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Using ODE45 to solve 2 dependent ODE: dX/dt (mass transfer) and dT/dt (heat transfer) of a particle during fluidized bed drying process.

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jean dnc
jean dnc on 22 Oct 2017
Commented: jean dnc on 23 Oct 2017
My problem is using ODE45 to solve simultaneously the two differential equations of heat and mass transfer of a particle during fluidized bed drying process. ODE1: Equation of mass transfer:
dX/dt = -hm*A*(Ye-Yinf)*Rhoda/M
ODE2: Equation of heat transfer:
dT/dt = (h*A*(Tinf-T)+M*(lath-Cpw*T)*(dX/dt))/(M*(Cpw*X+Cps))
where:
M = 0.05; % weight of dry matter of particle, [kg]
hm = 0.02; % velocity of surrounding air, [m/s]
A = 1e-4; % surface area of particle, [m2]
Ye = 0.3; % equilibrium moisture content of air at surface of particle, based on dry air, [kg H2O/kg dry air]
Yinf = 0.05; % moisture content of air based on dry air, [kg H2O/kg dry air]
Rhoda = 1.2; % density of air, [kg/m3]
h = 350; % convective heat transfer coefficient, [W/m2.K]
Tinf = 80; % temperature of air using for drying, [degC]
lath = 2260000, % latent heat of water, [J/kg]
Cpw = 4200; % specific heat of water, [J/kg.K]
Cps = 1700; % specific heat of air, [J/kg.K]
X0 = 0.5; % initial moisture content of particle, in dry basis, fraction
T0 = 20; % initial temperature of particle, [degC]

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Answers (1)

Mischa Kim
Mischa Kim on 22 Oct 2017
jean, check out this answer.

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jean dnc
jean dnc on 23 Oct 2017
% M = 0.05; % Weight of dry matter of the particle, [kg of dry matter]
hm = 0.02; % Velocity of air surrounding the particle, [m/s]
A = 1e-4; % Surface area of the particle, [m2]
Ye = 0.3; % Equilibrium moisture content of air at the surface of the
% particle, based on dry air, [kg H2O/kg dry air]
Yinf = 0.05; % Moisture content of air based on dry air, [kg H2O/kg dry air]
rhoda = 1.2; % Density of air, [kg/m3]
h = 350; % Convective heat transfer coefficient, [W/m2.K]
Tinf = 80; % Temperature of air using for drying, [degC]
lath = 2260000; % Latent heat of water, [J/kg]
Cpw = 4200; % Specific heat of water, [J/kg.K]
Cps = 1700; % Specific heat of solid substance of particle, [J/kg.K]
X0 = 0.5; % Initial moisture content of particle, in dry basis,
%, [kg H2O/kg dry matter]
T0 = 20; % Initial temperature of the particle, [degC]
t0 = 0; % Initial moment of simulation duration, [s]
te = 500; % End moment of simulation duration, [s]
tint = 20; % Time step, [s]
%DYNAMIC
C1 = -hm*A*rhoda*(Ye-Yinf)/M; %Calculate dX/dt = C1 = constant
F = @(t,y) [C1; (h*A*(Tinf-y(2))+M*(lath-Cpw*y(2))*C1)/...
(M*(Cpw*y(1)+Cps))];
[t,y] = ode45(F, [t0:tint:te], [X0 T0]);
%PLOT RESULTS
title('Moiture Content & Temperature vs. Drying Time')
yyaxis left % y-axis on the left indicates Moiture Content
plot(t,y(:,1),'-*');
xlabel('Time [s]');
ylabel('Moisture content, [kg H2O/kg dry matter]');
yyaxis right % y-axis on the right indicates Partilce Temperature
plot(t,y(:,2),'->');
xlabel('Time [s]');
ylabel('Particle Temperature, [degC]');

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