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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 /536 | |
| 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 '536')
102 files changed, 2879 insertions, 0 deletions
diff --git a/536/CH1/EX1.1/Example_1_1.sce b/536/CH1/EX1.1/Example_1_1.sce new file mode 100755 index 000000000..756ec7a62 --- /dev/null +++ b/536/CH1/EX1.1/Example_1_1.sce @@ -0,0 +1,12 @@ +clc;
+
+printf("Example 1.1\n");
+// 1 Poise = 1g/cm s = ((1/453.6)lb)/((1/30.48)ft*1s)
+be=30.48/453.6*3600; //be->british engineering unit
+printf("\n 1 Poise = %.4f lb/ft s",be/3600);
+printf("\n = %.0f lb/ft h",be);
+
+// 1 Poise = 1g/cm s = ((1/1000)kg)/((1/100)m*1s)
+si=100/1000; //si->SI units
+printf("\n 1 Poise = %.1f kg/m s ",si);
+printf("\n = %.1f N s/m^2 ",si);
\ No newline at end of file diff --git a/536/CH1/EX1.2/Example_1_2.sce b/536/CH1/EX1.2/Example_1_2.sce new file mode 100755 index 000000000..eec901b57 --- /dev/null +++ b/536/CH1/EX1.2/Example_1_2.sce @@ -0,0 +1,13 @@ +clc; + +printf("Example 1.2\n"); +// 1 kW= 103 W = 103 J/s = 10^3 * (1 kg*1 m^2)/1 s^3 +// = (10^3 * (1/0.4536) lb x (1/0.3048)^2 ft^2)/1 s^3 +lfs=(10^3*(1/0.4536)*(1/.3048)^2); //lfs->lb ft^2/s^3 +printf("\n 1 kW = %.0f lb ft^2/s^3",lfs); +sfs=lfs/32.2; //sfs->slug ft^2/s^3 +printf("\n 1 kW = %.0f slug ft^2/s^3",sfs); +printf("\n 1 kW = %.0f lbf ft/s",sfs); +hp=sfs/550; +printf("\n 1 kW = %.2f h.p.",hp); +printf("\n 1 h.p.= %.3f kW",1/hp)
\ No newline at end of file diff --git a/536/CH10/EX10.1/Example_10_1.sce b/536/CH10/EX10.1/Example_10_1.sce new file mode 100755 index 000000000..0c763e8e7 --- /dev/null +++ b/536/CH10/EX10.1/Example_10_1.sce @@ -0,0 +1,22 @@ +clc;
+clear;
+
+printf("\n Example 10.1\n");
+
+x=1e-3; //Thickness of stagnant air film
+D=1.8e-5;//Difffusivity of ammonia
+R=8314; //Gas constant
+T=295; //Temperature
+P=101.3e3; //Total Pressure
+
+//If the subscripts 1 and 2 refer to the two sides of the stagnant layer and
+//the subscripts A and B refer to ammonia and air respectively,
+P_A1=.50*P;
+P_A2=0;
+P_B1=P-P_A1;
+P_B2=P-P_A2;
+P_BM=(P-P_A1)/log(P/P_A1);
+//Thus, substituting in equation 10.31 gives:
+N_A=(-D/(R*T*x))*(P/P_BM)*(P_A2-P_A1);
+printf("\n The rate of diffusion of ammonia through the layer = %.2f *10^-4 kmol/m^2*s",N_A*1e4);
+
diff --git a/536/CH10/EX10.10/Example_10_10.sce b/536/CH10/EX10.10/Example_10_10.sce new file mode 100755 index 000000000..67fa3ef18 --- /dev/null +++ b/536/CH10/EX10.10/Example_10_10.sce @@ -0,0 +1,13 @@ +clc;
+clear;
+
+printf("Example 10.10\n");
+
+//The mass transfer rate (moles/unit area and unit time) is given by equation
+//10.180, where denoting the original conditions by subscript 1 and the
+//conditions at the higher temperature by subscript 2 gives
+//N_A2=0.83*N_A1
+//Substituting the numerical values gives:
+n=2*(log(0.83/(1.35)^0.5)/log(0.8))-1;
+printf("\n n = %.2f ",n);
+printf("\n Thus the reaction is of second order");
diff --git a/536/CH10/EX10.11/Example_10_11.sce b/536/CH10/EX10.11/Example_10_11.sce new file mode 100755 index 000000000..4b4c05bb9 --- /dev/null +++ b/536/CH10/EX10.11/Example_10_11.sce @@ -0,0 +1,30 @@ +clear;
+clc;
+//Coulson and Richardson's Chemical Engineering Volume I
+//Chapter 10 Example 11
+//Page 630
+printf('Example 10.11');
+//What factor will the mass transfer rate across interface change
+
+k = 2.5*10^-6 //[s^-1] Rate constant
+E = 2.643*10^7 //[J/kmol] Energy of Activation
+R = 8314 //[J/kmol.K] Universal gas contss
+D = 10^-9 //[m^2/s] MOlecular diffuisivity
+L = .01 //[m] Film Thickness
+
+//At T =293K
+T = 293 //[K] temperature
+A = k/exp(-E/(R*T)); //[s^-1]
+e = exp(-2*L*sqrt(k/D));
+N = sqrt(k/D)*(1+e)/(1-e); //Consider relative Solubility at 293 K be unity
+
+//At T =313K
+T2 = 313 //[K] temperature
+k2 = A*exp(-E/(R*T2)); //[s^-1]
+e2 = exp(-2*L*sqrt(k2/D));
+N2 = .8*sqrt(k2/D)*(1+e2)/(1-e2); //Consider relative Solubility at 313 K be .8 wrt that of 293K
+
+Nr = N2/N;
+
+printf('\n\nChange in mass transfer rate is given by factor %.2f',Nr)
+//END
\ No newline at end of file diff --git a/536/CH10/EX10.12/Example_10_12.sce b/536/CH10/EX10.12/Example_10_12.sce new file mode 100755 index 000000000..32703a55a --- /dev/null +++ b/536/CH10/EX10.12/Example_10_12.sce @@ -0,0 +1,24 @@ +clc;
+clear;
+
+printf("\n Example 10.12\n");
+
+k=5e-4;//first order rate constant
+D_e=2e-9;//effective diffusivity of reactants in the pores of the particles
+
+lambda=(k/D_e)^0.5;
+// (i) For the platelet of thickness 8 mm,
+L=0.5*(8e-3);
+phi=lambda*L;//thiele modulus
+//From equation 10.202, the effectiveness factor 'eta' is given by:
+eta=(1/phi)*tanh(phi);
+printf("\n (i) Thiele modulus = %.1f",phi);
+printf("\n The effectiveness factor = %.3f",eta);
+
+//(ii) For the sphere of diameter 10 mm, r_o = 0.005 m^-1.
+r_o=5e-3;
+phi_o=lambda*r_o;//Thiele modulus
+//From equation 10.212, the effectiveness factor 'eta' is given by:
+eta_o=(3/phi_o)*(coth(phi_o)-(1/phi_o));
+printf("\n (i) Thiele modulus = %.1f",phi_o);
+printf("\n The effectiveness factor = %.3f",eta_o);
diff --git a/536/CH10/EX10.13/Example_10_13.sce b/536/CH10/EX10.13/Example_10_13.sce new file mode 100755 index 000000000..ada45c71f --- /dev/null +++ b/536/CH10/EX10.13/Example_10_13.sce @@ -0,0 +1,18 @@ +clc;
+clear;
+printf("\n Example 10.13\n");
+
+D_e=1e-5;//Effective diffusivity for the reactants in the catalyst particle
+k=14.4;//first order rate constant
+L=2.5e-3;
+
+lambda=(k/D_e)^0.5;
+phi=(k/D_e)^0.5*(L);//Thiele modulus
+//From equation 10.202, the effectiveness factor,
+eta=(1/phi)*tanh(phi);
+printf("\n (i) The effectiveness factor = %.3f",eta);
+//The concentration profile is given by equation 10.198
+y=1.25e-3;
+C_Ai=0.15;
+C_A=C_Ai*(cosh(lambda*y)/cosh(lambda*L));
+printf("\n (ii) The concentration of reactant at a position half-way between the centre and the outside of the\n\t pellet = %.3f kmol/m^3",C_A);
diff --git a/536/CH10/EX10.14/Example_10_14.sce b/536/CH10/EX10.14/Example_10_14.sce new file mode 100755 index 000000000..433664f67 --- /dev/null +++ b/536/CH10/EX10.14/Example_10_14.sce @@ -0,0 +1,18 @@ +clear;
+clc;
+
+printf("\n Example 10.14\n");
+
+R_r=8.2e-2;//reaction rate when concentration =0.011 kmol/m^3
+D_e=7.5e-8;//Effective diffusivity
+
+//Since the value of the first-order rate constant is not given, lambda and
+//phi_l cannot be calculated directly. The reaction rate per unit volume of
+//catalyst = eta*k*C_Ai (equation 10.217),
+//eta=phi_L^-1
+//It is assumed that the reactor is operating in this regime and the assumption
+//is then checked.Substituting numerical values in equation 10.217:
+k=(1.217*R_r/0.011)^2;
+phi_L=1.217*(k)^0.5;
+eta=phi_L^-1;
+printf("\n Effectiveness factor = %.4f",eta);
\ No newline at end of file diff --git a/536/CH10/EX10.2/Example_10_2.sce b/536/CH10/EX10.2/Example_10_2.sce new file mode 100755 index 000000000..5f223922a --- /dev/null +++ b/536/CH10/EX10.2/Example_10_2.sce @@ -0,0 +1,57 @@ +clear;
+clc;
+
+printf("Example 10.2\n");
+
+th=[0 0 3 7 22 32 46 55 80 106]; //Time in hours
+tm=[0 26 5 36 16 38 50 25 22 25]; //Time in min
+
+//Conversion to kilo seconds
+for i=1:10
+ tm(i)=tm(i)*60;
+ th(i)=th(i)*3600;
+ tim(i)=(tm(i)+th(i))/1000;
+end
+
+L=[0 2.5 12.9 23.2 43.9 54.7 67.0 73.8 90.3 104.8]; //in mm
+
+Lo=L(1);
+
+//Calculations are done as indicated in the procedure
+//To obtain the values of x and y as below
+//For plotting x and t axis of graph
+x=L-Lo;
+
+y(1)=0;
+for j=2:10
+ y(j)=tim(j)/(L(j)-Lo);
+end
+
+plot2d(x,y);
+plot(x,y,'+');
+xtitle('t/(L-L0) vs (L-L0)','(L-L0) in mm','t/(L-L0) in ks/mm^2');
+
+//Calculation of slope
+s=(y(4)-y(3))/(x(4)-x(3))*10^3*10^6;
+printf("\nSlope is %.2e sec/m^2\n",s);
+
+Vl=22.4; //Molar volume in litres
+den=1540; //Density in kg/m^3
+T0=273;
+T=321;
+vp=37.6; //vapour pressure in kPa
+P0=101.3; //PRessue in kPa
+M=154;
+
+Ct=T0/(Vl*T);
+Ca=(vp*Ct)/P0;
+
+Cb1=Ct;
+Cb2=(P0-vp)*Ct/P0;
+Cbm=(Cb1-Cb2)/log(Cb1/Cb2);
+
+//Diffusivity calculation
+D=den*Cbm/(2*M*Ca*Ct*s);
+printf("\nDiffusivity is %.2e m^2/s\n",D);
+
+//End
diff --git a/536/CH10/EX10.3/Example_10_3.sce b/536/CH10/EX10.3/Example_10_3.sce new file mode 100755 index 000000000..96665184a --- /dev/null +++ b/536/CH10/EX10.3/Example_10_3.sce @@ -0,0 +1,17 @@ +clear;
+clc;
+
+printf("Example 10.3\n");
+
+P=101.3e3; //pressure of the operating column
+T=295; //Temperature of the operating column
+P_A=7e3; //partial pressure of ammonia
+x=1e-3; //=(y1-y2)Thickness of stationary gas film
+D=2.36e-5; //Diffusivity of ammonia
+
+C_A=(1/22.4)*(273/T)*(P_A/P);//=(C_A1-C_A2)Concentration of ammonia gas
+//X=C_T/C_BM
+X=P*log(P/(P-P_A))/(P-(P-P_A));
+//From equation 10.33
+N_A_=(D/x)*X*(C_A);
+printf("\n The transfer rate per unit area = %.2f *10^-5 kmol/m^2*s",N_A_*1e5)
\ No newline at end of file diff --git a/536/CH10/EX10.4/Example_10_4.sce b/536/CH10/EX10.4/Example_10_4.sce new file mode 100755 index 000000000..1a1224c23 --- /dev/null +++ b/536/CH10/EX10.4/Example_10_4.sce @@ -0,0 +1,22 @@ +clear;
+clc;
+
+printf("Example 10.4\n");
+
+Q=3e-6; //Flow rate of water
+Meu=1e-3; //Viscosity of water
+D=1.5e-9; //diffusivity of carbon dioxide in water
+rho=1e3; //Density of water
+
+//the mean velocity of flow is governed by equation 3.87 in which sin(phi)is
+//put equal to unity for a vertical surface:
+s=(Q*1e2*Meu*3/(rho*9.81))^(1/3);//Thickness of film
+
+//A=Ux/Us=0.95;
+A=0.95;
+y=s*(1-A)^0.5;//The distance below the free surface
+//using equation 10.108 and using tables of error fuctions
+t=(1.305/1.822)^2
+Us=rho*9.81*s^2/(2*Meu);//surface velocity
+L=Us*t;//The maximum lend=gth of column
+printf("\n The maximum length of column = %.2f m",L);
\ No newline at end of file diff --git a/536/CH10/EX10.5/Example_10_5.sce b/536/CH10/EX10.5/Example_10_5.sce new file mode 100755 index 000000000..d97d91f47 --- /dev/null +++ b/536/CH10/EX10.5/Example_10_5.sce @@ -0,0 +1,20 @@ +clear;
+clc;
+
+printf("Example 10.5\n");
+
+N_dot=50; //Initial maas transfer rate
+D=1.8e-9; //Diffusivity of gas in liquid phase
+
+C_bg=(1/22.4)*(273/293); //bulk gas concentration
+N_C=N_dot*C_bg; //Initial maas transfer rate in terms of cocentration
+h=N_C/0.04;// Effective Mass transfer coefficient
+R=1/h;//Equivalent resistance
+printf("\n Equivalent resistance = %.4f s/m",R);
+R_l=R*9; //Liquid phase resistance
+h_l=1/R_l; //Liquid phase coefficient
+
+//From equation 10.113 and using liquid phase resistance
+t=R_l^2/(%pi/D);
+
+printf("\n The required time is = %.2f *10^-11s",t*1e11)
\ No newline at end of file diff --git a/536/CH10/EX10.6/Example_10_6.sce b/536/CH10/EX10.6/Example_10_6.sce new file mode 100755 index 000000000..969aa707f --- /dev/null +++ b/536/CH10/EX10.6/Example_10_6.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+printf("Example 10.6\n");
+
+//Diffusivity of CO2 in ethanol
+D=4D-9; //in m^2/s
+t=100; //Time in sec
+
+//Solving all the integral as defined in the proces
+//as per described in the book
+//a useful result is obtained
+
+Cai=poly([0],'x');
+
+y=[0 10^-3];
+
+for i=1:2
+ mole(i)=((2*sqrt(D*t/%pi)*exp(-y(i)^2/(4*D*t)))-(y(i)*erfc(y(i)/(2*sqrt(D*t)))));
+end
+ret=(mole(1)-mole(2))/mole(1);
+
+printf("\nProportion retained is %.1f %%\n",ret*100);
+
+//End
\ No newline at end of file diff --git a/536/CH10/EX10.8/Example_10_8.sce b/536/CH10/EX10.8/Example_10_8.sce new file mode 100755 index 000000000..343f2a9a0 --- /dev/null +++ b/536/CH10/EX10.8/Example_10_8.sce @@ -0,0 +1,17 @@ +clc;
+clear;
+
+printf("Example 10.8\n")
+
+L=825e-3; //length of the tube
+d=15e-3; //diameter of the tube
+P_i=7.5e3; //Partial pressure of ammonia at inlet
+P_o=2e3; //Partial pressure of ammonia at inlet
+A_r=2e-5; //Air rate
+P=101.3e3; //Atmospheric pressure
+
+D_F_m=(P_i-P_o)/log(P_i/P_o);//Mean driving force
+A_absorbd=A_r*((P_i/(P-P_i))-(P_o/(P-P_o)));
+A_w=%pi*d*L;//Wetted surface
+K_G=(A_absorbd/(A_w*D_F_m));//Overall transfer coefficient
+printf("\n Overall Transfer coefficient = %.2f * 10^-9 kmol/[m^2 s (N/m^2)]",K_G*1e9)
\ No newline at end of file diff --git a/536/CH10/EX10.9/Example_10_9.sce b/536/CH10/EX10.9/Example_10_9.sce new file mode 100755 index 000000000..27e27862a --- /dev/null +++ b/536/CH10/EX10.9/Example_10_9.sce @@ -0,0 +1,25 @@ +clear;
+clc;
+
+printf("\n Example 10.9\n");
+
+//The proces is defined by
+// Ca = B1*exp(sqrt((k/D)y)) + B2*exp(-sqrt((k/D)y))
+
+//Boundary conditions as
+// Ca=Cai at y=0
+// Ca=Cai/2 at y=l
+
+//Using above 3 equations, final equation is derived as follows
+
+//Assuming
+// ratio = (Na)y=l / (Na)y=0
+// z=l*sqrt(k/D)
+
+z=0.693;
+
+ratio=(exp(sqrt(z))+exp(-sqrt(z))-4)/(2*(1-exp(-sqrt(z))-exp(sqrt(z))));
+
+printf("\n The final ratio is %.2f \n",ratio);
+
+//End
\ No newline at end of file diff --git a/536/CH11/EX11.1/Example_11_1.sce b/536/CH11/EX11.1/Example_11_1.sce new file mode 100755 index 000000000..2d85ab4c7 --- /dev/null +++ b/536/CH11/EX11.1/Example_11_1.sce @@ -0,0 +1,17 @@ +clc;
+clear;
+
+printf("\n Example 11.1\n");
+
+u_s=1; //Velocity of water
+w=0.6; //Width of plane surface
+l=1; //Length of plane surface
+A=0.6*1; //Area of the surface
+//Taking
+Meu=1e-3;//Viscosity of water
+rho=1000; //Density of water
+//Mean value of S/pw2 from equation 11.41
+//X=R/(rho*u^2)
+X=0.00214;
+F=X*rho*u_s^2*A;
+printf("\n Total drag force = %.2f N",F);
\ No newline at end of file diff --git a/536/CH11/EX11.2/Example_11_2.sce b/536/CH11/EX11.2/Example_11_2.sce new file mode 100755 index 000000000..d9a4f3752 --- /dev/null +++ b/536/CH11/EX11.2/Example_11_2.sce @@ -0,0 +1,18 @@ +clc;
+clear;
+
+printf("\n Example 11.2\n");
+
+x=150e-3; //Distance from leading edge where thicness is to be found
+Meu_o=0.05; //viscosity of oil
+rho_o=1000; //Density of oil
+u=0.3; //Velocity of flow
+
+Re_x=x*u*rho_o/Meu_o;
+//For streamline flow:
+//from equation 11.17
+del=4.64*x/Re_x^0.5;//thickness of the boundary layer
+printf("\n The thickness of the boundary layer = %.1f mm",del*1e3);
+//from equation 11.20
+del_star=0.375*del;
+printf("\n The displacement thickness = %.1f mm",del_star*1e3);
diff --git a/536/CH11/EX11.3/Example_11_3.sce b/536/CH11/EX11.3/Example_11_3.sce new file mode 100755 index 000000000..b2409eb96 --- /dev/null +++ b/536/CH11/EX11.3/Example_11_3.sce @@ -0,0 +1,23 @@ +clc;
+clear;
+
+printf("\n Example 11.3\n");
+
+D=50e-3; //Diameter of the pipe
+Q=2e-3; //Flow rate of benzene through pipe
+rho_b=870; //Density of benzene
+Meu_b=0.7e-3; //Viscosity of benzene
+
+G=Q*rho_b; //Mass flow rate of benzene
+Re=4*G/(Meu_b*%pi*D); //Reynolds number
+
+//From equation 11.49:
+del_b=62*D*Re^(-7/8);
+printf("\n The thickness of the laminar sub-layer = %.3f mm",del_b*1e3);
+
+area=%pi/4*D^2; //Cross sectional are of pipe
+u=G/(rho_b*area); //mean velocity
+
+//From equation 11.47:
+u_b=2.49*u*Re^(-1/8);
+printf("\n The velocity of the benzene at the edge of the laminar sub-layer = %.3f m/s",u_b);
diff --git a/536/CH12/EX12.1/Example_12_1.sce b/536/CH12/EX12.1/Example_12_1.sce new file mode 100755 index 000000000..3cb711638 --- /dev/null +++ b/536/CH12/EX12.1/Example_12_1.sce @@ -0,0 +1,33 @@ +clc;
+clear;
+
+printf("\n Example 12.1\n");
+
+d=250e-3; //internal diameter of pipe
+u=15; //Velocity of air through the pipe
+y1=50e-3; //First point where velocity is to be found out
+y2=5e-3; //Second point where velocity is to be found out
+rho_air=1.10; //Density of air
+Meu_air=20e-6; //Viscosity of air
+
+Re=d*u*rho_air/Meu_air;
+//Hence, from Figure 3.7: X=R/(rho*u^2)=0.0018
+X=0.0018;
+u_s=u/0.817;
+u_star=u*X^0.5;
+
+//At 50 mm from the wall:
+y1_r=2*y1/d;// y/r
+//Hence, from equation 12.34:
+u_x1=u_s+2.5*u_star*log(y1_r);
+printf("\n The fluid velocity at 50 mm from the wall = %.1f m/s",u_x1);
+
+//At 5 mm from the wall:
+y2_r=2*y2/d;// y/r
+//Hence, from equation 12.34:
+u_x2=u_s+2.5*u_star*log(y2_r);
+printf("\n The fluid velocity at 5 mm from the wall = %.1f m/s",u_x2);
+
+//The thickness of the laminar sub-layer is given by equation 12.54:
+del_b=5*d/(Re*X^0.5);
+printf("\n The thickness of the laminar sub-layer = %.3f mm",del_b*1e3);
\ No newline at end of file diff --git a/536/CH12/EX12.2/Example_12_2.sce b/536/CH12/EX12.2/Example_12_2.sce new file mode 100755 index 000000000..f0523fc01 --- /dev/null +++ b/536/CH12/EX12.2/Example_12_2.sce @@ -0,0 +1,19 @@ +clc;
+clear;
+
+printf("\n Example 12.2\n");
+
+u=10; //Velocity of air
+T=330; //Temperature of air
+d=25e-3; //Inner diameter of pipe
+T_p=415; //Temperature at which the pipe is maintained
