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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /1309 | |
| download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip | |
initial commit / add all books
Diffstat (limited to '1309')
172 files changed, 2830 insertions, 0 deletions
diff --git a/1309/CH1/EX1.1/Result1_1.pdf b/1309/CH1/EX1.1/Result1_1.pdf Binary files differnew file mode 100755 index 000000000..4a2d7d865 --- /dev/null +++ b/1309/CH1/EX1.1/Result1_1.pdf diff --git a/1309/CH1/EX1.1/ch1_1.sce b/1309/CH1/EX1.1/ch1_1.sce new file mode 100755 index 000000000..9c567fa6a --- /dev/null +++ b/1309/CH1/EX1.1/ch1_1.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+printf("\t\t\tChapter1_Example1\n\n\n");
+// determination of surface temperature on one side of a firewall
+k=9.4; // thermal conductivity in [BTU/hr.ft. ˚Rankine]
+q=6.3; // heat flux in [BTU/s. sq.ft]
+T1=350; // the outside surface temperature of one aide of the wall [˚F]
+// converting heat flux into BTU/hr sq.ft
+Q=6.3*3600 // [BTU/hr.sq.ft]
+printf("\nThe heat flux is %.2f BTU/hr. sq.ft",Q);
+dx=0.5; // thickness in [inch]
+//converting distance into ft
+Dx=0.5/12 // thickness in [ft]
+printf("\nThe thickness of firewall is %.2f ft",Dx);
+// solving for temeprature T2
+T2=T1-(Q*Dx/k); // [˚F]
+printf("\nThe required temperature on the other side of the firewall is %.1f degree Fahrenheit",T2);
diff --git a/1309/CH1/EX1.2/Result1_2.pdf b/1309/CH1/EX1.2/Result1_2.pdf Binary files differnew file mode 100755 index 000000000..f3ad05de8 --- /dev/null +++ b/1309/CH1/EX1.2/Result1_2.pdf diff --git a/1309/CH1/EX1.2/ch1_2.sce b/1309/CH1/EX1.2/ch1_2.sce new file mode 100755 index 000000000..83cb01171 --- /dev/null +++ b/1309/CH1/EX1.2/ch1_2.sce @@ -0,0 +1,12 @@ +clear;
+clc;
+printf("\t\t\tchapter1_example2\n\n\n");
+// determination of thermal conductivity of aluminium
+k_ss=14.4; // thermal conductivity of stainless steel in [W/m.K]
+printf("\nThe thermal conductivity of stainless steel is %.1f W/m.K",k_ss);
+dt_ss=40; // [K]
+dt_al=8.65; // [K]
+dz_ss=1; // [cm]
+dz_al=3; // [cm]
+k_al=k_ss*dt_ss*dz_al/(dt_al*dz_ss);// thermal conductivity of Al in [W/m.K]
+printf("\nThe thermal conductivity of aluminium is %d W/m.K",k_al);
diff --git a/1309/CH1/EX1.3/Result1_3.pdf b/1309/CH1/EX1.3/Result1_3.pdf Binary files differnew file mode 100755 index 000000000..720c31a67 --- /dev/null +++ b/1309/CH1/EX1.3/Result1_3.pdf diff --git a/1309/CH1/EX1.3/ch1_3.sce b/1309/CH1/EX1.3/ch1_3.sce new file mode 100755 index 000000000..91be46380 --- /dev/null +++ b/1309/CH1/EX1.3/ch1_3.sce @@ -0,0 +1,11 @@ +clear;
+clc;
+printf("\t\t\tchapter1_example3\n\n\n");
+// determination of heat transferred by convection
+h_c=3; // convective coefficient in [BTU/hr.ft^2
+A=30*18; // Cross sectional area in ft^2
+T_w=140; // Roof surface temperature in degree Fahrenheit
+T_inf=85; // Ambient temperature in degree Fahrenheit
+dT= (T_w-T_inf);
+Q_c=h_c*A*dT; // Convective heat transfer in BTU/hr
+printf("\nThe heat transferred by convection is %d BTU/hr",Q_c);
diff --git a/1309/CH1/EX1.4/Result1_4.pdf b/1309/CH1/EX1.4/Result1_4.pdf Binary files differnew file mode 100755 index 000000000..9e3585de3 --- /dev/null +++ b/1309/CH1/EX1.4/Result1_4.pdf diff --git a/1309/CH1/EX1.4/ch1_4.sce b/1309/CH1/EX1.4/ch1_4.sce new file mode 100755 index 000000000..ad120cb5b --- /dev/null +++ b/1309/CH1/EX1.4/ch1_4.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+printf("\t\t\tchapter1_example4\n\n\n");
+// determining average film conductance
+D=2.43/100; // diameter in meter
+L=20/100; // length in meter
+A=3.14*D*L; // cross-sectional area in sq.m
+cp=4200; // specific heat of water in J/kg.K
+T_b2=21.4; // temperature of bulk fluid in degree celsius
+T_in=20; // temperature of inlet water in degree celsius
+T_w=75; // temperature of wall in degree celsius
+Q=500; // volumetric flow rate in cc/s
+density=1000; // density of water in kg/cu.m
+m=Q*density/10^6; // mass flowa rate in kg/s
+printf("\nThe mass flow rate is %.1f kg/s",m);
+// using definition of specific heat and Newton's law of cooling
+hc=m*cp*(T_b2-T_in)/(A*(T_w-T_in));
+printf("\nThe average film conductance is %d W/sq.m. K",hc);
diff --git a/1309/CH1/EX1.5/Result1_5.pdf b/1309/CH1/EX1.5/Result1_5.pdf Binary files differnew file mode 100755 index 000000000..dde281729 --- /dev/null +++ b/1309/CH1/EX1.5/Result1_5.pdf diff --git a/1309/CH1/EX1.5/ch1_5.sce b/1309/CH1/EX1.5/ch1_5.sce new file mode 100755 index 000000000..c4d0e950f --- /dev/null +++ b/1309/CH1/EX1.5/ch1_5.sce @@ -0,0 +1,15 @@ +clear;
+clc;
+printf("\t\t\tchapter1_example5\n\n\n");
+// determination of heat loss rate by radiation
+W=14; // width in ft
+L=30; // length in ft
+A=W*L; // area in ft^2
+F_12=1; // view factor assumed to be 1
+T1=120+460; // driveway surface temperature in degree Rankine
+printf("\nThe driveway surface temperature is %d degree Rankine",T1);
+T2=0; // space temperature assumed to be 0 degree Rankine
+sigma=0.1714*10^(-8); // value of Stefan-Boltzmann's constant in BTU/(hr.ft^2.(degree Rankine)^4)
+e=0.9; // surface emissivity
+q=sigma*A*e*F_12*((T1)^4-(T2)^4);
+printf("\nThe heat loss rate by radiation is %d BTU/hr",q);
diff --git a/1309/CH1/EX1.6/Result1_6.pdf b/1309/CH1/EX1.6/Result1_6.pdf Binary files differnew file mode 100755 index 000000000..32a4b1bae --- /dev/null +++ b/1309/CH1/EX1.6/Result1_6.pdf diff --git a/1309/CH1/EX1.6/ch1_6.sce b/1309/CH1/EX1.6/ch1_6.sce new file mode 100755 index 000000000..f25808fc1 --- /dev/null +++ b/1309/CH1/EX1.6/ch1_6.sce @@ -0,0 +1,10 @@ +clear;
+clc;
+printf("\t\t\tchapter1_example6\n\n\n");
+// determination of radiation thermal conductance
+A=14*30; // area in sq.ft
+T1=120+460; // driveway surface temperature in degree Rankine
+T2=0; // surface temperature assumed to be 0 degree Rankine
+Qr=73320; // heat loss rate in BTU/hr
+hr=Qr/(A*(T1-T2)); // radiation thermal conductance in BTU/(hr.ft^2.(degree Rankine)
+printf("\nthe radiation thermal conductance is %.2f BTU/(hr. sq.ft.(degree Rankine))",hr);
diff --git a/1309/CH1/EX1.7/Result1_7.pdf b/1309/CH1/EX1.7/Result1_7.pdf Binary files differnew file mode 100755 index 000000000..13c21f68a --- /dev/null +++ b/1309/CH1/EX1.7/Result1_7.pdf diff --git a/1309/CH1/EX1.7/ch1_7.sce b/1309/CH1/EX1.7/ch1_7.sce new file mode 100755 index 000000000..07e10a8c1 --- /dev/null +++ b/1309/CH1/EX1.7/ch1_7.sce @@ -0,0 +1,27 @@ +clear;
+clc;
+printf("\t\t\tchapter1_example7\n\n\n");
+// Identification of all resistances and their values
+// Estimation of heat transfer per unit area
+// Determination of the inside and outside wall temperatures
+printf("\n\t\t\tSolution to part (b)\n");
+A=1; // assuming A=1 m^2 for convenience
+hc1_avg=(5+25)/2; // taking average of extreme values for hc [W/m^2.K]
+Rc1=1/(hc1_avg*A); // resistance on left side of wall [K/W]
+printf("\nThe resistance on left side of wall is %.3f K/W",Rc1);
+k=(0.38+0.52)/2; // thermal conductivity of common brick in W/M.k
+L=0.1; //10 cm converted into m
+Rk=(L/(k*A));// resistance of construction material, assume common brick
+printf("\nThe resistance of construction material of wall is %.3f K/W",Rk);
+Rc2=Rc1;
+printf("\nThe resistance on right side of wall is %.3f K/W",Rc2);
+printf("\n\n\t\t\tSolution to part (c)\n");
+T_inf1=1000; // temperature of exhaust gases in K
+T_inf2=283; // temperature of ambient air in K
+q=(T_inf1-T_inf2)/(Rc1+Rk+Rc2); // heat transferred per unit area
+printf("\nThe Heat transferred per unit area is %d W = %.3f kW",q,q/1000);
+printf("\n\n\t\t\tSolution to part (d)\n");
+T_in=T_inf1-Rc1*q; //
+T_out=T_inf2+Rc2*q;
+printf("\nThe inside wall temperature is %d K",T_in);
+printf("\nThe outside wall temperature is %d K",T_out);
diff --git a/1309/CH1/EX1.8/Result1_8.pdf b/1309/CH1/EX1.8/Result1_8.pdf Binary files differnew file mode 100755 index 000000000..5d538e1e0 --- /dev/null +++ b/1309/CH1/EX1.8/Result1_8.pdf diff --git a/1309/CH1/EX1.8/ch1_8.sce b/1309/CH1/EX1.8/ch1_8.sce new file mode 100755 index 000000000..a7f679141 --- /dev/null +++ b/1309/CH1/EX1.8/ch1_8.sce @@ -0,0 +1,28 @@ +clear;
+clc;
+printf("\t\t\tchapter1_example8\n\n\n");
+// determination of surface temperature
+k=0.604; // [BTU/(hr.ft.degree Rankine)]
+hc=3; // average value for natural convection in BTU/(hr.ft^2.degree Rankine)
+ew=0.93;
+f_wr=1; // shape factor
+sigma= 0.1714*10^(-8) // BTU/(hr.ft^2.degree Rankine).
+L=4/12; // length in ft
+T1=80+460; // temperature of side-walk in degree Rankine
+T_inf=20+460; // temperature of ambient air in degree Rankine
+T_r=0; // assuming space temperature to be 0 degree Rankine
+// LHS of the form a*Tw+b*Tw^4=c
+a=((k/L)+hc);
+b=(sigma*ew*f_wr);
+c=(k*T1/L)+(hc*T_inf)+(sigma*f_wr*ew*T_r^4);
+printf("\nRHS=%d",c);
+ Tw=[470;480;490;485;484.5];
+for i=1:5
+ LHS(i)=a*Tw(i)+b*Tw(i)^4;
+end
+printf("\nSolving by trial and error yields the following, where LHS is the left-hand side of the equation");
+printf("\n\tTw\tLHS");
+for i=1:5
+ printf("\n\t%.1f\t%d",Tw(i),LHS(i));
+end
+printf("\nThe Surface temperature is %.1f degree R = %.1f degree F",Tw(5),Tw(i)-460);
diff --git a/1309/CH10/EX10.1/Figure10_1.png b/1309/CH10/EX10.1/Figure10_1.png Binary files differnew file mode 100755 index 000000000..721ba5e28 --- /dev/null +++ b/1309/CH10/EX10.1/Figure10_1.png diff --git a/1309/CH10/EX10.1/Result10_1.pdf b/1309/CH10/EX10.1/Result10_1.pdf Binary files differnew file mode 100755 index 000000000..79df73832 --- /dev/null +++ b/1309/CH10/EX10.1/Result10_1.pdf diff --git a/1309/CH10/EX10.1/ch10_1.sce b/1309/CH10/EX10.1/ch10_1.sce new file mode 100755 index 000000000..fbb0d99e1 --- /dev/null +++ b/1309/CH10/EX10.1/ch10_1.sce @@ -0,0 +1,64 @@ +clc; +clear; +printf("\t\t\tChapter10_example1\n\n\n"); +// Calculation of the heat-transfer rate and the amount of steam condensed +// properties of engine oil at (328 + 325)/2 = 326.5 degree F = 320°F from appendix table C11 +rou_f= 0.909*62.4; // density in lbm/ft^3 +cp=1.037; // specific heat BTU/(lbm-degree Rankine) +v_f= 0.204e-5; // viscosity in ft^2/s +kf = 0.393; // thermal conductivity in BTU/(lbm.ft.degree Rankine) +a = 6.70e-3; // diffusivity in ft^2/hr +Pr = 1.099; // Prandtl Number +V_v=4.937; // specific volume in ft^3/lbm from superheated steam tables +rou_v=1/V_v; // vapour density +g=32.2; +hfg=888.8; // from saturated steam tables +Tg=327.81; +Tw=325; +L=2; // length in ft +W=3; // width in ft +z=0.2:0.2:2; // distance from entry of plate in ft +[n m]=size(z); +// film thickness is given as follows +for i=1:m +delta(i)=[(4*kf*v_f*z(i)*(Tg-Tw)/3600)/(rou_f*g*hfg*(1-(rou_v/rou_f)))]^(1/4); +hz(i)=(kf/delta(i)); +end +printf("\nGrowth of and Heat-Transfer Coefficient for the Condensate Film of Example 10.1 "); +printf("\nz, ft\tdelta, ft\tdelta, in\thz, BTU/(hr.sq.ft.degree Rankine)"); +for i=1:m +printf("\n%.1f\t%.2e\t%.4f\t\t%d\n",z(i),delta(i),12*delta(i),hz(i)); +end +hL=(4/3)*hz(m); // at plate end +mf=(hL*L*W*(Tg-Tw))/hfg; +printf("\nThe convective coefficient at the plate end is %d BTU/(hr.sq.ft. degree Rankine)",hL); +printf("\nThe amount of steam condensed is %.1f lbm/hr",mf); +q=mf*hfg; +printf("\nThe heat transfer rate is %.2e BTU/hr",q); +Re=(4*mf/3600)/(W*rou_f*v_f); +printf("\nThe Reynolds Number is %d",Re); +if Re<1800 then + printf("\nThe film is laminar and above equations apply"); + else printf("\nThe film is not laminar and above assumption is wrong"); +end +subplot(211); +plot(delta*12,z,'x-'); // our first figure +a1 = gca(); +a1.x_location="top"; +a1.axes_reverse=["off","on"]; +subplot(212) +plot(hz,z, 'o--'); // our second figure +a2 = gca(); +a2.axes_reverse=["off","on"]; +a2.x_location="bottom"; +a2.axes_visible = ["on", "on","on"]; +a2.y_location ="right"; +x_label1=a1.x_label; +x_label1.text="delta,in"; +x_label2=a2.x_label; +x_label2.text="hz, BTU/(hr.sq.ft.degree R)"; +y_label=a1.y_label; +y_label.text="z, ft"; +a1.axes_bounds=[0 0 1 1]; // modify the first figure to occupy the whole area +a2.axes_bounds=[0 0 1 1]; // modify the second figure to occupy the whole area too +a2.filled="off"; diff --git a/1309/CH10/EX10.2/Result10_2.pdf b/1309/CH10/EX10.2/Result10_2.pdf Binary files differnew file mode 100755 index 000000000..410bf5a58 --- /dev/null +++ b/1309/CH10/EX10.2/Result10_2.pdf diff --git a/1309/CH10/EX10.2/ch10_2.sce b/1309/CH10/EX10.2/ch10_2.sce new file mode 100755 index 000000000..ce45c7217 --- /dev/null +++ b/1309/CH10/EX10.2/ch10_2.sce @@ -0,0 +1,27 @@ +clc; +clear; +printf("\t\t\tChapter10_example2\n\n\n"); +// Determination of both the heat that the cooling fluid must remove and the condensation rate. +// properties of water at (100 + 60)/2 = 80°C from appendix table C11 +rou_f= 947; // density in kg/m^3 +cp_1= 4196; // specific heat in J/(kg*K) +v_1= 0.364e-6; // viscosity in m^2/s +Pr_1 =2.22; // Prandtl Number +kf= 0.668; // thermal conductivity in W/(m.K) +a_1 =1.636e-7; // diffusivity in m^2/s +Vv=1.9364; // specific volume in m^3/kg +rou_v=1/Vv; // vapor density; +g=9.81; +hfg=2257.06*1000; +Tg=100; +Tw=60; +L=1; +printf("\nThe vapor density is %.3f kg/cu.m",rou_v); +// specifications of 1 nominal schedule 40 pipe from appendix F1 +OD=.03340; +hD=0.782*[(g*rou_f*(1-(rou_v/rou_f))*(kf^3)*hfg)/(v_1*OD*(Tg-Tw))]^(1/4); +printf("\nThe average heat transfer coefficient is %.3e W/(sq.m.K)",hD); +q=hD*%pi*OD*L*(Tg-Tw); +printf("\nThe heat flow rate is %.1e W",q); +mf=q/hfg; +printf("\nThe rate at which steam condenses is %.2f kg/s = %d kg/hr",mf,.02*3600); diff --git a/1309/CH10/EX10.3/Result10_3.pdf b/1309/CH10/EX10.3/Result10_3.pdf Binary files differnew file mode 100755 index 000000000..77a85824d --- /dev/null +++ b/1309/CH10/EX10.3/Result10_3.pdf diff --git a/1309/CH10/EX10.3/ch10_3.sce b/1309/CH10/EX10.3/ch10_3.sce new file mode 100755 index 000000000..5a02cdc7d --- /dev/null +++ b/1309/CH10/EX10.3/ch10_3.sce @@ -0,0 +1,32 @@ +clc; +clear; +printf("\t\t\tChapter10_example3\n\n\n"); +// Calculation of (a) the power input to the water for boiling to occur, (b) the evaporation rate of water, and (c) the critical heat flux. +// properties of water at 100°C = 373 K from appendix table 10.3 +rou_f=958; // density in kg/m^3 +cp_f= 4217; // specific heat in J/(kg*K) +v_f= 2.91e-7; // viscosity in m^2/s +Pr_f =1.76; // Prandtl Number +rou_g=0.596; +sigma=0.0589; // surface tension in N/m +hfg=2257000; +Tw=120 +Tg=100; +D=.141; // diameter of pan in m +g=9.81; +gc=1; +// nucleate boiling regime +Cw=0.0132; // formechanically polished stainless steel from table 10.2 +q_A=(rou_f*v_f*hfg)*[(g*rou_f*(1-(rou_g/rou_f)))/(sigma*gc)]^(1/2)*[(cp_f*(Tw-Tg))/(Cw*hfg*Pr_f^1.7)]^3; +printf("\nThe heat transferred per unit area is %.2e W/sq.m",q_A); +A=%pi*D^2/4; +printf("\nThe area of the pan inside-bottom surface in contact with liquid is %.2e sq.m",A); +printf("\n\n\t\t\tSolution to part (a)"); +q=q_A*A; // power delivered to the water in W +printf("\nThe power delivered to the water is %.2f kW",q/1000); +printf("\n\n\t\t\tSolution to part (b)"); +mf=q/hfg; // water evaporation rate +printf("\nThe water evaporation rate is %.2e kg/s = %.2f kg/hr",mf,mf*3600); +printf("\n\n\t\t\tSolution to part (c)"); +q_cr=0.18*hfg*[sigma*g*gc*rou_f*rou_g^2]^(1/4); +printf("\nThe critical heat flux is %.2e W/sq.m",q_cr); diff --git a/1309/CH11/EX11.1/Result11_1.pdf b/1309/CH11/EX11.1/Result11_1.pdf Binary files differnew file mode 100755 index 000000000..1c47b1ba5 --- /dev/null +++ b/1309/CH11/EX11.1/Result11_1.pdf diff --git a/1309/CH11/EX11.1/ch11_1.sce b/1309/CH11/EX11.1/ch11_1.sce new file mode 100755 index 000000000..ba413e59a --- /dev/null +++ b/1309/CH11/EX11.1/ch11_1.sce @@ -0,0 +1,31 @@ +clc; +clear; +printf("\t\t\tChapter11_example1\n\n\n"); +// Calculation of the value of the solid angle subtended by surfaces dA2 and dA3 with respect to dA1 (b) the intensity of emission from dA, in the direction of the other areas (c) the rate at which radiation emitted by dA, is intercepted by the other areas +printf("\t\t\tSolution to Part (a)\n"); +// solid angle is calculate using the equation dw=dA*cos(Beta)/r^2 +// Beta is the angle between the surface normal of a receiver surface and the line connecting the two surfaces +// For area A2 +// dimensions are 1X1 in, so +dA2=(1*1)/144; +Beta1=40*%pi/180; +r=4; +dw2_1=dA2*cos(Beta1)/r^2; +printf("\nThe solid angle subtended by area dA2 with respect to dA1 is %.2e sr",dw2_1); +dA3=dA2; +Beta2=0; +dw3_1=dA3*cos(Beta2)/r^2; +printf("\nThe solid angle subtended by area dA3 with respect to dA1 is %.2e sr",dw3_1); +printf("\n\n\t\t\tSolution to Part (b)\n"); +theta2=%pi*50/180; +theta3=%pi*60/180; +I_theta2=2000*(1-0.4*(sin(theta2))^2); +I_theta3=2000*(1-0.4*(sin(theta3))^2); +printf("\n The intensity of radiation emitted from dA1 in the direction of dA2 is %d BTU/(hr.sq.ft.sr)",I_theta2); +printf("\n The intensity of radiation emitted from dA1 in the direction of dA3 is %d BTU/(hr.sq.ft.sr)",I_theta3); +printf("\n\n\t\t\tSolution to Part (c)\n"); +dA1=1/144; +dq1_2=I_theta2*dA1*cos(theta2)*dw2_1; +dq1_3=I_theta3*dA1*cos(theta3)*dw3_1; +printf("\nThe rate at which radiation emitted by dA1 is intercepted by dA2 is %.2e BTU/hr",dq1_2); +printf("\nThe rate at which radiation emitted by dA1 is intercepted by dA3 is %.2e BTU/hr",dq1_3); diff --git a/1309/CH11/EX11.2/Result11_2.pdf b/1309/CH11/EX11.2/Result11_2.pdf Binary files differnew file mode 100755 index 000000000..0e6ac25e3 --- /dev/null +++ b/1309/CH11/EX11.2/Result11_2.pdf diff --git a/1309/CH11/EX11.2/ch11_2.sce b/1309/CH11/EX11.2/ch11_2.sce new file mode 100755 index 000000000..5c6f2e5c0 --- /dev/null +++ b/1309/CH11/EX11.2/ch11_2.sce @@ -0,0 +1,22 @@ +clc; +clear; +printf("\t\t\tChapter11_example2\n\n\n"); +// Calculation of the value of the solid angle subtended by surfaces dA2 with respect to dA1 (b) the rate at which radiation emitted by dA1 is intercepted by dA2 (c) the irradiation associated with dA2 +printf("\t\t\tSolution to Part (a)\n"); +// solid angle is calculate using the equation dw=dA*cos(Beta)/r^2 +// The angle Beta is 0 because the surface normal of dA2 is directed at dA1 +dA2=0.02*0.02; +Beta=0; +r=1; +dw2_1=dA2*cos(Beta)/r^2; +printf("\nThe solid angle subtended by area dA2 with respect to dA1 is %.2e sr",dw2_1); +printf("\n\n\t\t\tSolution to Part (b)\n"); +dA1=dA2; +theta=%pi*30/180; +I_theta=1000;// The intensity of radiation leaving dA1 in any direction is 1 000 W/(m^2.sr +dq1_2=I_theta*dA1*cos(theta)*dw2_1; +printf("\nThe rate at which radiation emitted by dA1 is intercepted by dA2 is %.2e W",dq1_2); +printf("\n\n\t\t\tSolution to Part (c)\n"); +// The irradiation associated with dA2 can be found by dividing the incident radiation by the receiver area +dQ1_2=dq1_2/dA2; +printf("\nThe irradiation associated with dA2 is %.2e W/sq.m",dQ1_2); diff --git a/1309/CH11/EX11.3/Result11_3.pdf b/1309/CH11/EX11.3/Result11_3.pdf Binary files differnew file mode 100755 index 000000000..1f6b0349a --- /dev/null +++ b/1309/CH11/EX11.3/Result11_3.pdf diff --git a/1309/CH11/EX11.3/ch11_3.sce b/1309/CH11/EX11.3/ch11_3.sce new file mode 100755 index 000000000..df7ae08a2 --- /dev/null +++ b/1309/CH11/EX11.3/ch11_3.sce @@ -0,0 +1,24 @@ +clc; +clear; +printf("\t\t\tChapter11_example3\n\n\n"); +// (a) Calculation of the emissivity of the hole.