3
$\begingroup$

I'm developing a code to simulate thermal performance of automotive radiators.

My results are 40% under-predicted. But when I multiply them to 1.6, they match completely.

This is the graph that compares the results of program and experimental work.

Comparing experimental and numerical results

It should be noted, this number (1.6) differs when we use the other data sets.

Procedure of the program is as following :

  1. given air velocity, coolant flow rate, inlet air and coolant temperatures.

  2. assume outlet air and water temperatures (it should be a logical assumption).

  3. calculate air and fluid properties (using interpolation in existing tables based on mean temperatures)

  4. calculate heat transfer coefficient of water according to flow regime (using Petukhov or Gnielinski equations)

  5. calculate Colburn j factor (using Chang and Wang empirical correlation)

  6. calculate the total thermal resistance between air and coolant.

  7. calculate NTU and effectiveness

  8. Calculate heat rejection

  9. Calculate outlet temps and compare them with assumed temps. If it doesn't meet the specified error, go to step 3 and repeat calculations.


clc;clear all
% _____Jung Experiment_______
% Air Velocity m/s
    Va=4;
water_Flowrate=2e-3/3;% m3/s (40 L/min)
% Top tank and inlet air temperatures (deg C)
twih=85; t_ambient=25;
% Specifying the j-factor correlation (Chang & Wang)
cor='CW';
% Entering the Radiator characteristics
cas='C:\Users\Solid\Desktop\jung.xlsx';

% ___ Calculation ___
ITD=twih- t_ambient;
% Export Geometrical details from jung.xlsx file and calculate the needed
% Areas like AT(total heat transfer area etc).
[R,L,Afrt,Afrf,Ac,Afront,sigma,Af,Aw,AT...
    Lp,Fp, Fl, Ll, alpha, Fd, Tp, deltaf, Nlb,Td,LH,af,Dh,NT,kf...
    RH,RW,RD,N,Nct,Np,Nf,Yl,Dm,D_major,W,tt]=Area_Calc(cas);
% Radiator tubes are flat. So we need hydraulic diameter(dw). 
[dw,A_cross,Perimeter]=hydro2(Dm,L);
% ___ Pre_Procssing ___ 
t_top_tank=twih;
twi= t_top_tank;
tai = t_ambient;
%  defining an error;
error=.01;
% Assuming outlet temperatures of water(Two) and Air(Tao) logically.
Tao=50; Two=75; 
Y=1; 
% defining a counter;
ll=0;
        while Y
            Q=0;
        ll=ll+1;
%         At the end of the loop, calculated temperatures will be         compared
%         with the tepms of the past iteration.So we save the tepms of     the
% past iteration here.  
        Two1=Two; Tao1=Tao;
% Calculating the properties like viscosity(uw(coolant),ua(air)),Prandtel
% Number (Pra(air) ,Prw(coolant)) etc.
        [ uw,ua,rhow,rhoa,cw,ca,kw,ka,Prw,Pra,tam ]=...
            phyprop2(twi ,tai,Two,Tao);
%       Air side Reynolds number. ReLp is the reynolds number based on
% Louver pitch. 
        ReLp=rhoa*Va*Lp*1e-3/(ua);
        ReDh=rhoa*Va*Dh*1e-3/(ua);
%         Air mass flowrate
mair=Ac*rhoa*Va;

% --- Coolant side Reynolds number and mass flowrate ---
[Rew,mwater]=hydro(dw,A_cross,water_Flowrate,rhow,uw,...
    NT,'I');
% coolant heat transfer coff.
[hw]=hc2(Yl,dw,Rew,kw,Prw,1);
% Air heat transfer coff. and friction factor
% (friction factor(f) will be calculated but won't used in this program )
[ha,jcor,f]=ha2...
    (Lp,Fp,Fl, Ll, alpha,Fd, Tp, deltaf, Nlb,Td,LH,Dh,ua,rhoa,ca,Pra,...
    Va,cor,ReLp,ReDh,Dm,tt);
 % ----------- UA , NTU--------
 l=Fl*1e-3/2;
M=sqrt(2*ha/(kf*deltaf*1e-3));
% fin and Surface efficiency
nf=tanh(M*l)/(M*l);
no=1-Af*(1-nf)/(AT); Q=0;
% ----------Capacity Rate-----------
Ca=mair*ca; Cw=mwater*cw;