+DP_l=80; //Drop of static pressure along the pipe per unit length
+
+//From equations 12.98 and 3.18:
+//we get h=0.05*Cp
+//The heat taken up per unit time by the air dQ=0.0052*Cp*dT......(i)
+//The heat transferred through the pipe wall is also given by: = 0.039*Cp*(415-T)......(ii)
+//Equating (i) & (ii)
+//On integrating we get
+T_0=415-(85/exp(0.45))
+printf("\n The required air Temperature = %d K",T_0);
diff --git a/536/CH12/EX12.3/Example_12_3.sce b/536/CH12/EX12.3/Example_12_3.sce new file mode 100755 index 000000000..58a18d17d --- /dev/null +++ b/536/CH12/EX12.3/Example_12_3.sce @@ -0,0 +1,46 @@ +clc;
+clear;
+
+printf("\n Example 12.3\n");
+
+u=3.5; //Velocity of water
+d=25e-3; //Diameter of the pipe
+l=6; //Length of the pipe
+T1=300; //Temperature at enterance
+T2=330; //Temperature at exit
+rho=1000; //density of water at 310 K
+Meu=0.7e-3; //Viscosity of water at 310 K
+//Taking the fluid properties at 310 K and assuming that fully developed flow exists
+Cp=4.18e3; //heat capapcity
+k=0.65; //Thermal conductivity
+
+Re=d*u*rho/Meu;
+Pr=Cp*Meu/k;
+
+printf("\n (a) Reynolds analogy");
+h1=0.032*(Re^-0.25)*Cp*rho*u;//....Equation 12.139
+printf("\n h = %.2f kW/m^2 K",h1*1e-3);
+// on solving we get final equation as
+theta_dash1=330-10^(log10(30)-(0.0654*h1*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash1)
+
+printf("\n\n (b) Taylor Prandtl Equation");
+h2=0.032*(Re^-0.25)*(1+2*Re^(-1/8)*(Pr-1))^-1*Cp*rho*u;
+printf("\n h = %.2f kW/m^2 K",h2*1e-3);
+// on solving we get final equation as
+theta_dash2=330-10^(log10(30)-(0.0654*h2*1e-3/2.303));//....Equation 12.140
+printf("\n The outlet temperature = %.1f K",theta_dash2)
+
+printf("\n\n (c) Universal velocity profile equation");
+h3=0.032*(Re^-0.25)*(1+0.82*Re^(-1/8)*((Pr-1)+log(0.83*Pr+0.17)))^-1*Cp*rho*u;//...equation 12.141
+printf("\n h = %.2f kW/m^2 K",h3*1e-3);
+// on solving we get final equation as
+theta_dash3=330-10^(log10(30)-(0.0654*h3*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash3)
+
+printf("\n\n (d) Nu=0.023*Re^0.8*Pr^0.33");
+h4=k/d*0.023*Re^0.8*Pr^0.33;
+printf("\n h = %.2f kW/m^2 K",h4*1e-3);
+// on solving we get final equation as
+theta_dash4=330-10^(log10(30)-(0.0654*h4*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash4)
diff --git a/536/CH12/EX12.4/Example_12_4.sce b/536/CH12/EX12.4/Example_12_4.sce new file mode 100755 index 000000000..0c4aeacfb --- /dev/null +++ b/536/CH12/EX12.4/Example_12_4.sce @@ -0,0 +1,46 @@ +clc;
+clear;
+
+printf("\n Example 12.4\n");
+
+u=3.5; //Velocity of air
+d=25e-3; //Diameter of the pipe
+l=6; //Length of the pipe
+T1=290; //Temperature at enterance
+T2=350; //Temperature at exit
+rho=29/22.4*273/310; //density of air at 310 K
+Meu=0.018e-3; //Viscosity of air at 310 K
+//Taking the physical properties at 310 K and assuming that fully developed flow exists
+Cp=1.003e3; //heat capapcity
+k=0.024; //Thermal conductivity
+
+Re=d*u*rho/Meu;
+Pr=Cp*Meu/k;
+
+printf("\n (a) Reynolds analogy");
+h1=0.032*(Re^-0.25)*Cp*rho*u;//....Equation 12.139
+printf("\n h = %.2f W/m^2 K",h1);
+// on solving we get final equation as
+theta_dash1=350-10^(log10(60)-(239.88*h1*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash1)
+
+printf("\n\n (b) Taylor Prandtl Equation");
+h2=0.032*(Re^-0.25)*(1+2*Re^(-1/8)*(Pr-1))^-1*Cp*rho*u;
+printf("\n h = %.2f W/m^2 K",h2);
+// on solving we get final equation as
+theta_dash2=350-10^(log10(60)-(239.88*h2*1e-3/2.303));//....Equation 12.140
+printf("\n The outlet temperature = %.1f K",theta_dash2)
+
+printf("\n\n (c) Universal velocity profile equation");
+h3=0.032*(Re^-0.25)*(1+0.82*Re^(-1/8)*((Pr-1)+log(0.83*Pr+0.17)))^-1*Cp*rho*u;//...equation 12.141
+printf("\n h = %.2f W/m^2 K",h3);
+// on solving we get final equation as
+theta_dash3=350-10^(log10(60)-(239.88*h3*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash3)
+
+printf("\n\n (d) Nu=0.023*Re^0.8*Pr^0.33");
+h4=k/d*0.023*Re^0.8*Pr^0.33;
+printf("\n h = %.2f W/m^2 K",h4);
+// on solving we get final equation as
+theta_dash4=350-10^(log10(60)-(239.88*h4*1e-3/2.303));
+printf("\n The outlet temperature = %.1f K",theta_dash4)
diff --git a/536/CH13/EX13.1/Example_13_1.sce b/536/CH13/EX13.1/Example_13_1.sce new file mode 100755 index 000000000..6cf730384 --- /dev/null +++ b/536/CH13/EX13.1/Example_13_1.sce @@ -0,0 +1,26 @@ +clc;
+clear;
+
+printf("\n Example 13.1\n");
+
+P=101.3e3;
+T=297;
+R=8314; //gas constant
+RH=60; //Relative humidity
+p_b1=12.2e3;//Vapor pressure at 297 K
+p_b2=6e3; //Vapor pressure at 283 K
+M_w=78; //molecular weight of benzene
+M_a=28; //Mass of nitrogen
+
+//From the definition of percentage relative humidity (RH)
+P_w=(p_b1)*(RH/100);
+//In the benzene -nitrogen mixture:
+m_b=P_w*M_w/(R*T);//mass of benzene
+m_n=(P-P_w)*M_a/(R*T);//mass of nitrogen
+H=m_b/m_n; //Humidity at 297 K
+
+//In order to recover 80 per cent of the benzene, the humidity must be reduced to 20 per cent of the initial value
+H_o=H*.20;
+//Thus in equation 13.2
+P_r=p_b2+(p_b2/M_a*M_w)/H_o;
+printf("\n The required pressure is = %.0f kN/m^2",P_r*1e-3);
\ No newline at end of file diff --git a/536/CH13/EX13.2/Example_13_2.sce b/536/CH13/EX13.2/Example_13_2.sce new file mode 100755 index 000000000..d42daf09b --- /dev/null +++ b/536/CH13/EX13.2/Example_13_2.sce @@ -0,0 +1,32 @@ +clc;
+clear;
+
+printf("\n Example 13.2\n");
+
+P=101.3e3; //Given pressure
+T=300; //Given Temperature
+RH=25; //Percentage relative humidity of water
+P_wo=3.6e3; //partial pressure of water vapour when air is saturated with vapour
+M_w=18; //Molecular weight of water
+M_a=29; //Molecular weight of air
+R=8314; //gas constant
+
+printf("\n (a)\n The partial pressure of the water vapour in the vessel = ")
+P_w=P_wo*(RH/100);
+printf("%.1f kN/m^2",P_w*1e-3);
+
+printf("\n (b)");
+m_w=P_w*M_w/(R*T);//mass of water vapour
+m_a=(P-P_w)*M_a/(R*T);//mass of water air
+Vs_w=1/m_w;//specific volume of water vapour at 0.9 kN/m^2
+Vs_a=1/m_a;//specific volume of air at 100.4 kN/m^2
+printf("\n Specific volume of water vapour = %.0f m^3/kg",Vs_w);
+printf("\n Specific volume of air = %.3f m^3/kg",Vs_a);
+
+H=m_w/m_a;//Humidity
+printf("\n (a)\n Humidity of air = %.4f kg water/kg air",H);
+H_v=Vs_a;//Humid volume
+printf("\n Humid volume = %.3f m^3/kg",H_v);
+
+H_p=(P-P_wo)/(P-P_w)*RH; //Percentage humidity
+printf("\n (d)\n Percentage humidity = %.1f per cent",H_p)
\ No newline at end of file diff --git a/536/CH13/EX13.3/Example_13_3.sce b/536/CH13/EX13.3/Example_13_3.sce new file mode 100755 index 000000000..b321eb529 --- /dev/null +++ b/536/CH13/EX13.3/Example_13_3.sce @@ -0,0 +1,24 @@ +clc;
+clear;
+
+printf("\n Example 13.3\n");
+
+T=310; //Temperature of moist air
+T_w=300; //Wet bulb tempeature
+L=2440e3; //Latent heat of vapourisation of water at 300 K
+P=105e3; //Given total pressure
+P_wo1=3.6e3; //Vapour pressure of water vapour at 300 K
+P_wo2=6.33e3; //Vapour pressure of water vapour at 310 K
+M_w=18; //Molecular weight of water
+M_a=29; //Molecular weight of air
+
+H_w=(P_wo1/(P-P_wo1))*(M_w/M_a); //The humidity of air saturated at the wet-bulb temperature
+//Therefore, taking (h/hD*rho*A) as 1.0 kJ/kg K, in equation 13.8:
+H=H_w-(1e3/L)*(T-T_w);
+printf("\n The humidity of the air = %.3f kg/kg",H);
+
+//In equation 13.2:
+x=poly([0],'x');
+P_w=roots(H*(P-x)*M_a-M_w*x);
+RH=P_w/P_wo2*100;
+printf("\n The percentage relative humidity (RH)= %.1f per cent",RH);
diff --git a/536/CH13/EX13.4/Example_13_4.sce b/536/CH13/EX13.4/Example_13_4.sce new file mode 100755 index 000000000..3f0ae25ec --- /dev/null +++ b/536/CH13/EX13.4/Example_13_4.sce @@ -0,0 +1,31 @@ +clc;
+clear;
+
+printf('Example 13.4');
+
+//Refer HUMIDITY ENTHALPY PLOT Figure 13.5 Page 748 as Humidity Chart
+//According the given passes and situatuion
+T = [325 301 308 312 315] //[K]
+H = [.005 .015 .022 .027 .032] //[kg/kg]
+//From Humidity Chart on humidifying to 60 percent humidity
+Tw = [296 301 305 307] //[K]
+
+Hin = H(5)-H(1) //[kg/kg] Increase in Humidity
+
+printf('\n\n (a) The temperature of the material on each tray (in Kelvin)')
+disp(Tw);
+printf(' Thus the air leaving the system is at %i K and 60 per cent humidity.',T(5));
+
+//From Humitidy Chart at the obtained leaving conditions
+v = .893 //[m^3/kg] Specific Volume of dry air
+vs = .968 //[m^3/kg] Specific Volume of Saturated air
+vh = .937 //[m^3/kg] Humid Volume of air of 60 per cent humidity by Interpolation of Curve in Humidity Chart
+x = 5 //[m^3/s] Amount of moist air leaves the dryer in (b)
+m = x/vh //[kg/s] Mass of air passing through the dryer
+mw = m*Hin // [kg/s] Mass of water evaporated
+
+printf('\n\n (b) If 5 m^3/s moist air leaves the dryer, The amount of water removed is %.3f kg/s.',mw)
+Tb = 370 //[K] dry Bulb temperature corresponding to humidity of .005 kg/kg and wet-bulb temperature 307 K
+printf('\n\n (c) The Temperature to which the inlet air would have to be raised to carry out the drying in single stage is %i K.',Tb)
+
+//END
\ No newline at end of file diff --git a/536/CH13/EX13.5/Example_13_5.sce b/536/CH13/EX13.5/Example_13_5.sce new file mode 100755 index 000000000..ac0aad6b7 --- /dev/null +++ b/536/CH13/EX13.5/Example_13_5.sce @@ -0,0 +1,28 @@ +clc;
+clear;
+
+printf('Example 13.5\n')
+
+G1=1; //flow rate of air at 350 K
+PH1=10; //Percentage Humidity at 350 K
+G2=5; //flow rate of air at 300 K
+PH2=30; //Percentage Humidity at 300 K
+
+//from fig 13.4
+H1=0.043; //Humidity at 350 K and 10 percent humidity
+H2=0.0065; //Humidity at 300 K and 30 percent humidity
+//Thus, in equation 13.23:
+H=((G1*H1)+(G2*H2))/(G1+G2);
+printf("\n Humidity of final stream = %.4f kg/kg",H);
+
+//from fig 13.5
+H_1=192e3;//Entahlpy at 350 K and 10 percent humidity
+H_2=42e3;//Enthalpy at 300 K and 30 percent humidity
+x=poly([0],'x');
+H_=roots((G1*(x-H_1))-(G2*(H_2-x)));
+printf("\n Entahlpy of the resultant stream = %.0f kJ/kg",H_*1e-3);
+
+//From Figure 13.5:
+//at H_ (Enthalpy)= 67 kJ/kg and H(humidity) = 0.0125 kg/kg
+T=309;
+printf("\n Temperature of the resultant stream = %d K",T);
\ No newline at end of file diff --git a/536/CH13/EX13.6/Example_13_6.sce b/536/CH13/EX13.6/Example_13_6.sce new file mode 100755 index 000000000..7d8207a29 --- /dev/null +++ b/536/CH13/EX13.6/Example_13_6.sce @@ -0,0 +1,31 @@ +clc;
+clear;
+
+printf('Example 13.6\n')
+
+G_s=0.15; //Mass flow rate of steam
+T=400; //Temperature to which steam is superheated
+T_a=320;//Tremperature of air
+RH_a=20; //Percentage relative humidity of air
+G_a=5;//Mass flow rate of air
+L=2258e3; //latent heat of steam
+Cp=2e3; //Specific heat of superheated steam
+//Enthalpy of steam
+H_3=4.18*(373-273)+L+Cp*(T-373);
+//From Figure .13.5:
+//at T=320 K and 20 percent Relative Humidity
+H1=0.013;//Humidity
+H_1=83e3;//Enthalpy
+//By making required constructions we get
+H=0.043;
+printf("\n Relative humidity of stream= %.3f kg/kg",H);
+H_=165e3;
+printf("\n Entahlpy of stream = %d kJ/kg",H_*1e-3);
+T_s=324;
+printf("\n Temperature of stream = %d K",T_s);
+
+printf("\n\n When exit temperature = 330 K");
+//from chart and equation 13.28
+G_case2=0.41;
+printf("\n The required flow of steam = %.2f kg/s",G_case2);
+printf("\n humidity of this mixture = %.3f kg/kg",0.094);
diff --git a/536/CH2/EX2.1/Example_2_1.sce b/536/CH2/EX2.1/Example_2_1.sce new file mode 100755 index 000000000..64c2be42f --- /dev/null +++ b/536/CH2/EX2.1/Example_2_1.sce @@ -0,0 +1,38 @@ +clc;
+
+printf("Example 2.1\n");
+//For 1 kmol of methane
+
+//(a) PV = 1 * RT, where
+R=8314;
+P=60*10^6;
+T=320;
+Tc=191;
+Pc=4.64*10^6;
+printf("\n Given\n R=8314 J/kmol K.\n P=60*10^6 N/m^2\n T=320 K;")
+printf("\n Tc=191 K \n Pc=4.64*10^6 N/m^2")
+V1=8314*T/P;
+printf("\n(a)\n Volume of vessel (ideal gas law) = %.4f m^3",V1);
+
+
+//(b) In van der Waals equation (2.32), the constants may be taken as:
+a=27*R^2*Tc^2/(64*Pc);
+b=R*Tc/(8*Pc);
+printf("\n(b)\n a = %d (N/m^2)*(m^3)^2/(kmol)^2",a);
+printf("\n b = %.4f m^3/kmol",b);
+//Thus using equation 2.32:
+x=poly([0],'x');
+p=roots((60*10^6*x^2+a)*(x-0.0427)-(8314*320*x^2));
+printf(" \n Volume of vessel(van der waals eq.) = %.3f m^3",p(1,1));
+
+
+//(c) Tr=T/Tc ,Pr=P/Pc
+Tr=T/Tc;
+printf("\n(c)\n Tr = %.2f",Tr);
+Pr=P/Pc;
+printf("\n Pr = %.2f",Pr);
+//Thus from Figure 2.1,
+Z=1.33;
+//V = ZnRT/P (from equation 2.31)
+V3=Z*R*T/P;
+printf("\n Volume of vessel(generalised compressibility-factor chart) = %.4f m^3",V3);
\ No newline at end of file diff --git a/536/CH2/EX2.3/Example_2_3.sce b/536/CH2/EX2.3/Example_2_3.sce new file mode 100755 index 000000000..7a676df4d --- /dev/null +++ b/536/CH2/EX2.3/Example_2_3.sce @@ -0,0 +1,14 @@ +clc;
+
+printf("Example 2.3\n");
+//Mass rate of discharge of water, G = rho*u*A
+rho=1000; //Density of Water
+d=25*10^-3; //Diameter of nozzle
+u=25; //Velocity of water at nozzle
+printf("\n Given:\n Density of water = %d kg/m^3\n Nozzle diameter = %.3f m \n Velocity = %d m/s",rho,d,u);
+G=rho*u*%pi/4*d^2;
+printf("\n\n Mass rate of discharge of water, G = %.2f kg/s",G);
+//Momentum of fluid per second = Gu
+F=G*25;
+printf("\n Momentum of fluid per second = %.0f N",F);
+printf("\n Reaction force = Rate of change of momentum = %.0f N",F);
\ No newline at end of file diff --git a/536/CH2/EX2.4/Example_2_4.sce b/536/CH2/EX2.4/Example_2_4.sce new file mode 100755 index 000000000..35833dbaa --- /dev/null +++ b/536/CH2/EX2.4/Example_2_4.sce @@ -0,0 +1,12 @@ +clc;
+
+printf("Example 2.4\n");
+//Momentum per second of approaching liquid in Y-direction = rho*u^2*A
+rho=1000; //Density of water
+d=50*10^-3; //Diameter of pipe
+u=5; //Velocity of water in pipe
+printf("\n Given\n Density of water = %d kg/m^3\n Pipe diameter = %.3f m \n Velocity = %d m/s",rho,d,u);
+M=rho*u^2*%pi/4*d^2;
+printf("\n\n Momentum per second of approaching liquid in Y-direction = %.1f N",M);
+Rf=M*(cos(%pi/4)+sin(%pi/4));
+printf("\n The resultant force in direction of arm of bracket = %.1f N",Rf);
\ No newline at end of file diff --git a/536/CH2/EX2.5/Example_2_5.sce b/536/CH2/EX2.5/Example_2_5.sce new file mode 100755 index 000000000..ac22f1b50 --- /dev/null +++ b/536/CH2/EX2.5/Example_2_5.sce @@ -0,0 +1,19 @@ +clc;
+
+printf("Example 2.5\n");
+//From equation 2.68:
+// 0.5*((u2)^2-(u1)^2)=g*(z1-z2)+((P1-P2)/rho)
+//Suffix 1 to denote conditions in the pipe and suffix 2 to denote conditions in the jet
+//Symbols have their usual meaning
+u1=0;
+z1=0;
+z2=0;
+P1=250*10^3;
+P2=0;
+rho=1000;//Density of water
+printf("\n Suffix 1 to denote conditions in the pipe and suffix 2 to denote conditions in the jet")
+printf("\n Given:\n u1=%d m/s\n z1= %d m\n z2= %d m\n P1= %.3f kN/m^2\n P2= %d kN/m^2\n Density of water= %d kg/m^3",u1,z1,z2,P1,P2,rho);
+g=9.81;
+x=poly([0],'x');
+u2=roots((0.5*(x)^2)-((P1-P2)/rho));
+printf("\n\n Ans:\n Velocity of the jet, u2 = %.1f m/s",u2(1,1));
\ No newline at end of file diff --git a/536/CH2/EX2.6/Example_2_6.sce b/536/CH2/EX2.6/Example_2_6.sce new file mode 100755 index 000000000..a3a2a5707 --- /dev/null +++ b/536/CH2/EX2.6/Example_2_6.sce @@ -0,0 +1,13 @@ +clc;
+
+printf("Example 2.6\n");
+id=0.5; //internal diameter of pipe
+rs=50; //revolution speed
+ir=0.15; //internal radius of water
+rho=1000; //density of water
+printf("\n Given:\n Internal diameter = %.1f m\n Rotating speed = %d rev/s\n Inner radius of liquid = %.2f m\n Density of water= %d kg/m^3",id,rs,ir,rho);
+omega=2*%pi*rs;
+printf("\n\n Angular speed of rotation = %d rad/s",omega);
+//The wall pressure is given by equation 2.82 as:
+wall_pressure=rho*(omega)^2/2*((id/2)^2-ir^2);
+printf("\n The wall pressure is = %f N/m^2 \n\t\t\t=%.2f x 10^6 N/m^2",wall_pressure,wall_pressure/10^6);
\ No newline at end of file diff --git a/536/CH3/EX3.1/Example_3_1.sce b/536/CH3/EX3.1/Example_3_1.sce new file mode 100755 index 000000000..e44f55a71 --- /dev/null +++ b/536/CH3/EX3.1/Example_3_1.sce @@ -0,0 +1,22 @@ +clc;
+
+printf("Example 3.1\n\n");
+sap=1.25; //Sulphuric acid pumped
+d=25e-3; //Diameter of pipe
+l=30; //length of pipe
+meu=25e-3; //Viscosity of acid
+rho_a=1840; //Density of acid
+printf(" Given :\n Sulphuric acid pumped = %.2f kg/s\n Diameter of pipe = %.3f m\n length of pipe = %d m\n Viscosity of acid = %d x 10^-3 N s/m^2\n Density of acid = %d kg/m^3",sap,d,l,meu*1000,rho_a);
+Re=4*sap/(%pi*meu*d);
+printf("\n\n\n Reynolds number , Re=(u*d*rho)/meu = 4G/(pi*meu*d)= %d",Re);
+
+//For a mild steel pipe, suitable for conveying the acid, the roughness e will be between 0.05 and 0.5 mm (0.00005 and 0.0005 m).
