initial commit / add all books

6 files changed, 290 insertions, 0 deletions

diff --git a/534/CH11/EX11.1/11_1_Counterflow_tube_HeatX.sce b/534/CH11/EX11.1/11_1_Counterflow_tube_HeatX.sce
new file mode 100644
index 000000000..15c8fd824
--- /dev/null
+++ b/534/CH11/EX11.1/11_1_Counterflow_tube_HeatX.sce
@@ -0,0 +1,51 @@
+clear;

+clc;

+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 11.1   Page 680 \n'); //Example 11.1

+// Tube Length to achieve a desired hot fluid temperature

+

+//Operating Conditions

+Tho = 60+273    ;//[K] Hot Fluid outlet Temperature

+Thi = 100+273   ; //[K] Hot Fluid intlet Temperature

+Tci = 30+273    ;//[K] Cold Fluid intlet Temperature

+mh = .1          ;//[kg/s] Hot Fluid flow rate

+mc = .2          ;//[kg/s] Cold Fluid flow rate

+Do = .045        ;//[m] Outer annulus

+Di = .025        ;//[m] Inner tube

+

+//Table A.5 Engine Oil Properties T = 353 K

+cph = 2131              ;//[J/kg.K] Specific Heat

+kh = .138               ; //[W/m.K] Conductivity

+uh = 3.25*10^-2           ; //[N.s/m^2] Viscosity

+//Table A.6 Saturated water Liquid Properties Tc = 308 K

+cpc = 4178              ;//[J/kg.K] Specific Heat

+kc = 0.625               ; //[W/m.K] Conductivity

+uc = 725*10^-6           ; //[N.s/m^2] Viscosity

+Pr = 4.85                ;//Prandtl Number

+

+q = mh*cph*(Thi-Tho);

+

+Tco = q/(mc*cpc)+Tci;

+

+T1 = Thi-Tco;

+T2 = Tho-Tci;

+Tlm = (T1-T2)/(2.30*log10(T1/T2));

+

+//Through Tube

+Ret = 4*mc/(%pi*Di*uc);

+printf("\n Flow through Tube has Reynolds Number as %i. Thus the flow is Turbulent", Ret);

+//Equation 8.60

+Nut = .023*Ret^.8*Pr^.4;

+hi = Nut*kc/Di;

+

+//Through Shell

+Reo = 4*mh*(Do-Di)/(%pi*uh*(Do^2-Di^2));

+printf("\n Flow through Tube has Reynolds Number as %i. Thus the flow is Laminar", Reo);

+//Table 8.2

+Nuo = 5.63;

+ho = Nuo*kh/(Do-Di);

+

+U = 1/[1/hi+1/ho];

+L = q/(U*%pi*Di*Tlm);

+

+printf("\n Tube Length to achieve a desired hot fluid temperature is %.1f m",L);

+//END
\ No newline at end of file
diff --git a/534/CH11/EX11.2/11_2_Counterflow_plate_HeatX.sce b/534/CH11/EX11.2/11_2_Counterflow_plate_HeatX.sce
new file mode 100644
index 000000000..79317e1c3
--- /dev/null
+++ b/534/CH11/EX11.2/11_2_Counterflow_plate_HeatX.sce
@@ -0,0 +1,71 @@
+clear;

+clc;

+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 11.2   Page 683 \n'); //Example 11.2

+// Exterior Dimensions of heat Exchanger

+// Pressure drops within the plate-type Heat exchanger with N=60 gaps

+

+//Operating Conditions

+Tho = 60+273    ;//[K] Hot Fluid outlet Temperature

+Thi = 100+273    ;//[K] Hot Fluid intlet Temperature

+Tci = 30+273    ;//[K] Cold Fluid intlet Temperature

+mh = .1          ;//[kg/s] Hot Fluid flow rate

+mc = .2          ;//[kg/s] Cold Fluid flow rate

+Do = .045        ;//[m] Outer annulus

+Di = .025        ;//[m] Inner tube

+

+//Table A.5 Engine Oil Properties T = 353 K

+cph = 2131              ;//[J/kg.K] Specific Heat

+kh = .138                ;//[W/m.K] Conductivity

+uh = 3.25*10^-2           ; //[N.s/m^2] Viscosity

+rhoh = 852.1            ;//[kg/m^3] Density

+//Table A.6 Saturated water Liquid Properties Tc = 308 K

+cpc = 4178              ;//[J/kg.K] Specific Heat

+kc = 0.625                ;//[W/m.K] Conductivity

+uc = 725*10^-6            ;//[N.s/m^2] Viscosity

+Pr = 4.85                ;//Prandtl Number

+rhoc = 994              ;//[kg/m^3] Density

+

+q = mh*cph*(Thi-Tho);