(b) the rate of radiant emission from the hole +D=2.5/12; // diameter in ft +L=4.5/12; // length in ft +A=(2*%pi*D^2/4)+(%pi*D*L); +printf("\nThe inside surface area is %.3f sq.ft ",A); +A_hole=%pi*(1/(8*12))^2/4; +printf("\nThe area of a 1/8 inch hole is %.3e sq.ft",A_hole); +f=A_hole/A; // fraction of area removed +printf("\nThe fraction of area removed is %.3e ",f); +printf("\n\n\t\t\tSolution to Part (a)\n"); +// for rolled and polished aluminum, that emissivity = 0.039 from appendix table E1 +emissivity=0.039; +emissivity_hole=emissivity/(emissivity+(1-emissivity)*f); +printf("\nThe emissivity of the hole is %.4f",emissivity_hole); +printf("\n\n\t\t\tSolution to Part (b)\n"); +sigma=0.1714e-8; // stefan Boltzmann constant in BTU/(hr~ft^2 degree R) +T=150+460; // temperature in degree R +qe=emissivity_hole*sigma*T^4; +printf("\nThe heat lost per unit area of the hole is %d BTU/hr",qe); +Qe=A_hole*qe; +printf("\nThe heat lost by the hole is %.2e BTU/hr",Qe); diff --git a/1309/CH11/EX11.4/Result11_4.pdf b/1309/CH11/EX11.4/Result11_4.pdf Binary files differnew file mode 100755 index 000000000..8d25eb849 --- /dev/null +++ b/1309/CH11/EX11.4/Result11_4.pdf diff --git a/1309/CH11/EX11.4/ch11_4.sce b/1309/CH11/EX11.4/ch11_4.sce new file mode 100755 index 000000000..72c78b773 --- /dev/null +++ b/1309/CH11/EX11.4/ch11_4.sce @@ -0,0 +1,14 @@ +clc; +clear; +printf("\t\t\tChapter11_example4\n\n\n"); +// Determination of the percentage of total emitted energy that lies in the visible range. +T=2800; +lambda1=4e-7; +lambda2=7e-7; +hT=lambda1*T; +lambdaT=lambda2*T; +printf("\nhT=%.2e m.K and lambda2_T=%.2e m.K",hT,lambdaT); +I1=0.0051; //Fraction of Total Radiation Emitted for lower Wavelength-Temperature Product from Table 11.1 +I2=0.065; //Fraction of Total Radiation Emitted for upper Wavelength-Temperature Product from Table 11.1 +dI=I2-I1; +printf("\nThe percentage of total emitted energy that lies in the visible range is %.1f percent",dI*100); diff --git a/1309/CH11/EX11.5/Result11_5.pdf b/1309/CH11/EX11.5/Result11_5.pdf Binary files differnew file mode 100755 index 000000000..3dc562e60 --- /dev/null +++ b/1309/CH11/EX11.5/Result11_5.pdf diff --git a/1309/CH11/EX11.5/ch11_5.sce b/1309/CH11/EX11.5/ch11_5.sce new file mode 100755 index 000000000..d9237a5d1 --- /dev/null +++ b/1309/CH11/EX11.5/ch11_5.sce @@ -0,0 +1,12 @@ +clc; +clear; +printf("\t\t\tChapter11_example5\n\n\n"); +// Estimation of the surface temperature of the sun and the emitted heat flux +lambda_max=0.5e-6; // maximum wavelength in m +// From Wien’s Displacement Law we can write lambda_max*T=2.898e-3 m.K +T=2.898e-3/lambda_max; +printf("\nThe Surface Temperature of the Sun is %d K",T); +// The heat flux is given by the Stefan-Boltzmann Equation as q=sigma*T^4 +sigma=5.675e-8; // value of Stefan-Boltzmann constant in W/(m^2.K^4) +q=sigma*T^4; +printf("\nThe heat flux emitted is %.3e W/sq.m",q); diff --git a/1309/CH11/EX11.6/Result11_6.pdf b/1309/CH11/EX11.6/Result11_6.pdf Binary files differnew file mode 100755 index 000000000..8d80635eb --- /dev/null +++ b/1309/CH11/EX11.6/Result11_6.pdf diff --git a/1309/CH11/EX11.6/ch11_6.sce b/1309/CH11/EX11.6/ch11_6.sce new file mode 100755 index 000000000..56d5cd1e2 --- /dev/null +++ b/1309/CH11/EX11.6/ch11_6.sce @@ -0,0 +1,28 @@ +clc; +clear; +printf("\t\t\tChapter11_example6\n\n\n"); +// (a) Calculation of the rate at which the sun’s radiant energy is transmitted through the glass windshield. The interior of the car is considered to be a black body that radiates at 100°F. (b) Calculation of the rate at which radiant energy from the car interior is transmitted through the glass windshield. +printf("\t\t\tSolution to Part (a)\n"); +lambda1=300e-9; // lower limit of wavelength +lambda2=380e-9; // upper limit of wavelength +T=5800; +lambda1_T=lambda1*T; +lambda2_T=lambda2*T; +printf("\nThe Lower and Upper limits of Wavelength-Temperature Products are %.2e m.K and %.3e m.K respectively",lambda1_T,lambda2_T); +I1=0.101; //Fraction of Total Radiation Emitted for lower Wavelength-Temperature Product from Table 11.1 +I2=0.0334; //Fraction of Total Radiation Emitted for upper Wavelength-Temperature Product from Table 11.1 +dI=abs(I2-I1); +t=dI*0.68; // transmissivity +printf("\nThe Transmittivity is %.4f",t); +q=1100; // radiation received by car in W/sq.m +q_in=t*q; // energy transmitted from the sun through the glass +printf("\nThe energy transmitted from the sun through the glass is %.1f W/sq.m",q_in); +printf("\n\t\t\tSolution to Part (b)\n"); +Tb=311; // temperature of black body source in K +lambda1_Tb=lambda1*Tb; +lambda2_Tb=lambda2*Tb; +printf("\nThe Lower and Upper limits of Wavelength-Temperature Products are %.2e m.K and %.2e m.K respectively",lambda1_Tb,lambda2_Tb); +dI_b=0; // Table 11.1 gives negligibly small values of the corresponding integrals. +t_b=dI_b*0.68; // transmissivity +q_out=t_b*q; +printf("\nthe rate at which radiant energy from the car interior is transmitted through the glass windshield is %d W/sq.m",q_out); diff --git a/1309/CH12/EX12.3/Result12_3.pdf b/1309/CH12/EX12.3/Result12_3.pdf Binary files differnew file mode 100755 index 000000000..6a905460c --- /dev/null +++ b/1309/CH12/EX12.3/Result12_3.pdf diff --git a/1309/CH12/EX12.3/ch12_3.sce b/1309/CH12/EX12.3/ch12_3.sce new file mode 100755 index 000000000..06061dc96 --- /dev/null +++ b/1309/CH12/EX12.3/ch12_3.sce @@ -0,0 +1,37 @@ +clc; +clear; +printf("\t\t\tChapter12_example3\n\n\n"); +// Determination of the heat transferred by radiation from dA1 to A. +// The view factor Fd1_2 can be calculated as Fd1_2=Fd1_3+Fd1_4+Fd1_5 +// For Fd1_3 +a_13=100; +b_13=250; +c_13=100; +X_13=a_13/c_13; +Y_13=b_13/c_13; +printf("\nFor Fd1_3, the values of a/c=%.1f and b/c=%.1f",X_13,Y_13); +Fd1_3=0.17; // value for Fd1_3 corresponding to above calculated values of a/c and b/c +// For Fd1_4 +a_14=300; +b_14=50; +c_14=100; +X_14=a_14/c_14; +Y_14=b_14/c_14; +printf("\nFor Fd1_4, the values of a/c=%.1f and b/c=%.1f",X_14,Y_14); +Fd1_4=0.11; //value for Fd1_4 corresponding to above calculated values of a/c and b/c +// For Fd1_5 +a_15=100; +b_15=50; +c_15=100; +X_15=a_15/c_15; +Y_15=b_15/c_15; +printf("\nFor Fd1_5, the values of a/c=%.1f and b/c=%.1f",X_15,Y_15); +Fd1_5=0.09; //value for Fd1_3 corresponding to above calculated values of a/c and b/c +Fd1_2=Fd1_3+Fd1_4-Fd1_5; +printf("\nFd1_2=%.2f",Fd1_2); +printf("\n%d percent of the energy leaving dA1 reaches A",100*Fd1_2); +sigma=0.1714e-8; // Stefan-Boltzmann constant +T1=660; +T2=560; +q12_A1=sigma*Fd1_2*(T1^4-T2^4); +printf("\nThe net heat transferred is %.1f BTU/(hr.sq.ft)",q12_A1); diff --git a/1309/CH12/EX12.4/Result12_4.pdf b/1309/CH12/EX12.4/Result12_4.pdf Binary files differnew file mode 100755 index 000000000..49171e3c7 --- /dev/null +++ b/1309/CH12/EX12.4/Result12_4.pdf diff --git a/1309/CH12/EX12.4/ch12_4.sce b/1309/CH12/EX12.4/ch12_4.sce new file mode 100755 index 000000000..876b57d5b --- /dev/null +++ b/1309/CH12/EX12.4/ch12_4.sce @@ -0,0 +1,16 @@ +clc; +clear; +printf("\t\t\tChapter12_example4\n\n\n"); +// Determination of the heat transferred to the conveyed parts for the conditions given +L1=1; +angle=%pi*45/180; +L2=L1*sin(angle); +L3=L2; +printf("\nThe Widths are L1=%d m, L2=%.3f m and L3=%.3f m",L1,L2,L3); +T1=303; +T2=473; +sigma=5.67e-8; // Stefan-Boltzmann constant +q21_A2=sigma*(T2^4-T1^4)*((L1/L2)+1-(L3/L2))/2; +q31_A3=sigma*(T2^4-T1^4)*((L1/L2)-1+(L3/L2))/2; +printf("\nThe heat transferred from A2 to A1 is %.2e W/sq.m",q21_A2); +printf("\nThe heat transferred from A3 to A1 is %.2e W/sq.m",q31_A3); diff --git a/1309/CH12/EX12.5/Result12_5.pdf b/1309/CH12/EX12.5/Result12_5.pdf Binary files differnew file mode 100755 index 000000000..490d263b0 --- /dev/null +++ b/1309/CH12/EX12.5/Result12_5.pdf diff --git a/1309/CH12/EX12.5/ch12_5.sce b/1309/CH12/EX12.5/ch12_5.sce new file mode 100755 index 000000000..08cc42a57 --- /dev/null +++ b/1309/CH12/EX12.5/ch12_5.sce @@ -0,0 +1,22 @@ +clc; +clear; +printf("\t\t\tChapter12_example5\n\n\n"); +// Determination of the heat exchanged between the two plates +// The view factor can be found with the crossed-string method +// from figure 12.13(b) +ac=1; +bd=1; +ad=(9+1)^0.5; +bc=ad; +crossed_strings=ad+bc; +uncrossed_strings=ac+bd; +L1_F12=(1/2)*(crossed_strings-uncrossed_strings); +printf("\nThe Product L1F12=%.2f ft",L1_F12); +L1=3; +F12=L1_F12/L1; +printf("\nThe view factor F12=%.2f",F12); +sigma=5.67e-8; // Stefan-Boltzmann constant +T1=560; +T2=460; +q12_A1=sigma*(T1^4-T2^4)*F12; +printf("\nThe heat transfer rate is %.2e W/sq.m",q12_A1); diff --git a/1309/CH12/EX12.6/Result12_6.pdf b/1309/CH12/EX12.6/Result12_6.pdf Binary files differnew file mode 100755 index 000000000..59053a362 --- /dev/null +++ b/1309/CH12/EX12.6/Result12_6.pdf diff --git a/1309/CH12/EX12.6/ch12_6.sce b/1309/CH12/EX12.6/ch12_6.sce new file mode 100755 index 000000000..ce5bb6eaf --- /dev/null +++ b/1309/CH12/EX12.6/ch12_6.sce @@ -0,0 +1,26 @@ +clc; +clear; +printf("\t\t\tChapter12_example6\n\n\n"); +// Determination of the heat that must be supplied to each of the isothermal surfaces, and also the temperature of the insulated surface. +// we can apply the equations as follows +// q1=sigma*A1*[(T1^4-T2^4)F12+(T1^4-T3^4)F13]..... (1) +// q2=sigma*A2*[(T2^4-T1^4)F21+(T2^4-T3^4)F23]..... (2) +// q3=sigma*A3*[(T3^4-T1^4)F31+(T3^4-T2^4)F32]..... (3) +// given data: +T1=1000; +T3=500; +q2=0; +F12=1/2; +F13=1/2; +F21=1/2; +F23=1/2; +F31=1/2; +F32=1/2; +T2=[(T1^4+T3^4)/2]^(1/4); // using equation (2) +printf("\nThe temperature T2=%.1f degree R",T2); +sigma=0.1714e-8; // Stefan-Boltzmann constant +q1_A1=sigma*[(T1^4-T2^4)*F12+(T1^4-T3^4)*F13]; // using equation (1) +printf("\nThe heat flux through area A1 is %d BTU/(hr.sq.ft)",q1_A1); +q3_A3=sigma*[(T3^4-T1^4)*F31+(T3^4-T2^4)*F32]; // using equation (3) +printf("\nThe heat flux through area A3 is %d BTU/(hr.sq.ft)",q3_A3); +printf("\nThe results are logical in that the heat entering the system (the oven itself) must equal that which leaves under steady-state conditions."); diff --git a/1309/CH12/EX12.7/Result12_7.pdf b/1309/CH12/EX12.7/Result12_7.pdf Binary files differnew file mode 100755 index 000000000..4f328c0a4 --- /dev/null +++ b/1309/CH12/EX12.7/Result12_7.pdf diff --git a/1309/CH12/EX12.7/ch12_7.sce b/1309/CH12/EX12.7/ch12_7.sce new file mode 100755 index 000000000..f1bf55ff8 --- /dev/null +++ b/1309/CH12/EX12.7/ch12_7.sce @@ -0,0 +1,18 @@ +clc; +clear; +printf("\t\t\tChapter12_example7\n\n\n"); +// Determination of the heat lost by the oven through its top surface. +// all energy leaving A1 is intercepted by A2 and vice versa +F12=1; +F21=1; +F11=0; // the surfaces are flat +F22=0; +emissivity1=0.94; // for oxidized steel from appendix table E1 +emissivity2=0.94 +T1=533; +T2=323; +sigma=5.67e-8; // Stefan-Boltzmann constant +q1=(sigma*(T1^4-T2^4))/((1/emissivity1)+(1/emissivity2)-1); +printf("\nThe heat lost through bottom surface is %d W/sq.m",q1); +q2=-q1; +printf("\nThe heat lost through top surface is %d W/sq.m",q2); diff --git a/1309/CH12/EX12.8/Result12_8.pdf b/1309/CH12/EX12.8/Result12_8.pdf Binary files differnew file mode 100755 index 000000000..e729a7b95 --- /dev/null +++ b/1309/CH12/EX12.8/Result12_8.pdf diff --git a/1309/CH12/EX12.8/ch12_8.sce b/1309/CH12/EX12.8/ch12_8.sce new file mode 100755 index 000000000..ac436d2e1 --- /dev/null +++ b/1309/CH12/EX12.8/ch12_8.sce @@ -0,0 +1,24 @@ +clc; +clear; +printf("\t\t\tChapter12_example8\n\n\n"); +// Determination of the net heat exchanged between the dish and the surroundings by radiation at the instant the dish is removed from the oven. Perform the calculations (a) if the dish and surroundings behave like black bodies, and again (b) if the dish has an emissivity of 0.82 and the surroundings have an emissivity of 0.93. +D=12/12; // diameter in ft +L=6/12; // length in ft +A=2*%pi*D^2/4+%pi*D*L; +printf("\nThe Surface area is %.2f sq.ft",A); +printf("\n\t\t\tSolution to part (a)\n"); +F12=1; // the view factor between the dish and the surroundings is unity +T1=810; +T2=530; +sigma=0.1714e-8; // Stefan-Boltzmann constant +q1=sigma*A*(T1^4-T2^4)*F12; +printf("\nThe heat exchanged between the dish and the surroundings is %d BTU/hr",q1); +printf("\n\t\t\tSolution to part (b)\n"); +// For gray-surface behavior, we can apply the following Equation +// q1/(A1e1)-[F11*(q1/A1)*(1-e1)/e1+F12*(q2/A2)*(1-e2)/e2]=sigma*T1^4-(F11*sigma*T1^4+F12*sigma*T2^4)... equation (1) +F11=0; +e1=0.82; +e2=0.93; +// putting q2/A2=0 in equation (1) as A2 tends to infinity +q1_=A*e1*[sigma*T1^4-F12*sigma*T2^4]; +printf("\nThe heat exchanged between the dish and the surroundings for the second case is %d BTU/hr",q1_); diff --git a/1309/CH2/EX2.1/Result2_1.pdf b/1309/CH2/EX2.1/Result2_1.pdf Binary files differnew file mode 100755 index 000000000..7dee40f68 --- /dev/null +++ b/1309/CH2/EX2.1/Result2_1.pdf diff --git a/1309/CH2/EX2.1/ch2_1.sce b/1309/CH2/EX2.1/ch2_1.sce new file mode 100755 index 000000000..acd027382 --- /dev/null +++ b/1309/CH2/EX2.1/ch2_1.sce @@ -0,0 +1,23 @@ +clc;
+clear;
+printf("\t\t\tChapter2_example1\n\n\n");
+// determination of the heat flow through a composite wall
+T3=-10; // temperature of inside wall in degree Fahrenheit
+T0=70; // temperature of outside wall in degree Fahrenheit
+dT=T0-T3; // overall temperature difference
+// values of thermal conductivity in BTU/(hr.ft.degree Rankine) from appendix table B3
+k1=0.38; // brick masonry
+k2=0.02; // glass fibre
+k3=0.063; // plywood
+dx1=4/12; // thickness of brick layer in ft
+dx2=3.5/12; // thickness of glass fibre layer in ft
+dx3=0.5/12; // thickness of plywood layer in ft
+A=1; // cross sectional area taken as 1 ft^2
+R1=dx1/(k1*A); // resistance of brick layer in (hr.degree Rankine)/BTU
+R2=dx2/(k2*A); // resistance of glass fibre layer in (hr.degree Rankine)/BTU
+R3=dx3/(k3*A); // resistance of plywood layer in (hr.degree Rankine)/BTU
+printf("\nResistance of brick layer is %.3f (hr.degree Rankine)/BTU",R1);
+printf("\nResistance of glass fibre layer is %.1f (hr.degree Rankine)/BTU",R2);
+printf("\nResistance of plywood layer is %.3f (hr.degree Rankine)/BTU",R3);
+qx=(T0-T3)/(R1+R2+R3);
+printf("\nHeat transfer through the composite wall is %.2f BTU/hr",qx);
diff --git a/1309/CH2/EX2.10/Result2_10.pdf b/1309/CH2/EX2.10/Result2_10.pdf Binary files differnew file mode 100755 index 000000000..d01f81070 --- /dev/null +++ b/1309/CH2/EX2.10/Result2_10.pdf diff --git a/1309/CH2/EX2.10/ch2_10.sce b/1309/CH2/EX2.10/ch2_10.sce new file mode 100755 index 000000000..390cc3c5e --- /dev/null +++ b/1309/CH2/EX2.10/ch2_10.sce @@ -0,0 +1,25 @@ +clc;
+clear;
+printf("\t\t\tChapter2_example10\n\n\n");
+// determination of optimum fin length and heat transferred by fin
+k=8.32; // thermal conductivity of Type 304 stainless steel in BTU/(hr.ft.degree Rankine)from appendix table B2
+hc=400; // the convective heat transfer coefficient given in BTU/(hr.ft^2. degree Rankine)
+printf("\n\t\t\tSolution to part (a)\n");
+delta_opt=0.55/(12*2);
+// determination of dimension of one fin using the equation delta_opt=0.583*hc*Lc^2/k
+Lc=sqrt(delta_opt*k/(0.583*hc));
+printf("\nThe optimum length is %.3f in",Lc*12);
+printf("\n\n\t\t\tSolution to part (b)\n");
+A=Lc*delta_opt;
+// determination of parameter for finding out efficiency from graph
+parameter=Lc^1.5*sqrt(hc/(k*A));
+printf("\nThe parameter value for finding the efficiency is: %.2f",parameter);
+efficiency=0.6;
+printf("\nThe efficiency found from the graph in figure 2.36 is %.1f", efficiency);
+W=1/(2*12);// width in ft
+T_w=190; // wall temperature in degree fahrenheit
+T_inf=58; // ambient temperature in degree fahrenheit
+L=1; // length in ft
+delta=W/2;
+q_ac=efficiency*hc*2*W*sqrt(L^2+delta^2)*(T_w-T_inf);
+printf("\nThe actual heat transferred is %d BTU/hr",q_ac);
diff --git a/1309/CH2/EX2.11/Result2_11.pdf b/1309/CH2/EX2.11/Result2_11.pdf Binary files differnew file mode 100755 index 000000000..19a0f3a29 --- /dev/null +++ b/1309/CH2/EX2.11/Result2_11.pdf diff --git a/1309/CH2/EX2.11/ch2_11.sce b/1309/CH2/EX2.11/ch2_11.sce new file mode 100755 index 000000000..6309fe193 --- /dev/null +++ b/1309/CH2/EX2.11/ch2_11.sce @@ -0,0 +1,41 @@ +clc;
+clear;
+printf("\t\t\tChapter2_example11\n\n\n");
+// determination of heat transferred and fin effectiveness
+printf("\t\t\tSolution to part (a)\n");
+//parameters of the problem are
+N=9; // number of fins
+delta=0.003/2;
+L=0.025;
+Lc=L+delta;
+R=0.219/2;
+R2c=R+delta;
+R1=R-L;
+T_w=260; // root wall temperature in degree celsius
+T_inf=27; // ambient temperature in degree celsius
+hc=15;
+k=52; // thermal conductivity of cast iron in W/(m.K)from appendix table B2
+Ap=2*delta*Lc;
+As=2*3.14*(R2c^2-R1^2);
+radius_ratio=R2c/R1; // for finding efficiency from figure 2.38
+variable=Lc^1.5*sqrt(hc/(k*Ap));
+printf("\n\nThe value of R2c/R1 is %.2f",radius_ratio);
+printf("\n\nThe value of Lc^(3/2)(hc/kAp)^(1/2) is %.2f",variable);
+efficiency=0.93; // efficiency from figure 2.38
+printf("\n\nThe efficiency of the fin from figure 2.38 is %.2f",efficiency);
+qf=N*efficiency*As*hc*(T_w-T_inf);
+printf("\n\nThe heat transferred by the nine fins is %.1f w",qf);
+Sp=0.0127; // fin spacing
+Asw=2*3.14*R1*Sp*N; // exposed surface area
+qw=hc*Asw*(T_w-T_inf);
+printf("\n\nThe heat transferred by exposed surface of the cylinder is %d W",qw);
+q=qf+qw;
+printf("\n\nThe total heat transferred from the cylinder is %d W",q)
+printf("\n\n\t\t\tSolution to part (b)\n");
+H=N*(Sp+2*delta);// height of cylinder
+Aso=2*3.14*R1*H; // surface area without fins
+qo=hc*Aso*(T_w-T_inf);
+printf("\n\nThe Heat transferred without fins is %d W",qo)
+printf("\n\n\t\t\tSolution to part (c)\n");
+effectiveness=q/qo; // effectiveness defined as ratio of heat transferred with fins to heat transferred without fins
+printf("\nThe fin effectiveness is %.2f",effectiveness);
diff --git a/1309/CH2/EX2.2/Result2_2.pdf b/1309/CH2/EX2.2/Result2_2.pdf Binary files differnew file mode 100755 index 000000000..a2eee18bf --- /dev/null +++ b/1309/CH2/EX2.2/Result2_2.pdf diff --git a/1309/CH2/EX2.2/ch2_2.sce b/1309/CH2/EX2.2/ch2_2.sce new file mode 100755 index 000000000..a4dcbbe4c --- /dev/null +++ b/1309/CH2/EX2.2/ch2_2.sce @@ -0,0 +1,27 @@ +clc; +clear; +printf("\t\t\tChapter2_example2\n\n\n"); +// determination of heat transfer through composite wall for materials in parallel +// values of thermal conductivities in W/(m.K) from appendix table B3 +k1=0.45;// thermal conductivity of brick +k2a=0.15; // thermal conductivity of pine +k3=0.814; // thermal conductivity of plaster board +k2b=0.025; // thermal conductivity of air from appendix table D1 +// Areas needed fpor evaluating heat transfer in sq.m +A1=0.41*3; // cross sectional area of brick layer +A2a=0.038*3; // cross sectional area of wall stud +A2b=(41-3.8)*0.01*3; // cross sectional area of air layer +A3=0.41*3; // cross sectional area of plastic layer +dx1=0.1; // thickness of brick layer in m +dx2=0.089; // thickness of wall stud and air layer in m +dx3=0.013; // thickness of plastic layer in m +R1=dx1/(k1*A1); // Resistance of brick layer in K/W +R2=dx2/(k2a*A2a+k2b*A2b); // Resistance of wall stud and air layer in K/W +R3=dx3/(k3*A3); // Resistance of plastic layer in K/W +printf("\nResistance of brick layer is %.3f K/W",R1); +printf("\nResistance of wall stud and air layer is %.2f K/W",R2); +printf("\nResistance of plastic layer is %.3f K/W",R3); +T1=25; // temperature of inside wall in degree celsius +T0=0; // temperature of outside wall in degree celsius +qx=(T1-T0)/(R1+R2+R3); // heat transfer through the composite wall in W +printf("\nHeat transfer through the composite wall is %.1f W",qx); diff --git a/1309/CH2/EX2.3/Result2_3.pdf b/1309/CH2/EX2.3/Result2_3.pdf Binary files differnew file mode 100755 index 000000000..c83517040 --- /dev/null +++ b/1309/CH2/EX2.3/Result2_3.pdf diff --git a/1309/CH2/EX2.3/ch2_3.sce b/1309/CH2/EX2.3/ch2_3.sce new file mode 100755 index 000000000..c5697d5db --- /dev/null +++ b/1309/CH2/EX2.3/ch2_3.sce @@ -0,0 +1,31 @@ +clc;
+clear;
+printf("\t\t\tChapter2_example3\n\n\n");
+// determination of heat transfer rate and overall heat transfer coefficient
+k1=24.8; // thermal conductivity of 1C steel in BTU/(hr.ft.degree Rankine)from appendix table B2
+k2=0.02; // thermal conductivity of styrofoam steel in BTU/(hr.ft.degree Rankine)
+k3=0.09; // thermal conductivity of fibreglass in BTU/(hr.ft.degree Rankine)
+hc1=0.79; // convection coefficient between the air and the vertical steel wall in BTU/(hr.ft^2.degree Rankine)
+hc2=150; // the convection coefficient between the ice water and the fiberglass
+A=1; // calculation based on per square foot
+dx1=0.04/12; // thickness of steel in ft
+dx2=0.75/12; // thickness of styrofoam in ft
+dx3=0.25/12; // thickness of fiberglass in ft
+// Resistances in (degree Fahrenheit.hr)/BTU
+disp('Resistances in (degree Fahrenheit.hr)/BTU:');
+Rc1=1/(hc1*A); // Resistance from air to sheet metal
+printf("\nResistance from air to sheet metal: %.3f degree F.hr/BTU",Rc1);
+Rk1=dx1/(k1*A); // Resistance of steel layer
+printf("\nResistance of steel layer: %.4f degree F.hr/BTU",Rk1);
+Rk2=dx2/(k2*A); // Resistance of styrofoam layer
+printf("\nResistance of styrofoam layer: %.3f degree F.hr/BTU",Rk2);
+Rk3=dx3/(k3*A); // Resistance of fiberglass layer
+printf("\nResistance of fiberglass layer: %.3f degree F.hr/BTU",Rk3);
+Rc2=1/(hc2*A); // Resistance from ice water to fiberglass
+printf("\nResistance from ice water to fiberglass: %.4f degree F.hr/BTU",Rc2);
+U=1/(Rc1+Rk1+Rk2+Rk3+Rc2); // overall heat transfer coefficient in BTU/(hr.ft^2.degree Rankine)
+printf("\nThe overall heat transfer coefficient is %.3f BTU/(hr. sq.ft.degree Rankine)",U);
+T_inf1=90;// temperature of air in degree F
+T_inf2=32; // temperature of mixture of ice and water in degree F
+q=U*A*(T_inf1-T_inf2);
+printf("\nThe heat transfer rate is %.1f BTU/hr",q);
diff --git a/1309/CH2/EX2.4/Result2_4.pdf b/1309/CH2/EX2.4/Result2_4.pdf Binary files differnew file mode 100755 index 000000000..7b556f3f2 --- /dev/null +++ b/1309/CH2/EX2.4/Result2_4.pdf diff --git a/1309/CH2/EX2.4/ch2_4.sce b/1309/CH2/EX2.4/ch2_4.sce new file mode 100755 index 000000000..4bc174774 --- /dev/null +++ b/1309/CH2/EX2.4/ch2_4.sce @@ -0,0 +1,13 @@ +clc;
+clear;
+printf("\t\t\tChapter2_example4\n\n\n");
+// determination of the heat transfer through the pipe wall per unit length of pipe.