+//The relative roughness is thus: e/d = 0.002 to 0.02
+//From Figure 3.7: R/(rho*u^2) = 0.006 over this range of e/d
+u=sap/(rho_a*%pi/4*d^2);
+printf("\n Velocity is , u=G/(rho *A) = %.2f m/s",u);
+
+//calculating pressure drop from the energy balance equation and equation 3.19
+Dp=rho_a*((0.5+4*0.006*30/0.025)*u^2+9.81*12);
+printf("\n Pressure Drop = %.0f N/m^2",Dp);
+printf("\n Pressure drop = %.0f kN/m^2",(Dp/10^3));
\ No newline at end of file diff --git a/536/CH3/EX3.10/Example_3_10.sce b/536/CH3/EX3.10/Example_3_10.sce new file mode 100755 index 000000000..63d38c1e4 --- /dev/null +++ b/536/CH3/EX3.10/Example_3_10.sce @@ -0,0 +1,35 @@ +clc;
+
+printf("Example 3.10\n");
+
+k=10;
+n=0.2;
+//Using the power-law model (equation 3.121):
+printf("\n Given:\n Consistency coefficient k = %d N.s^n/m^-2",k);
+printf("\n Flow behaviour index = %.1f",n);
+Ucl=1; // centre line velocity
+printf("\n Centre line velocity = %d m/s",Ucl);
+l=200; // length of pipe
+printf("\n Length of pipe = %d m",l)
+r=.02; // radius of pipe
+printf("\n Radius of pipe = %.2f m",r);
+dux_dy_1=10;
+dux_dy_2=50;
+Ry_1=k*dux_dy_1^0.2;
+Ry_2=k*dux_dy_2^0.2;
+//Using the Bingham-plastic model (equation 3.125):
+A=[1 10;1 50]
+B=[15.85;21.87]
+C=inv(A)*B;
+Ry=C(1);
+Meu_p=C(2);
+printf("\n\n Plastic viscosity (Meu_p) = %.3f N s/m^2",C(2));
+printf("\n Yeild stress (Ry) = %.2f N s/m^2",C(1));
+// Using Equation 3.131
+DP=2*k*l*Ucl^n*((n+1)/n)^n*r^(-n-1);
+printf("\n Pressure drop (Bingham plastic model)= %.0f kN/m^2",DP/1e3);
+// For a Bingham-plastic fluid:
+// The centre line velocity is given by equation 3.145:
+X=(l*2*Ry)/(r*DP);
+Up=(DP*r^2*(2-4*X+2*X^2))/(8*Meu_p*l);
+printf("\n centre line velocity (Bingham plastic model) = %.2f m/s",Up);
\ No newline at end of file diff --git a/536/CH3/EX3.11/Example_3_11.sce b/536/CH3/EX3.11/Example_3_11.sce new file mode 100755 index 000000000..5727c6d5b --- /dev/null +++ b/536/CH3/EX3.11/Example_3_11.sce @@ -0,0 +1,27 @@ +clc;
+
+printf("Example 3.11\n");
+// given:
+Meu=0.1; // Viscosity of liquid
+printf("\n Given \n Viscosity of liquid = %.1f N s/m^2",Meu);
+d=25e-3; // Diameter of pipe
+printf("\n Diameter of pipe = %.3f m",d);
+l=20; // length of pipe
+printf("\n length of pipe = %d m",l);
+DP=1e5; // Pressure drop
+printf("\n Pressure drop = %d N/m^2",DP);
+n=1/3; // flow index of polymer solution
+printf("\n flow index = %.2f",n);
+dux_dy=1000;
+k=Meu;
+Meu_a=Meu;
+k_poly_sol=Meu_a/(dux_dy)^(n-1);
+Ry=10*(dux_dy)^n;
+//From equation 3.136:
+//For a power-law fluid:
+u2=((DP/(4*k_poly_sol*l))^3)*(n*(d^((n+1)/n)))/(2*(3*n+1));
+printf("\n\n Velocity for polymer solution = %.4f m/s",u2);
+u1=(DP/(4*k*l))*(d^2)/8
+printf("\n Velocity for original solution = %.3f m/s",u1);
+ratio=u2/u1;
+printf("\n Ratio of the volumetric flow rates of the two liquids = %.3f",ratio);
diff --git a/536/CH3/EX3.2/Example_3_2.sce b/536/CH3/EX3.2/Example_3_2.sce new file mode 100755 index 000000000..6abf2b0a1 --- /dev/null +++ b/536/CH3/EX3.2/Example_3_2.sce @@ -0,0 +1,20 @@ +clc; + +printf("Example 3.2\n\n"); +d=50e-3; //Diameter of pipe +l=100; //length of pipe +e=0.013; //Roughness of pipe +DPf=50e3; //Maximum pressure drop +rho=1000; //density of water +meu=1e-3; //viscosity of water +printf(" Given:\n Diameter of pipe = %.3f m\n length of pipe = %d m\n Roughness of pipe = %.3f \n Maximum pressure drop = %d kN/m^2\n Density of water = %d kg/m^3\n Viscosity of water = %.1f mN s/m^2",d,l,e,DPf/10e3,rho,meu*10e3); +//From Equation 3.23 +// phi*Re^2=R*Re^2/(rho*u^2)=-(DPf)*d^3*rho/(4*l*meu^2) +phi_re2=(DPf)*d^3*rho/(4*l*meu^2); +e_d=e/(d*1e3); +printf("\n\n phi*Re^2 = %.2f*10^7\n e/d = %.5f",phi_re2*1e-7,e_d); +//From Figure 3.8, for given phi*Re^2 = 1.56 x 10^7 and (e/d) = 0.00026, then: +//Re=7.9*10^4 +Re=7.9e4; +u=Re*meu/(rho*d); +printf("\n\n Ans \n The maximum allowable velocity is = %.1f m/s",u);
\ No newline at end of file diff --git a/536/CH3/EX3.3/Example_3_3.sce b/536/CH3/EX3.3/Example_3_3.sce new file mode 100755 index 000000000..c5d7a66e5 --- /dev/null +++ b/536/CH3/EX3.3/Example_3_3.sce @@ -0,0 +1,23 @@ +clc;
+
+printf("Example 3.3\n");
+Dia_tank=5; //Diameter of the tank
+len_pipe=100; //Length of pipe
+dia_pipe=225e-3; //Diameter of pipe
+printf("\n Given:\n Diameter of the tank = %d m\n Length of pipe = %d m\n Diameter of pipe = %.2f m",Dia_tank,len_pipe,dia_pipe);
+
+//If at time t the liquid level is D m above the bottom of the tank, then
+// designating point 1 as the liquid level and point 2 as the pipe outlet,
+// and applying the energy balance equation (2.67) for turbulent flow, then:
+
+// The equation becomes (u2^2/2)-D*g+(4*R*len_pipe*u2^2/(rho*u^2*dia_pipe))
+
+//As the level of liquid in the tank changes from D to (D + dD), the quantity
+// of fluid discharged = (pi/4)5^2(-dD) = -19.63dD m^3.
+// The time taken for the level to change by an amount dD is given by:
+//dt=19.63dD/((pi/4)0.225^2 x 4.43D^0.5/([l + 3552(/R/rho*u^2)])^0.5)
+
+// calculating the value of X=R/rho*u^2 as given in example (refer to book)
+X=.0020;
+t=integrate('111.5*(1+(3552*X))^0.5*D^-0.5','D',0.3,3);
+printf("\n\n The time taken for the level to fall is therefore about %d s",t);
\ No newline at end of file diff --git a/536/CH3/EX3.4/Example_3_4.sce b/536/CH3/EX3.4/Example_3_4.sce new file mode 100755 index 000000000..fc5c4254c --- /dev/null +++ b/536/CH3/EX3.4/Example_3_4.sce @@ -0,0 +1,31 @@ +clc;
+
+printf("Example 3.4\n");
+
+d1=0.3; //diameter of pipe from junction A to D or B to D
+l1=1.5e3; //length of pipe from junction A to D or B to D
+d2=0.5; // diameter of pipe from junction D to C
+l2=0.75e3; // length of pipe from junction D to C
+h_A=10; // height of tank A above C
+h_B=h_A+6; // height of tank A above C
+rho=870; // density of liquid
+Meu_l=0.7e-3; // viscosity of liquid
+
+//It may be assumed, as a first approximation, that R/(rho*u^2) is the same in each pipe and that the velocities in pipes AD, BD, and DC are u1,u2 and u3
+//respectively,
+//Taking the roughness of mild steel pipe e as 0.00005 m, e/d varies from
+//0.0001 to 0.00017. As a first approximation, R/(rho*u^2) may be taken as 0.002
+//Then applying the energy balance equation between D and the liquid level in
+//each of the tanks gives
+//On forming and solving the equations
+
+x=poly([0],'x');
+u2=roots(x^4-(7.38*x^2)+13.57);
+u1=(u2^2-1.47)^0.5;
+u3=(u1+u2)/2.78;
+//taking the positive values and which satisfy equation 7
+U1=u1(4);
+U2=u2(4);
+U3=u3(4);
+Q=%pi/4*d2^2*U3;
+printf("\n The volumetric flow rate = %.2f m^3/s",Q);
\ No newline at end of file diff --git a/536/CH3/EX3.5/Example_3_5.sce b/536/CH3/EX3.5/Example_3_5.sce new file mode 100755 index 000000000..df20ed43e --- /dev/null +++ b/536/CH3/EX3.5/Example_3_5.sce @@ -0,0 +1,19 @@ +clc;
+
+printf("Example 3.5\n");
+
+// Ux = Ucl*(y/r)^l/7 equation 3.59 (Prandtl one-seventh power law)
+//where UCL is the velocity at the centre line of the pipe, and r is the radius of the pipe.
+// Then total flow, Q = 49/60*pi*r^2*Ucl equation 3.62
+
+//When the flow in the central core is equal to the flow in the surrounding annulus, then taking a = y/r, the flow in the central core is:
+//Qc=pi*r^2*Ucl*(105*a^(8/7)-56*a^(15/7))/60
+//flow in the core = 0.5 (flow in the whole pipe)
+
+r=50;
+a=poly([0],'a');
+p=roots((a^8*(105-56*a)^7)-24.5^7);
+
+printf("\n a = %.2f",p(8));
+y=p(8)*r;
+printf("\n y = %.1f mm",y)
\ No newline at end of file diff --git a/536/CH3/EX3.6/Example_3_6.sce b/536/CH3/EX3.6/Example_3_6.sce new file mode 100755 index 000000000..575f23518 --- /dev/null +++ b/536/CH3/EX3.6/Example_3_6.sce @@ -0,0 +1,16 @@ +clc;
+
+printf("Example 3.6\n");
+
+Q=7.2;//Water flow rate
+d1=40e-3; //initial pipe diameter
+d2=50e-3; //diameter of pipe after enlargement
+g=9.81;
+printf("\n Given\n Water flow rate = %.1f m^3/h\n d1 = %d mm\n d1 = %d mm",Q,d1*1e3,d2*1e3);
+
+u1=(Q/3600)/(%pi/4*d1^2);//Velocity in 40 mm pipe
+u2=(Q/3600)/(%pi/4*d2^2);//Velocity in 50 mm pipe
+printf("\n Velocity in 40 mm pipe = %.2f m/s\n Velocity in 50 mm pipe = %.2f m/s",u1,u2);
+// The head lost is given by equation 3.77 as:
+hf=(u1-u2)^2/(2*g);
+printf("\n\n Ans\n Head lost = %.1f mm of water",hf*1e3);
\ No newline at end of file diff --git a/536/CH3/EX3.7/Example_3_7.sce b/536/CH3/EX3.7/Example_3_7.sce new file mode 100755 index 000000000..646f122ce --- /dev/null +++ b/536/CH3/EX3.7/Example_3_7.sce @@ -0,0 +1,35 @@ +clc;
+
+printf("Example 3.7\n");
+Q_h=2.27; // flow rate of water in m^3/h
+T=320; //Temperature of water to be pumped
+id=40e-3; //internal diameter of pipe
+l_h=150; //length of pipe horizontally
+l_v=10; //length of pipe vertically
+e=0.2e-3;
+g=9.81;
+rho=1000;
+printf("\n Given\n flow rate of water in m^3/h = %.2f m^3/h\n Temperature of water to be pumped = %d K\n internal diameter of pipe = %d mm\n length of pipe horizontally = %d m\n length of pipe vertically = %d m",Q_h,T,id*1e3,l_h,l_v);
+
+rel_rough=e/id; //Relative roughness
+printf("\n\n Relative roughness = %.3f",rel_rough);
+meu=0.65e-3; //Viscosity at 320 K
+Q_s=Q_h/3600; //flow rate of water in m^3/s
+area=%pi/4*id^2; // Area for flow
+printf("\n Area for flow = %.2f * 10^-3 m^2",area*1e3);
+u=Q_s/area; //Velocity
+printf("\n Velocity = %.2f m/s",u);
+Re=(id*u*rho)/meu;
+printf("\n Reynolds No. = %d",Re);
+
+//X=R/(rho*u^2)=0.004 (from Figure 3.7)
+X=.004;
+equi_len=l_h+l_v+(260*id); // Equivalent length of pipe
+printf("\n Equivalent length of pipe = %.1f m",equi_len);
+hf=4*X*equi_len*u^2/(id*g);
+tot_head=hf+1.5+10; // Total head to be developed
+printf("\n Total head to be developed = %.2f m",tot_head);
+mass_thr=Q_s*rho; //Mass throughput
+printf("\n Mass throughput = %.2f kg/s",mass_thr);
+power_reqd=(mass_thr*tot_head*g)/0.60;
+printf("\n\n Power required = %.1f W = %.3f kW",power_reqd,power_reqd*1e-3);
\ No newline at end of file diff --git a/536/CH3/EX3.8/Example_3_8.sce b/536/CH3/EX3.8/Example_3_8.sce new file mode 100755 index 000000000..06ab67ba3 --- /dev/null +++ b/536/CH3/EX3.8/Example_3_8.sce @@ -0,0 +1,19 @@ +clc;
+printf("Example 3.8\n");
+
+d=0.15; // diameter of pipe
+g=9.81;
+printf("\n Given\n Diameter of pipe = %.2f",d);
+// From equation 3.20, the head lost due to friction is given by:
+// hf = 4*phi*l*u^2/(d*g)m water
+// The total head loss is:
+// h=(u^2/(2*g))+hf+loss in fittings
+// From Table 3.2., the losses in the fittings are:From Table 3.2., the losses in the fittings are:
+//6.6*u^2/(2*g)
+//Taking
+phi=.0045;
+x=poly([0],'x');
+u=roots((7.6+4*phi*(105/.15))*x^2/(2*g)-10);
+printf("\n\n Velocity = %.2f m/s",u(1));
+rate_dis=u(1)*%pi*d^2/4;
+printf("\n Rate of discharge = %.3f m^3/s = %d kg/s",rate_dis,rate_dis*1e3);
\ No newline at end of file diff --git a/536/CH3/EX3.9/Example_3_9.sce b/536/CH3/EX3.9/Example_3_9.sce new file mode 100755 index 000000000..ba42a2778 --- /dev/null +++ b/536/CH3/EX3.9/Example_3_9.sce @@ -0,0 +1,14 @@ +clc;
+
+printf("Example 3.9\n");
+
+u1=1.5; // velocity
+D1=75e-3; //depth
+g=9.81;
+printf("\n Given\n velocity before jump= %.1f m/s\n depth before jump= %d mm",u1,D1*1e3);
+//The depth of fluid in the channel after the jump is given by:
+D2=0.5*(-D1+(D1^2+(8*u1^2*D1/g)^0.5)); //equation 3.113
+printf("\n\n The depth of fluid in the channel after the jump is = %.1f mm",D2*1e3);
+//If the channel is of uniform cross-sectional area, then:
+u2=u1*D1/D2;
+printf("\n The velocity of fluid in the channel after the jump is = %.2f m/s",u2);
\ No newline at end of file diff --git a/536/CH4/EX4.1/Example_4_1.sce b/536/CH4/EX4.1/Example_4_1.sce new file mode 100755 index 000000000..dc5a0e75f --- /dev/null +++ b/536/CH4/EX4.1/Example_4_1.sce @@ -0,0 +1,38 @@ +clc;
+clear;
+
+printf("\n Example 4.1\n");
+
+d=0.006;//Diameter of the cylinder
+Gamma=1.4;
+//The critical pressure ratio for discharge through the valve
+C_r=(2/(Gamma+1))^(Gamma/(Gamma-1));
+printf("\n The graphs are plotted between\n (1) Rate of discharge of air from the cylinder against cylinder pressure\n (2) For a constant pressure of 5 MN/m^2 in the cylinder, the discharge rate vs Downstream pressure.")
