NATURAL CONVECTION HEAT TRANSFER IN COMBUSTION SYSTEMS

In the last week we investigated heat trasnfer when combustion gases are flowing whith some velocities. If velocities of combustion gases are very low or non-existent, there are still heat trasnfer between gas and surface , or the fluid on the other side of the Wall or pipe. For this reason natural convection is also an important mechanism when considered heat transfer of a combustion gas.

NATURAL CONVECTION HEAT TRANSFER EQUATIONS IN VERTICAL PLATE

Grashoff Number: where is the volumetric thermal expansion coefficient. This term may be approximated as: . For a perfect gas: therefore: 1/K

Rayleigh Number:

Critical Rayleigh Number

Laminar free convection: Incropera-DeWitt equation[35] similarity solution

Laminar free convection: Coban-2 equation similarity solution

Churchill & Chu Equation for all Ra range[36]

Churchill & Chu Equation for Laminar Ra range[36]

PROBLEM : exhaust gas temperature of an oven is =226.85 C and pressure isnside oven P=101.325 kPa. Oven has a wall of heigh L=0.5 m and width W=1 m. Wall surface temperature is T_s=126.85 C. Find the heat transfer from natural convection.

Exhaust gas :

CO2	3	9.81%
H2O	4	13.08%
O2	1	3.27%
N2	22.57143	73.83%

public class HT_natural_vertical_plate_exhaust_gas

{ public Gmix g; //exhaust gas mix

public HT_natural_vertical_plate_exhaust_gas(String gname[],double n[])

{g=new Gmix("exhaust gas",gname,n);

g.mole=false;

// st=new steamIAPWS_IF97();

}

public double Nu(double Ts,double Tinf,double P,double L,int select)

{//Nusselt number

double T=(Tinf+Ts)/2.0;

double Ra=Ra(Ts,Tinf,P,L);

double Pr=g.Prandtl(Tinf);

double Gr=Ra/Pr;

double Nu=0.0;

double Racritic=1e9;

double Pr_05=Math.sqrt(Pr);

if(Ra>Racritic || select==4 ) {Nu=(0.825+0.387*Math.pow(Ra,(1.0/6.0))/Math.pow((1+Math.pow((0.492/Pr),(9.0/16.0))),(8.0/27.0)));

Nu=Nu*Nu;

}

else

{if(select==1) Nu=4.0/3.0*Math.pow((Gr/4.0),(1.0/4.0))*0.75*Pr_05/Math.pow((0.609+1.221*Pr_05+1.238*Pr),(1.0/4.0));

else if(select==2) Nu=4.0/3.0*Math.pow((Gr/4.0),(1.0/4.0))*(0.022830145+0.752624416*Pr_05+2.04e-4*Pr)/Math.pow((1.222236219+1.010421489*Pr_05+1.389905434*Pr),(1.0/4.0));

else if(select==3) Nu=(0.68+0.67*Math.pow(Ra,(1.0/4.0))/Math.pow((1+Math.pow((0.492/Pr),(9.0/16.0))),(4.0/9.0)));

}

return Nu;

}

public double Ra(double Ts,double Tinf,double P,double L)

{ double T=(Ts+Tinf)/2.0;

double mu=g.vis(T);

double ro=1.0/g.v(T,P);

double Cp=g.Cp(T);

double k=g.k(Tinf);

double beta=1/T;

double nu=mu/ro;

double alpha=k/(ro*Cp);

double g=9.806; //m/s^2

double Ra=g*beta*Math.abs(Tinf-Ts)*L*L*L/alpha/nu;

System.out.println("Ra="+Ra);

return Ra;

}

public static void main(String arg[])

{ String s1[]={"co2","h2o","o2","n2"};

double n1[]={0.0981,0.1308,0.0327,0.7383};

HT_natural_vertical_plate_exhaust_gas ht=new HT_natural_vertical_plate_exhaust_gas(s1,n1);

double Ts=126.85+273.15; // Wall surface temperature degree K

double Tinf=226.85+273.15; //exhaust gas temperature degree K;

double L=0.5; // m

double W=1.0;

double P=101.325; //kPa

double Nu1=ht.Nu(Ts,Tinf,P,L,1);

double Nu2=ht.Nu(Ts,Tinf,P,L,2);

double Nu3=ht.Nu(Ts,Tinf,P,L,3);

double Nu4=ht.Nu(Ts,Tinf,P,L,4);

