sci.math.numanalysis
SUNA40: Numerical Heat Exchange
===============================
Wrap a controlvolume 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 = massflow; 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 35: " ... 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 onedimensional elements,
or specializations of the 2D 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 midbundle 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 endresult 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 massflows is: Gp/Gs = (TsL  Ts0) / (TpL  Tp0) .
Remark: the massflow 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
* Email: Han.deBruijn@RC.TUDelft.NL  Fax: +31 15 78 37 87 * Blood