+//(i) Sonic velocity will occur until the
+P_c=101.3/C_r;//pressure at which sonic velocity will occur
+M=29;//molecular mass of air
+//The rate of discharge for cylinder pressures greater than 191.1 kN/m^2 is
+//given by equation 4.30: taking mean value for Gamma i.e. 1.47
+//we get
+//G_max=4.23e-8*P1 kg/s
+//For cylinder pressures below 191.1 kN/m2, the mass flowrate is given by equation 4.20
+P1a=[0.1 0.125 0.15 0.17 0.19 0.2 0.5 1.0 2.0 3.0 4.0 5.0 6.0]
+for i=5:13
+ G(i)=4.23e-2*P1a(i);
+end
+for j=1:4
+ G(j)=0.0314*P1a(j)^0.286*((1-0.519*P1a(j)^(-0.286)))^0.5;
+end
+xset('window',1)
+plot(P1a,G);
+xtitle('Rate of discharge of air vs Cylider Pressure','Cylinder pressure P1a (MN/m )','Mass flow (kg/s)');
+xset('window',2);
+P2a=[0 1 2 2.65 3 3.5 4 4.5 4.9 4.95 5];
+for j=5:11
+ G2(j)=0.2548*P2a(j)^0.714*((1-0.631*P2a(j)^0.286))^0.5;
+end
+for i=1:4
+ G2(i)=0.210;
+end
+plot(P2a,G2);
+xtitle('Rate of discharge of air vs Downstream Pressure','Downstream pressure P2a (MN/m )','Mass flow (kg/s)')
\ No newline at end of file diff --git a/536/CH4/EX4.2/Example_4_2.sce b/536/CH4/EX4.2/Example_4_2.sce new file mode 100755 index 000000000..5b57ce4dc --- /dev/null +++ b/536/CH4/EX4.2/Example_4_2.sce @@ -0,0 +1,18 @@ +clc;
+
+printf("\n Example 4.2\n");
+
+l=30;//Length of the tube
+d=150e-3;//Diameter of the tube
+P1=0.4e3;//Initial Pressure
+P2=0.13e3;//final Pressure
+//X=e/d, Relative roughness
+//Y=R/(rho*u^2) = 0.004
+X=0.003;
+Y=0.005;
+v1=21.15e1;
+
+G_A=poly([0],'G_A');
+f=(G_A^2*log(P1/P2))+((P2^2-P1^2)/(2*P1*v1))+(4*(Y*l/d)*G_A^2);
+A=roots(f);
+printf("\n The approximate flow rate = %.2f kg/m^2 s",A(1));
\ No newline at end of file diff --git a/536/CH4/EX4.3/Example_4_3.sce b/536/CH4/EX4.3/Example_4_3.sce new file mode 100755 index 000000000..3cd5a5913 --- /dev/null +++ b/536/CH4/EX4.3/Example_4_3.sce @@ -0,0 +1,47 @@ +clc;
+
+printf("\n Example 4.3\n");
+
+Q=50; //volumetric flow rate of methane
+P=101.3e3;//Given Pressure
+T1=288;//Given Temperature
+d=0.6;//Diameter of pipeline
+l=3e3;//length of the pipe line
+R_R=0.0001;//Relative roughness
+P2=170e3;//Pressure at which methane is to be discharged
+T2=297;//Temperature at which methane leaves the compressor
+M=16;//molecular mass of methane
+R=8314;//Gas constant
+Meu=1e-5;//Viscosity of methane at 293 K
+
+T=(T1+T2)/2;//Mean temperature
+P1_v1=R*T/(M);
+//At 288 K and 101.3 kN/m^2
+v=P1_v1/P*T1/T;
+G=Q/v;//Mass flow rate of methane
+A=%pi/4*d^2;//cross sectional area of pipeline
+G_A=G/A;
+Re=G_A*d/Meu;
+//Y=R/(rho*u^2) = 0.0015
+Y=0.0015;//(from fig 3.7)
+//The upstream pressure is calculated using equation 4.55:
+function[y]=pressure(P1)
+ y=G_A^(2)*log(P1/P2)+(P2^2-P1^2)/(2*1.5525e5)+4*Y*(l/d)*G_A^2;
+ funcprot(0);
+endfunction
+P1 = 1e5;
+z = fsolve(P1,pressure);
+printf("\n Pressure to be developed at the compressor in order to achieve this flowrate = %.2f * 10^5 N/m^2",z*1e-5);
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/536/CH4/EX4.4/Example_4_4.sce b/536/CH4/EX4.4/Example_4_4.sce new file mode 100755 index 000000000..3a3d6e7e2 --- /dev/null +++ b/536/CH4/EX4.4/Example_4_4.sce @@ -0,0 +1,29 @@ +clc;
+clear;
+
+printf("\n Example 4.4");
+
+A1=0.007;//cross sectional area of stack pipe
+A2=4000e-6;//flow area of ruptured disc
+P1=10e6;//Pressure of the gas in the vessel
+Gamma=1.4;
+M=40;//mean molecular weight of gas
+
+w_c=(2/(Gamma+1))^(Gamma/(Gamma-1));
+P_c=P1*w_c;
+V1=(22.4/M)*(500/273)*(101.3e3/P1);//Specific volume of the gas in the reactor
+V=V1*(1/w_c)^(1/Gamma);//Specific volume of gas at the throat
+u=(Gamma*P_c*V)^0.5;//velocity at the throat
+G=u*A2/V;//initial rate of discharge
+
+printf("\n (a)Initial rate of discharge of gas = %.1f kg/s",G);
+//obtaining the equations as given in book and solving for 'w' we get
+w=0.0057;//Pressure ratio
+P_u=P1*w;
+printf("\n (b)The pressure upstream from the shockwave = %.0f kN/m^2",P_u*1e-3);
+Mach_no=2.23*(w^(-0.29)-1)^0.5;
+printf("\n The mach number is = %.2f",Mach_no);
+P_s=56.3*w*(w^(-0.29)-1)*1e6;
+printf("\n (c)The pressure downstream from the shockwave = %.0f kN/m^2",P_s*1e-3);
+
+
diff --git a/536/CH5/EX5.1/Example_5_1.sce b/536/CH5/EX5.1/Example_5_1.sce new file mode 100755 index 000000000..f1476f0cf --- /dev/null +++ b/536/CH5/EX5.1/Example_5_1.sce @@ -0,0 +1,37 @@ +clc;
+clear;
+
+printf("Example 5.1\n");
+id=75e-3; // internal diameter of pipe
+printf("\n Given: \n Internal diameter of pipe = %d mm",id*1e3);
+f_r_s=0.05; // Flow rate of steam in (kg/s)
+printf("\n Flow rate of steam = %.2f kg/s",f_r_s);
+f_r_w=1.5; // Flow rate of water in (kg/s)
+printf("\n Flow rate of water = %.1f kg/s",f_r_w);
+T=330; // Mean Temperature
+printf("\n Mean Temperature = %d K",T);
+P=120; // Mean Pressure drop
+printf("\n Mean Pressure drop = %d kN/m^2",P);
+area=%pi*id^2/4; // Cross-sectional area for flow
+f_r_w_m3s=f_r_w/1000; // Flow of water
+wtr_vel=f_r_w_m3s/area; //Water velocity
+rho_steam=18*273*120/(22.4*330*101.3); // density of steam at 330 K and 120 kN/m^2
+f_r_s_m3s=f_r_s/rho_steam; //Flow of Steam
+steam_vel=f_r_s_m3s/area; //Steam velocity
+printf("\n\n Calculations:\n Cross-sectional area for flow = %.5f m^2\n Water velocity = %.3f m/s\n Steam velocity = %.2f m/s",area,wtr_vel,steam_vel);
+meu_steam=0.0113e-3;
+meu_water=0.52e-3;
+Rel=id*wtr_vel*1000/meu_water;
+Reg=id*steam_vel*rho_steam/meu_steam;
+printf("\n Reynolds no.(water) = %.2f *10^4",Rel*1e-4);
+printf("\n Reynolds no.(steam) = %.2f *10^4",Reg*1e-4);
+// That is, both the gas and liquid are in turbulent flow. From the friction chart (Figure 3.7), assuming e/d = 0.00015:
+// R/(rho*u^2) liq=0.0025 R/(rho*u^2) gas=0.0022
+// From equation 3.18:
+DPl=4*0.0025*(1000*wtr_vel^2)/id;
+DPg=4*0.0022*(rho_steam*steam_vel^2)/id;
+X=(DPl/DPg)^0.5;
+phi_l=4.35;
+phi_g=3.95;
+DP_tpf=phi_g^2*DPg;
+printf("\n Pressure drop per unit length of pipe = %.0f N/m^2",DP_tpf);
diff --git a/536/CH5/EX5.2/Example_5_2.sce b/536/CH5/EX5.2/Example_5_2.sce new file mode 100755 index 000000000..1a8c072ba --- /dev/null +++ b/536/CH5/EX5.2/Example_5_2.sce @@ -0,0 +1,37 @@ +clc;
+clear;
+
+printf("\n Example 5.2\n");
+
+M_p_d=0.2e-3; // Mean particle diameter
+printf("\n Given:\n Mean particle diameter = %.1f mm",M_p_d*1e3);
+f_r_w=0.5; //Flow rate of water
+printf("\n Flow rate of water = %.1f kg/s",f_r_w);
+id=25e-3; //Diameter of pipe
+printf("\n Diameter of pipe = %d mm",id*1e3);
+l=100; //length of pipe
+printf("\n length of pipe = %d m",l);
+t_vel=0.0239; //Terminal velocity of falling sand particles
+printf("\n Terminal velocity of falling sand particles = %.4f m/s",t_vel);
+//Assuming the mean velocity of the suspension is equal to the water velocity, that is, neglecting slip, then:
+Um=f_r_w/(1000*%pi*id^2/4);
+printf("\n\n Calculations:\n Mean velocity of suspension = %.2f m/s",Um);
+Re=id*Um*1000/0.001;
+printf("\n Reynolds no. of water alone = %d",Re);
+//Assuming e/d = 0.008, then, from Figure 3.7:
+phi=0.0046;
+f=0.0092;
+//From, equation 3.20, the head loss is:
+hf=4*phi*l*Um^2/(9.81*id);
+printf("\n Head loss = %.1f m water",hf);
+iw=hf/l;
+printf("\n Hydraulic gradient = %.3f m water/m",iw);
+i=300*1000/(1000*9.81*100);
+// Substituting in equation 5.20:
+C=(iw/(i-iw)*(1100*9.81*id*(2.6-1)*t_vel)/(Um^2*Um))^-1;
+printf("\n C = %.2f",C);
+//If G kg/s is the mass flow of sand, then:
+G=poly([0],'G');
+p=2600^-1*G-0.30*(2600^-1*G+.0005);
+printf("\n Mass flow of sand = %.2f kg/s",roots(p));
+printf("")
\ No newline at end of file diff --git a/536/CH5/EX5.3/Example_5_3.sce b/536/CH5/EX5.3/Example_5_3.sce new file mode 100755 index 000000000..ff60e517f --- /dev/null +++ b/536/CH5/EX5.3/Example_5_3.sce @@ -0,0 +1,42 @@ +clc;
+clear;
+
+printf("\n Example 5.3");
+
+p_s=1.25e-3; // Particle size of sand
+printf("\n Given:\n Particle size of sand = %.2f mm",p_s*1e3);
+rho_sand=2600; //Density of sand
+printf("\n Density of sand = %d kg/m^3",rho_sand);
+flow_sand=1; //flow rate of sand in air
+printf("\n flow rate of sand in air = %d kg/s",flow_sand);
+l=200; //length of pipe
+printf("\n length of pipe = %d m",l);
+// Assuming a solids:gas mass ratio of 5, then:
+flow_air=flow_sand/5;
+vol_flow_air=1*flow_air;
+printf("\n\n Calculations:\n Volumetric flow rate of air = %.2f m^3/s",vol_flow_air);
+//In order to avoid an excessive pressure drop, an air velocity of 30 m/s is acceptable
+d=100e-3; // taking nearest standard size of pipe
+// For sand of particle size 1.25 mm and density 2600 kg/m3, the free-falling velocity is given in Table 5.3 as:
+Uo=4.7;
+// In equation 5.37:
+area=%pi*d^2/4;
+printf("\n The cross-sectional area of a 100 mm ID. pipe = %.5f m^2",area);
+Ug=flow_air/area;
+Us=Ug-(Uo/(0.468+(7.25*(Uo/rho_sand)^0.5)));
+printf("\n Air velocity = %.1f m/s",Ug);
+printf("\n solids velocity = %.1f m/s",Us);
+//Taking
+Meu_air=1.7e-5; // viscosity of air
+rho_air=1; // Density of air
+Re=(d*Ug*rho_air/Meu_air);
+printf("\n Reynolds no. of air alone = %d",Re);
+phi=0.004;
+//Assuming isothermal conditions and incompressible flow, then, in equation 3.18:
+DP_air=(4*phi*l/d)*rho_air*Ug^2/2;
+printf("\n Pressure drop due to air = %.1f kN/m^2",DP_air*1e-3);
+//and in equation 5.38:
+DP_x=2805*DP_air/(Uo*Us^2);
+printf("\n Pressure drop due to sand particles = %.1f kN/m^2",DP_x*1e-3);
+DP=DP_air+DP_x;
+printf("\n The total pressure drop = %.1f kN/m^2",DP*1e-3);
\ No newline at end of file diff --git a/536/CH6/EX6.1/Example_6_1.sce b/536/CH6/EX6.1/Example_6_1.sce new file mode 100755 index 000000000..9ebfb7731 --- /dev/null +++ b/536/CH6/EX6.1/Example_6_1.sce @@ -0,0 +1,25 @@ +clc;
+clear;
+
+printf("\n Example 6.1");
+d_o=25e-3;//Diameter of orifice
+printf("\n\n Given:\n Diameter of orifice = %d mm",d_o*1e3);
+d_p=75e-3;//Diameter of pipe
+printf("\n Diameter of pipe = %d mm",d_p*1e3);
+flow_o=300e-6;//Flow rate through pipe
+printf("\n Flow rate through pipe = %d m^3/s",flow_o*1e6);
+Meu_watr=1e-3;//Viscosity of water
+printf("\n Viscosity of water = %d mN s/m^2",Meu_watr*1e3);
+area_o=%pi/4*d_o^2;//Area of orifice
+printf("\n\n Calculations:\n Area of orifice = %.2f * 10^-4 m^2",area_o*1e4);
+vel_o=flow_o/area_o;//Velocity of water through the orifice
+printf("\n Velocity of water through the orifice = %.2f m/s",vel_o);
+Re_o=d_o*vel_o*1000/Meu_watr;//Re at the orifice
+printf("\n Re at the orifice = %d",Re_o);
+//From Figure 6.16, the corresponding value of Cd = 0.61 (diameter ratio = 0.33)
+Cd=0.61;
+G=flow_o*1e3; //mass flow rate water
+//Equation 6.21 may therefore be applied:
+ho=poly([0],'ho');
+p=G^2-((Cd*area_o*1000)^2*2*9.81*ho);
+printf("\n Difference in level on a water manometer = %.0f mm of water",roots(p)*1e3);
\ No newline at end of file diff --git a/536/CH6/EX6.2/Example_6_2.sce b/536/CH6/EX6.2/Example_6_2.sce new file mode 100755 index 000000000..9af19b3c7 --- /dev/null +++ b/536/CH6/EX6.2/Example_6_2.sce @@ -0,0 +1,28 @@ +clc;
+clear;
+
+printf("\n Example 6.2\n");
+rho_sul=1300;//Density of sulphuric acid
+printf("\n Given:\n Density of sulphuric acid = %d kg/ m^3",rho_sul);
+id=50e-3;//Internal diameter of pipe
+printf("\n Internal diameter of pipe = %d mm",id*1e3);
+d_o=10e-3;//Diameter of orifice
+printf("\n Diameter of orifice = %d mm",d_o*1e3);
+h=.1;//Differential pressure shown on a mercury manometer
+printf("\n Differential pressure shown on a mercury manometer = %.1f m",h);
+Cd=0.61//Coeffecient of discharge
+printf("\n Coeffecient of discharge = %.2f",Cd);
+rho_merc=13550;//Density of mercury
+printf("\n Density of mercury = %d kg/m^3",rho_merc);
+rho_watr=1000;//Density of water
+printf("\n Density of water = %d kg/m^3",rho_watr);
+printf("\n\n Calculations:\n (a)");
+area_o=%pi/4*d_o^2;//area of orifice
+//The differential pressure is given by:
+h_sul=h*(rho_merc-rho_sul)/rho_sul;//
+//The mass flow-rate G is given by:
+//substituting in equation 6.21 gives the mass flowrate as:
+G_sul=Cd*area_o*rho_sul*(2*9.81*h_sul)^0.5;
+printf("\n The mass flow rate of acid = %.3f kg/s\n (b)",G_sul);
+DP=rho_sul*9.81*h_sul;
+printf("\n The drop in pressure = %.0f kN/m^2",DP*1e-3);
\ No newline at end of file diff --git a/536/CH6/EX6.3/Example_6_3.sce b/536/CH6/EX6.3/Example_6_3.sce new file mode 100755 index 000000000..a6c56f12b --- /dev/null +++ b/536/CH6/EX6.3/Example_6_3.sce @@ -0,0 +1,17 @@ +clc;
+clear;
+
+printf("\n Example 6.3\n");
+d=150e-3;//Diameter of pipe
+printf("\n Given:\n Diameter of pipe = %d mm",d*1e3);
+d_t=50e-3;//Throat diameter
+printf("\n Throat diameter = %d mm",d_t*1e3);
+hv=121e-3;//Pressure drop over the converging section
+printf("\n Pressure drop over the converging section = %d mm of water",hv*1e3);
+G=2.91; //Mass Flow rate of water
+printf("\n Mass Flow rate of water = %.2f kg/s",G);
+//From equation 6.32, the mass rate of flow
+A1=%pi*d^2/4;
+A2=%pi*d_t^2/4;
+Cd=G*(A1^2-A2^2)^0.5/(1000*A1*A2*(2*9.81*hv)^0.5);
+printf("\n\n Calculations:\n Coefficient for the converging cone of the meter at given flowrate = %.3f",Cd);
\ No newline at end of file diff --git a/536/CH6/EX6.4/Example_6_4.sce b/536/CH6/EX6.4/Example_6_4.sce new file mode 100755 index 000000000..8d7353860 --- /dev/null +++ b/536/CH6/EX6.4/Example_6_4.sce @@ -0,0 +1,36 @@ +clc;
+clear;
+
+printf("\n Example 6.4\n");
+l=0.3;//length of tube
+printf("\n Given:\n length of tube = %.1f m",l);
+id_t=25e-3;//Top internal diameter of tube
+printf("\n Top internal diameter of tube = %d mm",id_t*1e3);
+id_b=20e-3;//Bottom internal diameter of tube
+printf("\n Bottom internal diameter of tube = %d mm",id_b*1e3);
+d_f=20e-3;//Diameter of float
+printf("\n Diameter of float = %d mm",d_f*1e3);
+v_f=6e-6;//Volume of float
+printf("\n Volume of float = %d cm^3",v_f*1e6);
+Cd=0.7;//Coefficient of discharge
+printf("\n Coefficient of discharge = %.1f",Cd);
+rho=1000;//Density of water
+printf("\n Density of water = %d kg/m^3",rho);
+rho_f=4800;//Density of float
+printf("\n Density of float = %d kg/m^3",rho_f);
+area_t=%pi/4*id_t^2;//Cross-sectional area at top of tube
+printf("\n\n Calculations:\n Cross-sectional area at top of tube = %.2f *10^-4 m^2",area_t*1e4);
+area_b=%pi/4*id_b^2;//Cross-sectional area at bottom of tube
+printf("\n Cross-sectional area at bottom of tube = %.2f *10^-4 m^2",area_b*1e4);
+A_f=%pi/4*d_f^2;//Area of float
+printf("\n Area of float = %.2f *10^-4 m^2",A_f*1e4);
+//When the float is halfway up the tube, the area at the height of the float A1 is given by:
+A1=%pi/4*((id_t+id_b)/2)^2;
+printf("\n The area of the height of the float A1 is = %.2f *10^-4 m^2",A1*1e4)
+//The area of the annulus A2 is given by:
+A2=A1-A_f;
+printf("\n The area of the annulus A2 is = %.2f *10^-4 m^2",A2*1e4)
+//Substituting into equation 6.36:
+//The flow rate of water =
+G=Cd*A2*((2*9.81*v_f*(rho_f-rho)*rho)/(A_f*(1-(A2/A1)^2)))^0.5;
+printf("\n\n The flow rate of water = %.3f kg/s",G);
\ No newline at end of file diff --git a/536/CH6/EX6.5/Example_6_5.sce b/536/CH6/EX6.5/Example_6_5.sce new file mode 100755 index 000000000..4fe222eb2 --- /dev/null +++ b/536/CH6/EX6.5/Example_6_5.sce @@ -0,0 +1,13 @@ +clc;
+clear;
+
+printf("\n Example 6.5\n");
+
+L=0.5;// Length of the weir
+printf("\n Given\n Length of the weir = %.1f m",L);
+D=100e-3;//Height of water over the weir
+printf("\n Height of water over the weir = %d",D*1e3);
+n=0;
+//Using Francis formula:
+Q=1.84*(L-(0.1*n*D))*D^1.5;
+printf("\n\n Calculations:\n Volumetric flowrate of water = %.2f m^3/s",Q);
\ No newline at end of file diff --git a/536/CH6/EX6.6/Example_6_6.sce b/536/CH6/EX6.6/Example_6_6.sce new file mode 100755 index 000000000..c4090afc2 --- /dev/null +++ b/536/CH6/EX6.6/Example_6_6.sce @@ -0,0 +1,15 @@ +clc;
+clear;
+
+printf("\n Example 6.6\n");
+
+G=15; //Mass flow rate of organic liquid
+printf("\n Given:\n Mass flow rate of organic liquid = %d kg/s",G)
+L_ow=2;//Length of the weir
+printf("\n Length of the weir = %.1f m",L_ow);
+rho_l=650;
+printf("\n Density of liquid = %d kg/m^3",rho_l);
+Q=G/rho_l;
+//Use is made of the Francis formula (equation 6.43),
+h_ow=(2/3)*(Q/L_ow)^(2/3);
+printf("\n\n Calculations:\n Height of liquid flowing over the weir = %.2f mm",h_ow*1e3);
\ No newline at end of file diff --git a/536/CH7/EX7.2/Example_7_2.sce b/536/CH7/EX7.2/Example_7_2.sce new file mode 100755 index 000000000..4fd4bec43 --- /dev/null +++ b/536/CH7/EX7.2/Example_7_2.sce @@ -0,0 +1,25 @@ +clc;
+clear;
+
+printf("\n Example 7.2\n");
+
+rho_sol=1650;//Density of the solution
+printf("\n Given \n Density of the solution = %d kg/m^3",rho_sol);
+Meu_sol=50e-3;//Viscosity of the solution
+printf("\n Viscosity of the solution = %d mN s/m^2",Meu_sol*1e3);
+Dt=2.28;//Density of the tank
+printf("\n Density of the tank = %.2f m",Dt);
+D=0.5;//Diameter of the propeller mixer
+printf("\n Diameter of the propeller mixer = %.2f m",D);
+H=2.28;//Liquid depth
+printf("\n Liquid depth = %.2f m",H);
+Za=0.5;//Height of the propeller
+printf("\n Height of the propeller = %.1f m",Za);
+N=2;//Rotational speed
+//In this problem the geometrical arrangement corresponds with the configuration for which the curves in Figure 7.6 are applicable.