+k=14.4; // thermal conductivity of 304 stainless steel in W/(m.K) from appendix table B2
+// dimensions of steel pipes in cm from appendix table F1
+D2=32.39;
+D1=29.53;
+T1=40;
+T2=38;
+Qr_per_length=(2*3.14*k)*(T1-T2)/log(D2/D1);
+format(6);
+printf("\nThe heat transfer through the pipe wall per unit length of pipe is %.1f W/m = %.2f kW/m",Qr_per_length,Qr_per_length/1000);
diff --git a/1309/CH2/EX2.5/Result2_5.pdf b/1309/CH2/EX2.5/Result2_5.pdf Binary files differnew file mode 100755 index 000000000..3b9164fb2 --- /dev/null +++ b/1309/CH2/EX2.5/Result2_5.pdf diff --git a/1309/CH2/EX2.5/ch2_5.sce b/1309/CH2/EX2.5/ch2_5.sce new file mode 100755 index 000000000..1a6db96e3 --- /dev/null +++ b/1309/CH2/EX2.5/ch2_5.sce @@ -0,0 +1,24 @@ +clc;
+clear;
+printf("\t\t\tChapter2_example5\n\n\n");
+// determination of the heat gain per unit length
+k1=231; // thermal conductivity of copper in BTU/(hr.ft.degree Rankine)from appendix table B1
+k2=0.02; // thermal conductivity of insuLtion in BTU/(hr.ft.degree Rankine)
+// Specifications of 1 standard type M copper tubing from appendix table F2 are as follows
+D2=1.125/12; // outer diameter in ft
+D1=0.08792; // inner diameter in ft
+R2=D2/2;// outer radius
+printf("\nOuter radius is %.4f ft",R2);
+R1=D1/2; // inner radius
+printf("\nOuter radius is %.3f ft",R1);
+t=0.5/12; // wall thickness of insulation in ft
+R3=R2+t;
+printf("\nRadius including thickness is %.4f ft",R3);
+LRk1=(log(R2/R1))/(2*3.14*k1); // product of length and copper layer resistance
+printf("\nProduct of length and copper layer resistance is: %.1e",LRk1);
+LRk2=(log(R3/R2))/(2*3.14*k2); // product of length and insulation layer resistance
+printf("\nProduct of length and insulation layer resistance is: %.2f",LRk2);
+T1=40; // temperature of inside wall of tubing in degree fahrenheit
+T3=70; // temperature of surface temperature of insulation degree fahrenheit
+q_per_L=(T1-T3)/(LRk1+LRk2); // heat transferred per unit length in BTU/(hr.ft)
+printf("\nThe heat transferred per unit length is %.2f BTU/(hr.ft)",q_per_L);
diff --git a/1309/CH2/EX2.6/Result2_6.pdf b/1309/CH2/EX2.6/Result2_6.pdf Binary files differnew file mode 100755 index 000000000..56478a8c4 --- /dev/null +++ b/1309/CH2/EX2.6/Result2_6.pdf diff --git a/1309/CH2/EX2.6/ch2_6.sce b/1309/CH2/EX2.6/ch2_6.sce new file mode 100755 index 000000000..fccda732a --- /dev/null +++ b/1309/CH2/EX2.6/ch2_6.sce @@ -0,0 +1,18 @@ +clc;
+clear;
+printf("\t\t\tChapter2_example6\n\n\n");
+// Determination of the overall heat transfer coefficient
+k12=24.8; // thermal conductivity of 1C steel in BTU/(hr.ft.degree Rankine)from appendix table B2
+k23=.023; // // thermal conductivity of glass wool insulation in BTU/(hr.ft.degree Rankine)from appendix table B3
+// Specifications of 6 nominal, schedule 40 pipe (no schedule was specified, so the standard is assumed) from appendix table F1 are as follows
+D2=6.625/12; // outer diameter in ft
+D1=0.5054; // inner diameter in ft
+printf("\nOuter diameter is %.3f ft",D2);
+printf("\nInner diameter is %.4f ft",D1);
+t=2/12; // wall thickness of insulation in ft
+D3=D2+t;
+printf("\nDiameter including thickness is %.5f ft",D3);
+hc1=12; // convection coefficient between the air and the pipe wall in BTU/(hr. sq.ft.degree Rankine).
+hc2=1.5; // convection coefficient between the glass wool and the ambient air in BTU/(hr. sq.ft.degree Rankine).
+U=1/((1/hc1)+(D1*log(D2/D1)/k12)+(D1*log(D3/D2)/k23)+(D1/(hc2*D3)));
+printf("\nOverall heat transfer coefficient is %.3f BTU/(hr. sq.ft.degree Fahrenheit)",U);
diff --git a/1309/CH2/EX2.7/Result2_7.pdf b/1309/CH2/EX2.7/Result2_7.pdf Binary files differnew file mode 100755 index 000000000..52777abcf --- /dev/null +++ b/1309/CH2/EX2.7/Result2_7.pdf diff --git a/1309/CH2/EX2.7/ch2_7.sce b/1309/CH2/EX2.7/ch2_7.sce new file mode 100755 index 000000000..966447fd4 --- /dev/null +++ b/1309/CH2/EX2.7/ch2_7.sce @@ -0,0 +1,27 @@ +clc;
+clear;
+printf("\t\t\tChapter2_example7\n\n\n");
+// Determination of the thermal contact resistance
+k=14.4; // thermal conductivity of 304 stainless steel in W/(m.K)from appendix table B2
+T1=543; // temperature in K at point 1
+T2=460; // temperature in K at point 2
+dT=T1-T2; // temperature difference between point 1 and 2
+dz12=0.035; // distance between thermocouple 1 and 2 in cm
+qz_per_A=k*dT/dz12; // heat flow calculated in W/m^2 calculated using Fourier's law
+printf("\nHeat flow calculated is %.2f kW/sq.m",qz_per_A/1000);
+dz56=4.45; // distance between thermocouple 5 and 6 in cm
+dz6i=3.81; // distance between thermocouple 6 and interface in cm
+dz5i=dz56+dz6i; // distance between thermocouple 5 and interface in cm
+T5=374; // temperature in K at point 5
+T6=366; // temperature in K at point 6
+T_ial=T5-(dz5i*(T5-T6)/dz56); // temperature of aluminium interface in K
+printf("\nTemperature of aluminium interface is %.1f K",T_ial);
+dzi7=2.45; // distance between thermocouple 7 and interface in cm
+dz78=4.45; // distance between thermocouple 7 and 8 in cm
+dzi8=dzi7+dz78; // distance between thermocouple 8 and interface in cm
+T7=349; // temperature in K at point 7
+T8=337; // temperature in K at point 8
+T_img=dzi8*(T7-T8)/dz78+T8; // temperature of magnesium interface in K
+printf("\nTemperature of magnesium interface is %.1f K",T_img);
+Rtc=(T_ial-T_img)/qz_per_A;
+printf("\nThe required thermal contact resistance is %.2e K. sq.m/W",Rtc);
diff --git a/1309/CH2/EX2.8/Figure2_8.jpeg b/1309/CH2/EX2.8/Figure2_8.jpeg Binary files differnew file mode 100755 index 000000000..95230e645 --- /dev/null +++ b/1309/CH2/EX2.8/Figure2_8.jpeg diff --git a/1309/CH2/EX2.8/Result2_8.pdf b/1309/CH2/EX2.8/Result2_8.pdf Binary files differnew file mode 100755 index 000000000..76fc3ce27 --- /dev/null +++ b/1309/CH2/EX2.8/Result2_8.pdf diff --git a/1309/CH2/EX2.8/ch2_8.sce b/1309/CH2/EX2.8/ch2_8.sce new file mode 100755 index 000000000..323d0a7f5 --- /dev/null +++ b/1309/CH2/EX2.8/ch2_8.sce @@ -0,0 +1,52 @@ +clc; +clear; +printf("\t\t\tChapter2_example8\n\n\n"); +// determination of temperature profile, heat transferred, efficiency, effectiveness. +printf("\n\t\t\tSolution to part (a)"); +k=24.8; // thermal conductivity of 1C steel in BTU/(hr.ft.degree Rankine)from appendix table B2 +D=(5/16)/12; // diameter of the rod in ft +P=(3.14*D); // Circumference of the rod in ft +printf("\nThe perimeter is %.4f ft",P); +A=(3.14/4)*D^2; // Cross sectional area of the rod in sq.ft +printf("\nThe Cross sectional area is %.6f sq.ft",A); +hc=1; // assuming the convective heat transfer coefficient as 1 BTU/(hr. sq.ft. degree Rankine) +m=sqrt(hc*P/(k*A)); +printf("\nThe value of parameter m is: %.3f/ft",m); +L=(9/2)/12; // length of rod in ft +// using the equation (T-T_inf)/(T_w-T_inf)=(cosh[m(L-z)])/(cosh(mL)) for temperature profile +T_inf=70; +T_w=200; +dT=T_w-T_inf; +const=dT/cosh(m*L); +printf("\nThe temperature profile is:\t"); +printf("T=%d+%.2fcosh[%.3f(%.3f-z)]",T_inf,const,m,L); +z=0:.05:L; +T=T_inf+const*cosh(m*(L-z)); +x=linspace(0,4.5,8); +plot(x,T); +a=gca(); +a.data_bounds=[0,140;5,200]; +newticks=a.x_ticks; +newticks(2)=[0;1;2;3;4;5]; +newticks(3)=['0';'1';'2';'3';'4';'5']; +a.x_ticks=newticks; +newticks1=a.y_ticks; +newticks1(2)=[140;150;160;170;180;190;200]; +newticks1(3)=['140';'150';'160';'170';'180';'190';'200']; +a.y_ticks=newticks1; +xlabel('Rod length z, in'); +ylabel('Temperature T, degree fahrenheit'); +title('Temperature_distribution_within_the_rod'); +printf("\n\n\t\t\tSolution to part (b)\n"); +// the heat transferred can be calculated using the equation qz=k*A*m*(T_w-T_inf)*tanh(m*L) +qz=k*A*m*dT*tanh(m*L); +printf("\nThe heat transferred is %.2f BTU/hr",qz); +printf("\n\n\t\t\tSolution to part (c)\n"); +mL=m*L; +printf("\nThe value of mL is: %.3f",mL); +efficiency=0.78; +printf("\nThe efficiency found from the graph in figure 2.30 is: %.2f",efficiency); +printf('\n\n\t\t\tSolution to part (d)\n'); +// the effectiveness can be found using the equation effectiveness=sqrt(k*P/h*A)*tanh(mL) +effectiveness=sqrt(k*P/(hc*A))*tanh(mL); +printf("\nThe effectiveness is found to be: %.1f",effectiveness); diff --git a/1309/CH2/EX2.9/Result2_9.pdf b/1309/CH2/EX2.9/Result2_9.pdf Binary files differnew file mode 100755 index 000000000..fab77620c --- /dev/null +++ b/1309/CH2/EX2.9/Result2_9.pdf diff --git a/1309/CH2/EX2.9/ch2_9.sce b/1309/CH2/EX2.9/ch2_9.sce new file mode 100755 index 000000000..e1bac49c5 --- /dev/null +++ b/1309/CH2/EX2.9/ch2_9.sce @@ -0,0 +1,26 @@ +clc;
+clear;
+printf("\t\t\tChapter2_example9\n\n\n");
+// determination of heat transferred
+k=136; // thermal conductivity of aluminium in BTU/(hr.ft.degree Rankine)from appendix table B1
+L=9/(8*12);
+W=9/(4*12);
+delta=1/(32*12);
+printf("\nLength=%.5f ft, Width=%.4f ft, Delta=%.6f ft",L,W,delta);
+hc=0.8; // the convective heat transfer coefficient estimated as 1 BTU/(hr.ft^2. degree Rankine)
+T_w=1000;// the root temperature in degree fahrenheit
+T_inf=90; // the ambient temperature in degree fahrenheit
+m=sqrt(hc/(k*delta));
+printf("\nThe value of m is %.3f",m);
+P=2*W;
+A=2*delta*W;
+printf("\n\t\t\tSolution to part (a)\n");
+qz1=sqrt(hc*P*k*A)*(T_w-T_inf)*(sinh(m*L)+(hc/(m*k)*cosh(m*L)))/(cosh(m*L)+(hc/(m*k)*sinh(m*L)));
+printf("\nThe heat transferred is %.2f BTU/hr",qz1);
+printf("\n\n\t\t\tSolution to part (b)\n");
+qz2=sqrt(k*A*hc*P)*(T_w-T_inf)*tanh(m*L);
+printf("\nThe heat transferred is %.2f BTU/hr\n",qz2);
+printf("\n\t\t\tSolution to part (c)\n");
+Lc=L+delta;
+qz3=k*A*m*(T_w-T_inf)*tanh(m*L*(1+delta/Lc));
+printf("\nThe heat transferred is %.2f BTU/hr\n",qz3);
diff --git a/1309/CH3/EX3.1/Result3_1.pdf b/1309/CH3/EX3.1/Result3_1.pdf Binary files differnew file mode 100755 index 000000000..3ae292148 --- /dev/null +++ b/1309/CH3/EX3.1/Result3_1.pdf diff --git a/1309/CH3/EX3.1/ch3_1.sce b/1309/CH3/EX3.1/ch3_1.sce new file mode 100755 index 000000000..17ef2388f --- /dev/null +++ b/1309/CH3/EX3.1/ch3_1.sce @@ -0,0 +1,18 @@ +clc; +clear; +printf("\t\t\tChapter3_example1\n\n\n"); +// Determination of the heat-flow rate from one tube +// specifications of 1 standard type K from table F2 +OD=0.02858; // outer diameter in m +// from figure 3.11 +M=8; // total number of heat-flow lanes +N=6; // number of squares per lane +S_L=M/N; // conduction shape factor +printf("\nThe Conduction shape factor is %.3f",S_L); +k=0.128; // thermal conductivity in W/(m.K) for concrete from appendix table B3 +T1=85; // temperature of tube surface +T2=0; // temperature of ground beneath the slab +q_half=k*S_L*(T1-T2); +printf("\nThe heat flow per unit length from one half of one tube is %.1f W/m",q_half); +q=2*q_half; +printf("\nThe total heat flow per tube is %.1f W/m",q); diff --git a/1309/CH3/EX3.2/Result3_2.pdf b/1309/CH3/EX3.2/Result3_2.pdf Binary files differnew file mode 100755 index 000000000..521390feb --- /dev/null +++ b/1309/CH3/EX3.2/Result3_2.pdf diff --git a/1309/CH3/EX3.2/ch3_2.sce b/1309/CH3/EX3.2/ch3_2.sce new file mode 100755 index 000000000..dfa8e4e81 --- /dev/null +++ b/1309/CH3/EX3.2/ch3_2.sce @@ -0,0 +1,15 @@ +clc; +clear; +printf("\t\t\tChapter3_example2\n\n\n"); +// Determination of the heat transferred from the buried pipe per unit length +// shape factor number 8 is selected from table 3.1 +// specifications of 10 nominal, schedule 80 pipe from table F1 +OD=10.74/12; // diameter in ft +R=OD/2; +T1=140; +T2=65; +k=0.072; // thermal conductivity in BTU/(hr-ft. degree R) +d=18/12; // distance from centre-line +S_L=(2*%pi)/(acosh(d/R)); +q_L=k*S_L*(T1-T2); +printf("\nThe heat transferred from the buried pipe per unit length is %.1f BTU/(hr.ft)",q_L); diff --git a/1309/CH3/EX3.3/Result3_3.pdf b/1309/CH3/EX3.3/Result3_3.pdf Binary files differnew file mode 100755 index 000000000..a5f90bd5a --- /dev/null +++ b/1309/CH3/EX3.3/Result3_3.pdf diff --git a/1309/CH3/EX3.3/ch3_3.sce b/1309/CH3/EX3.3/ch3_3.sce new file mode 100755 index 000000000..215a007ae --- /dev/null +++ b/1309/CH3/EX3.3/ch3_3.sce @@ -0,0 +1,46 @@ +clc; +clear; +printf("\t\t\tChapter3_example3\n\n\n"); +// Determination of the heat lost through the walls, using the shape-factor method. (b) Repeat the calculations but neglect the effects of the corners; that is, assume only one-dimensional effects through all the walls. +k = 1.07; // thermal conductivity of silica brick from appendix table B3 in W/(m.K) +// Calculation of total shape factor +// From figure 3.12, for component A +S1_A=0.138*0.138/0.006; +nA=2; +St_A=nA*S1_A; // Total shape factor of component A +printf("\nThe Total shape factor of component A is %.3f ",St_A); +// For component B +S1_B=0.138*0.188/0.006; +nB=4; +St_B=nB*S1_B; // Total shape factor of component B +printf("\nThe Total shape factor of component B is %.3f ",St_B); +// For component C +S3_C=0.15*0.006; +nC=8; +St_C=nC*S3_C; // Total shape factor of component C +printf("\nThe Total shape factor of component C is %.4f ",St_C); +// For component D +S2_D=0.54*0.188; +nD=4; +St_D=nD*S2_D; // Total shape factor of component D +printf("\nThe Total shape factor of component D is %.5f ",St_D); +// For component E +S2_E=0.138*0.54; +nE=8; +St_E=nE*S2_E; // Total shape factor of component E +printf("\nThe Total shape factor of component E is %.5f ",St_E); +S=St_A+St_B+St_C+St_D+St_E; +printf("\nThe Total shape factor is %.2f",S); +printf("\n\t\t\tSolution to part (a)\n"); +T1=550; +T2=30; +q=k*S*(T1-T2); +printf("\nThe heat transferred through the walls of the furnace is %d W = %.1f kW",q,q/1000); +printf("\n\n\t\t\tSolution to part (b)\n"); +// Neglecting the effects of the edges and corners, the shape factor for all walls is found as +S=St_A+St_B; +printf("\nNeglecting the effects of the edges and corners, the shape factor for all walls is %.2f",S); +q_1=k*S*(T1-T2); +printf("\nNeglecting the effects of the edges and corners, the heat transferred is %d W = %.1f kW",q_1,q_1/1000); +Error=(q-q_1)/q; +printf("\nThe error introduced by neglecting heat flow through the edges and corners is %.1f percent",Error*100); diff --git a/1309/CH3/EX3.4/Result3_4.pdf b/1309/CH3/EX3.4/Result3_4.pdf Binary files differnew file mode 100755 index 000000000..ffefe1947 --- /dev/null +++ b/1309/CH3/EX3.4/Result3_4.pdf diff --git a/1309/CH3/EX3.4/ch3_4.sce b/1309/CH3/EX3.4/ch3_4.sce new file mode 100755 index 000000000..6bbd6f5af --- /dev/null +++ b/1309/CH3/EX3.4/ch3_4.sce @@ -0,0 +1,19 @@ +clc; +clear; +printf("\t\t\tChapter3_example4\n\n\n"); +// Determination of the conduction shape factor for the underground portion of the configuration +// specifications of 4 nominal, schedule 40 pipe from table F1 +OD=4.5/12; // diameter in ft +R=OD/2; +// For pipe A +L_A=4.5; // length in ft +// shape factor number 9 is selected from table 3.1 +S_A=(2*%pi*L_A)/(log(2*(L_A)/R)); +printf("\nThe Shape Factor of pipe A is %.1f",S_A); +// For pipe B +L_B=18; // length in ft +// shape factor number 9 is selected from table 3.1 +S_B=(2*%pi*L_B)/(acosh(L_A/R)); +printf("\nThe Shape Factor of pipe B is %.1f",S_B); +S=2*S_A+S_B; +printf("\nThe total conduction shape factor for the system is %.1f",S); diff --git a/1309/CH3/EX3.5/Figure3_5.jpg b/1309/CH3/EX3.5/Figure3_5.jpg Binary files differnew file mode 100755 index 000000000..db2db3106 --- /dev/null +++ b/1309/CH3/EX3.5/Figure3_5.jpg diff --git a/1309/CH3/EX3.5/Result3_5.pdf b/1309/CH3/EX3.5/Result3_5.pdf Binary files differnew file mode 100755 index 000000000..f97ea373c --- /dev/null +++ b/1309/CH3/EX3.5/Result3_5.pdf diff --git a/1309/CH3/EX3.5/ch3_5.sce b/1309/CH3/EX3.5/ch3_5.sce new file mode 100755 index 000000000..0ff8a7bd2 --- /dev/null +++ b/1309/CH3/EX3.5/ch3_5.sce @@ -0,0 +1,50 @@ +clc; +clear; +printf("\t\t\tChapter3_example5\n\n\n"); +// (a) Using the pin-fin equations for the case where the exposed tip is assumed insulated, graph the temperature distribution existing within the rod. (b) Use the numerical formulation of this section to obtain the temperature distribution. (c) Compare the two models to determine how well the numerical results approximate the exact results +h=1.1; // convective coefficient in BTU/(hr.ft^2. degree R) +Tw=200; +T_inf=68; // ambient temperature +printf("\n\t\t\tSolution to part (a)\n"); +k=0.47; // thermal conductivity in BTU/(hr.ft.degree R) from table B3 +D=0.25/12; // diameter in ft +A=%pi*D^2/4; // cross sectional area in ft^2 +P=%pi*D; // perimeter in ft +printf("\nThe cross sectional area is %.3e sq.ft and Perimeter is %.3e ft",A,P); +L=6/12; // length in ft +mL=L*((h*P)/(k*A))^0.5; +printf("\nThe value of Product mL is %.2f",mL); +z=0:1.5:6; +[n m]=size(z); +for i=1:m + T(i)=T_inf+(Tw-T_inf)*(cosh(mL*(1-(z(i)/6)))/(cosh(mL))); +end +printf("\n\n\t\t\tSolution to part (b)\n"); +d_zeta=1/4; +K=2+(mL*d_zeta)^2; +printf("\nThe value of K is %.4f",K); +T_(5)=T_inf+(Tw-T_inf)*(2/(K^4-4*K^2+2)); +T_(4)=T_inf+(Tw-T_inf)*(K/(K^4-4*K^2+2)); +T_(3)=T_inf+(Tw-T_inf)*((K^2-1)/(K^4-4*K^2+2)); +T_(2)=T_inf+(Tw-T_inf)*((K^3-3*K)/(K^4-4*K^2+2)); +T_(1)=200; +printf("\n\nA Comparison of Exact to Numerical Results for the Data of Example 3.5"); +printf("\nz,in\tExact (e) T\tNumerical (n) T\t Percent error (e - n)/e"); +for i=1:m +err(i)=(T(i)-T_(i))/T(i); +printf("\n%.1f\t%.2f\t\t%.2f\t\t%.2f\n",z(i),T(i),T_(i),err(i)*100); +end +plot(z,T,z,T_); +a=gca(); +newticks1=a.x_ticks; +newticks1(2)=[0;1.5;3.0;4.5;6]; +newticks1(3)=['0';'1.5';'3.0';'4.5';'6']; +a.x_ticks=newticks1; +newticks2=a.y_ticks; +newticks2(2)=[75;100;125;150;175;200]; +newticks2(3)=['75';'100';'125';'150';'175';'200']; +a.y_ticks=newticks2; +title('A comparison of the exact to the numerical temperature profiles for the pin fin of Example 3.5'); +xlabel("z, in"); +ylabel("T, degree F"); +hl=legend(['Exact Solution';'Numerical Solution']);