if min(Ca,Cw)==Ca
Cmin=Ca;
Cmax=Cw;
Q=1;
else
Cmin=Cw;
Cmax=Ca;
end
Cs=Cmin/Cmax;
% Calculation of overall heat transfer coefficient using thermal
% resistances between air and water(fouling and wall resistances are
% negligible)
UA=(( 1/(hw*Aw)+1/(no*ha*AT) ) )^-1;
NTU=UA/Cmin;
% -------------  E - NTU (calculation of effectiveness _epe_ using         existed 
% correlations) -----------------
epe=eps1(Q,N,Cs,NTU,Cmin,Ca);
% Air and water outlet temps
tao=tai+epe*Cmin*(twi-tai)/Ca;
two=twi-epe*Cmin*(twi-tai)/Cw;

Tao=tao;
Two=two;
% do these temps meet the errors ?
Y=abs(Tao-Tao1)/Tao>error2 ||...
    abs(Two-Two1)/Two>error2;
        end
%         Radiator Heat Rejection
        q=epe*Cmin*ITD;
        disp(q)

What's wrong?

$\endgroup$
  • 2
    $\begingroup$ Without seeing your mathematical model it would be impossible to determine what the issue is, short of guessing. If you run it on other, newer data points does it have a consistent systematic error? $\endgroup$ – wwarriner Jan 21 '16 at 2:21
  • $\begingroup$ I also just noticed step (2) says "assume..." Have you checked those assumptions? $\endgroup$ – wwarriner Jan 21 '16 at 2:21
  • 1
    $\begingroup$ $1.6\approx \frac{\pi}{2}$, coincidence? $\endgroup$ – Chris Mueller Jan 21 '16 at 12:57
  • $\begingroup$ How are you doing this without knowing the radiator characteristics? I don't see that anywhere, and I would expect it to be with step (1). Do you iterate at all? As @starrise points out, you assume output temperatures, and from there you calculate air and fluid properties?? It looks like you have a given starting point, you assume an end point, and then you base all of your calculations off of the assumed heat transfer and force the other values to meet that assumption. What are the calculations you're using? I would expect Q = hA($\Delta$T), but again, I don't see radiator specs anywhere. $\endgroup$ – Chuck Jan 21 '16 at 14:24
  • $\begingroup$ @Chuck 1- If I run it on other data points, it have a consistent systematic error. 2- 1.6= pi/2 is just a coincidence . When I use another data set, this number changes. 3- This is an iterative procedure, as it is mentioned above. 4- we use the radiator characteristics to calculate Colburn j-factor. Chang and Wang correlation uses these characteristics and gives Colburn j-factor. 5- We obtain the air and coolant properties using interpolation from the properties tables. $\endgroup$ – Alish Jan 30 '16 at 12:20
2
$\begingroup$

Either your definition of "completely" is a bit unusual, or there's something wrong with those charts, because in those charts, 1.6 x the model does not completely match any of the six observations you have (assuming the point values are at the centre of each shape).

Either way: your model outputs don't match your observations, with or without a 1.6 fudge factor.

So it's some combination of:

  • your assumptions don't match reality;
  • your inputs don't match reality;
  • your model doesn't match reality;
  • your observations don't match reality;
  • there are things outside the system boundary of your model that are crucial to the results, that you are not accounting for (this is kind of a subset of the first bullet point).

So, you need to go through, step-by-step, verifying or correcting each of those.

At a first guess: have you actually calibrated your model at any point: that is, taken a set of inputs that you can be absolutely sure of, and a set of observations that you can be absolutely sure of, and got a match. Or is what you're presenting to us, actually an uncalibrated model - in which case, why should it look like reality? The second-best way to tackle this: does your model reproduce the results of some other, calibrated model?

| improve this answer | |
$\endgroup$
  • $\begingroup$ I'm absolutely sure of the set of inputs and the set of observations . Because they've gotten from journal article. There are some journal articles that present a model and validate their model with experiment. But my model doesn't reproduce the results of any of them. $\endgroup$ – Alish Jan 22 '16 at 1:40
  • $\begingroup$ Cromwell's rule always applies, so you should never be absolutely sure of your inputs or your observations. And lots of things published in journal articles are wrong. $\endgroup$ – 410 gone Jan 22 '16 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.