+

+Tco = q/(mc*cpc)+Tci;

+

+T1 = Thi-Tco;

+T2 = Tho-Tci;

+Tlm = (T1-T2)/(2.30*log10(T1/T2));

+

+N = linspace(20,80,100);

+L = q/Tlm*[1/(7.54*kc/2)+1/(7.54*kh/2)]*(N^2-N)^-1;

+clf();

+plot(N,L);

+xtitle("Size of Heat Xchanger vs Number of gaps", "Number of Gaps (N)", "L (m)");

+

+N2 = 60;

+L = q/((N2-1)*N2*Tlm)*[1/(7.54*kc/2)+1/(7.54*kh/2)];

+a = L/N2;

+Dh = 2*a        ;//Hydraulic Diameter [m]

+//For water filled gaps

+umc = mc/(rhoc*L^2/2);

+Rec = rhoc*umc*Dh/uc;

+//For oil filled gaps

+umh = mh/(rhoh*L^2/2);

+Reh = rhoh*umh*Dh/uh;

+printf("\n Flow of the fluids has Reynolds Number as %.2f & %i. Thus the flow is Laminar for both", Reh,Rec);

+

+//Equations 8.19 and 8.22a

+delpc = 64/Rec*rhoc/2*umc^2/Dh*L        ;//For water

+delph = 64/Reh*rhoh/2*umh^2/Dh*L        ;//For oil

+

+//For example 11.1

+L1 = 65.9;

+Dh1c = .025;

+Dh1h = .02;

+Ret = 4*mc/(%pi*Di*uc);

+f = (.790*2.30*log10(Ret)-1.64)^-2        ;//friction factor through tube Eqn 8.21

+umc1 = 4*mc/(rhoc*%pi*Di^2);

+delpc1 = f*rhoc/2*umc1^2/Dh1c*L1;

+Reo = 4*mh*(Do-Di)/(%pi*uh*(Do^2-Di^2));

+umh1 = 4*mh/(rhoh*%pi*(Do^2-Di^2));

+delph1 = 64/Reo*rhoh/2*umh1^2/Dh1h*L1;

+

+printf("\n Exterior Dimensions of heat Exchanger L = %.3f m \n Pressure drops within the plate-type Heat exchanger with N=60 gaps\n For water = %.2f N/m^2    For oil = %.2f N/m^2\n Pressure drops tube Heat exchanger of example 11.1\n For water = %.1f kN/m^2    For oil = %.1f kN/m^2",L,delpc,delph,delpc1/1000,delph1/1000);

+//END
\ No newline at end of file
diff --git a/534/CH11/EX11.3/11_3_Crossflow_finned_tube_HeatX.sce b/534/CH11/EX11.3/11_3_Crossflow_finned_tube_HeatX.sce
new file mode 100644
index 000000000..ef5e9baa6
--- /dev/null
+++ b/534/CH11/EX11.3/11_3_Crossflow_finned_tube_HeatX.sce
@@ -0,0 +1,34 @@
+clear;

+clc;

+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 11.3   Page 692 \n'); //Example 11.3

+// Required gas side surface area

+

+//Operating Conditions

+Tho = 100+273    ;//[K] Hot Fluid outlet Temperature

+Thi = 300+273    ;//[K] Hot Fluid intlet Temperature

+Tci = 35+273    ;//[K] Cold Fluid intlet Temperature

+Tco = 125+273   ; //[K] Cold Fluid outlet Temperature

+mc = 1          ;//[kg/s] Cold Fluid flow rate

+Uh = 100        ;//[W/m^2.K] Coefficient of heat transfer

+//Table A.5 Water Properties T = 353 K

+cph = 1000       ;       //[J/kg.K] Specific Heat

+//Table A.6 Saturated water Liquid Properties Tc = 308 K

+cpc = 4197        ;      //[J/kg.K] Specific Heat

+

+Cc = mc*cpc;