\ No newline at end of file diff --git a/1309/CH4/EX4.1/Result4_1.pdf b/1309/CH4/EX4.1/Result4_1.pdf Binary files differnew file mode 100755 index 000000000..4f011f37a --- /dev/null +++ b/1309/CH4/EX4.1/Result4_1.pdf diff --git a/1309/CH4/EX4.1/ch4_1.sce b/1309/CH4/EX4.1/ch4_1.sce new file mode 100755 index 000000000..8205a1f31 --- /dev/null +++ b/1309/CH4/EX4.1/ch4_1.sce @@ -0,0 +1,19 @@ +clc; +clear; +printf("\t\t\tChapter4_example1\n\n\n"); +// determination of response time +k=12; // thermal conductivity in BTU/(hr.ft.degree Rankine) +c=0.1; // specific heat in BTU/(lbm.degree Rankine) +D=0.025/12; // diameter in ft +density=525; // density in lbm/cu.ft +hc=80; // convective coefficient in BTU/(hr. sq.ft. degree Rankine) +T_i=65; // intial temperature in degree fahrenheit +T_inf=140; // fluid temperature in degree fahrenheit +As=3.14*D^2; // surface area in sq.ft +Vs=3.14*D^3/6; // volume in cu.ft +reciprocal_timeconstant=(hc*As)/(density*Vs*c); +printf("\nThe reciprocal of time constant is %.1f /hr",reciprocal_timeconstant); +// selecting T=139 degree fahrenheit as T=140 gives an infinite time through the equation (T-T_inf)/(T_i-T_inf)=exp(-hc*As/density*Vs*c)t +T=139; +t=log((T-T_inf)/(T_i-T_inf))/(-reciprocal_timeconstant); +printf('\n\nThe response time of the junction is %.1f s",t*3600); diff --git a/1309/CH4/EX4.10/Result4_10.pdf b/1309/CH4/EX4.10/Result4_10.pdf Binary files differnew file mode 100755 index 000000000..0fdfaf9cd --- /dev/null +++ b/1309/CH4/EX4.10/Result4_10.pdf diff --git a/1309/CH4/EX4.10/ch4_10.sce b/1309/CH4/EX4.10/ch4_10.sce new file mode 100755 index 000000000..280a1dafc --- /dev/null +++ b/1309/CH4/EX4.10/ch4_10.sce @@ -0,0 +1,17 @@ +clc; +clear; +printf("\t\t\tChapter4_example10\n\n\n"); +// determination of time required to cool to a certain temperature +rou=7.817*62.4; +c=.110; +k=8.32; +alpha=0.417e-4; +dx=1/12; +// taking Fo=1 +Fo=1; +dt=Fo*dx^2/alpha; +printf("\nThe time increments is %.1f s",dt); +// We have to draw the Saul'ev plot to determine the number of time intervals +n=8; //Enter the number of time intervals from Saulev plot +time=n*dt; +printf("\nThe required time is %.2f hr",time/3600); diff --git a/1309/CH4/EX4.2/ch4_2.sce b/1309/CH4/EX4.2/ch4_2.sce new file mode 100755 index 000000000..7b85c7fdb --- /dev/null +++ b/1309/CH4/EX4.2/ch4_2.sce @@ -0,0 +1,28 @@ +clc; +clear; +printf("\t\t\tChapter4_example2\n\n\n"); +// Determination of temperature of metal and cumulative heat rate +// properties of aluminium from appendix table B1 +k=236; // thermal conductivity in W/(m.K) +Cp=896;// specific heat in J/(kg.K) +sp_gr=2.702; // specific gravity +density=2702; // density in kg/cu.m +D=0.05; // Diameter in m +L=0.60; // length in m +hc=550; // unit surface conductance between the metal and the bath in W/(K.sq.m) +Vs=(3.14*D^2*L)/4; // Volume in cu.m +As=(2*3.14*D^2/4)+(3.14*D*L); // surface area in sq.m +printf("\n\nThe volume of cylinder is %.5f cu.m",Vs); +printf("\n\nThe surface area of cylinder is %.3f sq.m",As); +Bi=(hc*Vs)/(k*As); // Biot Number +printf("\n\nThe Biot number is %.3f",Bi); +// Biot number is less than 1 hence lump capacitance equations apply +printf("\n\n\t\t\tSolution to part (a)\n"); +T_i=50; // initial temperature in degree celsius +T_inf=2; // temperature of ice water bath in degree celsius +t=60; // time=1 minute=60 s +T=T_inf+(T_i-T_inf)*exp(-(hc*As)/(density*Vs*Cp)*t); +printf("\nThe temperature of aluminium is %.1f degree celsius",T); +printf("\n\n\t\t\tSolution to part (b)\n"); +Q=density*Vs*Cp*(T_inf-T_i)*[1-exp(-(hc*As)/(density*Vs*Cp)*t)]; +printf("\nThe cumulative heat transferred is %d J =%.1f kJ",abs(Q),abs(-Q/1000)); diff --git a/1309/CH4/EX4.3/Figure4_3.jpeg b/1309/CH4/EX4.3/Figure4_3.jpeg Binary files differnew file mode 100755 index 000000000..ff1b509de --- /dev/null +++ b/1309/CH4/EX4.3/Figure4_3.jpeg diff --git a/1309/CH4/EX4.3/Result4_3.pdf b/1309/CH4/EX4.3/Result4_3.pdf Binary files differnew file mode 100755 index 000000000..2f20eac4b --- /dev/null +++ b/1309/CH4/EX4.3/Result4_3.pdf diff --git a/1309/CH4/EX4.3/ch4_3.sce b/1309/CH4/EX4.3/ch4_3.sce new file mode 100755 index 000000000..b78cb8e36 --- /dev/null +++ b/1309/CH4/EX4.3/ch4_3.sce @@ -0,0 +1,82 @@ +clc;
+clear;
+printf("\t\t\tChapter4_example3\n\n\n");
+hc=30;
+L=0.24;
+k=1.25;
+c=890;
+rou=550;
+Bi=hc*L/k;
+alpha=k/(rou*c);
+printf("The value of diffusivity is %.2e sq.m/s",alpha);
+Tc=150;
+T_inf=600;
+T_i=50;
+printf("\nThe Biot number is %.2f,",Bi);
+if Bi<0.1 then
+ n=0;
+else if Bi>0.1 then
+ n=1;
+ end
+end
+select n
+case 0 then
+ disp('The Lumped capacity approach is applicable');
+case 1 then
+ disp('Since value of Biot number is greater than 0.1, Lumped capacity approach would not give accurate results, so figure 4.6 is to be used');
+ reciprocal_Bi=1/Bi;
+ dimensionless_temp=(Tc-T_inf)/(T_i-T_inf);
+ Fo=0.4; //the value of Fourier Number from figure 4.6(a)
+ t=L^2*Fo/alpha;
+ printf("The required time is %d s = %.1f hr",t,t/3600);
+end
+// reading values of dimensionless temperature from figure 4.6(b) using reciprocal of Biot number
+x_per_L=[0 0.2 0.4 0.6 0.8 0.9 1.0];
+[n,m]=size(x_per_L);
+printf("\nThe choosen values of x/L are: \n");
+disp(x_per_L);
+printf("\n Values for dimensionless temperature for corresponding values of x/L:")
+dim_T=[1.0 .97 .86 .68 .48 .36 .24]; // value for dimensionless temperature for corresponding value of x/L
+disp(dim_T);
+printf("the temperature profile with distance is\n");
+printf("\tx/L\t\t");
+for j=1:m
+ printf("%.2f\t",x_per_L(1,j));
+
+end
+printf("\n");
+printf("(T-T_inf)/T_i-T_inf)\t");
+for i=1:m
+ printf("%.2f\t",dim_T(i));
+end
+T=zeros(1,m);
+x=zeros(1,m);
+for i=1:m
+ T(1,i)=dim_T(1,i)*dimensionless_temp*(T_i-T_inf)+T_inf;
+ x(1,i)=x_per_L(1,i)*L;
+end
+printf("\n\tx,cm\t\t");
+for i=1:m
+ X(1,i)=x(1,i)*100;
+ printf("%.1f\t",X(1,i));
+end
+printf("\nT, degree celsius\t");
+for i=1:m
+ printf("%d\t",T(1,i));
+end
+plot2d(X,T,rect=[0,0,24,600]);
+a=gca();
+newticks=a.x_ticks;
+newticks(2)=[0;4;8;12;16;20;24];
+newticks(3)=['0';'4';'8';'12';'16';'20';'24'];
+a.x_ticks=newticks;
+newticks1=a.y_ticks;
+newticks1(2)=[0;100;200;300;400;500;600];
+newticks1(3)=['0';'100';'200';'300';'400';'500';'600'];
+a.y_ticks=newticks1;
+xlabel('x,cm');
+ylabel('t,degree celsius');
+title('Temperature profile in the 24-cm slab after 2.5 hr.');
+filename='Temperature profile in the 24-cm slab after 2.5 hr.';
+xgrid(1);
+xs2jpg(0,filename);
diff --git a/1309/CH4/EX4.4/Result4_4.pdf b/1309/CH4/EX4.4/Result4_4.pdf Binary files differnew file mode 100755 index 000000000..0a465b779 --- /dev/null +++ b/1309/CH4/EX4.4/Result4_4.pdf diff --git a/1309/CH4/EX4.4/ch4_4.sce b/1309/CH4/EX4.4/ch4_4.sce new file mode 100755 index 000000000..5e73af129 --- /dev/null +++ b/1309/CH4/EX4.4/ch4_4.sce @@ -0,0 +1,44 @@ +clc; +clear; +printf("\t\t\tChapter4_example4\n\n\n"); +hc=6; +D=0.105; +k=0.431; +c=2000; +rou=998; +Vs=%pi*D^3/6; +As=%pi*D^2; +// calculating Biot Number for lumped capacitance approach +Bi_lumped=hc*Vs/(k*As); +printf("\nThe Biot number is %.3f,",Bi_lumped); +alpha=k/(rou*c); +printf("\nThe value of diffusivity is %.2e sq.m/s",alpha); +Tc=20; +T_inf=23; +T_i=4; +if Bi_lumped<0.1 then + n=0; +else if Bi_lumped>0.1 then + n=1; + end +end +select n +case 0 then + disp('The Lumped capacity approach is applicable'); +case 1 then + printf("\n\nSince value of Biot number is greater than 0.1,\nLumped capacity approach would not give accurate results, so figure 4.8 is to be used\n"); + // calculating Biot Number for using figure 4.8 + Bi_figure=hc*D/(2*k); + printf("\nThe Biot Number for using figure 4.8 is %.3f",Bi_figure); + reciprocal_Bi=1/Bi_figure; + dimensionless_temp=(Tc-T_inf)/(T_i-T_inf); + printf("\nThe dimensionless temperature is %.3f",dimensionless_temp); + Fo=1.05;//The corresponding value of Fourier Number from figure 4.8a + t=(D/2)^2*Fo/alpha; + printf("\nThe required time is %.2e s = %.1f hr",t,t/3600); +end +Bi2Fo=Bi_figure^2*Fo; +printf("\nBi^2Fo=%.1e",Bi2Fo); +Dimensionless_HeatFlow=0.7; // The corresponding dimensionless heat flow ratio from figure 4.8c +Q=Dimensionless_HeatFlow*rou*c*Vs*(T_i-T_inf); +printf("\nThe heat transferred is %.3e J",Q); diff --git a/1309/CH4/EX4.5/Result4_5.pdf b/1309/CH4/EX4.5/Result4_5.pdf Binary files differnew file mode 100755 index 000000000..861926659 --- /dev/null +++ b/1309/CH4/EX4.5/Result4_5.pdf diff --git a/1309/CH4/EX4.5/ch4_5.sce b/1309/CH4/EX4.5/ch4_5.sce new file mode 100755 index 000000000..884c7e043 --- /dev/null +++ b/1309/CH4/EX4.5/ch4_5.sce @@ -0,0 +1,23 @@ +clc; +clear; +printf("\t\t\tChapter4_example5\n\n\n"); +hc=6; +D=0.105; +k=0.3; +c=0.41; +sp_gr=2.1; +rou_water=62.4; +alpha=k/(sp_gr*rou_water*c); +printf("\nThe diffusivity of the soil is %.2e sq.ft/hr",alpha); +t=3*30*24; +printf("\nTime in hours is %d hr",t); +// Bi_sqrt(Fo) is infinite +T_inf=10; +Ts=10; +T=32; +T_i=70; +dimensionless_temp=(T-T_i)/(T_inf-T_i); +printf("\nThe dimensionless temperature is %.4f",dimensionless_temp); +variable_fig4_12=0.38; //The value of x/(2*(alpha*t)^0.5) from figure 4.12 +x=2*sqrt(alpha*t)*variable_fig4_12; +printf("\nThe depth of the freeze line in soil is %.2f ft",x); diff --git a/1309/CH4/EX4.6/Result4_6.pdf b/1309/CH4/EX4.6/Result4_6.pdf Binary files differnew file mode 100755 index 000000000..4f1b2c15d --- /dev/null +++ b/1309/CH4/EX4.6/Result4_6.pdf diff --git a/1309/CH4/EX4.6/ch4_6.sce b/1309/CH4/EX4.6/ch4_6.sce new file mode 100755 index 000000000..f4d48765a --- /dev/null +++ b/1309/CH4/EX4.6/ch4_6.sce @@ -0,0 +1,29 @@ +clc; +clear; +printf("\t\t\tChapter4_example6\n\n\n"); +// properties of aluminium from appendix table B1 +k_al=236; +p_al=2.7*1000; +c_al=896; +// properties of oak from appendix table B3 +k_oak=0.19; +p_oak=0.705*1000; +c_oak=2390; +sqrt_kpc_al=sqrt(k_al*p_al*c_al); +printf("\nThe square root of kpc product of aluminium is %.2e sq.W.s/(m^4.sq.K)",sqrt_kpc_al); +kpc_R=4; +T_Li=20; +T_Ri=37.3; +T_al=(T_Li*(sqrt_kpc_al)+T_Ri*sqrt(kpc_R))/(sqrt_kpc_al+sqrt(kpc_R)); +printf("\nThe temperature of aluminium is felt as %.1f degree celsius",T_al); +sqrt_kpc_oak=sqrt(k_oak*p_oak*c_oak); +printf("\nThe square root of kpc product of oak is %.2e sq.W.s/(m^4.sq.K)",sqrt_kpc_oak); +T_oak=(T_Li*(sqrt_kpc_oak)+T_Ri*sqrt(kpc_R))/(sqrt_kpc_oak+sqrt(kpc_R)); +printf("\nThe temperature of oak is felt as %.1f degree celsius",T_oak); +if (T_al>T_oak) then + printf("\nThe aluminium will feel warmer."); +elseif (T_al<T_oak) then + printf("\nThe oak will feel warmer."); +else + printf("\nBoth will be felt equally warm.") +end diff --git a/1309/CH4/EX4.7/Result4_7.pdf b/1309/CH4/EX4.7/Result4_7.pdf Binary files differnew file mode 100755 index 000000000..e57a8bc95 --- /dev/null +++ b/1309/CH4/EX4.7/Result4_7.pdf diff --git a/1309/CH4/EX4.7/ch4_7.sce b/1309/CH4/EX4.7/ch4_7.sce new file mode 100755 index 000000000..99d2b30c3 --- /dev/null +++ b/1309/CH4/EX4.7/ch4_7.sce @@ -0,0 +1,50 @@ +clc; +clear; +printf("\t\t\tChapter4_example7\n\n\n"); +// properties of water at 68 degree fahrenheit from appendix table C11 +rou=62.46; +cp=0.9988; +k=0.345; +alpha=k/(rou*cp); +printf("\nThe diffusivity at 68 degree fahrenheit is %.2e sq.ft/hr",alpha); +D=2.5/12; +L=4.75/12; +Vs=%pi*D^2*L/4; +As=(%pi*D*L)+(%pi*D^2)/2; +Lc=Vs/As; +printf("\nThe volume of the can is %.4f cu.ft",Vs); +printf("\nThe surface area of the can is %.3f sq.ft",As); +printf("\nThe characteristic length is %.3f ft",Lc); +hc=1.7; +Bi=hc*Lc/k; +printf("\nThe Biot number is %.3f",Bi); +t=4; +// for the cylinder solution +Fo_cylinder=alpha*t/(D/2)^2; +Bi_cylinder=hc*(D/2)/k; +printf("\nFor the cylinder, The Fourier number is %.2f and Biot Number is %.3f",Fo_cylinder,Bi_cylinder); +reciprocal_Bi_cylinder=1/Bi_cylinder; +printf("\nThe reciprocal for Biot number for cylinder is %.2f",reciprocal_Bi_cylinder); +dim_T_cylinder=0.175; //The value of dimensionless temperature of cylinder from figure 4.7a at corresponding values of Fo and 1/Bi +// for the infinite plate solution +Fo_plate=alpha*t/(L/2)^2; +Bi_plate=hc*L/(2*k); +printf("\nFor the infinite plate, The Fourier number is %.3f and Biot Number is %.2f",Fo_plate,Bi_plate); +reciprocal_Bi_plate=1/Bi_plate; +printf("\nThe reciprocal for Biot number for infinite plate is %.2f",reciprocal_Bi_plate); +dim_T_plate=0.55; //The value of dimensionless temperature of infinite plate from figure 4.7a at corresponding values of Fo and 1/Bi +// Table 4. I, for the short-cylinder problem, indicates that the solution is the product of the infinite-cylinder problem (Figure 4.7) and the infinite-plate problem (Figure 4.6). +// For short cylinder problem +dim_T_shortcylinder=dim_T_cylinder*dim_T_plate; +printf("\nThe value of dimensionless temperature for short cylinder is %.3f ",dim_T_shortcylinder); +T_inf=30; +T_i=72; +Tc=dim_T_shortcylinder*(T_i-T_inf)+T_inf; +printf("\nThe temperature at centre of can is %.1f degree celsius",Tc); +dim_Tw_cylinder=0.77; //The dimensionless temperature from figure 4.7b corresponding to the value of 1/Bi and r/R=1 +dim_Tw_plate=0.65; //The dimensionless temperature from figure 4.6b corresponding to the value of 1/Bi and x/L=1 +dim_Tw_shortcylinder=dim_Tw_cylinder*dim_Tw_plate; +printf("\nThe value of dimensionless temperature at the wall for short cylinder is %.2f ",dim_Tw_shortcylinder); +Tw=dim_Tw_shortcylinder*(Tc-T_inf)+T_inf; +printf("\nThe wall temperature is %.1f degree F",Tw); + diff --git a/1309/CH4/EX4.8/Result4_8.pdf b/1309/CH4/EX4.8/Result4_8.pdf Binary files differnew file mode 100755 index 000000000..b57cf729b --- /dev/null +++ b/1309/CH4/EX4.8/Result4_8.pdf diff --git a/1309/CH4/EX4.8/ch4_8.sce b/1309/CH4/EX4.8/ch4_8.sce new file mode 100755 index 000000000..827efcbc4 --- /dev/null +++ b/1309/CH4/EX4.8/ch4_8.sce @@ -0,0 +1,39 @@ +clc; +clear; +printf("\t\t\tChapter4_example8\n\n\n"); +rou=7817; +c=461; +k=14.4; +alpha=.387e-5; +L1=.03; +L2=0.03; +L3=0.04; +x=0.04; +T_i=95; +T_inf=17; +// for infinite plate +L=L1/2; +hc=50; +reciprocal_Bi_plate=k/(hc*L); +printf("\nThe value of 1/Bi for infinite plate is %.1f",reciprocal_Bi_plate); +T=50; +n=1; +t=[3000 1500 700 400 200 300 350]; +[n m]=size(t); +// parameter for infinite plate Fourier Number,Fo is named as parameter1 +for i=1:m + parameter1(i)=alpha*t(i)/L^2; +// parameters for semi-infinite solid Bi(Fo)^0.5 and x/(2*(alpha*t)^0.5) are named as parameter2 and parameter3 +parameter2(i)=hc*((alpha*t(i))^0.5)/k; +parameter3(i)=x/(2*(alpha*t(i))^0.5); +dim_T_plate=[0.085 0.34 0.55 0.7 0.8 0.8 0.7]; //the corresponding values of dimensionless temperature for infinite plate from figure 4.6a +dim_T_solid=[0.225 0.14 0.075 0.046 0.02 0.035 0.042]; // the corresponding values of dimensionless temperature for semi-infinite solid from figure 4.12 +dim_T_bar(i)=dim_T_plate(i)*dim_T_plate(i)*(1-dim_T_solid(i)); +T(i)=dim_T_plate(i)*dim_T_plate(i)*(1-dim_T_solid(i))*(T_i-T_inf)+T_inf; +end +printf("\nThe Results for different time instances:\n"); +printf("\n\tInfinite Plate\t\t\t\t\t\tSemi-Infinite Solid\t\t\t\tDimensionless Temperature\tTemperature"); +printf("\ntime t, s\t1/Bi\tFo\t(T-Tinf)/(Ti-Tinf)\tBi(Fo)^0.5\tx/(2*(at)^0.5)\t(T-Tinf)/(Ti-Tinf)\t(T-Tinf)/(Ti-Tinf)\t\tT"); +for i=1:m + printf("\n%d\t\t%.1f\t%.2f\t\t%.2f\t\t%.3f\t\t%.3f\t\t%.3f\t\t\t%.3f\t\t\t\t%.1f",t(i),reciprocal_Bi_plate,parameter1(i),dim_T_plate(i),parameter2(i),parameter3(i),dim_T_solid(i),dim_T_bar(i),T(i)); +end diff --git a/1309/CH4/EX4.9/Figure4_9.jpg b/1309/CH4/EX4.9/Figure4_9.jpg Binary files differnew file mode 100755 index 000000000..6bfaacbad --- /dev/null +++ b/1309/CH4/EX4.9/Figure4_9.jpg diff --git a/1309/CH4/EX4.9/Result4_9.pdf b/1309/CH4/EX4.9/Result4_9.pdf Binary files differnew file mode 100755 index 000000000..edc3b0b27 --- /dev/null +++ b/1309/CH4/EX4.9/Result4_9.pdf diff --git a/1309/CH4/EX4.9/ch4_9.sce b/1309/CH4/EX4.9/ch4_9.sce new file mode 100755 index 000000000..f10032d86 --- /dev/null +++ b/1309/CH4/EX4.9/ch4_9.sce @@ -0,0 +1,115 @@ +clc; +clear; +printf("\t\t\tChapter4_example9\n\n\n"); +rou=.5*1000; +cp=837; +k=0.128; +alpha=0.049e-5; +// let Fo=0.5 and dx=0.05 +dt=0.5*(0.05)^2/alpha; +printf("\nThe time increment is %.3f hr",dt/3600); +p=1; +m=6; +A=2*eye(6,6); +n=1; +N=1; +for j=1:n + for i=1:6 + T(i,j)=20; + end +end +for n=1:7 + for i=1:4 + B(i+1,n)=T(i+2,n)+T(i,n); + B(1,n)=T(i+1,n)+200; + B(6,n)=2*T(i+1,n); + end +Temp=inv(A)*B(:,n); // temperature at the different points +printf("\nThe temperature at different points after %d time interval are:",n); +T(:,n+1)=Temp; +disp(T(:,n+1)); +end +time=n*dt; +printf("\nThe required time is %.2f hr",time/3600); +x=0:5:30; +plot(x,[200;T(:,2)]); +a1=gca(); +a1.data_bounds=[0,0;30,200]; +xtitle('(a) After 0.709 hr','T degree C','x, cm'); +newticks=a1.x_ticks; +newticks(2)=[0;10;20;30]; +newticks(3)=['0';'10';'20';'30']; +a1.x_ticks=newticks; +newticks1=a1.y_ticks; +newticks1(2)=[0;50;100;150;200]; +newticks1(3)=['0';'50';'100';'150';'200']; +a1.y_ticks=newticks1; +plot(x,[200;T(:,3)]); +a2=gca(); +hl=legend(['After 2(0.709) hr ';'After (0.709) hr ']); +a2.data_bounds=[0,0;30,200]; +xtitle('(b) After 2(0.709) hr ','T degree C','x, cm'); +newticks=a2.x_ticks; +newticks(2)=[0;10;20;30]; +newticks(3)=['0';'10';'20';'30']; +a2.x_ticks=newticks; +newticks1=a2.y_ticks; +newticks1(2)=[0;50;100;150;200]; +newticks1(3)=['0';'50';'100';'150';'200']; +a2.y_ticks=newticks1; +filename='(b) After 2(0.709) hr '; +clf(); +plot(x,[200;T(:,4)],x,[200;T(:,3)]); +a3=gca(); +hl=legend(['After 3(0.709) hr ';'After 2(0.709) hr ']); +a3.data_bounds=[0,0;30,200]; +xtitle('(c) After 3(0.709) hr ','T degree C','x, cm'); +newticks=a3.x_ticks; +newticks(2)=[0;10;20;30]; +newticks(3)=['0';'10';'20';'30']; +a3.x_ticks=newticks; +newticks1=a3.y_ticks; +newticks1(2)=[0;50;100;150;200]; +newticks1(3)=['0';'50';'100';'150';'200']; +a3.y_ticks=newticks1; +clf(); +plot(x,[200;T(:,5)],x,[200;T(:,4)]); +a4=gca(); +hl=legend(['After 4(0.709) hr ';'After 3(0.709) hr ']); +a4.data_bounds=[0,0;30,200]; +xtitle('(d) After 4(0.709) hr ','T degree C','x, cm'); +newticks=a4.x_ticks; +newticks(2)=[0;10;20;30]; +newticks(3)=['0';'10';'20';'30']; +a4.x_ticks=newticks; +newticks1=a4.y_ticks; +newticks1(2)=[0;50;100;150;200]; +newticks1(3)=['0';'50';'100';'150';'200']; +a4.y_ticks=newticks1; +clf(); +plot(x,[200;T(:,6)],x,[200;T(:,5)]); +a5=gca(); +hl=legend(['After 5(0.709) hr ';'After 4(0.709) hr ']); +a5.data_bounds=[0,0;30,200]; +xtitle('(e) After 5(0.709) hr ','T degree C','x, cm'); +newticks=a5.x_ticks; +newticks(2)=[0;10;20;30]; +newticks(3)=['0';'10';'20';'30']; +a5.x_ticks=newticks; +newticks1=a5.y_ticks; +newticks1(2)=[0;50;100;150;200]; +newticks1(3)=['0';'50';'100';'150';'200']; +a5.y_ticks=newticks1; +clf(); +plot(x,[200;T(:,7)]); +a6=gca(); +a6.data_bounds=[0,0;30,200]; +xtitle('(f) After 