+Re=D^2*N*rho_sol/(Meu_sol);
+Fr=N^2*D/9.81;
+//From figure 7.6
+Np=0.5;
+P=Np*rho_sol*N^3*D^5;
+printf("\n\n Calculations:\n Power provided by propeller to the liquid = %.0f W",P);
\ No newline at end of file diff --git a/536/CH7/EX7.3/Example_7_3.sce b/536/CH7/EX7.3/Example_7_3.sce new file mode 100755 index 000000000..418dcca08 --- /dev/null +++ b/536/CH7/EX7.3/Example_7_3.sce @@ -0,0 +1,31 @@ +clc; +clear; + +printf('Example 7.3\n'); + +d=0.6;//Tank diameter +N1=4;//Rotor dpeed in Hertz +P1=0.15;//Power consumption +Re1=160000;//Reynold's number +//The correlation of power consumption and Reynolds number is given by: +//equation (7.13) +printf("\n For Constant impeller tip speed \n"); +D1=d/3; +D2=6*D1; +N2=%pi*N1*D1/(%pi*D2); +printf("\n The new rotor speed = %.2f Hz",N2); +//from eq(1) of the solution +P2=7.32*N2^3*D2^5; +printf("\n The new power required = %.2f kW",P2); +//For thermal similarity, that is the same temperature in both systems: +Re2=Re1*(N2*D2^2/(N1*D1^2)); +printf("\n The new reynolds number = %d",Re2); + +printf("\n\n For Constant power input per unit volume\n"); +V2=(%pi/4)*(6*d)^3; +P=V2*0.884; +printf("\n The new power required = %.2f kW",P); +N=(P/(7.32*(6*0.6/3)^5))^(1/3); +printf("\n The new rotor speed = %.2f Hz",N); +Re=Re1*(N*D2^2/(N1*D1^2)); +printf("\n The new reynolds number = %d",Re);
\ No newline at end of file diff --git a/536/CH8/EX8.1/Example_8_1.sce b/536/CH8/EX8.1/Example_8_1.sce new file mode 100755 index 000000000..73ed697a7 --- /dev/null +++ b/536/CH8/EX8.1/Example_8_1.sce @@ -0,0 +1,32 @@ +clc;
+clear;
+printf("\n Example 8.1\n");
+
+dia_cy=110e-3; //Cylinder diameter
+printf("\n Given\n Cylinder diameter = %d mm",dia_cy*1e3)
+str=230e-3; //stroke
+printf("\n Stroke length = %d mm",str*1e3);
+l_su=6; //Suction line length
+printf("\n Suction line length = %d m",l_su);
+d_su=50e-3; //Suction line diameter
+printf("\n Suction line diameter = %d mm",d_su*1e3);
+lvl_wtr=3; //level of the water in the suction tank
+printf("\n level of the water in the suction tank = %d m",lvl_wtr);
+atm_P=10.36;
+printf("\n Atmospheric pressure is equivalent to = %.2f m of water",atm_P);
+//If the maximum permissible speed of the pump is N Hz:
+//Angular velocity of the driving mechanism = 2*pi*N radians/s
+//Acceleration of piston = 0.5x0.230(2*pi*N)^2*cos(2*pi*N*t) m/s^2
+//Maximum acceleration(when t=0) = 4.54*N^2 m/s^2
+//Maximum acceleration of the liquid in the suction pipe
+// =(.110/.05)^2 x 4.54*N^2 = 21.91*N^2 m/s^2
+//Accelerating force acting on the liquid
+// = 21.97*N^2*pi/4*(0.050)^2*(6*1000) N
+//Pressure drop in suction line due to acceleration
+// = 21.97*N^2 *6*1000 N/m^2
+// = 1.32x 10^5*N^2 N/m^2
+// = 13.44*N^2 m water
+//Pressure head at cylinder when separation is about to occur,
+x=poly([0],'x');
+N=roots(1.20-(10.36-3.0-13.44*x^2));
+printf("\n\n Calculations:\n Maximum speed at which the pump can run = %.3f Hz",N(1));
\ No newline at end of file diff --git a/536/CH8/EX8.10/Example_8_10.sce b/536/CH8/EX8.10/Example_8_10.sce new file mode 100755 index 000000000..313bc7530 --- /dev/null +++ b/536/CH8/EX8.10/Example_8_10.sce @@ -0,0 +1,29 @@ +clc;
+clear;
+printf("\n Example 8.10\n");
+
+Meu_H2=0.009e-3; //Viscosity of hydrogen
+P2=2e6; //Downstream Pressure
+P1=2.5e6;//Upstream pressure
+P_m=(P1+P2)/2;//Mean Pressure
+T=295;//Temperature of the gas
+l=500;//Length of the pipe used
+d=50e-3;//diameter of pipe used
+rho_H2=2*P_m*273/(22.4*101.3e3*T);//Density of hydrogen at the mean pressure
+A=%pi*d^2/4;//Area of the pipe
+eta=0.60; //Efficiency of the pump
+v_m=1/rho_H2;
+//Firstly, an approximate value of G is obtained by neglecting the kinetic energy of the fluid
+//Using equation 4.56
+//phi*Re^2=7.02 * 10 ^8
+//Taking the roughness of the pipe surface, e as 0.00005 m
+//e/d= 0.001 and Re = 5.7 x 10^5 from Figure 3.8
+//G=.201(approximate value)
+//From Figure 3.7,
+phi=0.0024;
+//Taking the kinetic energy of the fluid into account, equation 4.56 may be used:
+x=poly([0],'x');
+G=roots((x/A)^2*log(P1/P2)+(P2-P1)*rho_H2+4*phi*l/d*(x/A)^2);
+printf("\n Mass flow rate = %.2f kg/s",G(1));
+P=G(1)*P_m*v_m*log(P1/P2)/eta;
+printf("\n Power required = %.1f kW",P*1e-3);
\ No newline at end of file diff --git a/536/CH8/EX8.2/Example_8_2.sce b/536/CH8/EX8.2/Example_8_2.sce new file mode 100755 index 000000000..440294e6b --- /dev/null +++ b/536/CH8/EX8.2/Example_8_2.sce @@ -0,0 +1,34 @@ +clc;
+clear;
+printf("\n Example 8.2\n");
+
+rho_l=800; //Density of liquid
+printf("\n Given\n Density of liquid = %d kg/m^3",rho_l);
+Meu_l=0.5e-3;//Viscosity of liquid
+printf("\n Viscosity of liquid = %.1f * 10^-3 N s/m^2",Meu_l*1e3);
+Q=0.0004;//Volumetric flow rate
+printf("\n Volumetric flow rate = %d m^3/s",Q*1e6);
+liq_depth=0.07;
+d=25e-3;//Diameter of pipe used
+printf("\n Diameter of pipe used = %d",d*1e3);
+p_v_r=1e3;//Pressure of vapor in reboiler
+printf("\n Pressure of vapor in reboiler = %d kN/m^2",p_v_r*1e-3);
+Z=2;//Net Positive Suction Head
+printf("\n Net Positive Suction Head = %d m",Z);
+A=%pi/4*d^2;//Cross sectional area of pipe
+printf("\n\n Calculations:\n Cross sectional area of pipe = %.5f m^2",A);
+u=Q/A;//Velocity in pipe
+printf("\n Velocity in pipe = %.3f m/s",u);
+Re=d*u*rho_l/Meu_l;//Reynolds no.
+printf("\n Reynolds no. = %d ",Re);
+//From Figure 3.7, the friction factor for a smooth pipe is:
+phi=0.0028;
+hf_l=(4*phi*u^2)/(d*9.81);
+printf("\n head loss due to friction per unit length = %.4f m/m of pipe",hf_l);
+//It should be noted that a slightly additional height will be required if the kinetic energy at the pump inlet cannot be utilised.
+//Thus the height between the liquid level in the reboiler and the pump, HQ, depends on the length of pipe between the reboiler and the pump. If this is say 10 m
+l=10;
+hf=hf_l*l;
+//equation 8.26 becomes:
+ho=Z+hf;
+printf("\n The minimum height required = %.1f m",ho);
diff --git a/536/CH8/EX8.3/Example_8_3.sce b/536/CH8/EX8.3/Example_8_3.sce new file mode 100755 index 000000000..30b27d752 --- /dev/null +++ b/536/CH8/EX8.3/Example_8_3.sce @@ -0,0 +1,36 @@ +clc;
+clear;
+printf("\n Example 8.3\n");
+
+Q=0.1;//Flow rate of air suppplied by compressor
+printf("\n Given:\n Flow rate of air suppplied by compressor = %.1f m^3/s",Q);
+T=273;//Temperature
+printf("\n Temperature = %d K",T);
+P=101.3e3;//Pressure
+printf("\n Pressure = %.1f kN/m^2",P*1e-3);
+P2=380e3;//Air compressed to a pressure
+printf("\n Air compressed to a pressure = %d kN/m^2",P2*1e-3);
+T2=289;//Suction Temperature
+printf("\n Suction Temperature = %d K",T2);
+l=0.25;//Length of the stroke
+printf("\n Length of the stroke = %.2f m",l);
+u=4;//Speed
+printf("\n Speed = %d Hz",u);
+c=4/100;//Cylinder clearance
+printf("\n Cylinder clearance = %.2f",c);
+Gamma=1.4;
+V=Q*T2/(u*T);//Volume per stroke
+printf("\n\n Calculations:\n Volume per stroke = %.4f m^3",V);
+R=P2/P;//Compression ratio
+printf("\n Compression ratio = %.2f",R);
+//The swept volume is given by equation 8.42
+Vs=V/(1+c-(c*(R)^(1/Gamma)));
+printf("\n The swept volume is = %.4f m^3",Vs);
+A=Vs/l;//Cross sectional Area of cylinder
+printf("\n Cross sectional Area of cylinder = %.3f m^2",A);
+d=(A/%pi*4)^0.5;//Diameter of cylinder
+printf("\n Diameter of cylinder = %.2f m",d);
+//From equation 8.41, work of compression per cycle
+W=P*V*(Gamma/(Gamma-1))*((R)^((Gamma-1)/Gamma)-1);
+printf("\n Work of compression per cycle = %.0f J",W);
+printf("\n Theoretical power requirements = %.1f kW",W*4/1e3);
\ No newline at end of file diff --git a/536/CH8/EX8.4/Example_8_4.sce b/536/CH8/EX8.4/Example_8_4.sce new file mode 100755 index 000000000..f902b846b --- /dev/null +++ b/536/CH8/EX8.4/Example_8_4.sce @@ -0,0 +1,43 @@ +clc;
+clear;
+printf("\n Example 8.4\n");
+
+T=290;//Temperature at which compression takes place
+printf("\n Given:\n Temperature at which compression takes place = %d K",T);
+P1=101.3e3;//Initial pressure
+P2=2065e3;//Final pressure
+printf("\n Compressed from a Pressure of %.1f kN/m^2 to %d kN/m^2",P1*1e-3,P2*1e-3);
+eta=.85;//Mechanical efficiecy
+printf("\n Mechanical efficiecy = %d percent",eta*1e2);
+c1=4/100;//Clearance in cylinder 1
+printf("\n Clearance in cylinder 1 = %d percent",c1*1e2);
+c2=5/100;//Clearance in cylinder 1
+printf("\n Clearance in cylinder 2 = %d percent",c2*1e2);
+R=P2/P1;//Overall compression ratio
+printf("\n\n Overall compression ratio = %.1f",R);
+V_spe=22.4/28.8*T/273;//Specific volume of air at 290 K
+printf("\n Specific volume of air at 290 K = %.3f m^3/kg\n (a)",V_spe);
+W=P1*V_spe*2*(1.25/(1.25-1))*(R^.1-1);
+//Energy supplied to the compressor, that is the work of compression
+W_act=W/0.85;
+printf("\n Energy supplied to the compressor, that is the work of compression = %.1f kJ/kg",W_act*1e-3);
+printf("\n (b)");
+//the work done in isothermal compression of 1 kg of gas
+W_it=P1*V_spe*log(R);
+//Isothermal efficiency
+eta_it=100*W_it/W_act;
+printf("\n Isothermal efficiency = %.0f percent",eta_it);
+printf("\n (c)");
+Gamma=1.4;
+//the work done in isentropic compression of 1 kg of gas
+W_ie=P1*V_spe*(Gamma/(Gamma-1))*((R)^((Gamma-1)/Gamma)-1);
+//Isentropic efficiency
+eta_ie=100*W_ie/W_act;
+printf("\n Isentropic efficiency = %d percent",eta_ie);
+printf("\n (d)")
+//From equation 8.47, volume swept out in first cylinder in compression of 1 kg of gas is given by:
+Vs1=V_spe/(1+c1-(c1*(R)^(1/(2*2.5))));
+//Similarly, the swept volume of the second cylinder is given by:
+Vs2=V_spe*(1/R)^0.5/(1+c2-(c2*(R)^(1/(2*2.5))));
+ratio=Vs1/Vs2;
+printf("\n the ratio of the swept volumes in the two cylinders = %.2f",ratio);
\ No newline at end of file diff --git a/536/CH8/EX8.5/Example_8_5.sce b/536/CH8/EX8.5/Example_8_5.sce new file mode 100755 index 000000000..2aafb3193 --- /dev/null +++ b/536/CH8/EX8.5/Example_8_5.sce @@ -0,0 +1,37 @@ +clc;
+clear;
+printf("\n Example 8.5\n");
+
+Q_l=7.5e-4;
+printf("\n Given:\n Volume flow rate of liquid = %.1f m^3/s",Q_l*1e4);
+rho_l=1200;
+printf("\n Density of liquid = %d kg/m^3",rho_l);
+h=20;
+printf("\n height to which liquid is raised = %d m",h);
+P=450e3;
+printf("\n Air is available at pressure = %d kN/m^2",P*1e-3);
+eta=30/100;
+printf("\n Efficiency = %d percent",eta*100);
+P_atm=101.3e3;
+Gamma=1.4;
+G=Q_l*rho_l;//Mass flow of liquid
+//Work per unit time done by the pump
+W=G*9.81*h;
+printf("\n\n Calculations:\n Work per unit time done by the pump = %.1f W",W);
+//Actual work of expansion of air per unit time
+W_act=W/eta;
+printf("\n Actual work of expansion of air per unit time = %.1f W",W_act);
+//Taking the molecular weight of air
+M=28.9;
+//the specific volume of air at 101.3 kN/m2 and 273 K
+va=22.4/M;
+//and in equation 8,49:
+x=poly([0],'x');
+Ga=roots(P_atm*va*x*log(P/P_atm)-W_act);
+Q=Ga*va;
+printf("\n volume flow rate of air = %.4f m^3/s",Q);
+//From equation 8.37
+//Power for compression
+Power=(P_atm*Q)*(Gamma/(Gamma-1))*((P/P_atm)^((Gamma-1)/Gamma)-1);
+Power_reqd=Power/1000;
+printf("\n power requirement of the pump = %.3f kW",Power_reqd);
\ No newline at end of file diff --git a/536/CH8/EX8.6/Example_8_6.sce b/536/CH8/EX8.6/Example_8_6.sce new file mode 100755 index 000000000..00425619e --- /dev/null +++ b/536/CH8/EX8.6/Example_8_6.sce @@ -0,0 +1,41 @@ +clc;
+clear;
+printf("\n Example 8.6\n");
+
+P1=101.3e3;
+Q_watr=0.01;
+printf("\n Given:\n Flow rate of Water = %.2f m^3/s",Q_watr);
+depth=100;
+printf("\n Depth of well = %d m",depth);
+d=100e-3;
+printf("\n Diameter of pipe = %d mm",d*1e3);
+depth_watr=40;
+printf("\n Level of water below water = %d m",depth_watr);
+Q_air=0.1;
+printf("\n Flow rate of Air = %.2f m^3/s",Q_air);
+P2=800e3;
+Gamma=1.4;
+//V1=Q_air;
+G_watr=Q_watr*1000;//Mass flow of water
+W=G_watr*depth_watr*9.81;
+//The energy needed to compress 0. 1 m^3/s of air is given by:
+E=P1*Q_air*(1.4/0.4)*((P2/P1)^(0.4/1.4)-1);// equation 8.37
+printf("\n\n Calculations:\n The power required for this compression is = %d W",E);
+effi=W/E*100;
+printf("\n Efficiency = %.1f per cent",effi);
+//the mean pressure
+P=345e3;
+printf("\n The mean pressure = %d kN/m^2",P);
+v1=8314*273/(29*P);
+printf("\n The specific volume v of air at 273 K and given pressure is = %.3f m^3/kg",v1);
+v2=8314*273/(29*P1);
+printf("\n The specific volume v of air at 273 K and 101.3 kN/m^2 is = %.3f m^3/kg",v2);
+G_air=Q_air/v2; //mass flowrate of the air is:
+Q_mean=G_air*v1;//Mean volumetric flowrate of air
+Q_tot=Q_watr+Q_mean;//Total volumetric flowrate
+A=%pi/4*d^2;//Area of pipe
+v_mean=Q_tot/A;
+printf("\n Mean velocity of the mixture = %.2f m/s",v_mean);
+
+
+
diff --git a/536/CH8/EX8.7/Example_8_7.sce b/536/CH8/EX8.7/Example_8_7.sce new file mode 100755 index 000000000..5d83c0b95 --- /dev/null +++ b/536/CH8/EX8.7/Example_8_7.sce @@ -0,0 +1,29 @@ +clc; +clear; +printf("\n Example 8.7\n"); + +d=40e-3;//Internal Diameter of the pipe +l_p=150; //Lendth of pipe +Q_watr=600e-6;//Flow of water +h1=10; //Vertical Height +h2=2; //head lost across heat exchanger +eta=60/100; //Efficiency of pump + +A=%pi/4*d^2; //Area for flow +printf("\n Area for flow = %.4f m^2",A); +u=Q_watr/A; //Velocity +//At 320 K, +Meu=0.65e-3; +rho=1000; +Re=d*u*rho/Meu; +printf("\n Reynolds no. = %d",Re); +phi=0.004; //for a relative roughness of 0.005 +l=l_p+h1+(260*d); +printf("\n Equivalent length of pipe = %.1f m",l); +hf=4*phi*l*u^2/(d*9.81); +h_tot=hf+h1+h2;//Total head to be developed +printf("\n Total head to be developed = %.2f m",h_tot); +G=Q_watr*rho;// Mass flow of water +P_r=G*h_tot*9.81; //Power Required +P_s=P_r/eta; //Power Supplied +printf("\n Power Required = %.0f W",P_s);
\ No newline at end of file diff --git a/536/CH8/EX8.8/Example_8_8.sce b/536/CH8/EX8.8/Example_8_8.sce new file mode 100755 index 000000000..1321290c1 --- /dev/null +++ b/536/CH8/EX8.8/Example_8_8.sce @@ -0,0 +1,27 @@ +clc;
+clear;
+printf("\n Example 8.8\n");
+
+eta=0.50;
+Q=[0.0028 0.0039 0.0050 0.0056 0.0059]
+h=[23.2 21.3 18.9 15.2 11.0]
+plot(Q,h,'o-');
+//The head to be developed, h=10+4.12*u^2 m water
+//h=10+2.205e5*Q^2
+Q1=0.0015:0.0001:0.0060
+h1=10+2.205e5*Q1^2;
+plot2d(Q1,h1,style=1);
+xtitle("Data for Example 8.8","Discharge (Q m^3/s)","Head (m water)");
+legend("Pump characteristics","h=10+2.205e5*Q^2");
+//showing the intersection point
+x1=[0 0.0054];
+y1=[16.43 16.43];
+x2=[0.0054 0.0054];
+y2=[0 16.43];
+plot(x1,y1,x2,y2);
+Q_r=0.0054;
+printf("\n The discharge at the point of intersection between\n the purnp characteristic equation = %.4f m^3/s",Q_r);
+h_r=10+2.205e5*Q_r^2;
+printf("\n The total head developed = %.2f m",h_r);
+P=Q_r*1000*h_r*9.81/eta;
+printf("\n Power required = %.0f W = %.2f kW",P,P*1e-3);
\ No newline at end of file diff --git a/536/CH9/EX9.1/Example_9_1.sce b/536/CH9/EX9.1/Example_9_1.sce new file mode 100755 index 000000000..9524d342d --- /dev/null +++ b/536/CH9/EX9.1/Example_9_1.sce @@ -0,0 +1,29 @@ +clc;
+clear;
+
+printf("\n Example 9.1\n");
+
+M_dot1=20; //rate of mass to be cooled
+M_dot2=25; //rate of cooling water
+Cp=4.18e3; //Heat capacity
+T1=360; //Initial temp.