System.out.println("Nu1="+Nu1+" Nu2="+Nu2+" Nu3="+Nu3+" Nu4="+Nu4);

double T=(Tinf+Ts)/2.0;

double k1=ht.g.k(Tinf);

double h1=Nu1*k1/L;

double h2=Nu2*k1/L;

double h3=Nu3*k1/L;

double h4=Nu4*k1/L;

System.out.println("h1="+h1+" h2="+h2+" h3="+h3+" h4="+h4);

double A=W*L;

double Q1=h1*A*(Tinf-Ts);

double Q2=h2*A*(Tinf-Ts);

double Q3=h3*A*(Tinf-Ts);

double Q4=h4*A*(Tinf-Ts);

System.out.println("Q1="+Q1+" Q2="+Q2+" Q3="+Q3+"Q4 = "+Q4);

}}

---------- Capture Output ----------

> "E:\co\java\bin\javaw.exe" HT_natural_vertical_plate_exhaust_gas

Ra=7.305303667489275E7

Nu1=26.15818484714037 Nu2=27.733820597755294 Nu3=26.882730874865885 Nu4=27.919353448371727

h1=1.9774365660927675 h2=2.096547267630974 h3=2.0322088607842446 h4=2.1105726843471677

Q1=98.87182830463837 Q2=104.8273633815487 Q3=101.61044303921223Q4 = 105.52863421735839

> Terminated with exit code 0.

PROBLEM : Exhaust gas in a furnace is =326.85 C and furnace pressure is P=101.325 kPa. Furnace has a vertical pipe of length L=10 m and diameter D=0.01 m. Pipe surface temperature is constant and T_s=126.85 C. Find the heat transfer from natural convection.

Exhaust gas :

CO2	3	9.81%
H2O	4	13.08%
O2	1	3.27%
N2	22.57143	73.83%

public class HT_natural_vertical_plate_exhaust_gas

{ public Gmix g; //exhaust gas mix

public HT_natural_vertical_plate_exhaust_gas(String gname[],double n[])

{g=new Gmix("exhaust gas",gname,n);

g.mole=false;

// st=new steamIAPWS_IF97();

}

public double Nu(double Ts,double Tinf,double P,double L,int select)

{//Nusselt number

double T=(Tinf+Ts)/2.0;

double Ra=Ra(Ts,Tinf,P,L);

double Pr=g.Prandtl(Tinf);

double Gr=Ra/Pr;

double Nu=0.0;

double Racritic=1e9;

double Pr_05=Math.sqrt(Pr);

if(Ra>Racritic || select==4 ) {Nu=(0.825+0.387*Math.pow(Ra,(1.0/6.0))/Math.pow((1+Math.pow((0.492/Pr),(9.0/16.0))),(8.0/27.0)));

Nu=Nu*Nu;

}

else

{if(select==1) Nu=4.0/3.0*Math.pow((Gr/4.0),(1.0/4.0))*0.75*Pr_05/Math.pow((0.609+1.221*Pr_05+1.238*Pr),(1.0/4.0));

else if(select==2) Nu=4.0/3.0*Math.pow((Gr/4.0),(1.0/4.0))*(0.022830145+0.752624416*Pr_05+2.04e-4*Pr)/Math.pow((1.222236219+1.010421489*Pr_05+1.389905434*Pr),(1.0/4.0));

else if(select==3) Nu=(0.68+0.67*Math.pow(Ra,(1.0/4.0))/Math.pow((1+Math.pow((0.492/Pr),(9.0/16.0))),(4.0/9.0)));

}

return Nu;

}

public double Ra(double Ts,double Tinf,double P,double L)

{ double T=(Ts+Tinf)/2.0;

double mu=g.vis(T);

double ro=1.0/g.v(T,P);

double Cp=g.Cp(T);

double k=g.k(Tinf);

double beta=1/T;

double nu=mu/ro;

double alpha=k/(ro*Cp);

double g=9.806; //m/s^2

double Ra=g*beta*Math.abs(Tinf-Ts)*L*L*L/alpha/nu;