7(0.709) hr ','T degree C','x, cm'); +newticks=a6.x_ticks; +newticks(2)=[0;10;20;30]; +newticks(3)=['0';'10';'20';'30']; +a6.x_ticks=newticks; +newticks1=a6.y_ticks; +newticks1(2)=[0;50;100;150;200]; +newticks1(3)=['0';'50';'100';'150';'200']; +a6.y_ticks=newticks1; diff --git a/1309/CH5/EX5.1/Result5_1.pdf b/1309/CH5/EX5.1/Result5_1.pdf Binary files differnew file mode 100755 index 000000000..b0beb54ae --- /dev/null +++ b/1309/CH5/EX5.1/Result5_1.pdf diff --git a/1309/CH5/EX5.1/ch5_1.sce b/1309/CH5/EX5.1/ch5_1.sce new file mode 100755 index 000000000..be744c630 --- /dev/null +++ b/1309/CH5/EX5.1/ch5_1.sce @@ -0,0 +1,14 @@ +clc; +clear; +printf("\t\t\tChapter5_example1\n\n\n"); +// properties of CO at 300K from appendix table D2 +Cp=871; +Gamma=1.3; +Cv=Cp/Gamma; +printf("\nThe specific heat at constant volume is %d J/(kg.K)",Cv); +dT=20; +m=5; +Qp=m*Cp*dT; +Qv=m*Cv*dT; +printf("\n The heat required at constant pressure is %.1f kJ",Qp/1000); +printf("\nThe heat required at constant volume is %d kJ",Qv/1000); diff --git a/1309/CH5/EX5.2/Result5_2.pdf b/1309/CH5/EX5.2/Result5_2.pdf Binary files differnew file mode 100755 index 000000000..b8a7c90d1 --- /dev/null +++ b/1309/CH5/EX5.2/Result5_2.pdf diff --git a/1309/CH5/EX5.2/ch5_2.sce b/1309/CH5/EX5.2/ch5_2.sce new file mode 100755 index 000000000..c1a51e1ca --- /dev/null +++ b/1309/CH5/EX5.2/ch5_2.sce @@ -0,0 +1,20 @@ +clc; +clear; +printf("\t\t\tChapter5_example2\n\n\n"); +// properties of Freon-12 from appendix table C3 +T1_Fr=-50; +T2_Fr=-40; +rou1_Fr=1.546*1000; +rou2_Fr=1.518*1000; +beta_Fr=-(rou1_Fr-rou2_Fr)/(rou1_Fr*(T1_Fr-T2_Fr)); +printf("\nThe volumetric thermal expansion coefficient calculated for Freon-12 is %.3e /K",beta_Fr); +beta_acc_Fr=2.63e-3; // the accurate value of volumetric thermal expansion coefficient for Freon-12 +error_Fr=(beta_acc_Fr-beta_Fr)*100/beta_acc_Fr; +printf("\nThe error introduced in the case of Freon-12 is %d percent",error_Fr); +// properties of helium from appendix table D3 +T1_He=366; +T2_He=477; +rou1_He=0.13280; +rou2_He=0.10204; +beta_He=-(rou1_He-rou2_He)/(rou1_He*(T1_He-T2_He)); +printf("\nThe volumetric thermal expansion coefficient calculated for Freon-12 is %.3e /K",beta_He); diff --git a/1309/CH6/EX6.1/Result6_1.pdf b/1309/CH6/EX6.1/Result6_1.pdf Binary files differnew file mode 100755 index 000000000..82c83933e --- /dev/null +++ b/1309/CH6/EX6.1/Result6_1.pdf diff --git a/1309/CH6/EX6.1/ch6_1.sce b/1309/CH6/EX6.1/ch6_1.sce new file mode 100755 index 000000000..96510720a --- /dev/null +++ b/1309/CH6/EX6.1/ch6_1.sce @@ -0,0 +1,36 @@ +clc; +clear; +printf("\t\t\tChapter6_example1\n\n\n"); +// Determination of the fluid outlet tetnperature and the tube-wall temperature at the outlet. +// properties of ethylene glycol at 20 degree celsius from appendix table C5 +Cp_20=2382; +rou_20=1.116*1000; +v_20=19.18e-6; +kf_20=.249; +a_20=.939e-7; +Pr_20=204; +// specifications of 1/2 standard type M seamless copper water tubing from appendix table F2 +OD=1.588/100; +ID=1.446/100; +A=1.642e-4; +Q=3.25e-6; +V=Q/A; +printf("\nThe average flow velocity is %.1f m/s",V*100); +// calculation of Reynold's Number to check flow regime +Re=V*ID/v_20; +printf("\nThe Reynolds Number is %.1f",Re); +// since Re>he 2100, the flow regime is laminar and the hydrodynamic length can be calculated as +Z_h=0.05*ID*Re; +printf("\nThe hydrodynamic length is %.1f cm",Z_h*100); +Tbi=20; // bulk-fluid inlet temperature in degree celsius +qw=2200; // incident heat flux in W/m^2 +L=3; // Length of copper tube in m +R=ID/2; // inner radius in m +Tbo=Tbi+(2*qw*a_20*L)/(V*kf_20*R); +printf("\nThe bulk-fluid outlet temperature is %.1f degree celsius",Tbo); +// This result is based on fluid properties evaluated at 20°C. taken as a first approximation +Z_t=0.05*ID*Re*Pr_20; +printf("\nThe thermal entry length is %.1f m",Z_t); +Two=Tbo+(11*qw*ID)/(48*kf_20); // The wall temperature at outlet in degree celsius +printf("\nThe wall temperature at outlet is %.1f degree celsius",Two); +//The result is based on first approximation based on flow properties evaluated at the fluid inlet temperature. diff --git a/1309/CH6/EX6.2/Result6_2.pdf b/1309/CH6/EX6.2/Result6_2.pdf Binary files differnew file mode 100755 index 000000000..af48d23bf --- /dev/null +++ b/1309/CH6/EX6.2/Result6_2.pdf diff --git a/1309/CH6/EX6.2/ch6_2.sce b/1309/CH6/EX6.2/ch6_2.sce new file mode 100755 index 000000000..28135818c --- /dev/null +++ b/1309/CH6/EX6.2/ch6_2.sce @@ -0,0 +1,26 @@ +clc; +clear; +printf("\t\t\tChapter6_example2\n\n\n"); +// determination of average convection coefficient +T_avg=(140+70)/2; +printf("\nThe average bulk temperature is %d degree celsius",T_avg); +// properties of water at average bulk temperature from appendix table C11 +rou=.994*62.4; +kf=.363; +cp=.9980; +a=5.86e-3; +v=0.708e-5; +Pr=4.34; +// specifications of 1 standard type M copper tube from appendix table F2 +OD=1.125/12; // outer diameter in ft +ID=0.8792; // inner diameter in ft +A=0.006071 // cross sectional area in sq.ft +m_flow=1.5; // mass flow rate in lbm/s +V=m_flow*3600/(rou*A); // velocity in ft/hr +printf("\nThe velocity is %d ft/hr",V); +L=20; +Tw=240; +Tbo=140; +Tbi=70; +hL=-(rou*V*ID*cp*log((Tw-Tbo)/(Tw-Tbi)))/(4*L); +printf("\nThe average convective coefficient is %d BTU/(hr. sq.ft.degree Rankine)",hL); diff --git a/1309/CH6/EX6.3/Figure6_3.jpg b/1309/CH6/EX6.3/Figure6_3.jpg Binary files differnew file mode 100755 index 000000000..02b73a4d5 --- /dev/null +++ b/1309/CH6/EX6.3/Figure6_3.jpg diff --git a/1309/CH6/EX6.3/Result6_3.pdf b/1309/CH6/EX6.3/Result6_3.pdf Binary files differnew file mode 100755 index 000000000..b1e59a41f --- /dev/null +++ b/1309/CH6/EX6.3/Result6_3.pdf diff --git a/1309/CH6/EX6.3/ch6_3.sce b/1309/CH6/EX6.3/ch6_3.sce new file mode 100755 index 000000000..87c08de9b --- /dev/null +++ b/1309/CH6/EX6.3/ch6_3.sce @@ -0,0 +1,80 @@ +clc; +clear; +printf("\t\t\tChapter6_example3\n\n\n"); +// Determination of the variation of wall temperature with length up to the point where the flow becomes fully developed. +// properties of milk +kf=0.6; // thermal conductivity in W/(m-K) +cp=3.85*1000; // specific heat in J/(kg*K) +rou=1030; // density in kg/m^3 +mu=2.12e3; // viscosity in N s/m^2 +// specifications of 1/2 standard type K tubing from appendix table F2 +OD=1.588/100; // outer diameter in m +ID=1.340/100; // inner diameter in m +A=1.410e-4 // cross sectional area in m^2 +rou=1030; +V=0.1; +mu=2.12e-3 +// determination of flow regime +Re=rou*V*ID/(mu); +printf("\nThe Reynolds Number is %d",Re); +// The flow being laminar, the hydrodynamic entry length is calculated as follows +ze=0.05*ID*Re; +printf("\nThe hydrodynamic entry length is %.1f cm",ze*100); +Tbo=71.7; // final temperature in degree celsius +Tbi=20; // initial temperature in degree celsius +L=6; // heating length in m +qw=rou*V*ID*cp*(Tbo-Tbi)/(4*L); +printf("\nThe heat flux is %d W/sq.m",qw); +q=qw*%pi*ID*L; +printf("\nThe power required is %.1f W",q); +printf("\nA 3000 W heater would suffice"); +Pr=(cp*mu)/kf; // Prandtl Number +printf("\nThe Prandtl Number is %.1f",Pr); +zf=0.05*ID*Re*Pr; +printf("\nThe length required for flow to be thermally developed is %.1f m",zf); +// calculations of wall temperature of the tube +reciprocal_Gz=[0.002 0.004 0.01 0.04 0.05];// values of 1/Gz taken +[n m]=size(reciprocal_Gz); +Nu=[12 10 7.5 5.2 4.5]; //Enter the corresponding value of Nusselts Number from figure 6.8 +for i=1:m + z(i)=ID*Re*Pr*reciprocal_Gz(i); + h(i)=kf*Nu(i)/ID; + Tbz(i)=20+(8.617*z(i)); + Twz(i)=Tbz(i)+(11447/h(i)); +end +printf("\nSummary of Calculations to Find the Wall Temperature of the Tube"); +printf("\n\t1/Gz\t\tNu\t\tz (m)\t\th W/(sq.m.K)\t\tTbz (degree celsius)\t\tTwz (degree celsius)"); +for i=1:m +printf("\n\t%.3f\t\t%.1f\t\t%.3f\t\t%d\t\t\t%.1f\t\t\t\t%.1f",reciprocal_Gz(i),Nu(i),z(i),h(i),Tbz(i),Twz(i)); +end +subplot(211); +plot(z,Tbz,'r--d',z,Twz,'r-'); // our first figure +a1 = gca(); +h1=legend(["Tbz";"Twz"]); +subplot(212) +plot(z,h, 'o--'); // our second figure +hl=legend(['h'],2); +title('Variation of temperature and local convection coefficient with axial distance for the constant- wall-flux tube'); +a2 = gca(); +a2.axes_visible = ["off", "on","on"]; +a2.y_location ="right"; + +a1.axes_bounds=[0 0 1 1]; // modify the first figure to occupy the whole area +a2.axes_bounds=[0 0 1 1]; // modify the second figure to occupy the whole area too + +a1.data_bounds=[0,0;6,140]; +a2.data_bounds=[0,0;6,700]; + +a1.x_ticks = tlist(["ticks", "locations", "labels"], (0:6)', ["0";"1";"2";"3";"4";"5";"6"]); +a1.x_label +a1.y_label +x_label=a1.x_label; +x_label.text=" z,m" +a2.x_label +a2.y_label +y_label=a1.y_label; +y_label.text="T, degree celsius" +y_label=a2.y_label; +y_label.text="h, W/(sq.m.K)" +xgrid(1); +a2.filled = "off"; diff --git a/1309/CH6/EX6.4/Figure6_4.jpg b/1309/CH6/EX6.4/Figure6_4.jpg Binary files differnew file mode 100755 index 000000000..3922573f9 --- /dev/null +++ b/1309/CH6/EX6.4/Figure6_4.jpg diff --git a/1309/CH6/EX6.4/Result6_4.pdf b/1309/CH6/EX6.4/Result6_4.pdf Binary files differnew file mode 100755 index 000000000..91671f065 --- /dev/null +++ b/1309/CH6/EX6.4/Result6_4.pdf diff --git a/1309/CH6/EX6.4/ch6_4.sce b/1309/CH6/EX6.4/ch6_4.sce new file mode 100755 index 000000000..3fb1a5d5a --- /dev/null +++ b/1309/CH6/EX6.4/ch6_4.sce @@ -0,0 +1,105 @@ +clc; +clear; +printf("\t\t\tChapter6_example4\n\n\n"); +// The average bulk temperature of the Freon-12 is [-4O +(-4)]/2 = -22°F +// properties of Freon-12 at average bulk temperature +kf=0.04; // thermal conductivity in BTU/(hr.ft.°R) +cp=0.2139; // specific heat in BTU/(lbm-°R) +rou= 1.489*(62.4); // density in lbm/cu.ft +v=0.272e-5; // viscosity in sq.ft/s +a=2.04e-3; // diffusivity in sq.ft/hr +Pr=4.8; // Prandtl Number +// specifications of 3/8 standard type K copper tubing from appendix table F2 +OD=0.5/12; // outer diameter in ft +ID=0.03350; // inner diameter in ft +A=0.0008814 // cross sectional area in sq.ft +// Laminar conditions are asssumed +z=5; +Tw=32; +Tbo=-4; +Tbi=-40; +L=5; +i=1; +V_assumed(i)=100; //assumed value for velocity +for i=1:6 + inv_Gz(i)=(z*a)/(V_assumed(i)*ID^2); + Nu=[4.7 5.8 6.2 6.3 6.4 6.4]; // corresponding Nusselt numbers from fig. 8.8: + hL(i)=Nu(i)*kf/ID; + V(i)=-(2*a*L*hL(i))/((kf*ID/2)*log((Tw-Tbo)/(Tw-Tbi))); + V_assumed(i+1)=V(i); +end +printf("\nSummary of Results\n"); +printf("Assmued V (ft/hr)\t1/Gz\tNu(fig 8.8)\thL BTU/(hr. sq.ft. degree R)\tV (ft/hr)\n"); +for j=1:6 +printf("\t%d\t\t%.4f\t%.1f\t\t%.2f\t\t\t\t%d\n",V_assumed(j),inv_Gz(j),Nu(j),hL(j),V(j)); +end +V_final=V(i-1); +hL_final=hL(i-1); +printf("\nThe final velocity is %d ft/hr = %.4f ft/s",V_final,V_final/3600); +printf("\nThe final convective coefficient is %.2f BTU/(hr. sq.ft. degree R)",hL_final); +// checking the laminar-flow assumption by calculating the Reynolds number +Re=(V_final/3600)*ID/v; +printf("\nThe Reynolds number is %d",Re); +// The flow is laminar +m_Fr=rou*A*V_final/3600; +printf("\nThe mass flow rate of Freon-12 is %.2e lbm/s = %.2f lbm/hr",m_Fr,m_Fr*3600); +As=%pi*ID*L; +q=hL_final*As*[(Tw-Tbo)-(Tw-Tbi)]/(log((Tw-Tbo)/(Tw-Tbi))); +printf("\nThe heat gained by Freon-12 is %.1f BTU/hr",q); +q_check=m_Fr*cp*(Tbo-Tbi); +printf("\nOn checking the heat transferred we find almost equal to the heat gained by Freon-12"); +rou_water=1.002*62.4; // density of water in lbm/ft^3 from appendix table C11 +m_water=rou_water*L*(2/12)*(3/12); +printf("\nThe mass of water in the prescribed volume is %.1f lbm",m_water); +// to remove 144 BTU/lbm of water, the time required is caalculated as below +t=144*m_water/q; +printf("\nThe required time is %.1f hr",t); +inv_Gz1=[0.001 0.004 0.01 0.015 0.02 0.0271]; // guess values of 1/Gz +Nu_D=[19.3 12.1 8.9 7.7 7.1 6.4]; //corresponding Nusselt number from fig. 6.8 +[n m]=size(inv_Gz1); +for j=1:m + Z(j)=ID*Re*Pr*(inv_Gz1(j)); + hz(j)=Nu_D(j)*kf/ID; + Tbz(j)=32-72*exp(-0.01812*Z(j)*hz(j)); +end +printf("\nSummary of Data for Example 6.4 "); +printf("\n\t1/Gz\tNu_D\tz (ft)\thz, BTU/(hr. sq.ft.degree R)\tTbz,degree F\n"); +for p=1:m + printf("\t%.4f\t%.1f\t%.2f\t%.2f\t\t\t\t%.1f\n",inv_Gz1(p),Nu_D(p),Z(p),hz(p),Tbz(p)); +end +subplot(211); +plot(Z,Tbz,'r--d',Z,Tw,'r-'); // your first figure +a1 = gca(); +hl=legend(['Tbz';'Tw'],4); +subplot(212) +plot(Z,hz, 'o--'); // your second figure +a2 = gca(); +hl=legend(['hz'],1); +a2.axes_visible = ["off", "on","on"]; +a2.y_location ="right"; + +a1.axes_bounds=[0 0 1 1]; // modify the first figure to occupy the whole area +a2.axes_bounds=[0 0 1 1]; // modify the second figure to occupy the whole area too +a2.filled = "off"; +a1.data_bounds=[-2,-40;5,40]; +a2.data_bounds=[-2,0;5,30]; +x_label1=a1.x_label; +x_label1.text="z, ft"; +y_label2=a2.y_label; +y_label2.text="hz, BTU/(hr.sq.ft.degree R)"; +y_label=a1.y_label; +y_label.text="T, degree F"; +newticks1=a1.y_ticks; +newticks1(2)=[-40;-30;-20;-10;0;10;20;30;40]; +newticks1(3)=['-40';'-30';'-20';'-10';'0';'10';'20';'30';'40']; +a1.y_ticks=newticks1; +newticks2=a2.y_ticks; +newticks2(2)=[0;5;10;20;30]; +newticks2(3)=['0';'5';'10';'20';'30']; +a2.y_ticks=newticks2; +newticks=a1.x_ticks; +newticks(2)=[-2;-1;0;1;2;3;4;5]; +newticks(3)=['-2';'-1';'0';'1';'2';'3';'4';'5']; +a1.x_ticks=newticks; + +title('Graphical summary of the solution to the constant-wall-temperature tube of Example 6.4'); diff --git a/1309/CH6/EX6.5/Result6_5.pdf b/1309/CH6/EX6.5/Result6_5.pdf Binary files differnew file mode 100755 index 000000000..dbd235233 --- /dev/null +++ b/1309/CH6/EX6.5/Result6_5.pdf diff --git a/1309/CH6/EX6.5/ch6_5.sce b/1309/CH6/EX6.5/ch6_5.sce new file mode 100755 index 000000000..c15d94b29 --- /dev/null +++ b/1309/CH6/EX6.5/ch6_5.sce @@ -0,0 +1,42 @@ +clc; +clear; +printf("\t\t\tChapter6_example5\n\n\n"); +// Determination for the power required for heating and the wall temperature at the outlet. +// The liquid properties are evaluated at the mean temperature of (80 + 20)/2 = 50°C. +// specifications of 1 standard type K copper water tubing from appendix table F2 +OD = 2.858/100; // outer diameter in m +ID = 2.528/100; // inner diameter in m +A = 5.019e-4; // cross sectional area in sq.m +// 1 oz = 2.957e-5 m^3 +Q=80*2.957e-5/120; // The volume flow rate of water (at 20°C) in cu.m/s +printf("\nThe volume flow rate of water (at 20°C) is %.2e cu.m/s",Q); +p_20= 1.000*1000; // density of water at 20°C in kg/cu.m +// properties of water at 50°C from appendix table C11 +p_50= 0.990*(1000); // density in kg/m3 +cp= 4181; // specific heat in J/(kg*K) +v = 0.586e-6; // viscosity in sq.m/s +kf = 0.640; // thermal conductivity in W/(m.K) +a = 1.533e-7; // diffusivity in sq.m/s +Pr = 3.68; // Prandtl number +mass_flow=p_20*Q; // mass flow rate through the tube in kg/s +printf("\nmass flow rate through the tube is %.4f kg/s",mass_flow); +L=3; // length of tube in m +As=%pi*ID*L; +Tbo=80; // final temperature in °C +Tbi=20; // initial temperature in °C +qw=mass_flow*cp*(Tbo-Tbi)/(As); +q=qw*As; +A=%pi*(ID/2)^2; +printf("\nThe power required in %.3e W/sq.m = %d W",qw,q); +V=mass_flow/(p_50*A); // average velocity at 50 °C +printf("\nThe average velocity at 50°C is %.2e m/s",V); +Re=(V*ID)/v; // Reynold's Number +printf("\nThe Reynolds Number for the flow is %d",Re); +// The flow is laminar so we can use Figure 6.12 to obtain the information needed on Nusselt number and to find hz +inv_Gz=L/(Re*ID*Pr); // The inverse Graetz number at tube end, based on 50°C conditions +printf("\nThe inverse Graetz number at tube end, based on 50°C conditions is %.4f",inv_Gz); +Nu=6.9; //value of corresponding Nusselts Number from figure 6.12 +hz=(Nu*kf)/ID; +printf("\nThe local convection coefficient is %.1f W/(sq.m.K)",hz); +Two=(qw/hz)+Tbo; // The outlet wall temperature in °C +printf("\nThe outlet wall temperature is %d °C",Two); diff --git a/1309/CH6/EX6.6/Result6_6.pdf b/1309/CH6/EX6.6/Result6_6.pdf Binary files differnew file mode 100755 index 000000000..707b482bf --- /dev/null +++ b/1309/CH6/EX6.6/Result6_6.pdf diff --git a/1309/CH6/EX6.6/ch6_6.sce b/1309/CH6/EX6.6/ch6_6.sce new file mode 100755 index 000000000..7a2076bf7 --- /dev/null +++ b/1309/CH6/EX6.6/ch6_6.sce @@ -0,0 +1,38 @@ +clc; +clear; +printf("\t\t\tChapter6_example6\n\n\n"); +// determibation of heat gained +// air properties to be calculated at T=(72+45)/2=58.5 degree Fahrenheit +// properties at T=58.5 degree fahrenheit from appendix table D1 +p = 0.077; // density in lbm/ft^3 +cp = 0.240; // specific heat in BTU/(lbm.degree Rankine) +v = 15.28e-5; // viscosity in ft^2/s +kf = 0.0146; // thermal conductivity in BTU/(hr.ft."R) +a = 0.776; // diffusivity in ft^2/hr +Pr = 0.711; // prandtl number +D=7/12; // diameter in ft +L=40; // length in ft +Tbo=72; // outlet temperature in degree Fahrenheit +Tbi=45; // inlet temperature in degree Fahrenheit +A=%pi*(D^2)/4; // cross sectional area of duct in ft^2 +// density at outlet temperature in lbm/ft^3 +rou_o=.0748; +V=10; // average velocity in ft/s +mass_flow=rou_o*A*V; +printf("\nThe mass flow rate is %.1f lbm/s",mass_flow); +// average velocity evaluated by using the average bulk temperature +V_avg=mass_flow/(p*A); +printf("\nThe average velocity evaluated by using the average bulk temperature is %.2f ft/s",V_avg); +Re=(V_avg*D)/v; +printf("\nThe Reynolds number for the flow is %.3e ",Re); +// the flow is in turbulent regime +q=mass_flow*cp*(Tbo-Tbi); +printf("\nThe heat gained by air is %.3f BTU",q); +hc=1; // convection coefficient between the outside duct wall and the attic air in BTU/(hr. sq.ft.degree Rankine). +T_inf=105; // The temperature of attic air surrounding the duct in degree Fahrenheit +hz=(0.023*Re^(4/5)*Pr^0.4)*kf/D; // The local coefficient at the duct end is %.2f BTU/(hr. sq.ft.degree Rankine) +printf("\nThe local coefficient at the duct end is %.2f BTU/(hr. sq.ft.degree Rankine)",hz); +qw=(T_inf-Tbo)/((1/hc)+(1/hz)); // wall flux in BTU/(hr. sq.ft.degree Rankine) +printf("\nThe wall flux is %.1f BTU/(hr. sq.ft.degree Rankine)",qw); +Two=qw*(1/hz)+Tbo; // The wall temperature at exit in degree Fahrenheit +printf("\nThe wall temperature at exit is %.1f degree Fahrenheit",Two); diff --git a/1309/CH7/EX7.1/Figure7_1.jpeg b/1309/CH7/EX7.1/Figure7_1.jpeg Binary files differnew file mode 100755 index 000000000..1fd5736f2 --- /dev/null +++ b/1309/CH7/EX7.1/Figure7_1.jpeg diff --git a/1309/CH7/EX7.1/Result7_1.pdf b/1309/CH7/EX7.1/Result7_1.pdf Binary files differnew file mode 100755 index 000000000..6ce8becb5 --- /dev/null +++ b/1309/CH7/EX7.1/Result7_1.pdf diff --git a/1309/CH7/EX7.1/ch7_1.sce b/1309/CH7/EX7.1/ch7_1.sce new file mode 100755 index 000000000..e82146a44 --- /dev/null +++ b/1309/CH7/EX7.1/ch7_1.sce @@ -0,0 +1,64 @@ +clc;