+//Equation 11.6b and 11.7b

+Ch  = Cc*(Tco-Tci)/(Thi-Tho);

+// Equation 11.18

+qmax = Ch*(Thi-Tci);

+//Equation 11.7b

+q = mc*cpc*(Tco-Tci);

+

+e = q/qmax;

+ratio = Ch/Cc;

+

+printf("\n As effectiveness is %.2f with Ratio Cmin/Cmax = %.2f, It follows from figure 11.14 that NTU = 2.1",e,ratio);

+NTU = 2.1;

+A = 2.1*Ch/Uh;

+

+printf("\n Required gas side surface area = %.1f m^2",A);

+//END
\ No newline at end of file
diff --git a/534/CH11/EX11.4/11_4_Crossflow_finned_HeatX2.sce b/534/CH11/EX11.4/11_4_Crossflow_finned_HeatX2.sce
new file mode 100644
index 000000000..b9730f03d
--- /dev/null
+++ b/534/CH11/EX11.4/11_4_Crossflow_finned_HeatX2.sce
@@ -0,0 +1,36 @@
+clear;

+clc;

+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 11.4   Page 695 \n'); //Example 11.4

+// Heat Transfer Rate and Fluid Outlet Temperatures

+

+//Operating Conditions

+Thi = 250+273    ;//[K] Hot Fluid intlet Temperature

+Tci = 35+273    ;//[K] Cold Fluid intlet Temperature

+mc = 1          ;//[kg/s] Cold Fluid flow rate

+mh = 1.5        ;  //[kg/s] Hot Fluid flow rate

+Uh = 100        ;//[W/m^2.K] Coefficient of heat transfer

+Ah = 40         ; //[m^2] Area

+//Table A.5 Water Properties T = 353 K

+cph = 1000       ;       //[J/kg.K] Specific Heat

+//Table A.6 Saturated water Liquid Properties Tc = 308 K

+cpc = 4197        ;      //[J/kg.K] Specific Heat

+

+Cc = mc*cpc;

+Ch  = mh*cph;

+Cmin = Ch;

+Cmax = Cc;

+

+NTU = Uh*Ah/Cmin;

+ratio = Cmin/Cmax;

+

+printf("\n As Ratio Cmin/Cmax = %.2f and Number of transfer units NTU = %.2f, It follows from figure 11.14 that e = .82",ratio,NTU);

+e = 0.82;

+qmax = Cmin*(Thi-Tci);

+q = e*qmax;

+

+//Equation 11.6b

+Tco = q/(mc*cpc) + Tci;

+//Equation 11.7b

+Tho = -q/(mh*cph) + Thi;

+printf("\n Heat Transfer Rate  = %.2e W \n Fluid Outlet Temperatures Hot Fluid (Tho) = %.1f degC    Cold Fluid (Tco) = %.1f degC",q,Tho-273,Tco-273);

+//END
\ No newline at end of file
diff --git a/534/CH11/EX11.5/11_5_Shell_n_Tube_HeatX.sce b/534/CH11/EX11.5/11_5_Shell_n_Tube_HeatX.sce
new file mode 100644
index 000000000..3faf9cf71
--- /dev/null
+++ b/534/CH11/EX11.5/11_5_Shell_n_Tube_HeatX.sce
@@ -0,0 +1,39 @@
+clear;

+clc;

+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 11.5   Page 696 \n'); //Example 11.5

+// Outlet Temperature of cooling Water

+// Tube length per pass to achieve required heat transfer

+

+//Operating Conditions

+q = 2*10^9        ;//[W] Heat transfer Rate

+ho = 11000        ;//[W/m^2.K] Coefficient of heat transfer for outer surface

+Thi = 50+273    ;//[K] Hot Fluid Condensing Temperature

+Tho = Thi    ;//[K] Hot Fluid Condensing Temperature

+Tci = 20+273    ;//[K] Cold Fluid intlet Temperature

+mc = 3*10^4     ;     //[kg/s] Cold Fluid flow rate

+m =  1          ;//[kg/s] Cold Fluid flow rate per tube

+D = .025        ;//[m] diameter of tube

+//Table A.6 Saturated water Liquid Properties Tf = 300 K

+rho = 997        ;    //[kg/m^3] Density

+cp = 4179        ;      //[J/kg.K] Specific Heat

+k = 0.613        ;        //[W/m.K] Conductivity

+u = 855*10^-6    ;        //[N.s/m^2] Viscosity

+Pr = 5.83        ;        // Prandtl number

+

+//Equation 11.6b

+Tco = q/(mc*cp) + Tci;