+T2=340; //Final temp.
+theta_1=300; //Temperature of cooing water entering
+U=2e3; //Overall heat transfer coefficient
+
+Q=M_dot1*Cp*(T1-T2); //Heat load
+printf("\n Heat load = %.0f kW",Q*1e-3);
+//The cooling water outlet temperature is given by
+x=poly([0],'x');
+theta_2=roots(Q-(M_dot2*Cp*(x-300)));
+printf("\n The cooling water outlet temperature is = %.0f K",theta_2);
+printf("\n (a) Counter flow")
+//In equation 9.9:
+theta_m1=((T1-theta_2)-(T2-theta_1))/(log((T1-theta_2)/(T2-theta_1)));
+A1=Q/(U*theta_m1)
+printf("\n The surface area required %.2f m^2",A1);
+printf("\n (b) Co-current flow")
+//In equation 9.9:
+theta_m2=((T1-theta_1)-(T2-theta_2))/(log((T1-theta_1)/(T2-theta_2)));
+A2=Q/(U*theta_m2)
+printf("\n The surface area required %.2f m^2",A2);
diff --git a/536/CH9/EX9.10/Example_9_10.sce b/536/CH9/EX9.10/Example_9_10.sce new file mode 100755 index 000000000..a88f68a00 --- /dev/null +++ b/536/CH9/EX9.10/Example_9_10.sce @@ -0,0 +1,30 @@ +clc;
+clear;
+
+printf("\n Example 9.10\n");
+
+G=15;//Mass flow rate of benzene
+d_s=1; //Internal diameter of Heat Exchanger
+l=5; //Length of tubes
+od=19e-3; //Outer diameter of tubes
+C=6e-3; //Clearance
+l_b=0.25; //Baffle spacing
+Meu=.5e-3;
+Y=25e-3; //dimension of square pitch
+N=19; //no. of Baffles
+
+As=d_s*l_b*C/Y; //Cross-flow area
+printf("\n Cross-flow area = %.2f m^2",As);
+G_dash_s=G/As; //Mass flow
+printf("\n Mass flow = %d kg/m^2 s",G_dash_s);
+d_e=4*(Y^2-(%pi*od^2/4))/(%pi*od);//Equivalent Diameter
+printf("\n Equivalent Diameter = %.4f m",d_e);
+Re=G_dash_s*d_e/Meu;
+//From Figure 9.29:
+f_dash=.280;
+rho_b=881;//density of benzene
+DPf=f_dash*G_dash_s^2*(N+1)*d_s/(2*rho_b*d_e);
+printf("\n The pressure drop over the tube bundle = %.0f N/m^2",DPf);
+printf("\n\t\t\t\t\t= %.0f m of Benzene",DPf/(rho_b*9.81));
+
+
diff --git a/536/CH9/EX9.11/Example_9_11.sce b/536/CH9/EX9.11/Example_9_11.sce new file mode 100755 index 000000000..1efc802f0 --- /dev/null +++ b/536/CH9/EX9.11/Example_9_11.sce @@ -0,0 +1,21 @@ +clc;
+clear;
+
+printf("\n Example 9.11\n");
+
+d=0.15; //Diameter of pipe
+Ts=400; //Surface temperature
+Ta=294; //Air temperture
+//Over a wide range of temperature, k^4*(beta*g*rho^2*Cp/(Meu*k)) = 36.0
+//For air at a mean temperature i.e. 347 K
+k=0.0310; //Thermal conductivity ---Table 6, Appendix A1
+//X=beta*g*rho^2*Cp/(Meu*k)
+X=36/k^4;
+//From Equation 9.102:
+GrPr=X*(Ts-Ta)*d^3;
+//From Table 9.5:
+n=0.25;
+C_dd=1.32;
+//Thus, in Equation 9.104:
+h=C_dd*(Ts-Ta)^n*d^(3*n-1);
+printf("\n The heat transfer coefficient = %.2f W/m^2 K",h);
diff --git a/536/CH9/EX9.12/Example_9_12.sce b/536/CH9/EX9.12/Example_9_12.sce new file mode 100755 index 000000000..f0c5d590e --- /dev/null +++ b/536/CH9/EX9.12/Example_9_12.sce @@ -0,0 +1,18 @@ +clc;
+clear;
+
+printf("\n Example 9.12\n");
+
+lambda=1e-6;//Wavelength
+E_l_b=1e9; //Emissive power at given lambda
+
+//From equation 9.108
+C2=1.439e-2;
+C1=3.742e-16;
+T=C2/lambda/log(C1/(E_l_b*lambda^5));
+printf("\n The temperature of surface = %d K",T);
+//With an error of +2 per cent, the correct value is given by:
+E_l_b_n=(100-2)*E_l_b/100;
+//In equation 9.108:
+T_n=C2/lambda/log(C1/(E_l_b_n*lambda^5));
+printf("\n The temperature of surface with +2 per cent error= %.0f K",T_n);
\ No newline at end of file diff --git a/536/CH9/EX9.13/Example_9_13.sce b/536/CH9/EX9.13/Example_9_13.sce new file mode 100755 index 000000000..cc1798cb8 --- /dev/null +++ b/536/CH9/EX9.13/Example_9_13.sce @@ -0,0 +1,20 @@ +clc;
+clear;
+
+printf("\n Example 9.13\n");
+
+d=10e-3; //Diameter of carbide elements
+l=0.5; //Length of carbide elements
+Ts=1750; //Maximun surface temperature of carbide
+P=500e3; //Thermal power output required
+sigma=5.67e-8;
+
+//From equation 9.112, the total emissive power is given by:
+Eb=sigma*Ts^4;
+printf("\n The total emissive power is = %.2f *10^5 W/m^2",Eb*1e-5);
+A=%pi*d*l;
+printf("\n The area of one element = %.3f *10^-2 m^2",A*1e2);
+P1=Eb*A;//Power dissipated by one element
+printf("\n Power dissipated by one element = %.3f *10^3 W",P1*1e-3);
+n=P/P1; //Number of elements required
+printf("\n Number of elements required = %.0f ",n);
diff --git a/536/CH9/EX9.14/Example_9_14.sce b/536/CH9/EX9.14/Example_9_14.sce new file mode 100755 index 000000000..2a4570135 --- /dev/null +++ b/536/CH9/EX9.14/Example_9_14.sce @@ -0,0 +1,18 @@ +clc;
+clear;
+
+printf("\n Example 9.14\n");
+
+A=10; //Area of the surface
+P_r=1000e3; //Power radiated
+T1=1500; //First Temperature
+T2=1600; //Second Temperatue
+sigma=5.67e-8;
+
+E=P_r/A; //The emissive Power
+printf("\n The emissive Power when T=1500 K = %d kW",E*1e-3);
+//From equation 9,118:
+e=E/(sigma*T1^4);
+printf("\n Emissivity when T=1500 K = %.3f",e);
+E2=e*sigma*T2^4;
+printf("\n The Emissive power when T=1600 K = %.1f kW",E2*1e-3);
\ No newline at end of file diff --git a/536/CH9/EX9.15/Example_9_15.sce b/536/CH9/EX9.15/Example_9_15.sce new file mode 100755 index 000000000..224724116 --- /dev/null +++ b/536/CH9/EX9.15/Example_9_15.sce @@ -0,0 +1,16 @@ +clc;
+clear;
+
+printf("\n Example 9.15\n");
+
+A1=2; //Area of rectangle(Surface 1)
+A2=%pi*1^2/4; //Area of disc (Surface 2)
+T1=1500; //Temperature of Surface 1
+T2=750; //Temperature of Surface 2
+F12=0.25; //View factor
+sigma=5.67e-8;
+//From equation 9. 1 26:
+F21=A1*F12/A2;
+printf("\n View factor, F12 = %.3f",F21);
+Q12=sigma*A1*F12*(T1^4-T2^4);
+printf("\n The net radiation transfer = %.0f kW",Q12*1e-3);
\ No newline at end of file diff --git a/536/CH9/EX9.16/Example_9_16.sce b/536/CH9/EX9.16/Example_9_16.sce new file mode 100755 index 000000000..8dbc04a35 --- /dev/null +++ b/536/CH9/EX9.16/Example_9_16.sce @@ -0,0 +1,28 @@ +clc;
+clear;
+
+printf("\n Example 9.16\n");
+
+printf("\n (a)")
+//Using the nomenclature in Figure 9.40 iii;
+X=4;//width of horizontal plate and length vertical plate
+Y=6;//length of horizontal plate
+Z=3;//height of verical plate
+W=Y/X;
+H=Z/X;
+A1=Z*X;//Area of plate 1
+A2=X*Y;//Area of plate 2
+F12=0.12;
+printf("\n View Factor, F12= %.2f",F12);
+//From equation 9.126:
+F21=A1*F12/A2;
+printf("\n View Factor, F21= %.2f",F21);
+printf("\n (b)");
+//For the two spheres
+r1=1; //Diameter of sphere 1
+r2=2; //Diameter of sphere 2
+F12b=1;
+F21b=(r1/r2)^2;
+printf("\n View Factor, F21= %.2f",F21b);
+F22b=1-F21b;
+printf("\n View Factor, F22= %.2f",F22b);
\ No newline at end of file diff --git a/536/CH9/EX9.17/Example_9_17.sce b/536/CH9/EX9.17/Example_9_17.sce new file mode 100755 index 000000000..11efc9260 --- /dev/null +++ b/536/CH9/EX9.17/Example_9_17.sce @@ -0,0 +1,19 @@ +clc;
+clear;
+
+printf("\n Example 9.17\n");
+
+ri_u=0.2;// Inner radius of the upper ring
+ro_u=0.3;// Outer radius of the upper ring
+ri_l=0.3;// Inner radius of the lower ring
+ro_l=0.4;// Outer radius of the lower ring
+//F23 = ((A12/A2)*(F12_34))-F12_4-((A1/A2)*(F1_34 - F14))
+//Laying out the data in tabular form and obtaining F from Figure 9.40 ii, y,then
+F12_34=0.4;
+F12_4=0.22;
+F1_34=0.55;
+F14=0.30;
+A12_A2=ro_l^2/(ro_l^2-ri_l^2);
+A1_A2=ro_u^2/(ro_l^2-ri_l^2);
+F23=((A12_A2)*(F12_34-F12_4))+((A1_A2)*(F1_34-F14));
+printf("\n F23 = %.2f",F23);
\ No newline at end of file diff --git a/536/CH9/EX9.18/Example_9_18.sce b/536/CH9/EX9.18/Example_9_18.sce new file mode 100755 index 000000000..fd205f4ef --- /dev/null +++ b/536/CH9/EX9.18/Example_9_18.sce @@ -0,0 +1,37 @@ +clc;
+clear;
+
+printf("\n Example 9.18\n");
+
+d=1; //Diameter of plate
+r1=0.5;
+r4=r1; //Radius of the imaginary disc sealing the hemisphere
+L=r1; //The distance between the plate and the bottom of the dome
+
+A1=%pi*d^2/4; //Area of the plate
+A2=2*%pi*d^2/4; //Area of the underside of the Hemisphere
+A4=%pi*r4^2/4;//Area of an imaginary disc sealing the hemisphere and parallel
+ //to the plate
+T1=750;//Temperature of the plate
+T2=1200;//Temperature of hemispherical cone
+T3=290;//Temperature of the surroundings
+sigma=5.67e-8;
+//from equation 9.134, the net radiation to the surface of the plate 1 is
+//given by:
+//Q1=sigma*A2*F21*(T2^4-T1^4)+sigma*A3*F31(T3^4-T1^4)
+//using the reciprocity rule:
+//Q1=sigma*A1*F12*(T2^4-T1^4)+sigma*A3*F31(T3^4-T1^4)
+//All radiation from the disc 1 to the dome 2 is intercepted by the imaginary
+// disc 4 and hence F\2 = F\4, which may be obtained from Figure 9.39ii, with
+//i and j representing areas 1 and 4 respectively
+R1=r1/L;
+R4=r4/L;
+S=1+(1+R4^2)/(R1^2);
+F14=0.5*(S-(S^2-4*(r4/r1)^2)^0.5);
+F12=F14;
+//The summation rule states that
+//F11+F12+F13=1
+//F11=0
+F13=1-F12;
+Q1=sigma*A1*F12*(T2^4-T1^4)+sigma*A1*F13*(T3^4-T1^4);
+printf("\n the net rate of heat transfer by radiation to the plate = %.1f kW",Q1*1e-3);
\ No newline at end of file diff --git a/536/CH9/EX9.19/Example_9_19.sce b/536/CH9/EX9.19/Example_9_19.sce new file mode 100755 index 000000000..3534352b8 --- /dev/null +++ b/536/CH9/EX9.19/Example_9_19.sce @@ -0,0 +1,36 @@ +clc;
+clear;
+
+printf("\n Example 9.19\n");
+d=2; //Diameter of the cylinder
+h=1; //Depth of insulated cylinder
+A1=%pi*d^2/4; //Radiant heater surface
+A2=A1; //Under-Surface of the vessel
+A_R=%pi*d*h;
+T1=1500;
+T2=373;
+//From Figure 9.40ii, with i = 1, j = 2
+r1=1;
+r2=1;
+L=1;
+//The view factor may also be obtained from Figure 9.39ii as follows:
+//Using the nomenclature of Figure 9.39
+R1=r1/L;
+R2=r2/L;
+S=1+(1+R2^2)/(R1^2);
+F12=0.5*(S-(S^2-4*(r2/r1)^2)^0.5);
+sigma=5.67e-8;
+//Using the summation rule
+//F11=0
+F1R=1-F12;;
+F2R=F1R;
+Q2=(A1*F12+((1/(A1*F1R)+(1/(A2*F2R))))^-1)*sigma*(T1^4-T2^4);
+printf("\n The rate of radiant heat transfer to the vessel = %d kW",Q2*1e-3);
+//If the surroundings without insulation are surface 3 at
+T3=290;
+F23=F2R;
+//from equation 9.135
+Q2_d=sigma*A1*F12*(T1^4-T2^4)+sigma*A2*F23*(T3^4-T2^4);
+printf("\n The rate of radiant heat transfer to the vessel\n if the insulation were removed = %.0f kW",Q2_d*1e-3);
+red=(Q2-Q2_d)/Q2*100; //Percentage Reduction
+printf("\n\n Reduction percentage = %.0f per cent",red);
diff --git a/536/CH9/EX9.2/Example_9_2.sce b/536/CH9/EX9.2/Example_9_2.sce new file mode 100755 index 000000000..1799e8366 --- /dev/null +++ b/536/CH9/EX9.2/Example_9_2.sce @@ -0,0 +1,12 @@ +clc
+clear;
+
+printf("\n Example 9.2\n");
+dx=0.5; //Thickness of wall
+T1=400; //Temperartue of inner surface
+T2=300; //Temperature of outer surface
+K=0.7; //Thermal conductivity
+A=1; //Area of heat transfer
+//From equation 9.12:
+Q=K*A*(T1-T2)/dx;
+printf("\n The heat loss per square metre of surface = %.0f w/m^2",Q);
\ No newline at end of file diff --git a/536/CH9/EX9.20/Example_9_20.sce b/536/CH9/EX9.20/Example_9_20.sce new file mode 100755 index 000000000..a8d3ee8fd --- /dev/null +++ b/536/CH9/EX9.20/Example_9_20.sce @@ -0,0 +1,26 @@ +clc;
+clear;
+
+printf("\n Example 9.20\n");
+
+e=0.75; //Emissivity of grey surface
+r=1-e; //reflectivity of surface
+Ts=400; //Temperature of surface
+T_amb=295;
+sigma=5.67e-8;
+q1=3e3; //Rate of radiation arriving at grey surface
+//From equation 9.118
+Eb=sigma*Ts^4;
+printf("\n Emissive Power = %.0f W/m^2",Eb);
+//From equation 9.138
+qo=e*Eb+r*q1;
+printf("\n Radiosity = %.0f W/m^2",qo);
+//From equation 9.140
+Q_A=e/r*(Eb-qo);
+q=Q_A;
+printf("\n The net rate of radiation trasfer = %.0f W/m^2",q);
+printf("\n where the negative value indicates heat transfer to the surface.");
+//For convective heat transfer from the surface
+qc=-1*q;
+hc=qc/(Ts-T_amb);
+printf("\n Coefficient of heat transfer = %.1f W/m^2 K",hc);
\ No newline at end of file diff --git a/536/CH9/EX9.21/Example_9_21.sce b/536/CH9/EX9.21/Example_9_21.sce new file mode 100755 index 000000000..c5db3fee0 --- /dev/null +++ b/536/CH9/EX9.21/Example_9_21.sce @@ -0,0 +1,63 @@ +clc;
+clear;
+
+printf("\n Example 9.21\n");
+
+sigma=5.67e-8;
+T=[1000 500 300];//tempertaure of surfaces
+//Taking surface 1 as the heater, surface 2 as the heated plate and surface 3
+//as an imaginary enclosure
+A=[1.07 1.07 0.628];//Array of area of surfaces
+e=[0.75 0.50 1.0];//Array of emissivity of the surfaces
+r=[0.250 0.50];// Array of radius of two surfaces
+//X is ratio of area to radius(A/r)
+//Y = A*e/r
+L=0.2; //distance between two discs
+for i=1:2
+ X(i)=A(i)/r(i);
+ Y(i)=A(i)*e(i)/r(i);
+ R(i)=r(i)/L;
+end
+
+F11=0;
+F22=0;
+S=1+(1+R(2)^2)/(R(2)^2);
+F12=0.5*(S-(S^2-4*(r(2)/(2*r(1)))^2)^0.5);
+A1_F11=0;
+A2_F22=0;
+A1_F12=A(1)*F12;
+A1_F13=A(1)-(A(1)*F11+A(2)*F12);
+//for surface 2:
+A2_F21=A1_F12;
+A2_F23=A1_F13;
+//for surface 3:
+//By reciprocity rule
+A3_F31=A1_F13;
+A3_F32=A2_F23;
+A3_F33=A(3)-(A3_F31+A3_F32);
+
+//From equation 9.112:
+for i=1:3
+ E_b(i)=sigma*T(i)^4/1000;
+end
+
+//Since surface 3 is a black body
+q_o3=E_b(3);
+//From equations 9.157 and 9.158:
+//we get
+
+function [f]=F(x)
+ f(1)=(A1_F11-A(1)/r(1))*x(1)+A2_F21*x(2)+A3_F31*q_o3+E_b(1)*A(1)*e(1)/r(1);
+ f(2)=(A1_F12*x(1))+((A2_F22-A(2)/r(2))*x(2))+E_b(2)*A(2)*e(2)/r(2);
+ funcprot(0);
+endfunction
+
+x=[0 0];
+q_o=fsolve(x,F);
+
+//From equation 9.140:
+Q1=(A(1)*e(1)/r(1))*(E_b(1)-q_o(1));
+Q2=(A(2)*e(2)/r(2))*(E_b(2)-q_o(2));
+printf("\n Power input to the heater = %.1f kW",Q1);
+printf("\n The rate of heat transfer to the plate = %.2f kW",Q2);