+clear;
+printf("\t\t\tChapter7_example1\n\n\n");
+printf("\t\t\tSolution to part (a)\n");
+// determination of boundary layer growth with length
+// properties of air at 27 degree celsius from appendix table D.1
+rou=1.177; // density in kg/cu.m
+v=15.68e-6; // viscosity in sq.m/s
+L=0.5; // length in m
+V_inf=1; // air velocity in m/s
+Re= (V_inf*L)/v; // Reynolds Number
+printf("The Reynolds Number is %.2e ",Re);
+// Reynolds Number is less than 5e5 hence the flow is laminar and Blasius Solution applies
+x=[0 0.125 0.25 0.375 0.5]; // distances in m where boundary layer growth is determined
+[n,m]=size(x);
+for i=1:m
+ delta(i)=5*x(i)^0.5/(V_inf/v)^0.5;
+end
+subplot(211);
+plot(x,delta);
+a=gca();
+newTicks=a.x_ticks;
+newTicks(2)=[0;0.125;0.25;0.375;0.5];
+newTicks(3)=['0';'0.125';'0.25';'0.375';'0.50'];
+a.x_ticks=newTicks;
+title('Boundary-layer growth with distance');
+xlabel('x, m');
+ylabel('delta, m^(1/2)');
+printf("\n\t\t\tSolution to part (b)\n");
+// produce graph of velocity distribution at x=0.25 m
+eta=0:5;
+[p,q]=size(eta);
+f=[0 0.32979 0.62977 0.84605 0.95552 0.99155];//value for f for corresponding eta value from Table 7.1
+for j=1:q
+ y(j)=eta(j)*(v*0.25)^0.5;
+end
+printf("\n\t\t\tResults of Calculations for Example 7.1\n");
+printf("\teta\t\ty,m\t\t\tf=vx, m/s\n");
+for i=1:q
+printf("\t%d\t\t%.2e\t\t%.5f\n",eta(i),y(i),f(i));
+end
+subplot(212);
+plot(f,y);
+b=gca();
+newTicks1=b.x_ticks;
+newTicks1(2)=[0;0.25;0.5;0.75;1.0];
+newTicks1(3)=['0';'0.25';'0.5';'0.75';'1.0'];
+b.x_ticks=newTicks1;
+newTicks2=b.y_ticks;
+newTicks2(2)=[0;0.0025;0.005;0.0075;0.010];
+newTicks2(3)=['0';'0.0025';'0.005';'0.0075';'0.010'];
+b.y_ticks=newTicks2;
+title('Velocity Distribution at x=0.25 m');
+xlabel('Vx, m/s');
+ylabel('y, m');
+printf("\t\t\tSolution to part (c)\n");
+// calculation of absolute viscosity
+gc=1;
+mu=rou*v/gc;
+printf("\nThe absolute viscosity is %.3e N.s/sq.m",mu);
+b=1; // width in m
+Df=0.664*V_inf*mu*b*(Re)^0.5;
+printf("\nThe skin-drag is %.2e N",Df);
+printf("\nThe skin-drag including both sides of plate is %.2e N",2*Df);
diff --git a/1309/CH7/EX7.10/Result7_10.pdf b/1309/CH7/EX7.10/Result7_10.pdf Binary files differnew file mode 100755 index 000000000..dd875dd62 --- /dev/null +++ b/1309/CH7/EX7.10/Result7_10.pdf diff --git a/1309/CH7/EX7.10/ch7_10.sce b/1309/CH7/EX7.10/ch7_10.sce new file mode 100755 index 000000000..438f14e50 --- /dev/null +++ b/1309/CH7/EX7.10/ch7_10.sce @@ -0,0 +1,52 @@ +clc; +clear; +printf("\t\t\tChapter7_example10\n\n\n"); +// Calculation of the pressure drop for the air passing over the tubes and the heat transferred to the air. +// properties of air at 70 + 460 = 530 degree R = 540 degree R from appendix table D1 +rou= 0.0735; // density in Ibm/cu.ft +cp=0.240; // specific heat BTU/(lbm-degree Rankine) +v= 16.88e-5; // viscosity in sq.ft/s +kf = 0.01516 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a = 0.859; // diffusivity in sq.ft/hr +Pr = 0.708; // Prandtl Number +// specifications of 3/4 standard type K copper tubing from appendix table F2 +OD=0.875/12; // outer diameter in ft +ID=0.06208; // inner diameter in ft +A=0.003027; // cross sectional area in sq.ft +L=2; +sL=1.5/12; +sT=1.3/12; +V_inf=12; // velocity of air in ft/s +V1=(sT*V_inf)/(sT-OD); // velocity at area A1 in ft/s +printf("\nVelocity at area A1 is %.1f ft/s",V1); +sD=((sL)^2+(sT/2)^2)^0.5; // diagonal pitch in inch +printf("\nThe diagonal pitch is %.2f in",sD*12); +V2=(sT*V_inf)/(2*(sD-OD)); +printf("\nVelocity at area A2 is %.1f ft/s",V2); +if V1>V2 then + Vmax=V1; + else Vmax=V2; +end +Re_D=Vmax*OD/v; // Reynolds Number +printf("\nThe Reynolds number is %.2e ",Re_D); +sT_OD=1.3/0.875; +sT_sL=1.3/1.5; +printf("\nThe values of parameters are sT/Do=%.2f and sT/sL=%.2f",sT_OD,sT_sL); +f1=0.35; //value of f1 for above values of sT/Do and Re +f2=1.05; //Corresponding value of f2 for above values of sT/sL and Re +gc=32.2; +N=7; +dP=N*f1*f2*(rou*Vmax^2/(2*gc)); +printf("\nThe pressure drop is %.2f lbf/ft^2 = %.4f psi",dP, dP/147); +sL_Do=sL/OD; +C1=0.438; //value of C1 for above values of sT/Do and sL/Do +C2=0.97; //value of C2 for above values of sT/Do and sL/Do +m=0.565; //value of m for above values of sT/Do and sL/Do +hc=kf*1.13*C1*C2*(Re_D)^m*(Pr)^(1/3)/OD; // The convection coefficient +printf("\nThe convection coefficient is %.1f BTU/(hr.sq.ft.degree Rankine)",hc); +As=70*%pi*OD*L; // outside surface area of 70 tubes +printf("\nThe outside surface area of 70 tubes is %.1f sq.ft",As); +Tw=200; // outside surface temeperature in degree F +T_inf=70; // air temperature in degree F +q=hc*As*(Tw-T_inf);// heat transferred +printf("\nThe heat transferred is %.2e BTU/hr",q); diff --git a/1309/CH7/EX7.2/Figure7_2.jpeg b/1309/CH7/EX7.2/Figure7_2.jpeg Binary files differnew file mode 100755 index 000000000..561b2a3c4 --- /dev/null +++ b/1309/CH7/EX7.2/Figure7_2.jpeg diff --git a/1309/CH7/EX7.2/Result7_2.pdf b/1309/CH7/EX7.2/Result7_2.pdf Binary files differnew file mode 100755 index 000000000..dd7f5ef0f --- /dev/null +++ b/1309/CH7/EX7.2/Result7_2.pdf diff --git a/1309/CH7/EX7.2/ch7_2.sce b/1309/CH7/EX7.2/ch7_2.sce new file mode 100755 index 000000000..9dc973828 --- /dev/null +++ b/1309/CH7/EX7.2/ch7_2.sce @@ -0,0 +1,52 @@ +clc; +clear; +printf("\t\t\tChapter7_example2\n\n\n"); +// determination of temperature profile +// properties of water at (40 + 100)/2 = 70°F = 68°F from appendix table C11 +rou= 62.4; // density in Ibm/ft^3 +cp=0.9988; // specific heat BTU/(lbm-degree Rankine) +v= 1.083e-5; // viscosity in sq.ft/s +kf = 0.345 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a = 5.54e-3; // diffusivity in sq.ft/hr +Pr = 7.02; // Prandtl Number +V=1.2; // velocity in ft/s +x=[1 2]; // distances from plate entry in ft +for i=1:2 +Re(i)=(V*x(i))/v; // Reynolds Number at x=1 ft +printf("\nThe Reynolds Number at x=%d ft is %.3e",i,Re(i)); +// since Reynolds Number is less than 5*10^5, the flow is laminar +hL(i)=0.664*Pr^(1/3)*Re(i)^0.5*kf/x(i); +printf("\nThe average convection coefficient at x=%d is %.1f BTU/(hr. sq.ft. degree Rankine)",i,hL(i)); +Tw=100; // temperature of metal plate in degree fahrenheit +T_inf=40; // temperature of water in degree fahrenheit +A(i)=x(i)*18/12; // cross sectional area for 1 ft length +q(i)=hL(i)*A(i)*(Tw-T_inf); +printf("\nThe heat transferred to water over the x=%d ft is %.3e BTU/hr",i,q(i)); +end +eta=0:0.2:1.2; +[n m]=size(eta); +theta=[1 .75 .51 .31 .17 .08 0.01]; // values of dimensionless temperature from figure 7.7 corresponding to eta value taken +for i=1:m +y(i)=eta(i)*(v*x(1)/V(1))^0.5; +T(i)=theta(i)*(Tw-T_inf)+T_inf; +end +printf("\nSolution Chart for example 7.2\n"); +printf("\teta\t\ttheta\t\ty, ft\t\t\tT, degree F\n"); +for i=1:m +printf("\t%.1f\t\t%.2f\t\t%.1e\t\t\t%.1f\n",eta(i),theta(i),y(i),T(i)); +end +plot(T,y); +a=gca(); +newTicks=a.x_ticks; +newTicks(2)=[100; 90; 80; 70; 60;50; 40]; +newTicks(3)=['100'; '90'; '80'; '70'; '60';'50'; '40']; +a.x_ticks=newTicks; +newTicks1=a.y_ticks; +newTicks1(2)=[0; 0.001; 0.002; 0.003; 0.004]; +newTicks1(3)=['0'; '0.001'; '0.002'; '0.003'; '0.004']; +a.y_ticks=newTicks1; +a.axes_reverse=["on","off"]; +xgrid(1); +title('Temperature variation (at x = 1 ft) within the boundary layer for the water'); +xlabel('T, degree Fahrenheit'); +ylabel('y, ft'); diff --git a/1309/CH7/EX7.3/Result7_3.pdf b/1309/CH7/EX7.3/Result7_3.pdf Binary files differnew file mode 100755 index 000000000..5736cec8d --- /dev/null +++ b/1309/CH7/EX7.3/Result7_3.pdf diff --git a/1309/CH7/EX7.3/ch7_3.sce b/1309/CH7/EX7.3/ch7_3.sce new file mode 100755 index 000000000..a6a759a5b --- /dev/null +++ b/1309/CH7/EX7.3/ch7_3.sce @@ -0,0 +1,27 @@ +clc; +clear; +printf("\t\t\tChapter7_example3\n\n\n"); +// Determination of the average convection coefficient and the total drag force exerted on the plate. +// properties of air at (300 + 400)/2 = 350 K from appendix table D1 +rou= 0.998; // density in kg/cu.m +cp= 1009; // specific heat in J/(kg*K) +v= 20.76e-6; // viscosity in sq.m/s +Pr = 0.697; // Prandtl Number +k= 0.03003; // thermal conductivity in W/(m.K) +a = 0.2983e-4; // diffusivity in sq.m/s +L=1; // Length of plate in m +V=5; // velocity of air in m/s +b=0.5; // width in m +Re=V*L/v; // Reynolds number at plate end +printf("\nThe Reynolds number is %.2e",Re); +// Since the flow is laminar at plate end, The average convection coefficient is calculated with Equation Nu=h*L/k= 0.664 Re^(1/2)Pr^(1/3) +h=k*0.664*Re^(1/2)*Pr^(1/3)/L; // The average convection coefficient in W/(sq.m.K) +printf("\nThe average convection coefficient is %.2f W/(sq.m.K)",h); +Df=0.664*V*rou*v*b*(Re)^0.5; // drag force in N +printf("\nThe drag force is %.2e N",Df); +hx=(1/2)*h; // local convective coefficient +printf("\nThe local convective coefficient is %.2f W/(sq.m.K)",hx); +delta=5*L/(Re)^0.5; // The boundary-layer thickness at plate end +printf("\nThe boundary-layer thickness at plate end is %.2f cm",delta*100); +delta_t=delta/(Pr)^(1/3); +printf("\nThe thermal-boundary-layer thickness is %.2f cm",delta_t*100); diff --git a/1309/CH7/EX7.4/Result7_4.pdf b/1309/CH7/EX7.4/Result7_4.pdf Binary files differnew file mode 100755 index 000000000..41b03e603 --- /dev/null +++ b/1309/CH7/EX7.4/Result7_4.pdf diff --git a/1309/CH7/EX7.4/ch7_4.sce b/1309/CH7/EX7.4/ch7_4.sce new file mode 100755 index 000000000..62f2e98f5 --- /dev/null +++ b/1309/CH7/EX7.4/ch7_4.sce @@ -0,0 +1,21 @@ +clc; +clear; +printf("\t\t\tChapter7_example4\n\n\n"); +// Determination of the maximum heater-surface temperature for given conditions +// fluid properties at (300 degree R + 800 degree R)/2 = 550 degree R=540degree R from Appendix Table D.6 +rou= 0.0812; // density in Ibm/ft^3 +cp=0.2918; // specific heat BTU/(lbm-degree Rankine) +v= 17.07e-5; // viscosity in ft^2/s +kf = 0.01546 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a = 0.8862; // diffusivity in ft^2/hr +Pr = 0.709; // Prandtl Number +qw=10/(1.5*10.125)*(1/.2918)*144; // The wall flux +printf("\nThe wall flux is %d BTU/(hr. sq.ft)",qw); +V_inf=20; // velocity in ft/s +L=1.5/12; // length in ft +Re_L=V_inf*10*L/v; // Reynolds number at plate end +printf("\nThe Reynolds number at plate end is %.2e",Re_L); +// So the flow is laminar and we can find the wall temperature at plate end as follows +T_inf=300; // free stream temperature in degree Rankine +Tw=T_inf+(qw*L*10/(kf*0.453*(Re_L)^0.5*(Pr)^(1/3))); +printf("\nThe maximum heater surface temperature is %d degree Rankine",Tw); diff --git a/1309/CH7/EX7.5/Result7_5.pdf b/1309/CH7/EX7.5/Result7_5.pdf Binary files differnew file mode 100755 index 000000000..bcb325e28 --- /dev/null +++ b/1309/CH7/EX7.5/Result7_5.pdf diff --git a/1309/CH7/EX7.5/ch7_5.sce b/1309/CH7/EX7.5/ch7_5.sce new file mode 100755 index 000000000..1b68cdd89 --- /dev/null +++ b/1309/CH7/EX7.5/ch7_5.sce @@ -0,0 +1,25 @@ +clc; +clear; +printf("\t\t\tChapter7_example5\n\n\n"); +// validation of the equation st.(Pr)^(2/3)=Cd/2 where St: Stanton Number Pr:Prandtl Number Cd: Drag Coefficient +// values of parameters from example 7.4 +rou= 0.0812; // density in Ibm/ft^3 +cp=0.2918; // specific heat BTU/(lbm-degree Rankine) +v= 17.07e-5; // viscosity in ft^2/s +kf = 0.01546 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a = 0.8862; // diffusivity in ft^2/hr +Pr = 0.709; // Prandtl Number +Tw=469; // maximum heater temperature in degree Rankine +T_inf=300; // free-stream temperature in degree Rankine +qw=324; // The wall flux in BTU/(hr.ft^2) +V_inf=20; // velocity in ft/s +hx=qw/(Tw-T_inf); // The convection coefficient +printf("\nThe convection coefficient is %.2f BTU/(hr.sq.ft.degree R)",hx); +LHS=(hx/3600)*(Pr)^(2/3)/(rou*cp*V_inf); +printf("\nThe value of left hand side of the equation is %.2e",LHS); +Re_L=1.46e+005; // Reynolds number at plate end +RHS=0.332*(Re_L)^(-0.5); +printf("\nThe value of left hand side of the equation is %.2e",RHS); +err=(LHS-RHS)*100/LHS; +printf("\nThe error is %d percent",err); +printf("\nSince the error is only %d percent, the agreement is quite good",err); diff --git a/1309/CH7/EX7.6/Result7_6.pdf b/1309/CH7/EX7.6/Result7_6.pdf Binary files differnew file mode 100755 index 000000000..75f95318b --- /dev/null +++ b/1309/CH7/EX7.6/Result7_6.pdf diff --git a/1309/CH7/EX7.6/ch7_6.sce b/1309/CH7/EX7.6/ch7_6.sce new file mode 100755 index 000000000..0be0224fd --- /dev/null +++ b/1309/CH7/EX7.6/ch7_6.sce @@ -0,0 +1,17 @@ +clc; +clear; +printf("\t\t\tChapter7_example6\n\n\n"); +// Estimation of the drag due to skin friction +// properties of water at 68°F from Appendix Table C.11 +rou= 62.4; // density in Ibm/cu.ft +v= 1.083e-5; // viscosity in sq.ft/s +V_inf=5*.5144/.3048; // barge velocity in ft/s using conversion factors from appendix table A1 +printf("\nThe barge velocity is %.2f ft/s",V_inf); +L=20; // Length of barge in ft +Re_L=V_inf*L/v; // Reynolds number at plate end +printf("\nThe Reynolds number at plate end is %.2e",Re_L); +Cd=0.003; //value of Cd corresponding to the Reynolds number from figure 7.11 +gc=32.2; +b=12; // width in ft +Df=(Cd*rou*V_inf^2*b*L)/(2*gc); +printf("\nThe drag force is %d lbf",Df); diff --git a/1309/CH7/EX7.7/Result7_7.pdf b/1309/CH7/EX7.7/Result7_7.pdf Binary files differnew file mode 100755 index 000000000..5eb3546c4 --- /dev/null +++ b/1309/CH7/EX7.7/Result7_7.pdf diff --git a/1309/CH7/EX7.7/ch7_7.sce b/1309/CH7/EX7.7/ch7_7.sce new file mode 100755 index 000000000..22a800526 --- /dev/null +++ b/1309/CH7/EX7.7/ch7_7.sce @@ -0,0 +1,42 @@ +clc; +clear; +printf("\t\t\tChapter7_example7\n\n\n"); +// Determination of wattage requirement +// properties of carbon dioxide at a film temperature of (400+600)/2 = 500 K from appendix table D2 +rou= 1.0732; // density in kg/m^3 +cp= 1013; // specific heat in J/(kg*K) +v= 21.67e-6; // viscosity in m^2/s +Pr = 0.702; // Prandtl Number +k= 0.03352; // thermal conductivity in W/(m.K) +a = 0.3084e-4; // diffusivity in m^2/s +V_inf=60; // carbon dioxide velocity in m/s +x_cr=(5e5)*v/V_inf; // The transition length in m +printf("\nThe transition length is %.1f cm",x_cr*100); +w=4; // width of each heater in cm +b=.16; // effective heating length in m +Tw=600; // temperature of heater surface in K +T_inf=400; // temperature of carbon dioxide in K +r=pmodulo(x_cr*100,w); +n=(x_cr*100+r)/w; // number of heater where transition occurs +printf("\nThe transition thus occur at %dth heater",n); +m=6; // number of heater strips +q=zeros(m+1,1); +x=[0.04 0.08 0.12 0.16 0.20 0.24]; +for i=1:n-1 // transition occurs at 5th heater, so laminar zone equation is followed till then + h(i)=(0.664*k)*(V_inf/v)^0.5*(Pr)^(1/3)/x(i)^0.5; + printf("\n\nThe convective coefficient for heater no. %d is %d W/(sq.m.K)",i,h(i)); + q(i+1)=h(i)*x(i)*b*(Tw-T_inf); + dq(i)=q(i+1)-q(i); + printf("\nThe heat transferred by heater no. %d is %d W",i,dq(i)); +end +// Turbulent zone exists from 5th heater onwards so the following equation is followed Nu=h*x/kf=[0.0359*(Re_L)^(4/5)-830]*(Pr)^(1/3) +for i=5:6 + Re_L(i)=V_inf*x(i)/v; + h(i)=(k/x(i))*[0.0359*(Re_L(i))^(4/5)-830]*(Pr)^(1/3) + printf("\n\nThe Reynolds number for heater no. %d is %.2e",i,Re_L(i)); + printf("\nThe convective coefficient for heater no. %d is %.1f W/(sq.m.K)",i,h(i)); + q(i+1)=h(i)*x(i)*b*(Tw-T_inf); + dq(i)=q(i+1)-q(i); + printf("\nThe heat transferred by heater no. %d is %d W",i,dq(i)); +end + diff --git a/1309/CH7/EX7.8/Result7_8.pdf b/1309/CH7/EX7.8/Result7_8.pdf Binary files differnew file mode 100755 index 000000000..ce33b8313 --- /dev/null +++ b/1309/CH7/EX7.8/Result7_8.pdf diff --git a/1309/CH7/EX7.8/ch7_8.sce b/1309/CH7/EX7.8/ch7_8.sce new file mode 100755 index 000000000..ca71853dd --- /dev/null +++ b/1309/CH7/EX7.8/ch7_8.sce @@ -0,0 +1,30 @@ +clc; +clear; +printf("\t\t\tChapter7_example8\n\n\n"); +// Estimation of force exerted on the pole +// properties of air at given conditions from appendix table D1 +rou= 0.0735; // density in Ibm/ft^3 +v= 16.88e-5; // viscosity in ft^2/s +V=20*5280/3600; // flow velocity in ft/s +printf("\nThe flow velocity is %.1f ft/s",V); +D=12/12; // diameter of pole in ft +L=30;// length of pole in ft +gc=32.2; +Re_D=V*D/v; // Reynolds Number for flow past the pole +printf("\nThe Reynolds Number for flow past the pole is %.2e ",Re_D); +Cd_cylinder=1.1; // value of Cd for smooth cylinder from figure 7.22 +A_cylinder=D*L; // frontal area of pole +printf("\nThe frontal area of pole is %d sq.ft",A_cylinder); +Df_cylinder=Cd_cylinder*(1/2)*rou*V^2*A_cylinder/gc; +printf("\nThe Drag force exerted on only the pole is %.1f lbf",Df_cylinder); +D_square=2/12; // length of square part of pole +L_square=4; +Re_square=V*D_square/v; // Reynolds Number for square part of pole +printf("\nThe Reynolds Number for square part of pole is %.1e",Re_square); +Cd_square=2; // Corresponding value of Cd for square part from figure 7.23 +A_square=D_square*L_square; // projected frontal area of square part +printf("\nThe frontal area of square part of pole is %.3f sq.ft",A_square); +Df_square=Cd_square*(1/2)*rou*V^2*A_square/gc; +printf("\nThe Drag force exerted on cross piece of the pole is %.2f lbf",Df_square); +Df_total=Df_cylinder+Df_square; +printf("\nThe total drag force on the pole is %.1f lbf",Df_total); diff --git a/1309/CH7/EX7.9/Result7_9.pdf b/1309/CH7/EX7.9/Result7_9.pdf Binary files differnew file mode 100755 index 000000000..05273af03 --- /dev/null +++ b/1309/CH7/EX7.9/Result7_9.pdf diff --git a/1309/CH7/EX7.9/ch7_9.sce b/1309/CH7/EX7.9/ch7_9.sce new file mode 100755 index 000000000..0c440c728 --- /dev/null +++ b/1309/CH7/EX7.9/ch7_9.sce @@ -0,0 +1,30 @@ +clc; +clear; +printf("\t\t\tChapter7_example9\n\n\n"); +// determination of required current +// properties of air at film temperature (300 + 500)/2 = 400 K from