System.out.println("Ra="+Ra);

return Ra;

}

public static void main(String arg[])

{ String s1[]={"co2","h2o","o2","n2"};

double n1[]={0.0981,0.1308,0.0327,0.7383};

HT_natural_vertical_plate_exhaust_gas ht=new HT_natural_vertical_plate_exhaust_gas(s1,n1);

double Ts=126.85+273.15; // Wall surface temperature degree K

double Tinf=326.85+273.15; //exhaust gas temperature degree K;

double L=10.0; // m

double W=0.031415926;

double P=101.325; //kPa

double Nu1=ht.Nu(Ts,Tinf,P,L,1);

double Nu2=ht.Nu(Ts,Tinf,P,L,2);

double Nu3=ht.Nu(Ts,Tinf,P,L,3);

double Nu4=ht.Nu(Ts,Tinf,P,L,4);

System.out.println("Nu1="+Nu1+" Nu2="+Nu2+" Nu3="+Nu3+" Nu4="+Nu4);

double T=(Tinf+Ts)/2.0;

double k1=ht.g.k(Tinf);

double h1=Nu1*k1/L;

double h2=Nu2*k1/L;

double h3=Nu3*k1/L;

double h4=Nu4*k1/L;

System.out.println("h1="+h1+" h2="+h2+" h3="+h3+" h4="+h4);

double A=W*L;

double Q1=h1*A*(Tinf-Ts);

double Q2=h2*A*(Tinf-Ts);

double Q3=h3*A*(Tinf-Ts);

double Q4=h4*A*(Tinf-Ts);

System.out.println("Q1="+Q1+" Q2="+Q2+" Q3="+Q3+"Q4 = "+Q4);

}}

---------- Capture Output ----------

> "E:\co\java\bin\javaw.exe" HT_natural_vertical_plate_exhaust_gas

Ra=6.782784982644967E11

Nu1=452.2023920878429 Nu2=452.2023920878429 Nu3=452.2023920878429 Nu4=452.2023920878429

h1=2.0064086885485484 h2=2.0064086885485484 h3=2.0064086885485484 h4=2.0064086885485484

Q1=126.06637377039647 Q2=126.06637377039647 Q3=126.06637377039647Q4 = 126.06637377039647

> Terminated with exit code 0.

14.3 INCLINED PLATES NATURAL CONVECTION

Grashoff Number:

Rayleigh Number:

14.4 HORIZONTAL PLATES NATURAL CONVECTION

where A_s is the plate surface area and P is the perimeter

Upper surface of Hot Plate or Lower surface of Cold Plate

Lower surface of Hot Plate or Upper surface of Cold Plate

14.5 LONG HORIZONTAL CYLINDER NATURAL CONVECTION

Grashoff Number:

Rayleigh Number:

Morgan equation[37]

	C	n
10^-10-10^-2	0.675	0.058
10^-2-10²	1.02	0.148
10²-10⁴	0.850	0.188
10⁴-10⁷	0.480	0.250
10⁷-10¹²	0.125	0.333

Churchill & Chu equation[38]

PROBLEM : Exhaust gas in a furnace is =326.85 C and furnace pressure is P=101.325 kPa. Furnace has a horizontal pipe of length L=10 m and diameter D=0.01 m. Pipe surface temperature is constant and T_s=126.85 C. Find the heat transfer from natural convection.

