diff options
Diffstat (limited to '599/CH2')
30 files changed, 712 insertions, 0 deletions
diff --git a/599/CH2/EX2.1.a/example2_1_a.sce b/599/CH2/EX2.1.a/example2_1_a.sce new file mode 100755 index 000000000..dbf700887 --- /dev/null +++ b/599/CH2/EX2.1.a/example2_1_a.sce @@ -0,0 +1,23 @@ + +clear; +clc; +printf("\t Example 2_1_a\n"); +// let A denote nitrogen and B denote carboondioxide +rA=.3798; +rB=.3941; +rAB=(rA+rB)/2; //molecular seperation at collision +ebyk_A=71.4; +ebyk_B=195.2; +ebyk_AB=(ebyk_A/ebyk_B)^.5; //energy of molecular attraction +pt=1.013*10^5; //absolute total pressure in pascal +T=298; //absolute temperature in kelvin +s=T/ebyk_AB; //collision function + //from chart f(T/ebyk_AB) = 0.5 let it be = x +x=.5; //collision function +MA=28; //molecular weight of nitrogen +MB=44; //molecular weight of carbondioxide +Mnew=((1/MA)+(1/MB))^.5; +Dab= 10^-4*(1.084-.249*(Mnew))*T^1.5*((Mnew))/(pt*x*rAB^2); +printf("\n the diffisivity of nitrogen-carbondioxide is :%f *10^-5 m^2/s",Dab/10^-5) + +//end
\ No newline at end of file diff --git a/599/CH2/EX2.1.b/example2_1_b.sce b/599/CH2/EX2.1.b/example2_1_b.sce new file mode 100755 index 000000000..5d6a3bf37 --- /dev/null +++ b/599/CH2/EX2.1.b/example2_1_b.sce @@ -0,0 +1,25 @@ +
+clear;
+clc;
+printf("\t Example 2_1_b\n");
+// let A denote Hydrogen chloride and B denote air
+
+//"part(ii)"
+rA=.3339;
+rB=.3711;
+rAB=(rA+rB)/2; //molecular seperation at collision
+ebyk_A=344.7;
+ebyk_B=78.6;
+ebyk_AB=(ebyk_A/ebyk_B)^.5; //energy of molecular attraction
+pt=200*10^3; //absolute total pressure in pascal
+T=298; //absolute temperature in kelvin
+s=T/ebyk_AB; //collision function
+ //from chart f(T/ebyk_AB) = 0.62 let it be = x
+x=0.62; //collision function
+MA=36.5; //molecular weight of hydrogen chloride
+MB=29; //molecular weight of air
+Mnew=((1/MA)+(1/MB))^.5;
+Dab=10^-4*(1.084-.249*(Mnew))*T^1.5*((Mnew))/(pt*x*rAB^2);
+printf("\n the diffisivity of hydrogen chloride-air is :%f *10^-6 m^2/s",Dab/10^-6)
+
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.10.a/example2_10_a.sce b/599/CH2/EX2.10.a/example2_10_a.sce new file mode 100755 index 000000000..3493b722f --- /dev/null +++ b/599/CH2/EX2.10.a/example2_10_a.sce @@ -0,0 +1,19 @@ +
+clear;
+clc;
+printf("\t Example 2.10.a\n");
+ //part (i)
+r=(50/2)*10^-3; //radius pf circular tube
+pa1=(190); //vapour pressure of ammonia at pt.1
+pa2=(95); //vapour pressure of ammonia at pt.2
+Dab=2.1*10^-5 //molecular diffusivity in m^2/s
+z=1;
+R=760*22.414/273; //universal gas constant in mmHg*m^3*K*kmol
+ //carbondioxide and oxygen experiences equimolar counter diffusion
+T=298; //temperature in kelvin
+pt=(10/780)*1.013*10^5; //total pressure in pascal
+Na=Dab*(pa1-pa2)/(z*R*T); //flux in kmol/m^2*S
+rate=Na*(3.14*r^2); //rate of mass transfer..(3.14*r^2)-is the area
+printf("\n the rate of mass transfer.:%f *10^-10 kmol/s",rate/10^-10);
+
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.10.b/example2_10_b.sce b/599/CH2/EX2.10.b/example2_10_b.sce new file mode 100755 index 000000000..676ec13e3 --- /dev/null +++ b/599/CH2/EX2.10.b/example2_10_b.sce @@ -0,0 +1,23 @@ +
+clear;
+clc;
+printf("\t Example 2.10.b\n");
+ //part (i)
+r=(50/2)*10^-3; //radius pf circular tube
+pa1=(190); //vapour pressure of ammonia at pt.1
+pa2=(95); //vapour pressure of ammonia at pt.2
+Dab=2.1*10^-5 //molecular diffusivity in m^2/s
+R=760*22.414/273; //universal gas constant in mmHg*m^3*K*kmol
+ //carbondioxide and oxygen experiences equimolar counter diffusion
+T=298; //temperature in kelvin
+pt=(10/780)*1.013*10^5; //total pressure in pascal
+
+ //part (ii)
+//(ya-ya1)/(ya2-ya1)=(z-z1)/(z2-z1);
+z2=1; //diffusion path in m at pos.2
+z1=0; //diffusion path in m at pos.1
+z=.75; //diffusion at general z