appendix table D1 +rou= 0.883; // density in kg/cu.m +cp= 1014; // specific heat in J/(kg*K) +v= 25.90e-6; // viscosity in sq.m/s +Pr = 0.689; // Prandtl Number +kf= 0.03365; // thermal conductivity in W/(m.K) +a = 0.376e-4; // diffusivity in sq.m/s +V_inf=1; // velocity in m/s +D=0.00013; // diameter in m +L=1/100; // length of wire in cm +Re_D=V_inf*D/v; // The Reynolds number of flow past the wire +printf("\nThe Reynolds number of flow past the wire is %.3f",Re_D); +C=0.911; //value of C for cylinder from table 7.4 +m=0.385; //value of m for cylinder from table 7.4 +hc=kf*C*(Re_D)^m*(Pr)^(1/3)/D; // the convection coefficient in W/(m^2.K) +printf("\nThe convection coefficient is %d W/(sq.m.K)",hc); +Tw=500; // air stream temperature in K +T_inf=300; // wire surface temperature in K +As=%pi*D*L; // cross sectional area in sq.m +qw=hc*As*(Tw-T_inf); // The heat transferred to the air from the wire +printf("\nThe heat transferred to the air from the wire is %.3f W",qw); +resistivity=17e-6; // resistivity in ohm cm +Resistance=resistivity*(L/(%pi*D^2)); // resistance in ohm +printf("\nThe resistance is %.3f ohm",Resistance/100); +i=(qw*100/Resistance)^0.5; // current in ampere +printf("\nThe current is %.1f A",i); diff --git a/1309/CH8/EX8.1/Result8_1.pdf b/1309/CH8/EX8.1/Result8_1.pdf Binary files differnew file mode 100755 index 000000000..8b9737dc4 --- /dev/null +++ b/1309/CH8/EX8.1/Result8_1.pdf diff --git a/1309/CH8/EX8.1/ch8_1.sce b/1309/CH8/EX8.1/ch8_1.sce new file mode 100755 index 000000000..bf1f1f58b --- /dev/null +++ b/1309/CH8/EX8.1/ch8_1.sce @@ -0,0 +1,26 @@ +clc; +clear; +printf("\t\t\tChapter8_example1\n\n\n"); +// Determination of the heat transferred to the wall. +// air properties at (400+120)/2 =260 degree F = 720 degree R from Appendix Table D1 +rou= 0.0551; // density in Ibm/cu.ft +cp=0.2420; // specific heat BTU/(lbm-degree Rankine) +v= 27.88e-5; // viscosity in sq.ft/s +kf = 0.01944 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a = 1.457; // diffusivity in sq.ft/hr +Pr = 0.689; // Prandtl Number +T_inf=120+460; // wall temperature in degree R +Tw=400+460; // inside wall temperature in degree R +Beta=1/T_inf; +printf("\nThe volumetric thermal expansion coefficient is %.5f/degree R",Beta); +gc=32.2; +L=1; // length of wall in ft +W=2; // width in ft +Gr=(gc*Beta*(Tw-T_inf)*L^3)/v^2;// Grashof Number +printf("\nThe Grashof number is %.2e",Gr); +temperature_slope=0.505; //value of temperature slope from table 8.1 corresponding to Pr=.72 +hL=(kf/L)*(4/3)*(Gr/4)^(1/4)*temperature_slope; // The convection coefficient in BTU/(hr.ft^2.degree R) +printf("\nThe convection coefficient is %.2f BTU/(hr.sq.ft.degree R)",hL); +A=L*W; // cross sectional area in sq.ft +qw=hL*A*(Tw-T_inf); +printf("\nThe heat transferred is %d BTU/hr",qw); diff --git a/1309/CH8/EX8.2/Result8_2.pdf b/1309/CH8/EX8.2/Result8_2.pdf Binary files differnew file mode 100755 index 000000000..24bb63631 --- /dev/null +++ b/1309/CH8/EX8.2/Result8_2.pdf diff --git a/1309/CH8/EX8.2/ch8_2.sce b/1309/CH8/EX8.2/ch8_2.sce new file mode 100755 index 000000000..97c932aba --- /dev/null +++ b/1309/CH8/EX8.2/ch8_2.sce @@ -0,0 +1,47 @@ +clc; +clear; +printf("\t\t\tChapter8_example2\n\n\n"); +// Determination of heat lost through the glass per unit area +// properties of air at 22 + 273 = 295 K = 300 K(approx) and 273 K from appendix table D1 +rou= [1.177 1.295]; // density in kg/cu.m +cp= [1005 1005.5]; // specific heat in J/(kg*K) +v= [15.68e-6 12.59e-6]; // viscosity in sq.m/s +Pr = [0.708 0.713]; // Prandtl Number +kf= [0.02624 0.02426]; // thermal conductivity in W/(m.K) +a = [0.22160e-4 0.17661e-4]; // diffusivity in sq.m/s +T_inf=[22 0]// inside and outside temperature in K +Beta=[1/(T_inf(1)+273) 1/(T_inf(2)+273)]; // volumetric thermal expansion coefficient at 295 K and 273 K +printf("\nThe volumetric thermal expansion coefficients at 295 K and 273 K are respectively %.5f and %.5f",Beta(1),Beta(2)); +g=9.81; +t=0.005; // thickness of glass +L=0.60; // window length in m +k=0.81; // thermal conductivity of glass from appendix table B3 +// for first guess +Tw=[18 4]; +printf("\nFor first guess, the results are:\n"); +for i=1:2 + Ra(i)=(g*Beta(i)*(Tw(i)-T_inf(i))*L^3)/(v(i)*a(i)); + hL(i)=(kf(i)/L)*(0.68+((0.67*(abs(Ra(i)))^(1/4))/(1+(0.492/Pr(i))^(9/16))^(4/9))); +end +printf("\nThe Rayleigh Numbers are %.3e and %.3e",-Ra(1),Ra(2)); +printf("\nThe convective coefficients are %.2f W/(sq.m.K) and %.2f W/(sq.m.K)",hL(1),hL(2)); +q=(T_inf(1)-T_inf(2))/((1/hL(2))+(t/k)+(1/hL(1))); +printf("\nThe heat flux is %.1f W/sq.m",q); +for i=1:2 + Tw_final(i)=T_inf(i)-q*(1/hL(i)); + printf("\nThe wall temperature calculated is %.1f",abs(Tw_final(i))); + Tw(i)=abs(Tw_final(i)); // second guess +end +printf("\nFor second guess, the results are:\n"); +for i=1:2 + Ra(i)=(g*Beta(i)*(Tw(i)-T_inf(i))*L^3)/(v(i)*a(i)); + hL(i)=(kf(i)/L)*(0.68+((0.67*(abs(Ra(i)))^(1/4))/(1+(0.492/Pr(i))^(9/16))^(4/9))); +end +printf("\nThe Rayleigh Numbers are %.3e and %.3e",-Ra(1),Ra(2)); +printf("\nThe convective coefficients are %.2f W/(sq.m.K) and %.2f W/(sq.m.K)",hL(1),hL(2)); +q=(T_inf(1)-T_inf(2))/((1/hL(2))+(t/k)+(1/hL(1))); +printf("\nThe heat flux is %.1f W/sq.m",q); +for i=1:2 + Tw_final(i)=T_inf(i)-q*(1/hL(i)); + printf("\nThe wall temperature calculated is %.1f degree celsius",abs(Tw_final(i))); +end diff --git a/1309/CH8/EX8.3/Result8_3.pdf b/1309/CH8/EX8.3/Result8_3.pdf Binary files differnew file mode 100755 index 000000000..5d42a9460 --- /dev/null +++ b/1309/CH8/EX8.3/Result8_3.pdf diff --git a/1309/CH8/EX8.3/ch8_3.sce b/1309/CH8/EX8.3/ch8_3.sce new file mode 100755 index 000000000..c36c647ce --- /dev/null +++ b/1309/CH8/EX8.3/ch8_3.sce @@ -0,0 +1,23 @@ +clc; +clear; +printf("\t\t\tChapter8_example3\n\n\n"); +// determination of heat loss through the side. +rou= 0.0735; // density in Ibm/cu.ft +cp=0.240; // specific heat BTU/(lbm-degree Rankine) +v= 16.88e-5; // viscosity in sq.ft/s +kf = 0.01516 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a = 0.859; // diffusivity in sq.ft/hr +Pr = 0.708; // Prandtl Number +Tw=90; +T_inf=70; +g=32.2; +L=5.5; // length in ft +W=2+(4/12); // width in ft +Beta=1/(Tw+460); // volumetric thermal expansion coefficient in per degree Rankine +printf("\nThe volumetric thermal expansion coefficient is %.5f /degree R",Beta); +Ra=(g*Beta*(Tw-T_inf)*L^3)/(v*a/3600); +printf("\nThe Rayleigh Number is %.2e ",Ra); +hc=(kf/L)*(0.825+((0.387*(Ra)^(1/6))/(1+(0.492/Pr)^(9/16))^(8/27)))^2; +printf("\nThe value of convection coefficient is %.3f BTU/(hr.sq.ft.degree R)",hc); +q=hc*L*W*(Tw-T_inf); +printf("\nThe heat gained is %d BTU/hr",q); diff --git a/1309/CH8/EX8.4/Result8_4.pdf b/1309/CH8/EX8.4/Result8_4.pdf Binary files differnew file mode 100755 index 000000000..95f3709c4 --- /dev/null +++ b/1309/CH8/EX8.4/Result8_4.pdf diff --git a/1309/CH8/EX8.4/ch8_4.sce b/1309/CH8/EX8.4/ch8_4.sce new file mode 100755 index 000000000..f8203d12f --- /dev/null +++ b/1309/CH8/EX8.4/ch8_4.sce @@ -0,0 +1,41 @@ +clc; +clear; +printf("\t\t\tChapter8_example4\n\n\n"); +// Determination of the variation of average convection coefficient with distance +// properties of air at (65 + 20)/2 = 42.5 degree C =315 K. from appendix table D1 +rou= 1123; // density in kg/m^3 +cp= 1006.7; // specific heat in J/(kg*K) +v= 17.204e-6; // viscosity in m^2/s +Pr =0.703; // Prandtl Number +kf= 0.02738; // thermal conductivity in W/(m.K) +a = 0.2446e-4; // diffusivity in m^2/s +g=9.81; +L=5; +theta=45; +T_inf=20; // ambient air temperature in degree C +Tw=65; // roof surface temperature in degree C +Beta=1/(T_inf+273); // volumetric thermal expansion coefficient in per K +printf("\nThe volumetric thermal expansion coefficient is %.5f /K",Beta); +// determination of Laminar-turbulent transition length by Vliet equation Ra=3x10^5xexp(0.1368cos(90-theta)) +x=((3e5*exp(0.1368*cos(90-theta))*v*a)/(g*cos(theta)*Beta*(Tw-T_inf)))^(1/3); +printf("\nThe Laminar-turbulent transition length by Vliet equation is %.3f m",x); +i=1; +N=1; +n=0; +X=[0.02 0.04 0.05 0.051 0.1 1.0 3.0 5.0]; // entering values for length(m) +[n m]=size(X); +for i=1:m + if X(i)<=x then + // Laminar Flow regime exists + Ra(i)=(g*cos(%pi*45/180)*Beta*(Tw-T_inf)*X(i)^3)/(v*a); + hc(i)=(kf/X(i))*(0.68+(0.670*Ra(i)^(1/4))/(1+(0.492/Pr)^(9/16))^(4/9)); + else + // Turbulent Flow regime exists + Ra(i)=(g*Beta*(Tw-T_inf)*X(i)^3)/(v*a); + hc(i)=(0.02738/X(i))*(0.825+0.324*Ra(i)^(1/6))^2; + end +end +printf("\n\tx,m\t\tRa\t\thc,W/(sq.m.K)\n"); +for i=1:m + printf("\t%.2f\t\t%.2e\t%.2f\n",X(i),Ra(i),hc(i)); +end diff --git a/1309/CH8/EX8.5/Result8_5.pdf b/1309/CH8/EX8.5/Result8_5.pdf Binary files differnew file mode 100755 index 000000000..68197655b --- /dev/null +++ b/1309/CH8/EX8.5/Result8_5.pdf diff --git a/1309/CH8/EX8.5/ch8_5.sce b/1309/CH8/EX8.5/ch8_5.sce new file mode 100755 index 000000000..b1a8474fc --- /dev/null +++ b/1309/CH8/EX8.5/ch8_5.sce @@ -0,0 +1,38 @@ +clc; +clear; +printf("\t\t\tChapter8_example5\n\n\n"); +// determine if heat is lost lose more heat through its upper surface or one of its vertical sides +// properties of air at (100 + 60)/2 = 80°F = 540 degree R from appendix table D1 +rou= 0.0735; // density in lbm/cu.ft +cp=0.240; // specific heat BTU/(lbm-degree Rankine) +v= 16.88e-5; // viscosity in sq.ft/s +kf = 0.01516 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a = 0.859; // diffusivity in sq.ft/hr +Pr = 0.708; // Prandtl Number +Tw=100; // temperature of outside surface temperature of oven in degree F +T_inf=60; // ambient temperature in degree F +g=32.2; +L=2; // length in ft +W=2; // width in ft +Beta=1/(T_inf+460); // volumetric thermal expansion coefficient in per degree Rankine +printf("\nThe volumetric thermal expansion coefficient is %.5f /degree R",Beta); +Ra=(g*Beta*(Tw-T_inf)*L^3)/(v*a/3600); +printf("\nThe Rayleigh Number is %.2e ",Ra); +hc=(kf/L)*(0.68+(0.670*Ra^(1/4))/(1+(0.492/Pr)^(9/16))^(4/9)); +printf("\nThe value of convection coefficient is %.3f BTU/(hr.sq.ft.degree R)",hc); +q1side=hc*L*W*(Tw-T_inf); +printf("\nThe heat transferred from one side is %.1f BTU/hr",q1side); +// For the top, we have a heated surface facing upward, The characteristic length is determined as follows +Lc=(2*2)/(2+2+2+2); +Ra_L=(g*Beta*(Tw-T_inf)*Lc^3)/(v*a/3600); // Rayleigh number based on characteristic length +printf("\nThe Rayleigh Number based on characteristic length is %.2e ",Ra_L); +hc_L=(kf/Lc)*0.54*(Ra_L)^(1/4); +printf("\nThe convective coefficient based on characteristic length is %.3f BTU/(hr.sq.ft.degree R)",hc_L); +qtop=hc_L*L*W*(Tw-T_inf); +printf("\nThe heat transferred from top is %d BTU/hr",qtop); +if qtop>q1side then + printf("\nThe top transfers heat at a higher rate"); +elseif qtop<q1side + printf("\nThe side transfers heat at a higher rate"); + else printf("\nThe top and side transfer heat at equal rates"); +end diff --git a/1309/CH8/EX8.6/Result8_6.pdf b/1309/CH8/EX8.6/Result8_6.pdf Binary files differnew file mode 100755 index 000000000..b0fc1b239 --- /dev/null +++ b/1309/CH8/EX8.6/Result8_6.pdf diff --git a/1309/CH8/EX8.6/ch8_6.sce b/1309/CH8/EX8.6/ch8_6.sce new file mode 100755 index 000000000..81c659432 --- /dev/null +++ b/1309/CH8/EX8.6/ch8_6.sce @@ -0,0 +1,33 @@ +clc; +clear; +printf("\t\t\tChapter8_example6\n\n\n"); +// determination of heat lost from the insulation by convection +// properties of air at (50 + 5)/2 = 27.5 degree C = 300 K from appendix table D1 +rou= 1.177; // density in kg/cu.m +cp= 1005.7; // specific heat in J/(kg*K) +v= 15.68e-6; // viscosity in sq.m/s +Pr =0.708; // Prandtl Number +kf= 0.02624; // thermal conductivity in W/(m.K) +a = 0.22160e-4; // diffusivity in sq.m/s +g=9.81; +L=4; // length in m +D=15/100; // diameter in m +T_inf=5; // ambient air temperature in degree C +Tw=50; // outside surface temperature in degree C +Beta=1/(T_inf+273); // volumetric thermal expansion coefficient in per K +printf("\nThe volumetric thermal expansion coefficient is %.5f /K",Beta); +Ra=(g*Beta*(Tw-T_inf)*D^3)/(v*a); +printf("\nThe Rayleigh Number is %.2e ",Ra); +// for horizontal pipe, the convective coefficient is determined as follows +hc_h=(kf/D)*(0.60+(0.387*Ra^(1/6))/(1+(0.559/Pr)^(9/16))^(8/27))^2; +printf("\nThe convection coefficient for horizontal length is %.2f W/(sq.m.K)",hc_h); +As=%pi*D*L; +q_hor=hc_h*As*(Tw-T_inf); +printf("\nThe heat transferred from the horizontal length of 4 m is %d W",q_hor); +// for vertical pipe, the convective coefficient is determined as follows +hc_v=(kf/D)*0.6*(Ra*(D/L))^(1/4); +printf("\nThe convection coefficient for vertical length is %.2f W/(sq.m.K)",hc_v); +q_ver=hc_v*As*(Tw-T_inf); +printf("\nThe heat transferred from the vertical length of 4 m is %d W",q_ver); +q=q_ver+q_hor; +printf("\nThe total heat lost from the pipe is %d W",q); diff --git a/1309/CH8/EX8.7/Result8_7.pdf b/1309/CH8/EX8.7/Result8_7.pdf Binary files differnew file mode 100755 index 000000000..23b1e7ddc --- /dev/null +++ b/1309/CH8/EX8.7/Result8_7.pdf diff --git a/1309/CH8/EX8.7/ch8_7.sce b/1309/CH8/EX8.7/ch8_7.sce new file mode 100755 index 000000000..f31f4c697 --- /dev/null +++ b/1309/CH8/EX8.7/ch8_7.sce @@ -0,0 +1,23 @@ +clc; +clear; +printf("\t\t\tChapter8_example7\n\n\n"); +// Determinion of the convection coefficient about the ice cube +// properties of air at (0 + 70)/2 = 35°F == 495 degree R from appendix table D1 +rou= 0.0809; // density in lbm/cu.ft +cp=0.240; // specific heat BTU/(lbm-degree Rankine) +v= 13.54e-5; // viscosity in sq.ft/s +kf = 0.01402 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a = 0.685; // diffusivity in sq.ft/hr +Pr = 0.712; // Prandtl Number +Tw=0; // temperature of outside surface temperature of oven in degree F +T_inf=70; // ambient temperature in degree F +g=32.2; +Beta=1/(T_inf+460); // volumetric thermal expansion coefficient in per degree Rankine +printf("\nThe volumetric thermal expansion coefficient is %.5f /degree R",Beta); +// The characteristic length is found by using King Equation +Lc=1/((1/1)+(1/1.2)); +printf("\nThe characteristic length is %.3f ft",Lc); +Ra=(g*Beta*abs(Tw-T_inf)*Lc^3)/(v*a/3600); +printf("\nThe Rayleigh Number is %.2e ",Ra); +hc=(kf/Lc)*0.6*(Ra)^(1/4); +printf("\nThe value of convection coefficient is %.2f BTU/(hr.sq.ft.degree R)",hc); diff --git a/1309/CH8/EX8.8/Result8_8.pdf b/1309/CH8/EX8.8/Result8_8.pdf Binary files differnew file mode 100755 index 000000000..48748beb0 --- /dev/null +++ b/1309/CH8/EX8.8/Result8_8.pdf diff --git a/1309/CH8/EX8.8/ch8_8.sce b/1309/CH8/EX8.8/ch8_8.sce new file mode 100755 index 000000000..26e571626 --- /dev/null +++ b/1309/CH8/EX8.8/ch8_8.sce @@ -0,0 +1,35 @@ +clc; +clear; +printf("\t\t\tChapter8_example8\n\n\n"); +// determination of the maximum amount of heat that fins can transfer +// properties of air at (100 + 35)/2 = 67.5 degree C from appendix table D1 +rou= 0.998; // density in kg/cu.m +cp= 1009.0; // specific heat in J/(kg*K) +v= 20.76e-6; // viscosity in sq.m/s +Pr =0.697; // Prandtl Number +kf= 0.03003; // thermal conductivity in W/(m.K) +a = 0.2983e-4; // diffusivity in sq.m/s +g=9.81; +T_inf=35; // ambient air temperature in degree C +Tw=100; // surface temperature in degree C +Beta=1/(T_inf+273); // volumetric thermal expansion coefficient in per K +printf("\nThe volumetric thermal expansion coefficient is %.5f /K",Beta); +// properties of aluminium from appendix table B1 +rou_Al=2702; // density in kg/cu.m +k_Al=236; // thermal conductivity in W/(m.K) +cp_Al=896;// specific heat in J/(kg*K) +a_Al=97.5e-6; // diffusivity in sq.m/s +b=46/100; +w=24/100; +// Applying the Bar-Cohen Equations +zeta=((w*v^2)/(g*Beta*(Tw-T_inf)*Pr))^(1/4); +printf("\nThe value of zeta is %.2e ",zeta); +L=1.54*(k_Al/kf)^(1/2)*zeta; +printf("\nThe fin length is %.3f m",L); +S=2.89*zeta; +printf("\nThe fin spacing is %.5f m",S); +q=(b*w*(Tw-T_inf)*1.3*(k_Al*kf)^(1/2))/(6*zeta); +printf("\nThe heat transfer rate is %d W",q); +N=b/(2*S); +printf("\nThe number of fins can be atmost %d",N); + diff --git a/1309/CH9/EX9.1/Result9_1.pdf b/1309/CH9/EX9.1/Result9_1.pdf Binary files differnew file mode 100755 index 000000000..902f3b2f0 --- /dev/null +++ b/1309/CH9/EX9.1/Result9_1.pdf diff --git a/1309/CH9/EX9.1/ch9_1.sce b/1309/CH9/EX9.1/ch9_1.sce new file mode 100755 index 000000000..ff9cd87a7 --- /dev/null +++ b/1309/CH9/EX9.1/ch9_1.sce @@ -0,0 +1,16 @@ +clc; +clear; +printf("\t\t\tChapter9_example1\n\n\n"); +// determination of counterflow and parallel-flow configurations. +// temperatures of hot fluid in degree C +T1=100; +T2=75; +// temperatures of cold fluid in degree C +t1=5; +t2=50; +// for counterflow +LMTD_counter=((T1-t2)-(T2-t1))/(log((T1-t2)/(T2-t1))); +printf("\nThe LMTD for counter flow configuration is %.1f degree C",LMTD_counter); +// for parallel flow +LMTD_parallel=((T1-t1)-(T2-t2))/(log((T1-t1)/(T2-t2))); +printf("\nThe LMTD for parallel flow configuration is %.1f degree C",LMTD_parallel);
