## Optimization: Need to to optimize qr

Asked by Devdatt Thengdi

### Devdatt Thengdi (view profile)

on 6 Mar 2018
Latest activity Answered by Xavier

### Xavier (view profile)

on 13 Mar 2018
I need to optimize qr wrt to do, L, Tg, nb, uo. Do I need to substitute all variables into the formula for qr or is there any other way to solve it?
%Fuel flowrate
Ti = 251.9 + 273.15; %Fluid in degK
T0 = 386.1 + 273.15; %Fluid/gas out degK
cpL = 0.32; %Cp fluid KJ/kgdegK
flowL = 500; %flowrate fluid kg/s
THD = flowL*cpL*(T0 - Ti);%Total Heat Duty
Ef = 0.9320; %Fuel efficiency
qfuel = THD/Ef;%Heat released by fuel Kw
LHV = 41.05; %Fuel value KJ/kg
mfuel = qfuel/LHV;
%absorptivity
hp = 6.626*(10^-34); %planck's constant m2kg/s
cl = 3*(10^8); %speed of light m2/s
lamda = hp*cl/Qfuel; %wavelength
Avog = 6.023*10^24; %Avogadro's number
x1 = 0.9;
M1 = 16;
x2 = 0.1;
M2 = 29;
fatmass = x1*M1 + x2*M2; %fuel molecular mass
nA = mfuel*Avog*fatmass; %number of atoms
pi = 22/7;
alpha = 4*pi*nA/lamda;
%Exchange factor
x = pi*d0/2; %ctc
F = (sqrt(x^2 - 4) - x + 2*sinh(2/x))/2*pi;
%Heat transfer coefficient
k = 0.25*10^-3; %Thermal conductivity
rof = 1320; %density
Gm = 920880; %mass flowrate
mu = 130; %Viscosity Pa-s
A0 = pi*d0*d0/4;
h0 = (d0/k)*(0.023*((Gm*d0/mu*A0)^0.8)*((cpL*mu/k))^0.3);
%Tube wall temperature
Tref = 41; %Ambient
Tw = Tref + (Ti - T0)/L;
%Air
ea = 0.15; %excess air
cpair = 0.171; %Kcal/kgdegK
Tairin = 114;
atfr = 1.344; %in notebook
mair = mfuel*atfr*(1+ea);
%Flame calculation
cpsteam = 4.1855;
cpco2 = 0.918;
cpn2 = 1.044;
xco2 = 0.054;
xsteam = 0.034; %Total should be 8.8
xn2 = 0.66;
cpgas = xsteam*cpsteam + xco2*cpco2 + xn2*cpn2;
kgas = 0.25; %Thermal conductivity
mgas = mfuel + mair;
Lf = 0.0042*(qfuel^0.478); %flame length
tf = sqrt(pi*uo*Lf*cpgas/mgas*kgas); %flame size
db = 1.5*tf; %Burner tile diameter
dbc = db/(sin(180/nb) + (pi/nb)); %Burner circle diameter
%Cold plane area
cbt = (qfuel*1.055/4*10^6) + 1.5;
dtc = cbt + dbc; %tube circle diameter
ntube = pi*dtc/x;
Ar = L*ntubes*x;
sig = 5.67*10^-8; %Stefan-Boltzmann constant
qr = sig*alpha*F*(Tg^4 - Tw^4) + h0*(Tg - Tw);
%Heat leaving
Qr = qr * Ar; %total radiant heat absorbed
qair = mair*cpair*(Tairin - Tref);
qloss = 0.05*qfuel;
qout = qfuel + qair - (Qr + qloss);

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Devdatt Thengdi

### Devdatt Thengdi (view profile)

on 7 Mar 2018
Sorry for the ambiguity. I am designing a fired heater. Thus, I need to reduce qout(heat leaving the heater) i.e. I need to increase Qr which is dependent upon qr(flux) and Ar(area) which in turn comes to a conclusion that we need to maximize qr(32000<qr<45000) as a function of do, L, Tg, nb, uo. (take +infinity range for these variables) I've a;so, attached the flowsheet for you to understand the process.
Devdatt Thengdi

