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October 9, 2018, 09:10 
Pure Diffusion  Energy Balance

#1 
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Khan
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I try to model pure diffusion case, considered with solving of the onedimensional convectiondiffusion equation (considering no flow, v=0 so no convection but only diffusion!). The Matlab code is given below. Accordingly, the initial condition is with 'the top water mass of the tank is at 90 °C while the bottom water mass of the tank is at 40 °C' fiftyfifty. You can see the temperature profile changing through time, below (you can see that stratification expands with time):
The problem is that when I calculate the energy content/enthalpy through the tank height at each time step; I found that the overall tank enthalpy increases through time by only the diffusion effect. The energy content shall stay at the same level through time since there is only diffusion (no heat loss or gain considered). The calculation procedure is as the sum of all node enthalpies (each node enthalpy calculated by node mass x specific enthalpy of the node temperature) at each time step. Would you please tell me why this energy content (enthalpy) increase occurs with time? Can the reason be the finite difference method (Shall I accept such error, the enthalpy increase, due to numerical matters?)? Code:
function [T,H] = CDR_FTCS_Diffusion(t_final,n) %% 1D ConvectionDiffusionReaction (CDR) equation  (ForwardTime CentralSpace) % solves a d2u/dx2 + b du/dx + c u = du/dt %% INPUT % t_final : final time [s] % n : Node number [] %% OUTPUT % T : Temperature Matrix [degC] % H : Enthalpy Matrix [kJ] if mod(n,2)==0 % Checkinf if node number is even or not! error('Node number  n  must be odd number  Check function input!') end %% INPUT % Tank Dimensions height=1.05; % [m] Tank height diameter=0.45; % [m] Tank diameter % v=5.2397e4; % [m/s] Water velocity dx=height/n; % [m] mesh size dt=1; % [s] time step maxk=round(t_final/dt,0); % Number of time steps % Constant CDR coefficients  a d2u/dx2 + b du/dx + c u = du/dt % a_const=0.15e6; % [m2/s]Thermal diffusivity b=0;%v; % [m/s] Water velocity % c=0; %!!! Coefficient for heat loss calculation %% Initial condition T=zeros(n+1,maxk); % Case Half 90 degC Half 40 degC T(1:(n+1)/2,1)=90; T((n+1)/2+1:end,1)=40; %% Fasten Your Seat Belts  Calculation of Temperature Profile at each time step for j=2:maxk % Time Loop Tpre=T(:,j1); a(:,1)=2.16308E12.*Tpre.^2 + 5.67273E10.*Tpre + 1.32439E07; % [m2/s] Thermal diffusivity for i=2:n % Space Loop T(i,j)=Tpre(i1)*(a(i+1)*dt/(4*dx^2)+a(i)*dt/dx^2+a(i1)*dt/(4*dx^2)b*dt/(2*dx))... +Tpre(i)*(12*a(i)/(dx^2)*dt)... +Tpre(i+1)*((a(i+1)a(i1))/(4*dx^2)+a(i)/(dx^2)+b/(2*dx))*dt; end T(n+1,j)=T(n,j); % Neumann boundary condition with backward Euler T(1,j)=T(2,j); end %% Calculate Energy Content for each time step Tt=T; % Tt(1,:)=[]; % Tt(end,:)=[]; rhoNode=3.16677E03.*Tt.^2  1.22228E01.*Tt + 1.00199E+03; % [kg/m3] Density hlTNode= 4.18601E+00.*Tt + 1.32103E01; % [kJ/kg] Specific Enthalpy H_node=(pi*diameter^2/4)*dx.*rhoNode.*hlTNode; % [kJ] Enthalpy of each Node H=sum(H_node); % [kJ] Total enthalpy at each time step end 

October 9, 2018, 11:59 

#2 
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Filippo Maria Denaro
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I do not consider the matlab code now but start the question showing all the equations of your model that you want to solve. Describe the PDE and the discretization.


October 10, 2018, 07:27 

#3 
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Khan
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October 10, 2018, 11:19 

#4 
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Filippo Maria Denaro
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Why do you discretize this way the spatial derivative?
It is simply (uniform mesh) (T(i1)2*T(i)+T(i+1))/h^2 

October 10, 2018, 12:05 

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Khan
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Quote:
I carried out the same simulation with CrankNicolson scheme (this time no mixing characterization so alpha is constant). Still, I have this increase even with CrankNicolson. I need to indicate that the increase ratio at pure diffusion case (both by CrankNicolson and FTCS) is at around 0.6% (after some reasonably large time). Is this increase reasonable when using finite difference methods? 