\ No newline at end of file diff --git a/1309/CH9/EX9.2/Result9_2.pdf b/1309/CH9/EX9.2/Result9_2.pdf Binary files differnew file mode 100755 index 000000000..36f659907 --- /dev/null +++ b/1309/CH9/EX9.2/Result9_2.pdf diff --git a/1309/CH9/EX9.2/ch9_2.sce b/1309/CH9/EX9.2/ch9_2.sce new file mode 100755 index 000000000..535d0e11c --- /dev/null +++ b/1309/CH9/EX9.2/ch9_2.sce @@ -0,0 +1,15 @@ +clc; +clear; +printf("\t\t\tChapter9_example2\n\n\n"); +// Determination of the LMTD for both counterflow and parallel-flow configurations. +// temperatures of hot fluid in degree F +T1=250; +T2=150; +// temperatures of cold fluid in degree F +t1=100; +t2=150; +// for counterflow +LMTD_counter=((T1-t2)-(T2-t1))/(log((T1-t2)/(T2-t1))); +printf("\nThe LMTD for counter flow configuration is %.1f degree C",LMTD_counter); +// for parallel flow +printf("\nFor a finite heat-transfer rate and a finite overall heat-transfer coefficient,\nif parallel flow is to give equal outlet temperatures,\nthen the area needed must be infinite which is not feasible economically."); diff --git a/1309/CH9/EX9.3/Result9_3.pdf b/1309/CH9/EX9.3/Result9_3.pdf Binary files differnew file mode 100755 index 000000000..09defc366 --- /dev/null +++ b/1309/CH9/EX9.3/Result9_3.pdf diff --git a/1309/CH9/EX9.3/ch9_3.sce b/1309/CH9/EX9.3/ch9_3.sce new file mode 100755 index 000000000..e4c44d035 --- /dev/null +++ b/1309/CH9/EX9.3/ch9_3.sce @@ -0,0 +1,70 @@ +clc; +clear; +printf("\t\t\tChapter9_example3\n\n\n"); +// Determination of the outlet temperature of the ethylene glycol for counterflow. +// properties of air at (195 + 85)/2 = 140°F. from appendix table CII +rou_1= 0.985*62.4; // density in lbm/ft^3 +cp_1=0.9994; // specific heat BTU/(lbm-degree Rankine) +v_1= 0.514e-5; // viscosity in ft^2/s +kf_1 = 0.376 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a_1 = 6.02e-3; // diffusivity in ft^2/hr +Pr_1 = 3.02; // Prandtl Number +m_1=5000; // mass flow rate in lbm/hr +T_1=195; // temperature in degree F +// properties of ethylene glycol at 140 degree F from Appendix Table C.5 +rou_2= 1.087*62.4; // density in lbm/ft^3 +cp_2=0.612; // specific heat BTU/(lbm-degree Rankine) +v_2= 5.11e-5; // viscosity in ft^2/s +kf_2 = 0.150 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a_2 = 3.61e-3; // diffusivity in ft^2/hr +Pr_2 = 51; // Prandtl Number +m_2=12000; // mass flow rate in lbm/hr +T_2=85; // temperature in degree F +// specifications of seamless copper water tubing (subscripts: a = annulus, p = inner pipe or tube) from appendix table F2 +ID_a=0.1674; +ID_p=0.1076; +OD_p=1.375/12; +// Flow Areas +A_p=%pi*ID_p^2/4; +A_a=%pi*((ID_a)^2-(OD_p)^2)/4; +printf("\nThe area of annulus is %.5f sq.ft",A_a); +printf("\nThe area of inner pipe is %.5f sq.ft",A_p); +if A_a>A_p then + printf("\nAir flows through annulus"); + else printf("\ncarbon dioxide flows through annulus"); +end +// Annulus Equivalent Diameters +D_h=ID_a-OD_p; +D_e=(ID_a^2-OD_p^2)/(OD_p); +printf("\nThe Annulus Equivalent Diameter for friction is %.4f ft",D_h); +printf("\nThe Annulus Equivalent Diameter for heat transfer is %.4f ft",D_e); +// Reynolds Numbers +Re_1=(m_1/3600)*(ID_p)/(v_1*rou_1*A_p); +printf("\nThe Reynolds Number for water is %.1e",Re_1); +Re_2=(m_2/3600)*(D_e)/(v_2*rou_2*A_a); +printf("\nThe Reynolds Number for ethylene glycol is %.2e",Re_2); +// Nusselt numbers +Nu_1=0.023*(Re_1)^(4/5)*(Pr_1)^0.3; +Nu_2=0.023*(Re_2)^(4/5)*(Pr_2)^0.4; +printf("\nThe Nusselt number for water is %d",Nu_1); +printf("\nThe Nusselt number for ethylene glycol is %d",Nu_2); +// Convection Coefficients +h_1i=Nu_1*kf_1/ID_p; +h_1o=h_1i*ID_p/OD_p; +h_2=Nu_2*kf_2/D_e; +printf("\nThe convective coefficient for water based on inner diameter is %d BTU/(hr.ft^2.degree R)",h_1i); +printf("\nThe convective coefficient for water based on outer diameter is %d BTU/(hr.sq.ft.degree R)",h_1o); +printf("\nThe convective coefficient for ethylene glycol is %d BTU/(hr.sq.ft.degree R)",h_2); +// Exchanger Coefficient +Uo=1/((1/h_1o)+(1/h_2)); +printf("\nThe overall exchanger coefficient is %d BTU/(hr.sq.ft.degree R)",Uo); +R=(m_2*cp_2)/(m_1*cp_1); +L=20; +A=%pi*OD_p*L; +printf("\nThe ratio is %.2f and area is %.1f sq.ft",R,A); +T1=195; +t1=85; +T2=((T1*(R-1))-(R*t1*(1-exp((Uo*A*(R-1))/(m_2*cp_2)))))/(R*exp(Uo*A*(R-1)/(m_2*cp_2))-1); +printf("\nThe temperature T2=%d degree F",T2); +t2=t1+(T1-T2)/R; +printf("\nThe outlet temperature of Ethylene glycol is %.1f degree F",t2); diff --git a/1309/CH9/EX9.4/Result9_4.pdf b/1309/CH9/EX9.4/Result9_4.pdf Binary files differnew file mode 100755 index 000000000..6e271fccd --- /dev/null +++ b/1309/CH9/EX9.4/Result9_4.pdf diff --git a/1309/CH9/EX9.4/ch9_4.sce b/1309/CH9/EX9.4/ch9_4.sce new file mode 100755 index 000000000..756c2e166 --- /dev/null +++ b/1309/CH9/EX9.4/ch9_4.sce @@ -0,0 +1,99 @@ +clc; +clear; +printf("\t\t\tChapter9_example4\n\n\n"); +// Determination of (a) no. of exchangers required, (b) the overall coefficient of (all) the exchanger(s), and (c) the pressure drop for each stream. +// assuming counterflow arrangement +// properties of air at 323 K. from appendix table D1 +rou_1= 1.088; // density in kg/m^3 +cp_1= 1007; // specific heat in J/(kg*K) +v_1= 18.2e-6; // viscosity in m^2/s +Pr_1 =0.703; // Prandtl Number +kf_1= 0.02814; // thermal conductivity in W/(m.K) +a_1 = 0.26e-4; // diffusivity in m^2/s +m_1=100; // mass flow rate in kg/hr +// temperatures in K +t1_air=20+273; +t2_air=80+273; +// properties of carbon dioxide at [600 + (20 + 273)]/2 = 480 = 500 K. from appendix table D2 +rou_2= 1.0732; // density in kg/m^3 +cp_2= 1013; // specific heat in J/(kg*K) +v_2= 21.67e-6; // viscosity in m^2/s +Pr_2 =0.702; // Prandtl Number +kf_2= 0.03352; // thermal conductivity in W/(m.K) +a_2 = 0.3084e-4; // diffusivity in m^2/s +m_2=90; // mass flow rate in kg/hr +// temperatures in K +T1_CO2=600; +// specifications of seamless copper tubing from appendix table F2 +ID_a=.098; +ID_p=.07384; +OD_p=.07938; +// Flow Areas +A_p=%pi*ID_p^2/4; +A_a=%pi*((ID_a)^2-(OD_p)^2)/4; +printf("\nThe area of annulus is %.2e sq.m",A_a); +printf("\nThe area of inner pipe is %.2e sq.m",A_p); +if A_a>A_p then + printf("\nAir flows through annulus"); + else printf("\nair flows through inner pipe"); +end +// Heat Balance +q_air=(m_1/3600)*(cp_1)*(t2_air-t1_air); +printf("\nThe heat transferred is %.2e W",q_air); +T2_CO2=T1_CO2-(q_air/(m_2*cp_2/3600)); +printf("\nThe low temperature of carbon dioxide is %d K",T2_CO2); +// Log-Mean Temperature Difference +LMTD_counter=((T1_CO2-t2_air)-(T2_CO2-t1_air))/(log((T1_CO2-t2_air)/(T2_CO2-t1_air))); +printf("\nThe LMTD for counter flow configuration is %d degree C",LMTD_counter); +// Annulus Equivalent Diameters +D_h=ID_a-OD_p; +D_e=(ID_a^2-OD_p^2)/(OD_p); +printf("\nThe Annulus Equivalent Diameter for friction is %.5f m",D_h); +printf("\nThe Annulus Equivalent Diameter for heat transfer is %.4f m",D_e); +// Reynolds Numbers +Re_1=(m_1/3600)*(ID_p)/(v_1*rou_1*A_p); +printf("\nThe Reynolds Number for air is %.2e",Re_1); +Re_2=(m_2/3600)*(D_e)/(v_2*rou_2*A_a); +printf("\nThe Reynolds Number for carbon dioxide is %.2e",Re_2); +// Nusselt numbers +Nu_1=0.023*(Re_1)^(4/5)*(Pr_1)^0.3; +Nu_2=0.023*(Re_2)^(4/5)*(Pr_2)^0.4; +printf("\nThe Nusselt number for air is %.1f",Nu_1); +printf("\nThe Nusselt number for carbon dioxide is %.1f",Nu_2); +// Convection Coefficients +h_1i=Nu_1*kf_1/ID_p; +h_1o=h_1i*ID_p/OD_p; +h_2=Nu_2*kf_2/D_e; +printf("\nThe convective coefficient for air based on inner diameter is %.1f W/(sq.m.K)",h_1i); +printf("\nThe convective coefficient for air based on outer diameter is %.1f W/(sq.m.K)",h_1o); +printf("\nThe convective coefficient for carbon dioxide is %.1f W/(sq.m.K)",h_2); +// Fouling Factors in (m^2.K)/W +Rd_air=.0004; +Rd_CO2=0.002; +// exchanger coefficients +Uo=1/((1/h_1o)+(1/h_2)); +Uo=1/((1/Uo)+Rd_air+Rd_CO2); +printf("\nThe overall exchanger coefficient is %.1f W/(sq.m.K)",Uo); +// area required +A=q_air/(Uo*LMTD_counter); +printf("\nThe area required is %.2f sq.m",A); +// surface area of one exchanger is A=%pi*OD*L, so +L=(A/(%pi*OD_p)); // length of each exchanger +L_available=2; // available exchanger length +N=L_available/L; // no. of exchangers +printf("\nThe number of exchangers is %d",N); +//friction factors +fp=0.0245; //friction factor for air fom figure 6.14 corresponding to Reynolds Number calculated above +fa=0.033; //friction factor for carbon dioxide fom figure 6.14 corresponding to Reynolds Number calculated above +// Velocities +V_air=(m_1/3600)/(rou_1*A_p); +V_CO2=(m_2/3600)/(rou_2*A_a); +printf("\nThe velocity of air is %.2f m/s",V_air); +printf("\nThe velocity of carbon dioxide is %.2f m/s",V_CO2); +// pressure drops +dP_p=(fp*L_available*rou_1*V_air^2)/(ID_p*2); +dP_a=((rou_2*V_CO2^2)/2)*((fa*L_available/D_h)+1); +printf("\nThe pressure drop for tube side is %.2f Pa",dP_p); +printf("\nThe pressure drop for shell side is %d Pa",dP_a); +printf("\n\t\t\tSummary of Requested Information\n"); +printf("(a) Exchanger required: %d\n(b)Overall exchanger coefficient = %.1f W/(sq.m.K)\n(c)Air pressure drop = %.2f Pa\nDiesel exhaust pressure drop = %d Pa",N,Uo,dP_p,dP_a); diff --git a/1309/CH9/EX9.5/Result9_5.pdf b/1309/CH9/EX9.5/Result9_5.pdf Binary files differnew file mode 100755 index 000000000..9cc296a2c --- /dev/null +++ b/1309/CH9/EX9.5/Result9_5.pdf diff --git a/1309/CH9/EX9.5/ch9_5.sce b/1309/CH9/EX9.5/ch9_5.sce new file mode 100755 index 000000000..1630de0cb --- /dev/null +++ b/1309/CH9/EX9.5/ch9_5.sce @@ -0,0 +1,93 @@ +clc; +clear; +printf("\t\t\tChapter9_example5\n\n\n"); +// Determination of the outlet temperature of the distilled water and the pressure drop for each stream. +// properties of (distilled) water at 104°F from appendix table CII +rou_1= 0.994*62.4; // density in lbm/ft^3 +cp_1=0.998; // specific heat BTU/(lbm-degree Rankine) +v_1= 0.708e-5; // viscosity in ft^2/s +kf_1 = 0.363 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a_1 = 5.86e-3; // diffusivity in ft^2/hr +Pr_1 = 4.34; // Prandtl Number +m_1=170000; // mass flow rate in lbm/hr +T1=110; // temperature in degree F +// properties of (raw) water at 68°F from Appendix Table C11 +rou_2= 62.4; // density in lbm/ft^3 +cp_2=0.9988; // specific heat BTU/(lbm-degree Rankine) +v_2= 1.083e-5; // viscosity in ft^2/s +kf_2 = 0.345 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a_2 = 5.54e-3; // diffusivity in ft^2/hr +Pr_2 = 7.02; // Prandtl Number +m_2=150000; // mass flow rate in lbm/hr +t1=65; // temperature in degree F +// specifications of 3/4-in-OD, 18-BWG tubes, from table 9.2 +OD=3/(4*12); +ID=0.652/12; +OD_p=1.375/12; +Nt=224; // from table 9.3 +Np=2; // no. of tube passes +// Shell dimensions and other miscellaneous data +Ds=17.25/12; +Nb=15; // no. of baffles +B=1; +sT=15/(16*12); +C=sT-OD; +// flow areas +At=(Nt*%pi*ID^2)/(4*Np); +As=(Ds*C*B)/sT; +printf("\nThe areas are %.3f sq.ft and %.3f sq.ft",At,As); +if At>As then + printf("\nThe distilled water flows through the tubes"); + else printf("\nThe raw water flows through the tubes"); +end +// Shell Equivalent Diameter +De=4*[(sT/2)*(0.86*sT)-(%pi*OD^2/8)]/(%pi*OD/2); +printf("\nThe equivalent diameter is %.4f ft",De); +// Reynolds Numbers +Re_s=(m_1/3600)*(De)/(v_1*rou_1*As); +printf("\nThe Reynolds Number for raw water is %.2e",Re_s); +Re_t=(m_2/3600)*(ID)/(v_2*rou_2*At); +printf("\nThe Reynolds Number for distilled water is %.2e",Re_t); +// Nusselt numbers +Nu_t=0.023*(Re_t)^(4/5)*(Pr_2)^0.4; +Nu_s=0.36*(Re_s)^(.55)*(Pr_1)^(1/3); +printf("\nThe Nusselt number for raw water is %.1f",Nu_t); +printf("\nThe Nusselt number for distilled water is %.1f",Nu_s); +h_ti=Nu_t*kf_2/ID; +h_to=h_ti*ID/OD; +h_s=Nu_s*kf_1/De; +printf("\nThe convective coefficient for raw water based on inner diameter is %d BTU/(hr.sq.ft.degree R)",h_ti); +printf("\nThe convective coefficient for raw water based on outer diameter is %d BTU/(hr.sq.ft.degree R)",h_to); +printf("\nThe convective coefficient for distilled water is %d BTU/(hr.sq.ft.degree R)",h_s); +// Exchanger Coefficient +Uo=1/((1/h_to)+(1/h_s)); +printf("\nThe overall exchanger coefficient is %d BTU/(hr.sq.ft.degree R)",Uo); +R=(m_2*cp_2)/(m_1*cp_1); +L=16; +Ao=Nt*%pi*OD*L; +printf("\nThe ratio is %.3f and area is %.1f sq.ft",R,Ao); +UoAo_mccp=(Uo*Ao)/(m_2*cp_2); +printf("\n(UoAo)/(McCpc)=%.2f",UoAo_mccp); +S=0.58; //value of S from fig. 9.13 Ten Broeck graph corresponding to the value of (UoAo)/(McCpc) +t2=S*(T1-t1)+t1; +T2=T1-R*(t2-t1); +printf("\nt2=%.1f degree F\nT2=%.1f degree F",t2,T2); +//friction factors +ft=0.029; //friction factor for raw water fom figure 6.14 corresponding to Reynolds Number calculated above +printf("\nFriction factor for raw water fom figure 6.14 corresponding to Reynolds Number calculated above is %.3f",ft); +fs=0.281; //friction factor for distilled water fom figure 6.14 corresponding to Reynolds Number calculated above +printf("\nFriction factor for distilled water fom figure 6.14 corresponding to Reynolds Number calculated above is %.3f",fs); +// Velocities +V_t=(m_2/3600)/(rou_2*At); +V_s=(m_1/3600)/(rou_1*As); +printf("\nThe velocity of raw water is %.2f ft/s",V_t); +printf("\nThe velocity of distilled water is %.2f ft/s",V_s); +// pressure drops +gc=32.2; +dP_t=(rou_2*V_t^2)*((ft*L*Np/ID)+4*Np)/(2*gc); +dP_s=((rou_1*V_s^2)*(fs*Ds*(Nb+1)))/(2*gc*De); +printf("\nThe pressure drop for tube side is %.1f lbf/sq.ft = %.1f psi",dP_t,dP_t/147); +printf("\nThe pressure drop for shell side is %.1f lbf/sq.ft = %.1f psi",dP_s,dP_s/147); +printf("\n\t\t\tSummary of Requested Information\n"); +printf("\nOutlet Temperatures:\n\tRaw Water: %.1f degree F\n\tDistilled Water: %.1f degree F\n",t2,T2); +printf("\nPressure Drops:\n\tRaw Water: %.1f ddegree F\n\tDistilled Water: %.1f degree F\n",dP_t/147,dP_s/147); diff --git a/1309/CH9/EX9.6/Result9_6.pdf b/1309/CH9/EX9.6/Result9_6.pdf Binary files differnew file mode 100755 index 000000000..ad113232f --- /dev/null +++ b/1309/CH9/EX9.6/Result9_6.pdf diff --git a/1309/CH9/EX9.6/ch9_6.sce b/1309/CH9/EX9.6/ch9_6.sce new file mode 100755 index 000000000..6e0435f4c --- /dev/null +++ b/1309/CH9/EX9.6/ch9_6.sce @@ -0,0 +1,42 @@ +clc; +clear; +printf("\t\t\tChapter9_example6\n\n\n"); +// Using the effectiveness-NTU method to calculate the outlet temperatures of the fluids +// Data from Example 9.5 +// properties of (distilled) water at 104°F +m_1=170000; // mass flow rate in lbm/hr +T1=110; // temperature in degree F +cp_1=0.998; // specific heat BTU/(lbm-degree Rankine) +// properties of (raw) water at 68°F +m_2=150000; // mass flow rate in lbm/hr +t1=65; // temperature in degree F +cp_2=0.9988; // specific heat BTU/(lbm-degree Rankine) +Uo=350; // exchanger coefficient +Ao=703.7; +// The effectiveness-NTU approach is used when the overall heat transfer coefficient is known +// determining the capacitances +mcp_raw=m_2*cp_2; +mcp_distilled=m_1*cp_1; +printf("\nThe capacitance value of raw water is %d BTU/(hr. degree R)",mcp_raw); +printf("\nThe capacitance value of distilled water is %d BTU/(hr. degree R)",mcp_distilled); +if mcp_raw>mcp_distilled then + mcp_max=mcp_raw; + mcp_min=mcp_distilled; + printf("\nDistilled water has minimum capacitance"); + else mcp_max=mcp_distilled; + mcp_min=mcp_raw; + printf("\nRaw water has minimum capacitance"); +end +// determination of parameters for determining effectiveness +mcp_min_max=mcp_min/mcp_max; +UA_mcpmin=(Uo*Ao)/(mcp_min); +printf("\nThe required parameters are mcp_min/mcp_max=%.3f and (UoAo/mcp_min)=%.2f",mcp_min_max,UA_mcpmin); +effectiveness=0.58; //value of effectiveness from figure 9.15 corresponding to the above calculated values of capacitance ratio and (UoAo/mcp_min) +qmax=mcp_min*(T1-t1); +printf("\nThe maximum heat transfer is %.2e BTU/hr",qmax); +q=effectiveness*qmax; // actual heat transfer +printf("\nThe actual heat transfer is %.2e BTU/hr",q); +t2=(q/mcp_raw)+t1; +T2=T1-(q/mcp_distilled); +printf("\nThe Outlet temperatures are:\n\tRaw Water:%.1f degree F\n\tDistilled Water:%.1f degree F\n",t2,T2); + diff --git a/1309/CH9/EX9.7/Result9_7.pdf b/1309/CH9/EX9.7/Result9_7.pdf Binary files differnew file mode 100755 index 000000000..8c9dbdd87 --- /dev/null +++ b/1309/CH9/EX9.7/Result9_7.pdf diff --git a/1309/CH9/EX9.7/ch9_7.sce b/1309/CH9/EX9.7/ch9_7.sce new file mode 100755 index 000000000..2503fc00f --- /dev/null +++ b/1309/CH9/EX9.7/ch9_7.sce @@ -0,0 +1,72 @@ +clc; +clear; +printf("\t\t\tChapter9_example7\n\n\n"); +// (a) Determine the UA product for the exchanger. (b) Calculate the exit temperatures for the exchanger, assuming that only the inlet temperatures are known +// properties of engine oil at (190 + 158)/2 = 174°F = 176 degree F from appendix table C4 +rou_1= 0.852*62.4; // density in lbm/ft^3 +cp_1=0.509; // specific heat BTU/(lbm-degree Rankine) +v_1= 0.404e-3; // viscosity in ft^2/s +kf_1 = 0.08; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a_1 = 2.98e-3; // diffusivity in ft^2/hr +Pr_1 = 490; // Prandtl Number +m_1=39.8; // mass flow rate in lbm/min +// temperatures in degree F +T1=190; +T2=158; +// properties of air at (126 + 166)/2 = 146°F = 606 degree R from appendix table D1 +rou_2= 0.0653; // density in lbm/ft^3 +cp_2=0.241; // specific heat BTU/(lbm-degree Rankine) +v_2= 20.98e-5; // viscosity in ft^2/s +kf_2 = 0.01677 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) +a_2 = 1.066; // diffusivity in ft^2/hr +Pr_2 = 0.706; // Prandtl Number +m_2=67; // mass flow rate in lbm/min +// temperatures in degree F +t1=126; +t2=166; +// Heat Balance +q_air=m_2*cp_2*60*(t2-t1); +q_oil=m_1*cp_1*60*(T1-T2); +printf("\nThe heat gained by air is %.2e BTU/hr",q_air); +printf("\nThe heat lost by oil is %.2e BTU/hr",q_oil); +// for counterflow +LMTD=((T1-t2)-(T2-t1))/(log((T1-t2)/(T2-t1))); +printf("\nThe LMTD for counter flow configuration is %.1f degree F",LMTD); +// Frontal Areas for Each Fluid Stream +Area_air=(9.82*8)/144; +Area_oil=(3.25*9.82)/144; +printf("\nThe Core frontal area on the air side is %.3f sq.ft\nThe Core frontal area on the oil side is %.3f sq.ft ",Area_air,Area_oil); +// Correction Factors (parameters calculated first) +S=(t2-t1)/(T1-t1); +R=(T1-T2)/(t2-t1); +F=0.87; //value of correction factor from figure 9.21a corresponding to above calculated values of S and R +// Overall Coefficient (q = U*A*F*LMTD) +UA=q_air/(F*LMTD); +printf("\nThe Overall Coefficient is %.2e BTU/(hr. degree R)",UA); +// determining the capacitances +mcp_air=m_2*cp_2*60; +mcp_oil=m_1*cp_1*60; +printf("\nThe capacitance value of air is %d BTU/(hr. degree R)",mcp_air); +printf("\nThe capacitance value of engine oil is %d BTU/(hr. degree R)",mcp_oil); +if mcp_air>mcp_oil then + mcp_max=mcp_air; + mcp_min=mcp_oil; + printf("\nEngine Oil has minimum capacitance"); + else mcp_max=mcp_oil; + mcp_min=mcp_air; + printf("\nAir has minimum capacitance"); +end +// determination of parameters for determining effectiveness +mcp_min_max=mcp_min/mcp_max; +NTU=(UA/mcp_min); +printf("\nThe required parameters are mcp_min/mcp_max=%.3f and (UoAo/mcp_min)=%.2f",mcp_min_max,NTU); +effectiveness=0.62; //value of effectiveness from figure 9.21b corresponding to the above calculated values of capacitance ratio and (UoAo/mcp_min):'); +t2_c=(T1-t1)*effectiveness+t1; +T2_c=T1-(mcp_min_max)*(t2_c-t1); +printf("\n\t\t\tSummary of Requested Information\n"); +printf("\n(a) UA = %.2e BTU/(hr. degree R)",UA); +printf("\n(b) The Outlet temperatures (degree F)"); +printf("\n\tCalculated\tGiven in Problem Statement"); +printf("\nAir\t\t%d\t%d",t2_c,t2); +printf("\nEngine Oil\t%d\t%d",T2_c,T2); + |