+

+Re = 4*m/(%pi*D*u);

+printf("\n As the Reynolds number of tube fluid is %i. Hence the flow is turbulent. Hence using Diettus-Boetllor Equation 8.60", Re);

+Nu = .023*Re^.8*Pr^.4;

+hi = Nu*k/D;

+U = 1/[1/ho + 1/hi];

+N = 30000            ;//No of tubes

+T1 = Thi-Tco;

+T2 = Tho-Tci;

+Tlm = (T1-T2)/(2.30*log10(T1/T2));

+L2 = q/(U*N*2*%pi*D*Tlm);

+

+

+printf("\n Outlet Temperature of cooling Water = %.1f degC\n Tube length per pass to achieve required heat transfer = %.2f m",Tco-273,L2);

+//END
\ No newline at end of file
diff --git a/534/CH11/EX11.6/11_6_Finned_Compact_HeatX.sce b/534/CH11/EX11.6/11_6_Finned_Compact_HeatX.sce
new file mode 100644
index 000000000..d6c7ba472
--- /dev/null
+++ b/534/CH11/EX11.6/11_6_Finned_Compact_HeatX.sce
@@ -0,0 +1,59 @@
+clear;

+clc;

+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 11.6   Page 702 \n'); //Example 11.6

+// Gas-side overall heat transfer coefficient. Heat exchanger Volume

+

+//Operating Conditions

+hc = 1500        ;//[W/m^2.K] Coefficient of heat transfer for outer surface

+hi = hc;

+Th = 825    ;//[K] Hot Fluid Temperature

+Tci = 290    ;//[K] Cold Fluid intlet Temperature

+Tco = 370    ;//[K] Cold Fluid outlet Temperature

+mc = 1          ;//[kg/s] Cold Fluid flow rate

+mh =  1.25          ;//[kg/s] Hot Fluid flow rate

+Ah = .20            ;//[m^2] Area of tubes

+Di = .0138        ;//[m] diameter of tube

+Do = .0164        ;//[m] Diameter

+//Table A.6 Saturated water Liquid Properties Tf = 330 K

+cpw = 4184         ;     //[J/kg.K] Specific Heat

+//Table A.1 Aluminium Properties T = 300 K

+k = 237             ;   //[W/m.K] Conductivity

+//Table A.4 Air Properties Tf = 700 K

+cpa = 1075         ;     //[J/kg.K] Specific Heat

+u = 33.88*10^-6    ;        //[N.s/m^2] Viscosity

+Pr = .695          ;      // Prandtl number

+

+//Geometric Considerations

+si = .449;

+Dh = 6.68*10^-3        ;//[m] hydraulic diameter

+G = mh/si/Ah;

+Re = G*Dh/u;

+//From Figure 11.16

+jh = .01;

+hh = jh*G*cpa/Pr^.66667;

+

+AR = Di*2.303*log10(Do/Di)/(2*k*(.143));

+//Figure 11.16

+AcAh = Di/Do*(1-.830);

+//From figure 3.19

+nf = .89;

+noh = 1-(1-.89)*.83;

+

+U = [1/(hc*AcAh) + AR + 1/(noh*hh)]^-1;

+

+Cc = mc*cpw;

+q = Cc*(Tco-Tci);

+Ch = mh*cpa;

+qmax = Ch*(Th-Tci);

+e = q/qmax;

+ratio = Ch/Cc;

+

+printf("\n As effectiveness is %.2f with Ratio Cmin/Cmax = %.2f, It follows from figure 11.14 that NTU = .65",e,ratio);

+NTU = .65;

+A = NTU*Ch/U;

+//From Fig 11.16

+al = 269;            //[m^-1] gas side area per unit heat wxchanger volume

+V = A/al;

+

+printf("\n Gas-side overall heat transfer coefficient.r = %i W/m^2.K\n Heat exchanger Volume = %.3f m^3",U,V);

+//END;
\ No newline at end of file