+printf("\n where the negative sign indicates heat transfer to the plate")
\ No newline at end of file diff --git a/536/CH9/EX9.22/Example_9_22.sce b/536/CH9/EX9.22/Example_9_22.sce new file mode 100755 index 000000000..696c89bc2 --- /dev/null +++ b/536/CH9/EX9.22/Example_9_22.sce @@ -0,0 +1,63 @@ +clc;
+clear;
+
+printf("\n Example 9.22\n");
+
+d=0.5;//diameter of chamber
+l=2;//Length of chamber
+e=0.5;//Emissivity
+T_s=750;//Temperature at which the chamber is maintained
+P=150e3;
+T_g=1250;
+sigma=5.67e-8;
+
+//The partial pressures of carbon dioxide (P_c) and of water (P_w) are:
+P_c=0.1*P;
+P_w=P_c;
+//From Table 9.7:
+V=%pi/4*d^2*l;//Volume of the chamber
+A_s=(2*%pi/4*d^2)+(%pi*d*l);//total surface are of chamber
+
+L_e=3.6*(V/A_s);
+//FOR WATER VAPOUR
+//and from Figure 9.44, e_w = 0.075
+//Since P_w*L_e = 0.06 bar m, then from Figure 9.44:
+C_w=1.4;
+e_w1=C_w*0.075;
+//FOR CARBON DIOXIDE
+//Since P = 1.5 bar, Pc = 0.15 bar and P_c*L_e = 0.06 bar m, then, from
+//Figure 9.38:
+//and from Figure 9.45, e_c = 0.037
+C_c1=1.2;
+e_c1=(C_c1*0.037);
+A=(P_w+P_c)*L_e;
+B=P_c/(P_c+P_w)
+//Thus, from Figure 9.45 for T_g > 1203 K, De = 0.001
+//and, from equation 9.160:
+De=0.001;
+e_g=e_w1+0.044-De;
+
+//FOR WATER VAPOUR
+//Since 0.5*(P_w+P)=0.825 bar and P_w*L_e*(Ts/Tg)=P_c*L_e(Ts/Tg)=0.036 bar m,
+//then, from Figure 9.44:C_w=1.4
+e_w2=(0.12*C_w);
+//and the absorptivity, from equation 9.163 is:
+a_w=e_w2*(T_g/T_s)^0.65;
+//FOR CARBON DIOXIDE
+//From Figure 9.45 at 750 K, e_c=0.08
+//From Figure 9.45 at P=1.5 bar and P_c*L_e*(Ts/Tg)= 0.036 bar m:
+C_c2=1.02;
+e_c2=(0.08*C_c2);
+//and the absorptivity, from equation 9.164 is:
+a_c=e_c2*(T_g/T_s)^0.65;
+//P_w/(P_c+Pw)=0.5 and (P_c+P_w)*L_e*(T_s/T_g)=(0.036+0.036)=0.072 bar m
+//Thus, from Figure 9.46, for Tg=813 K, De=Da <0.01 and this may be neglected
+a_g=a_w+a_c;
+//If the surrounding surface is black, then:
+Q=sigma*A_s*(e_g*T_g^4-a_g*T_s^4);
+printf("\n Radiation to the walls if the surface is black = %.1f kW",Q*1e-3);
+//For grey walls, the correction factor allowing for multiple reflection of
+//incident radiation is:
+C_g=0.5/(1-(1-0.326)*(1-0.5));
+Q_w=(Q*C_g);
+printf("\n Net radiation to the walls = %.1f kW",Q_w*1e-3);
\ No newline at end of file diff --git a/536/CH9/EX9.23/Example_9_23.sci b/536/CH9/EX9.23/Example_9_23.sci new file mode 100755 index 000000000..7a5d36078 --- /dev/null +++ b/536/CH9/EX9.23/Example_9_23.sci @@ -0,0 +1,105 @@ +clear;
+clc;
+printf("\t Example 9.23\n");
+function[n]=mole(w,m)
+ n = w/m;
+ funcprot(0);
+endfunction
+
+function[p]=partial(n1)
+ p = 308*(n1/total);
+ funcprot(0);
+endfunction
+
+w_steam = 0.57; //mass flow rate of steam entering in [kg/sec]
+w_CO2 = 0.20; //mass flow rate of CO2 entering in [kg/sec]
+m_water = 18; //molecular mass of water in kg
+m_CO2 = 44; //molecular mass of CO2 in kg
+n_steam = mole(w_steam,m_water); //number of moles in kmol
+n_CO2 = mole(w_CO2,m_CO2); //number of moles in kmol
+printf("\n At the entrance there is %.3f kmol steam",n_steam);
+printf("\n At the entrance there is %.4f kmol water",n_CO2);
+total = n_steam + n_CO2;
+printf("\n Total number of moles fed to the condenser per second = %.4f kmol",total);
+p_steam = partial(n_steam);
+p_CO2 = partial(n_CO2);
+printf("\n At the inlet partial pressure is %d kN/m^2 watre",p_steam);
+printf("\n At the inlet there is %d kN/m^2 CO2",p_CO2);
+printf("\n From the Table 11A in the appendix Dew point = %d K",404);
+mean_mol = (0.57 + 0.20)/total; //mean molecular weight of the mixture in kg/kmol
+
+outlet_steam = 11.7; //partial pressure of water in kN/m^2
+outlet_CO2 = 308 - outlet_steam; //partial pressure of water in kN/m^2
+printf("\n At the outlet partial pressure is %.1f kN/m^2 water",outlet_steam);
+printf("\n At the outlet there is %.1f kN/m^2 CO2",outlet_CO2);
+n_s = n_CO2*outlet_steam/outlet_CO2;
+steam_condensed = n_steam - n_s;
+printf("\n steam codensed = %.5f kmol",steam_condensed);
+
+printf("\n\n For the interval 404 to 401 K");
+p_steam_401K = 252.2; //[kN/m^2]
+p_CO2_401K = 308 - 252.2;//[kN/m^2]
+steam_remaining = 0.0045*p_steam_401K/p_CO2_401K;
+s_c = n_steam - steam_remaining; //[kmol]
+Heat_cond = s_c*18*(2180 + 1.93*(404-401)); //[kW]
+Heat_uncondensed_steam = 0.0203*18*1.93*(404-401); //[kW]
+Heat_CO2 = 0.020*0.92*(404-401);
+total_heat = Heat_cond + Heat_uncondensed_steam + Heat_CO2;
+printf("\n Heat of condensation = %d kW",Heat_cond);
+printf("\n Heat of uncondensed steam = %.1f kW",Heat_uncondensed_steam);
+printf("\n Heat from CO2 = %.1f kW",Heat_CO2);
+printf("\n Total = %.1f kW",total_heat);
+
+printf("\n\n For other intervals simiarly");
+printf("\n Interval(K) Heat Load(kW)");
+printf("\n 404 - 401 %.1f ",total_heat);
+printf("\n 401 - 397 %.1f ",323.5);
+printf("\n 397 - 380 %.1f ",343.5);
+printf("\n 380 - 339 %.1f ",220.1);
+printf("\n 339 - 322 %.1f ",57.9);
+printf("\n total %.1f ",total_heat+323.5+343.5+220.1+57.9);
+flow_water = 1407.3/(4.187*(319-300)); //[kg/sec]
+hi = 6.36; //[kW/m^2 K]Based on flow velocity of 1425 kg/m^2 sec
+ho = 5.25; //[kW/m^2 K]Based on outside area
+Cp = (0.20*0.92 + 0.57*1.93)/0.77; //[kJ/kg K]
+printf("\n Mean specific heat, Cp = %.3f kJ/kg K",Cp);
+k_mean = 0.025; //[kW/m K]
+a = 0.0411; //[m^2]
+mass_velocity = (0.20+0.57)/0.0411; //[kg/m^2 sec]
+printf("\n Mass velocity = %.1f kg/m^2 sec",mass_velocity);
+hg = 107; //[W/m^2 K] at Re = 29,800 at equivalent diameter = 0.024m
+u_pD = 0.62; //(u/pD)^0.67 = 0.62
+Cpu_k = 1.01; //(Cp*u/k)^0.67
+Psf = (122.6 - 38)/log(122.6/38);
+kG = hg*(Cpu_k)/(1000*Cp*Psf*u_pD);
+printf("\n\n kG = %.4f",kG);
+
+printf("\n Point Ts Tc UT UTow Q A = Q/(UT) T Tow (Q/T)ow");
+printf("\n 1 404 378 309 - - - 84.4 - - ");
+printf("\n 2 401 356 228 268.5 468.4 1.75 88.1 86.3 5.42");
+printf("\n 3 397 336 145 186.5 323.5 1.74 88.6 8.4 3.66");
+printf("\n 4 380 312 40.6 88.1 343.5 3.89 76.7 82.7 4.15");
+printf("\n 5 339 302 5.4 17.5 220.1 12.58 38.1 55.2 4.00");
+printf("\n 6 322 300 2.1 3.5 51.9 14.83 22.2 29.6 1.75");
+
+printf("\n Assuming no scale resistance, the overall coefficient = %.3f W/m^ K",1407.3/(34.8*74.2));
+printf("\n The available surface area on the outside of the tubes = 0.060 m^2 or %.1f m^2",246*3.65*0.060);
+printf("\n Actual coeficient = %.3f kW/m^2 K",1407.3/(53.9*74.2));
+printf("\n Dirt factor = %.2f m^2 K/kW",(0.545-0.352)/(0.545*0.352));
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
diff --git a/536/CH9/EX9.24/Example_9_24.sce b/536/CH9/EX9.24/Example_9_24.sce new file mode 100755 index 000000000..e443e2ea8 --- /dev/null +++ b/536/CH9/EX9.24/Example_9_24.sce @@ -0,0 +1,37 @@ +clc;
+clear;
+
+printf("\n Example 9.24");
+
+d_v=1;//diameter of the vessel
+L=0.3;//diameter of propeller agitator
+N=2.5;//rotating speed of propeller agitator
+T=310;//Temperature
+G=0.5;//circulation speed of cooling water
+d_o=25e-3;//outer diameter of stainless steel coil
+d=22e-3;//inner diameter of stainless steel coil
+d_w=(d_o+d)/2;
+d_c=0.8;//diameter of helix
+T_m=290;//mean temperature
+k1=0.59;
+Meu1=1.08e-3;
+C_p1=4.18e3;
+x_w=1.5e-3;
+
+//From equations 9.202 and 9.203, the inside film coefficient for the water
+//is given by:
+h_i=(k1/d)*(1+3.5*(d/d_c))*0.023*(d*1315/Meu1)^0.8*(C_p1*Meu1/k1)^0.4;
+//The external film coefficient is given by equation 9.204:
+C_p2=1.88e3;//Specefic heat capacity
+Meu2=6.5e-3;//viscosity
+k2=0.40;
+rho=1666;
+Meu_s=8.6e-3;
+h_o=0.87*(C_p2*Meu2/k2)^(1/3)*(L^2*N*rho/Meu2)^0.62*(Meu2/Meu_s)^0.14*k2/d_v;
+
+k_w=15.9;
+R_o=0.0004;
+R_i=0.0002;
+U_o=((1/h_o)+(x_w*d_o/(k_w*d_w))+(d_o/(h_i*d))+(R_o)+(R_i*d_o/d))^-1;
+printf("\n\n The overall coeffecient of heat transfer = %.0f W/m^2.K",U_o);
+
diff --git a/536/CH9/EX9.25/Example_9_25.sce b/536/CH9/EX9.25/Example_9_25.sce new file mode 100755 index 000000000..caaf845d2 --- /dev/null +++ b/536/CH9/EX9.25/Example_9_25.sce @@ -0,0 +1,22 @@ +clc;
+clear;
+
+printf("\n Example 9.25")
+
+C_p=4e3;
+
+//If T K is the temperature of the liquid at time / s, then a heat balance on
+//the vessel gives:
+x=poly([0],'x');
+T_max=roots((600*0.5)*(393-x)-(10*6)*(x-293));
+printf("\n\n Maximum temperature to which it can be heated = %.1f K",T_max)
+//solving the equation finally we get
+
+t1=integrate('11111*(1/(376.3-T))','T',293,353);
+printf("\n Time taken to heat the liquid from 293 K to 353 K = %.0f s",t1);
+//The steam is turned off for 7200 s and during this time a heat balance gives:
+//on solving as given in book we get
+T=346.9;
+//The time taken to reheat the liquid to 353 K is then given by:
+t2=integrate('11111*(1/(376.3-T))','T',346.9,353);
+printf("\n Time taken to reheat the liquid to 353 K = %.0f s",t2);
\ No newline at end of file diff --git a/536/CH9/EX9.26/Example_9_26.sce b/536/CH9/EX9.26/Example_9_26.sce new file mode 100755 index 000000000..ad8aa046c --- /dev/null +++ b/536/CH9/EX9.26/Example_9_26.sce @@ -0,0 +1,20 @@ +clc;
+clear;
+
+printf("\n Example 9.26\n");
+
+//As in Example 9.1, the heat load = 1672 kW
+Q=1672;
+//With reference to Figure 9.71:
+T1=360;
+T2=340;
+theta1=300;//Temperature of cooling water entering
+theta2=316;
+X=(theta2-theta1)/(T1-theta1);
+Y=(T1-T2)/(theta2-theta1);
+//from Figure 9.58
+F=0.97;
+theta_m=41.9;
+//and hence:
+A=Q/(2*F*theta_m);//the heat transfer area
+printf("\n The heat transfer area is = %.1f m^2",A);
\ No newline at end of file diff --git a/536/CH9/EX9.27/Example_9_27.sce b/536/CH9/EX9.27/Example_9_27.sce new file mode 100755 index 000000000..e6b8fc5c4 --- /dev/null +++ b/536/CH9/EX9.27/Example_9_27.sce @@ -0,0 +1,31 @@ +clc;
+clear;
+
+printf("\n Example 9.27\n");
+
+//As in Example 9.1, the heat load = 1672 kW
+Q=1672;
+//With reference to Figure 9.71:
+T1=360;
+T2=340;
+theta1=300;//Temperature of cooling water entering
+theta2=316;
+F_theta_m=40.6;//corrected mean temperature difference
+T=(T1+T2)/2;
+d=1.9e-3;//Tube diameter
+u=1;//Water velocity
+//then, in equation 9,221:
+h_i=4.28*(0.00488*T-1)*u^0.8/d^0.2;
+//From Table 9.18, an estimate of the shell-side film coefficient is:
+h_o=(1700+11000)/2000;
+//For steel tubes of a wall thickness of 1.6 mm, the thermal resistance of the wall, from Table 9.15 is:
+xw_kw=0.025;
+//the thermal resistance for treated water, from Table 9.16, is 0.26 m2K/kW
+Ri=0.26;
+Ro=Ri;
+U=((1/h_o)+xw_kw+Ri+Ro+(1/h_i))^-1;
+A=Q/(F_theta_m*U);
+printf("\n The heat transfer area = %.1f m^2",A);
+
+
+
diff --git a/536/CH9/EX9.28/Example_9_28.sce b/536/CH9/EX9.28/Example_9_28.sce new file mode 100755 index 000000000..059799cb9 --- /dev/null +++ b/536/CH9/EX9.28/Example_9_28.sce @@ -0,0 +1,109 @@ +clear;
+clc;
+
+printf('Example 9.28');
+//Using Kern's method design Shell n tube heat exchanger
+
+mh = 30; //[kg/s] Hot fluid flow rate
+Thi = 370; //[K] Hot Fluid Inlet Temperature
+Tho = 315; //[K] Hot Fluid outlet Temperature
+Tci = 300; //[K] Cold Fluid Inlet Temperature
+Tco = 315; //[K] Cold Fluid Outlet Temperature
+cpc = 4.18*10^3; //[J/kg.K] Thermal Conductivity of Cold Fluid
+//From table A1.3 at mean temperature 343 K
+cph = 2.9*10^3; //[J/kg.K] Thermal Capacity of Hot fluid
+
+q = mh*cph*(Thi-Tho); //[kW] Heat load
+mc = q/(cpc*(Tco-Tci)); //[kg/s] Flow of cooling water
+
+Tln = ((Thi-Tho)-(Tco-Tci))/(log((Thi-Tho)/(Tco-Tci))); //[K] Logarithm mean temperature difference
+
+//for one-shell side pass and two-tube side pass Equation 9.213
+X = (Thi - Tho)/(Tco-Tci);
+Y = (Tco-Tci)/(Thi - Tci);
+
+//From Figure 9.757
+F = .85;
+//From Table 9.17
+U = 500 //[W/m^2.K]
+A = q/(F*Tln * U);
+//Thus COnvenient tubes to bo used
+od = .02 //[m] outer dia
+id = .016 //[m] inner dia
+l = 4.83 //[m] effective tube length
+
+s = %pi*od*l;
+N = A/s;
+//From equation 9.211
+db = (1210/.249)^(2.207)^-1*20/1000; //[m]
+//From figure 9.71
+dc = .068 //[m] diametric clearance between shell and tubes
+ds = db+dc //[m] Shell dia
+
+//Tube-Side Coefficient
+//From equation 9.218
+Ac = %pi/4*id^2 //[m^2] Cross sectional area
+Ntp = N/2;
+
+Af = N/2*Ac //[m^2] Tube side flow area
+mw = 76.3/Af //[kg/m^2.s] Mass velocity of water
+rho = 995 //[kg/m^3] mas density of water
+u = mw/rho //[m/s] water velocity
+
+//At mean temperature 308 K
+vu = .8*10^-3 //[N.s/m^2] viscosity
+k = .59 //[W/m.K]
+
+Re = id*u*rho/vu;
+Pr = cpc*vu/k;
+ld = l/id;
+//from figure 9.77
+jh = 3.7*10^-3
+hi = jh*Re*Pr^.3334*.59/id; //[W/m^2.K]
+
+//Shell-Side Coefficient
+//Baffle packing will be taken as 20percent of shell dia
+dbf = .20*ds; //[m] Baffle Dia
+tb = 1.25*20*10^-3 //[mm] Tube Pitch
+//From equation 9.226
+As = (25-20)/25*10^3*(ds*Ac) //[m^2]
+Gs = 30/As; //[kg/m^2.s]
+//From Equation 9.228
+de = 1.1*(.025^2-.917*od^2)/od; //[m]
+//At mean temperature 343 K Butyl Alcohol
+rho2 = 780 //[kg/m^3] density
+vu2 = .8*10^-3 //[N.s/m^2] viscosity
+Cp2 = 3.1*10^3 //[J/kg.K] Heat capacity
+k2 = .16 //[W/m.K]
+//Equation 9.229
+Re2 = Gs*de/vu2;
+Pr2 = Cp2*vu2/k2;
+//From figure 9.81
+jh2 = 5*10^-3;
+//Equation 9.230
+hs = jh2*Re2*Pr2^.334*k2/de;
+
+//Overall Coefficient
+//from Table 9.16
+k3 = 50 //[W/m.K] Thermal Conductivity
+Rw = .00020 //[m^2.K/W] Scale Resistances
+Ro = .00015 //[m^2.K/W] Resistance for organic
+
+U = [1/hs + Rw + .5*(od-id)/k3 + Ro*od/id+od/(id*hi)]^-1
+
+//From figure 9.78
+jf = 4.5*10^-3;