Exhaust gas :

CO2	3	9.81%
H2O	4	13.08%
O2	1	3.27%
N2	22.57143	73.83%

public class HT_natural_horizontal_cylinder_exhaust_gas

{

public Gmix g; //exhaust gas mix

public HT_natural_horizontal_cylinder_exhaust_gas(String gname[],double n[])

{g=new Gmix("exhaust gas",gname,n);

g.mole=false;

}

public double Nu(double Ts,double Tinf,double P,double D,int select)

{//Nusselt number

double T=(Tinf+Ts)/2.0;

double Ra=Ra(Ts,Tinf,P,D);

double Pr=g.Prandtl(T);

double Gr=Ra/Pr;

double Nu=0.0;

double Racritic=1e9;

double C=0,n=0;

double Pr_05=Math.sqrt(Pr);

if(select==1) {Nu=(0.60+0.387*Math.pow(Ra,(1.0/6.0))/Math.pow((1+Math.pow((0.559/Pr),(9.0/16.0))),(8.0/27.0)));

Nu=Nu*Nu;

}

else if(select==2)

{ if(Ra>=1e-10 && Ra<1e-2) {C=0.675;n=0.058;}

else if(Ra>=1e-2 && Ra<1e2) {C=1.02;n=0.148;}

else if(Ra>=1e2 && Ra<1e4) {C=0.85;n=0.188;}

else if(Ra>=1e4 && Ra<1e7) {C=0.480;n=0.250;}

else if(Ra>=1e7 && Ra<=1e12){C=0.125;n=0.333;}

Nu=C*Math.pow(Ra,n);

System.out.println("C="+C+"n="+n);

}

return Nu;

}

public double Ra(double Ts,double Tinf,double P,double D)

{ double T=(Ts+Tinf)/2.0;

double mu=g.vis(T);

double ro=1.0/g.v(T,P);

double Cp=g.Cp(T);

double k=g.k(Tinf);

double beta=1/T;

double nu=mu/ro;

double alpha=k/(ro*Cp);

double g=9.806; //m/s^2

double Ra=g*beta*Math.abs(Ts-Tinf)*D*D*D/alpha/nu;

System.out.println("Ra="+Ra);

return Ra;

}

public static void main(String arg[])

{ String s1[]={"co2","h2o","o2","n2"};

double n1[]={0.0981,0.1308,0.0327,0.7383};

HT_natural_horizontal_cylinder_exhaust_gas ht=new HT_natural_horizontal_cylinder_exhaust_gas(s1,n1);

double Tinf=326.85+273.15; //air temp. degree K;

double Ts=126.85+273.15;

double D=0.01; // m

double L=10.0; //m

double P=101.325; //kPa

double Nu1= ht.Nu(Ts,Tinf,P,D,1);

double Nu2= ht.Nu(Ts,Tinf,P,D,2);

System.out.println("Nu1="+Nu1+" Nu2="+Nu2);

double k1=ht.g.k(Tinf);

double h1=Nu1*k1/D;

double T=(Tinf+Ts)/2.0;

double k2=ht.g.k(T);

double h2=Nu2*k2/D;

System.out.println("h1="+h1+" h2="+h2);

double A=Math.PI*D*L;

double Q1=h1*A*(Tinf-Ts);

double Q2=h2*A*(Tinf-Ts);

System.out.println("Q1="+Q1+" Q2="+Q2);

}}

---------- Capture Output ----------

> "E:\co\java\bin\javaw.exe" HT_natural_horizontal_cylinder_exhaust_gas

Ra=678.2784982644968

C=0.85n=0.188

Nu1=1.52449915094878 Nu2=2.895500465604656

h1=6.764157810015154 h2=10.944315385956939

Q1=425.00456967331263 Q2=687.6516163018412

> Terminated with exit code 0.

HOMEWORK PROBLEMS

PROBLEM 1: Inside of a vertical pipe has saturated steam of T=100 C, pipe diameter outside diameter is D=0.05 m,inside diameter is Di=0.045m and pipe length L=5 m. Pipe is made of commercial steel (k=30 W/(mK). Outside of pipe exhaust gases at P=101.325 kPa and the following mixture existed.

Exhaust gas :

CO2	10%
H2O	14%
O2	2%
N2	74%

Calculate the heat transfer

PROBLEM 2: Inside of a horizontal pipe has saturated steam of T=100 C, pipe diameter outside diameter is D=0.05 m,inside diameter is Di=0.045m and pipe length L=5 m. Pipe is made of commercial steel (k=30 W/(mK). Outside of pipe exhaust gases at P=101.325 kPa and the following mixture existed.

Exhaust gas :

CO2	10%
H2O	14%
O2	2%
N2	74%

Calculate the heat transfer