October 10, 2018, 12:16 

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Filippo Maria Denaro
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well, the PDE you wrote is not congruent to what you are doing.... If alpha is variable, you wrote a wrong PDE since you should write d/dx(alpha(x,t)*dT/dx) A general convectiondiffusion equation would write in divergence form as dT/dt + d/dx (u*T  alpha(x,t)*dT/dx)=0 If you discretize on a compact stencil your formula is over 2h not over 4h 

October 10, 2018, 12:57 

#7  
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Khan
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Quote:
and the relative error (still there is this increase, killing me ): Code:
function [T,H] = CDR_FTCS_Simple_Diffusion(t_final,n) %% 1D ConvectionDiffusionReaction (CDR) equation  (ForwardTime CentralSpace) % solves a d2u/dx2 + b du/dx + c u = du/dt tStart = tic; %% INPUT % Tank Dimensions height=1.05; % [m] Tank height diameter=0.45; % [m] Tank diameter % v=5.2397e4; % [m/s] Water velocity dx=height/n; % [m] mesh size dt=1; % [s] time step maxk=round(t_final/dt,0); % Number of time steps % Constant CDR coefficients  a d2u/dx2 + b du/dx + c u = du/dt a=0.15e6; % [m2/s]Thermal diffusivity % b=0;%v; % [m/s] Water velocity % c=0; %!!! Coefficient for heat loss calculation r=a*dt/(dx^2); %% Initial condition  Tank water at 40 ?C T=zeros(n+1,maxk); % Case Half 90 degC Half 40 degC T(1:(n+1)/2,1)=90; T((n+1)/2+1:end,1)=40; %% Fasten Your Seat Belts for j=2:maxk % Time Loop Tpre=T(:,j1); for i=2:n % Space Loop T(i,j)=r*Tpre(i1)+(12*r)*Tpre(i)+r*Tpre(i+1); end T(n+1,j)=T(n,j); % Neumann boundary condition with backward Euler T(1,j)=T(2,j); end tElapsed = toc(tStart); SimulationTime=sprintf('Simulation time is %f [s]',tElapsed) %% Calculate Energy Content for each time step Tt=T; % First Node T_f=Tt(1,:); rho_f=3.16677E03.*T_f.^2  1.22228E01.*T_f + 1.00199E+03; % [kg/m3] Density hlT_f= 4.18601E+00.*T_f + 1.32103E01; % [kJ/kg] Specific Enthalpy H_f=((pi*diameter^2/4)*dx.*rho_f.*hlT_f)./2; % End Node T_e=Tt(end,:); rho_e=3.16677E03.*T_e.^2  1.22228E01.*T_e + 1.00199E+03; % [kg/m3] Density hlT_e= 4.18601E+00.*T_e + 1.32103E01; % [kJ/kg] Specific Enthalpy H_e=((pi*diameter^2/4)*dx.*rho_e.*hlT_e)./2; % Inner nodes Tt(1,:)=[]; Tt(end,:)=[]; rhoNode=3.16677E03.*Tt.^2  1.22228E01.*Tt + 1.00199E+03; % [kg/m3] Density hlTNode= 4.18601E+00.*Tt + 1.32103E01; % [kJ/kg] Specific Enthalpy H_node=(pi*diameter^2/4)*dx.*rhoNode.*hlTNode; % [kJ] Enthalpy of each Node H=sum(H_node)+H_e+H_f; % [kJ] Total enthalpy at each time step end 

October 10, 2018, 13:37 

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It seems that you change the density with temperature.
Neither your PDE nor your energy integration is taking that into account. 

October 10, 2018, 13:45 

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Filippo Maria Denaro
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How do you compute the curve? Be careful, your equation describes a pure diffusion problem but, depending on the BCs., you could be prescribing an entering heat flux!
What are exactly your BCs.? 

October 11, 2018, 05:13 

#10 
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Khan
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How shall I consider the density while calculating the energy content/enthalpy? Shall I assume a constant value? I just assigned 1000 kg/m3 still the curve has an increasing trend (the increasing rate changed under constant density).


October 11, 2018, 05:15 

#11  
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Khan
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Quote:
Code:
T(n+1,j)=T(n,j); % Neumann boundary condition with backward Euler T(1,j)=T(2,j); 

October 11, 2018, 05:35 

#12  
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Filippo Maria Denaro
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Quote:
Compute the integral of the temperature over the domain 

October 11, 2018, 06:07 

#13  
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Khan
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Quote:
(ii) and (iii) Can you please give more details about these points? I didn't get the idea. 

October 11, 2018, 07:52 

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Filippo Maria Denaro
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Let us start first by a theoretical result: the equation can be integrated over the total volume and can be rewritten as d/dt Int[V] T dx = alpha [ dT/dxBC_right dT/dxBC_left] = 0 Therefore the solution must ensure Int[V] T dx= constant, being such constant equal to that at the initial condition. If you adopt a FV method, the equation above can be used to define the flux function alpha*dT/dx on each section. The "telescopic" property ensures that, numerically, the temperature is conserved, depending only on the values of the flux at the boundaries. Therefore, the numerical check requires only to compute the difference between Int[V] T dx at any time and the value computed at t=0. Ideally such difference is zero, check if numerically is of the order of the adopted (single/double) precision. 

October 11, 2018, 10:51 

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Khan
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Quote:


Tags 
convectiondiffusion, enthalpy, finite difference, matlab, stratification 
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