+n = 2;
+delP = n*[4*jf*(4.830/id) + 1.25]*(rho*u^2);
+u2 = Gs/rho2;
+jf2 = 4.6*10^-2;
+N2 = 1;
+delP2 = N2*[4*jf2*(4.830/od)*(1005/14.2)]*(rho2*u2^2);
+
+//Increasing the baffle spacing pressure drop is reduced one-fourth
+delPs = delP2/4;
+
+ho = hs*(.5)^.8; //[W/m^2.K]
+U2 = [1/ho + Rw + .5*(od-id)/k3 + Ro*od/id+od/(id*hi)]^-1
+printf('\n\n Overall Coefficient of %i W/m^2.K \n Number of tubes/pass = %i\n Number of tubes required = %i ',U2,Ntp,N)
+//END
\ No newline at end of file diff --git a/536/CH9/EX9.29/Example_9_29.sce b/536/CH9/EX9.29/Example_9_29.sce new file mode 100755 index 000000000..41458f3e0 --- /dev/null +++ b/536/CH9/EX9.29/Example_9_29.sce @@ -0,0 +1,20 @@ +clc;
+clear;
+
+printf("\n Example 9.29\n");
+
+G=1; //Flow rate of organic liquid
+Cp=2e3//Heat capacity of organic liquid
+T1=350;
+T2=330;
+theta1=290;
+theta2=320;
+
+Q=G*Cp*(T2-T1);//heat load
+G_cool=Q/(4187*(theta1-theta2));//flow of water
+GCp_hot=(G*Cp);//for organic
+GCp_cold=(G_cool*4187);
+
+//From equation 9.235:
+eta=GCp_hot*(T1-T2)/(GCp_cold*(T1-theta1));
+printf("\n Effectiveness of the given double pipe heat exchanger = %.2f",eta);
\ No newline at end of file diff --git a/536/CH9/EX9.3/Example_9_3.sce b/536/CH9/EX9.3/Example_9_3.sce new file mode 100755 index 000000000..eb23dbf00 --- /dev/null +++ b/536/CH9/EX9.3/Example_9_3.sce @@ -0,0 +1,24 @@ +clc;
+clear;
+
+printf("\n Example 9.3\n");
+
+dx1=0.20; //thickness of firebrick
+dx2=0.10; //thickness of insulating brick
+dx3=0.20; //thickness of building brick
+k1=1.4; //Thermal conductivity of firebrick
+k2=0.21; //Thermal conductivity of insulating brick
+k3=0.7; //Thermal conductivity of building brick
+T1=1200; //Temperature at junction 1
+T4=330; //Temperature at junction 4
+
+//From equation 9.19:
+Q=(T1-T4)/((dx1/k1)+(dx2/k2)+(dx3/k3));
+printf("\n Heat loss per unit area = %d W/m^2",Q);
+//The ratio (Temperature drop over firebrick)/(Total temperature drop)
+R=(dx1/k1)/((dx1/k1)+(dx2/k2)+(dx3/k3));
+//Temperature drop over firebrick
+dT=(T1-T4)*R;
+printf("\n Temperature drop over firebrick = %.0f K",dT);
+T2=(T1-dT);
+printf("\n The temperature at the firebrick-insulating brick interface = %.0f K",T2);
\ No newline at end of file diff --git a/536/CH9/EX9.30/Example_9_30.sce b/536/CH9/EX9.30/Example_9_30.sce new file mode 100755 index 000000000..5a79575cd --- /dev/null +++ b/536/CH9/EX9.30/Example_9_30.sce @@ -0,0 +1,32 @@ +clear;
+clc;
+
+printf('Example 9.30');
+
+Tci = 320 //[K] Cold Fluid Initial Temperature
+Tce = 340 //[K] Cold Fluid Final Temperature
+mc = 4 //[kg/s] Flow rate of cold fluid
+mh = 8 //[kg/s] Flow rate of hot fluid
+Thi = [380 370 360 350] //[K] Hot fluid initial temperature
+Cp = 4.18 //[kJ/kg.K] mean heat capacity
+U = 1.5 //[W/m^2.K] Overall heat transfer coefficient
+
+GCpu= mh*Cp; //[kW/K]
+GCpp= mc*Cp; //[kW/K]
+if(GCpu<GCpp)
+ GCpmin = GCpu; //[kW/K]
+ ratio = GCpmin/GCpp;
+else
+ GCpmin = GCpp; //[kW/K]
+ ratio = GCpmin/GCpu;
+
+//Equation 9.235
+n = mc*Cp*(Tce-Tci)*(mc*Cp*(Thi - Tci))^-1;
+//From Figure 9.85b Number of transfer Units
+N = [.45 .6 .9 1.7]; //[NTU]
+A = N*GCpmin/U; //Area of required [m^2]
+
+format('v',4)
+printf('\n\n Thi(K) n N A (m^2)');
+disp([Thi(4) n(4) N(4) A(4)],[Thi(3) n(3) N(3) A(3)],[Thi(2) n(2) N(2) A(2)],[Thi(1) n(1) N(1) A(1)])
+//END
\ No newline at end of file diff --git a/536/CH9/EX9.31/Example_9_31.sce b/536/CH9/EX9.31/Example_9_31.sce new file mode 100755 index 000000000..352962e8d --- /dev/null +++ b/536/CH9/EX9.31/Example_9_31.sce @@ -0,0 +1,16 @@ +clc;
+clear;
+
+printf("\n Example 9.31\n");
+
+o_d=10e-3;//outer diameter of the tube
+i_d=8.2e-3; //inner diameter of the tube
+h=140;//coeffecient of heat transfer between gas and copper tube
+k=350;//Thermal conductivity of copper tube
+L=0.075;
+
+b=%pi*o_d;//perimeter of tube
+A=%pi/4*(o_d^2-i_d^2);//cross sectional area of the metal
+m=((h*b)/(k*A))^0.5;
+T_g=((475*cosh(m*L))-365)/(cosh(m*L)-1);
+printf("\n The gas temperature is = %.0f K",T_g)
\ No newline at end of file diff --git a/536/CH9/EX9.32/Example_9_32.sce b/536/CH9/EX9.32/Example_9_32.sce new file mode 100755 index 000000000..8f631f4db --- /dev/null +++ b/536/CH9/EX9.32/Example_9_32.sce @@ -0,0 +1,26 @@ +clc;
+clear;
+
+printf("\n Example 9.32\n");
+
+d2=54e-3; //outer diameter of the tube
+d1=70e-3; // fin diameter
+w=2e-3; //fin thickness
+n=230;// number of fins per metre run
+T_s=370; //Surface temperature
+T=280; //Temperature of surroundings
+h=30; //Heat transfer coeffecient between gas and fin
+k=43; //Thermal conductivity of steel
+L=(d1-d2)/2;
+
+theta1=T_s-T;
+//Assuming that the height of the fin is small compared with its circumference
+//and that it may be treated as a straight fin of length
+l=(%pi/2)*(d1+d2);
+b=2*l;//perimeter
+A=l*w;//the average area at right-angles to the heat flow
+m=((h*b)/(k*A))^0.5;
+//From equation 9.254, the heat flow is given for case (b) as:
+Qf=m*k*A*theta1*(%e^(2*m*L)-1)/(1+%e^(2*m*L));
+Q=Qf*n;//Heat loss per meter run of tube
+printf("\n The heat loss per metre run of tube = %.2f kW/m",Q*1e-3);
diff --git a/536/CH9/EX9.33/Example_9_33.sce b/536/CH9/EX9.33/Example_9_33.sce new file mode 100755 index 000000000..3bf476268 --- /dev/null +++ b/536/CH9/EX9.33/Example_9_33.sce @@ -0,0 +1,31 @@ +clc;
+clear;
+
+printf("\n Example 9.33\n");
+
+d=150e-3;//Internal diameter of tube
+d_o=168e-3;//outer diameter of tube
+d_w=159e-3;
+d_s=268e-3;
+d_m=(d_s-d_o)/log(d_s/d_o);//log mean of d_o and d_s
+h_i=8500;//The coefficient for condensing steam together with that for any scale
+k_w=45;
+k_l=0.073;
+x_l=50e-3;
+x_w=9e-3;
+DT=444-294;
+sigma=5.67e-8;
+//The temperature on the outside of the lagging is estimated at 314 K and (hr + hc) will be taken as 10 W/m2 K.
+//total thermal resisitance
+R=(h_i*%pi*d)^-1+(10*%pi*d_s)^-1+(k_w*%pi*d_w/x_w)^-1+(k_l*%pi*d_m/x_l)^-1;
+Q_l=DT/R; //The heat loss per metre of length(from eq 9.261)
+DT_lagging=((k_l*%pi*d_m/x_l)^-1/R)*DT;
+//Taking an emissivity of 0.9, from equation 9.119:
+
+h_r=(0.9*sigma*(310^4-294^4))/(310-294);
+C=1.32;
+//Substituting in equation 9.105 (putting l = diameter = 0.268 m):
+h_c=C*((310-294)/d_s)^0.25;
+//If the pipe were unlagged,(hc+hr)for DT=150 K would be about 20 W/m2 K and the heat loss would then be:
+Q_l=20*%pi*d_o*150;
+printf("\n The heat loss to the air = %.2f kW/m",Q_l*1e-3);
\ No newline at end of file diff --git a/536/CH9/EX9.34/Example_9_34.sce b/536/CH9/EX9.34/Example_9_34.sce new file mode 100755 index 000000000..10e8842ba --- /dev/null +++ b/536/CH9/EX9.34/Example_9_34.sce @@ -0,0 +1,23 @@ +clc;
+clear;
+printf('Example 9.34\n');
+
+
+T1=420;//temperature of steam
+k=0.1;//Thermal conductivity
+T2=285; //Ambient temperature
+h=10;//the coefficient of heat transfer from the outside of the lagging to
+ //the surroundings
+//determining Q/l from equation 9.21 and equating it to heat loss from the
+//outside of the lagging we get
+//(Q/l)=84.82/(log(d_o/0.1)+(0.02/d_o)) W/m
+//using various equations we finally get an equation in terms of d_o and we
+// will solve it by using fsolve
+function [f]=F(d_o)
+ f(1)=(1/(log(d_o/0.1)+(0.02/d_o))^2)-(2.35*(d_o^3)/(d_o-0.02));
+ funcprot(0);
+endfunction
+d_o=1;
+ans=fsolve(d_o,F);
+E_t=(ans-0.1)/2;
+printf("\n Economic thickness of lagging = %d mm",E_t*1e3);
\ No newline at end of file diff --git a/536/CH9/EX9.4/Example_9_4.sce b/536/CH9/EX9.4/Example_9_4.sce new file mode 100755 index 000000000..25b04c48f --- /dev/null +++ b/536/CH9/EX9.4/Example_9_4.sce @@ -0,0 +1,18 @@ +clc;
+clear;
+
+printf("\n Example 9.4\n");
+T=295; //initial temperature of surfaces
+T2f=375; //Final temperature of far surface
+dT1=900; //Temperature of near face raised
+//The temperature at any distance x from the near face at time t is given by equation 9.37
+//Choosing the temperature scale so that the initial temperature is everywhere zero, then:
+R=(T2f-T)/(2*(dT1-T)); //ratio of theta to twice of theta dash
+
+//An approximate solution is obtained by taking the first term only, to give:
+// R=erfc(346*t^-0.5)
+//erfc(1.30)=R
+//solving above equation
+x=poly([0],'x');
+t=roots((1.30^2*x)-346^2);
+printf("\n Time taken to rise from 295 to 375 K = %.1f h",t/3600);
\ No newline at end of file diff --git a/536/CH9/EX9.5/Example_9_5.sce b/536/CH9/EX9.5/Example_9_5.sce new file mode 100755 index 000000000..3f827b9cc --- /dev/null +++ b/536/CH9/EX9.5/Example_9_5.sce @@ -0,0 +1,23 @@ +clc;
+clear;
+
+printf("\n Example 9.5\n");
+
+T=295; //initial temperature of surfaces
+T2f=375; //Final temperature of far surface
+dT1=900; //Temperature of near face raised
+DH=4.2e-7; //Thermal diffusivity
+//The development of the temperature profile is shown in Figure 9.12
+//The problem will be solved by taking relatively large intervals for dx.
+//Choosing dx = 50 mm, the construction shown in Figure 9.12
+dx=50e-3;
+//Because the second face is perfectly insulated, the temperature gradient must
+// be zero at this point.
+//It is seen that the temperature is
+//less than 375 K after time 23dt and greater than 375 K after time 25dt
+//Thus:
+//t=24*dt
+//from equation 9.43
+dt=dx^2/(2*DH);
+t=24*dt;
+printf("\n The time taken to rise from 295 to 375 K = %.1f h",t/3600);
\ No newline at end of file diff --git a/536/CH9/EX9.6/Example_9_6.sce b/536/CH9/EX9.6/Example_9_6.sce new file mode 100755 index 000000000..377318ebf --- /dev/null +++ b/536/CH9/EX9.6/Example_9_6.sce @@ -0,0 +1,27 @@ +clc;
+clear;
+
+printf("\n Example 9.6\n");
+
+d=25e-3; //Diameter of copper sphere
+l=25e-3; //Side length of a copper cube
+h=75; //External heat transfer coefficient
+rho_cu=8950; //Density of copper at mean temperature
+Cp=0.38e3; //Heat capacity of copper at mean temperature
+k=385; // Thermal conductivity of copper at mean temperature
+Tf=923; //Temperature of the furnace
+Ta=368; //Temperature at which they are annealed
+t=5*60; // time taken
+
+V_Ae_S=(d/6); //V/Ae tor the sphere
+printf("\n V/Ae tor the sphere = %.2f * 10^-3 m",V_Ae_S*1e3);
+V_Ae_C=(l/6); //V/Ae tor the cube
+printf("\n V/Ae tor the cube = %.2f * 10^-3 m",V_Ae_C*1e3);
+Bi=h*(V_Ae_S)/k;
+//The use of a lumped capacity method is therefore justified
+tao=rho_cu*Cp*V_Ae_S/h;
+//Then using equation 9.49
+//theta=T
+x=poly([0],'x');
+T=roots(((x-Ta)/(Tf-Ta))-%e^(-t/tao));
+printf("\n Temperature of the sphere and of the cube at the end of 5 minutes = %.0f degree C",T-273);
\ No newline at end of file diff --git a/536/CH9/EX9.7/Example_9_7.sce b/536/CH9/EX9.7/Example_9_7.sce new file mode 100755 index 000000000..81e5dd046 --- /dev/null +++ b/536/CH9/EX9.7/Example_9_7.sce @@ -0,0 +1,22 @@ +clc;
+clear;
+
+printf("\n Example 9.7\n");
+
+k=2.5; //Thermal conductivity
+DH=2e-7;//Thermal diffusivity of the surrounding fluid
+h=100; //External heat transfer coefficient
+To=293; //Initial Temperature
+T_dash=373; //Oven Temperture
+Tc=353; //temperature throughout the whole of the sheet reaches a minimum
+l=10e-3; //thickness of sheet
+L=l/2;
+
+//For the given process, the Biot number
+Bi=h*L/k;
+Bi_1=1/Bi;
+lim_val=(T_dash-Tc)/(T_dash-To);
+//From Figure 9.17, the Fourier number
+Fo=7.7;
+t=Fo*L^2/DH
+printf("\n The minimum time for which the sheet must be heated = %.0f s or %.0f min approx.",t,t/60);
\ No newline at end of file diff --git a/536/CH9/EX9.8/Example_9_8.sce b/536/CH9/EX9.8/Example_9_8.sce new file mode 100755 index 000000000..90219026b --- /dev/null +++ b/536/CH9/EX9.8/Example_9_8.sce @@ -0,0 +1,12 @@ +clc;
+clear;
+
+printf("\n Example 9.8\n");
+
+l=5;//Length of the channel of uranium reactor
+Q=.25e6; //Heat release from uranium reactor
+k=33; //Thermal conductivity of the uranium
+Q_m=Q/l; //Heat release rate
+//Thus, from equation 9.52:
+dT=Q_m/(4*%pi*k);
+printf("\nThe temperature difference between the surface and the centre of the uranium element = %.0f deg K",dT);
\ No newline at end of file diff --git a/536/CH9/EX9.9/Example_9_9.sce b/536/CH9/EX9.9/Example_9_9.sce new file mode 100755 index 000000000..983be1631 --- /dev/null +++ b/536/CH9/EX9.9/Example_9_9.sce @@ -0,0 +1,35 @@ +clc;
+clear;
+
+printf("\n Example 9.9\n");
+
+Cp=2380;//specific heat capacity of nitrobenzene
+k=0.15;
+Meu=0.70e-3;//Viscosity of nitrobenzene
+d_i=15e-3;//internal diameter of tube
+d_o=19e-3;//external diameter of the tube
+d_s=0.44;//shell diameter
+b_s=0.150;//baffle spacing
+p=0.025;//pitch
+c=0.006;//clearance
+//(i)Tube side coefficient
+h_i=1000;//based on inside area
+h_io=1000*d_i/d_o;//based on outside area
+//(ii) Shell side coefficient.
+A=d_s*b_s*c/p;//Area for flow
+G_s_=4/A;
+//Taking Meu/Meu_s=1 in equation 9.91
+d_e=4*((25e-3^2-(%pi*d_o^2/4))/(%pi/d_o));
+h_o=0.36*k/d_e*(d_e*G_s_/Meu)^0.55*(Cp*Meu/k)^0.33;
+//(iii) Overall coefficient
+//The logarithmic mean temperature difference is given by:
+Tm=(((400-345)-(315-305))/log((400-345)/(315-305)));
+//The corrected mean temperature difference is
+Tm_c=Tm*0.8;
+Q=4*Cp*(400-315);
+//The surface area of each tube
+A_t=0.0598;
+U_o=Q/(2*166*5*A_t*Tm_c);
+//(iv) Scale resistance.
+R_d=(1/U_o)-(1/750)-(1/1000);
+printf("\n Value of scale resistance that could be allowed = %.5f m^2 K/W",R_d);
\ No newline at end of file |