tube plates c.Gp.v.dTp/dz + a.(Tp - Ts ) = 0 : vertical streamlines primary sodium c.Gs.dTs/dz + a.(Ts - Tp ) = 0 inner and outer shells Analytical elements ------------------- In addition, special analytical elements can be developed for the region in the middle of the bundle, where v = -1 . As follows: - c.Gp.dTp/dz + a.(Tp - Ts ) = 0 -> dTp/dz - a/(c.Gp).(Tp - Ts) = 0 c.Gs.dTs/dz + a.(Ts - Tp ) = 0 -> dTs/dz - a/(c.Gs).(Tp - Ts) = 0 ------------------------------ + ------------------------------- - - c.Gp.dTp/dz + c.Gs.dTs/dz = 0 : adding the former and ... substracting the latter: d(Tp - Ts)/dz - [a/(c.Gp) - a/(c.Gs)].(Tp - Ts) = 0 The first equation is integrated to: c.Gp.dTp = c.Gs.dTs c.Gp.(Tp2 - Tp1) = c.Gs.(Ts2 - Ts1) (a) And the second one: d(Tp - Ts)/(Tp - Ts) = a/c.(1/Gp - 1/Gs).dz Tp - Ts = C . exp [ a/c.(1/Gp - 1/Gs) ].z For z = 0 : Tp1 - Ts1 = C . Hence, if Gp <>0 and Gs <> 0 ... for z = L : Tp2 - Ts2 = (Tp1 - Ts1).exp[ a/c.(1/Gp - 1/Gs) ].L (b) The two equations (a) and (b) give an _exact_ solution for the analytical model of the mid-bundle region. With elements based upon it, the maximal coarse mesh can be adopted, which is of great computational advantage (especially with such a painfully slow XT type PC like mine). A special case occurs for Gp = Gs , giving: Tp2 - Tp1 = Ts2 - Ts1 (a) and essentially the same equation ........ : Tp2 - Ts2 = Tp1 - Ts1 (b) The two analytical equations are mutually dependent. Thus, an alternative must be found. Let's integrate the basic equations over the length of the element: (2) : z = L - c.Gp.(Tp2 - Tp1) \ / + a.| (Tp - Ts).dz = 0 - c.Gs.(Ts2 - Ts1) / / (1) : z = 0 The integral is worked out analytically with above theory, giving a term: a. (Tp1 - Ts1) / [ a/c.(1/Gp - 1/Gs) ] . { exp [ a/c.(1/Gp - 1/Gs).L ] - 1 } The special case for Gp = Gs = G is easily recognized now as a limit which is obtained by differentiating exp(h.L) to h = a/c.(1/Gp - 1/Gs) = 0 : Limit = a. (Tp1 - Ts1) . d/dh [ exp ( h.L ) ] = a. (Tp1 - Ts1) . L h=0 The end-result is: c.Gp.(Tp2 - Tp1) - a.L.(Tp1 - Ts1).TERM = 0 : equations now c.Gs.(Ts2 - Ts1) - a.L.(Tp1 - Ts1).TERM = 0 independent ! Define dimensionless quantities: P = (a.L)/(c.Gp) , S = (a.L)/(c.Gs) . Then: TERM = [ exp ( P - S ) - 1 ] / [ P - S ] forall P <> S . Special case if P = S ; then simply: TERM = 1 . Heat Balance ------------ The heat balance (a) is a thing that should be checked too for the problem as a whole, once it is solved: c.Gp.(TpL - Tp0) = c.Gs.(TsL - Ts0) Here: TpL = primary inlet temperature ; Tp0 = primary outlet temperature; TsL = secondary outlet temperature ; Ts0 = secondary inlet temperature . So the quotient of the two mass-flows is: Gp/Gs = (TsL - Ts0) / (TpL - Tp0) . Remark: the mass-flow rate Gp/Gs is also a dimensionless number (= P/Q) . To be continued ... - * Han de Bruijn; Applications&Graphics | "A little bit of Physics * No * TUD Computing Centre; P.O. Box 354 | would be NO idleness in * Oil * 2600 AJ Delft; The Netherlands. | Mathematics" (HdB). * for * E-mail: Han.deBruijn@RC.TUDelft.NL --| Fax: +31 15 78 37 87 ----* Blood