### Devdatt Thengdi (view profile)

on 7 Mar 2018
Anyway, I've made progress. I changed the target variable to qout. I am getting 'Not enough input arguments' as result. Any thoughts?
Function:
function qout = Lobo(d0, L, Tg, nb, uo)
clc
%Fuel flowrate
Ti = 251.9 + 273.15; %Fluid in degK
T0 = 386.1 + 273.15; %Fluid/gas out degK
cpL = 0.32; %Cp fluid KJ/kgdegK
flowL = 500; %flowrate fluid kg/s
THD = flowL.*cpL.*(T0 - Ti);%Total Heat Duty
Ef = 0.9320; %Fuel efficiency
qfuel = THD./Ef;%Heat released by fuel Kw
LHV = 41.05; %Fuel value KJ/kg
mfuel = qfuel./LHV;
%absorptivity
hp = 6.626*(10^-34); %planck's constant m2kg/s
cl = 3*(10^8); %speed of light m2/s
lamda = hp*cl/qfuel; %wavelength
Avog = 6.023*10^24; %Avogadro's number
x1 = 0.9;
M1 = 16;
x2 = 0.1;
M2 = 29;
fatmass = x1*M1 + x2*M2; %fuel molecular mass
nA = mfuel*Avog*fatmass; %number of atoms
pi = 22/7;
alpha = 4.*pi.*nA./lamda;
%Exchange factor
x = pi.*d0/2; %ctc
F = ((sqrt(x.^2 - 4) - x + 2.*sinh(2./x))/2.*pi);
%Heat transfer coefficient
k = 0.25*10^-3; %Thermal conductivity
%rof = 1320; %density
Gm = 920880; %mass flowrate
mu = 130; %Viscosity Pa-s
A0 = pi.*d0.*d0./4;
h0 = ((d0./k).*(0.023.*((Gm.*d0./mu.*A0).^0.8).*((cpL.*mu./k)).^0.3));
%Tube wall temperature
Tref = 41; %Ambient
Tw = Tref + (Ti - T0)/L;
%Air
ea = 0.15; %excess air
cpair = 0.171; %Kcal/kgdegK
Tairin = 114;
atfr = 1.344; %in notebook
mair = mfuel.*atfr.*(1+ea);
%Flame calculation
cpsteam = 4.1855;
cpco2 = 0.918;
cpn2 = 1.044;
xco2 = 0.054;
xsteam = 0.034; %Total should be 8.8
xn2 = 0.66;
cpgas = xsteam.*cpsteam + xco2.*cpco2 + xn2.*cpn2;
kgas = 0.25; %Thermal conductivity
mgas = mfuel + mair;
Lf = 0.0042.*(qfuel.^0.478); %flame length
tf = sqrt(pi.*uo.*Lf.*cpgas./mgas.*kgas); %flame size
db = 1.5.*tf; %Burner tile diameter
dbc = db./(sin(180./nb) + (pi./nb)); %Burner circle diameter
%Cold plane area
cbt = (qfuel.*1.055/4*10^6) + 1.5;
dtc = cbt + dbc; %tube circle diameter
ntubes = pi.*dtc./x;
Ar = L.*ntubes.*x;
stef = 5.67*10^-8; %Stefan-Boltzmann constant
qr = stef.*alpha.*F.*(Tg.^4 - Tw.^4) + ho.*(Tg - Tw);
%qr = stef*alpha*((sqrt((pi*d0)^2 - 16) - 2*x + 2*sinh(4/pi*d0))/4*pi)*(Tg^4 - (Tref + (Ti - T0)/L)^4) + ((d0/k)*(0.023*((Gm*d0/mu*pi*d0*d0/4)^0.8)*((cpL*mu/k))^0.3))*(Tg - (Tref + (Ti - T0)/L));
%Heat leaving
Qr = qr .* Ar; %total radiant heat absorbed
%Qr = qr * L*pi*(((qfuel*1.055/4*10^6) + 1.5) + ((1.5*(sqrt(pi*uo*Lf*cpgas/mgas*kgas)))/(sin(180/nb) + (pi/nb))));
qair = mair.*cpair.*(Tairin - Tref);
qloss = 0.05.*qfuel;
qout = qfuel + qair - ((stef.*alpha.*((sqrt((pi.*d0).^2 - 16) - 2.*x + 2*sinh(4/pi.*d0))/4*pi).*(Tg.^4 - (Tref + (Ti - T0)/L).^4) + ((d0/k).*(0.023.*((Gm.*d0/mu.*pi.*d0.*d0/4)^0.8).*((cpL.*mu./k)).^0.3)).*(Tg - (Tref + (Ti - T0)/L)).*L.*pi.*(((qfuel.*1.055/4*10^6) + 1.5) + ((1.5.*(sqrt(pi.*uo.*Lf.*cpgas./mgas.*kgas)))/(sin(180/nb) + (pi/nb))))) + qloss);
end
Solution:
%d0 = 0.01:1;
%L = 1:18;
%Tg = 1:1950;
%nb = 8:12;
%uo = 1:50;
x0 = [0.01, 1, 10, 8, 1];
options = optimset('PlotFcns',@optimplotfval);
X = fminsearch(@Lobo, x0, options);
qr = @Lobo;
plot (d0,qr)
xlabel('Tube circle diameter (d0)');
ylabel('flux (qr)');
Torsten

### Torsten (view profile)

on 7 Mar 2018
X = fminsearch(@(x)Lobo(x(1),x(2),x(3),x(4),x(5)), x0, options);
Best wishes
Torsten.