+pa=poly([0],'pa'); //calc. of conc. in gas phase
+x=roots((pa-pa1)/(pa2-pa1)-(z-z1)/(z2-z1));
+printf("\n partial pressure of co2 at o.75m from the end where partial pressure is 190mmhg is:%f mmHg",x);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.11.a/example2_11_a.sce b/599/CH2/EX2.11.a/example2_11_a.sce new file mode 100755 index 000000000..3632fa8f7 --- /dev/null +++ b/599/CH2/EX2.11.a/example2_11_a.sce @@ -0,0 +1,17 @@ +
+clear;
+clc;
+printf("\t Example 2_11_a\n");
+ya1=0.2; //initial mole fraction
+ya2=0.1; //final mole fraction
+T=(298); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=0.2*10^-2; //gas film thickness in m
+Dab=.215*10^-4; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+//part (i)when N2 is non diffusing
+
+Na=Dab*pt*log((1-ya2)/(1-ya1))/(z*R*T); //diffusion flux in kmol/m^2*s
+printf("\n diffusion flux if N2 is non diffusing :%f *10^-5 kmol/m^2*s ",Na/10^-5);
+
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.11.b/example2_11_b.sce b/599/CH2/EX2.11.b/example2_11_b.sce new file mode 100755 index 000000000..ef0fcce59 --- /dev/null +++ b/599/CH2/EX2.11.b/example2_11_b.sce @@ -0,0 +1,18 @@ +
+clear;
+clc;
+printf("\t Example 2_11_a\n");
+ya1=0.2;
+ya2=0.1;
+T=(298); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=0.2*10^-2; //gas film thickness in m
+Dab=.215*10^-4; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+
+//part (ii) equimolar counter diffusion
+
+Na=Dab*pt*(ya1-ya2)/(z*R*T) //diffusion flux in kmol/m^2*s
+printf("\n diffusion flux of oxygen during equimolar counter-diffusion :%f *10^-5 kmol/m^2*s ",Na/10^-5);
+
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.12/example2_12.sce b/599/CH2/EX2.12/example2_12.sce new file mode 100755 index 000000000..7b641863c --- /dev/null +++ b/599/CH2/EX2.12/example2_12.sce @@ -0,0 +1,31 @@ +
+clear;
+clc;
+printf("\t Example 2.12\n");
+// ammonia diffusing through inert air and air is non-diffusing
+ya1=0.1;
+ya2=0;
+T=(293); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=0.2*10^-2; //gas film thickness in m
+Dab=.185*10^-4; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+ //part (i)when air is assumed to be stagnant and non-diffusing
+
+Na=Dab*pt*log((1-ya2)/(1-ya1))/(z*R*T); //diffusion flux in kmol/m^2*s
+mw=17; //molecular weight of ammonia
+massflux=Na*mw; //mass flux of given NH3
+printf("\n diffusion flux when total presssure is 1atm and air is non-diffusing :%f *10^-4 kg/m^2*s ",massflux/10^-4);
+ //part (ii) when pressure is increased to 10atm
+
+//Dab_1/Dab_2=pt_2/pt_1
+pt_2=10; //final pressure in atm
+pt_1=1; //initially pressure was 1atm
+Dab_1=.185; //initially diffusion coefficient was.185
+Dab_2=Dab_1*pt_1/pt_2; //for gases Dab is proportional to 1/pt
+Dab=Dab_2*10^-4; //new diffusion coefficient
+pt=pt_2*1.013*10^5; //new total pressure
+Na=Dab*pt*log((1-ya2)/(1-ya1))/(z*R*T); //diffusion flux in kmol/m^2*s
+printf("\n diffusion flux when pressure is increased to 10atm :%f *10^-5 kmol/m^2*s ",Na/10^-5);
+printf("\n \n so the rate of diffusion remains same on increasing the pressure");
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.13/example2_13.sce b/599/CH2/EX2.13/example2_13.sce new file mode 100755 index 000000000..3f5ab1c4d --- /dev/null +++ b/599/CH2/EX2.13/example2_13.sce @@ -0,0 +1,35 @@ +
+clear;
+clc;
+printf("\t Example 2.13\n");
+//position 1 moles molefraction weight
+// acetic acid 0.15 0.0288 9
+// water 5 0.9712 91
+
+//position 2 moles molefraction weight
+// aceitic acid 0.05 0.0092 4
+// water 5.389 0.9908 96
+T=290; //temperature in kelvin
+z=2*10^-3; //film thickness sorrounding the water
+xa2=0.0092; //mole fraction of ethanol at pos.2
+xa1=0.0288; //mole fraction of ethanol at pos.1
+w1=60; //molecular weight of acetic acid
+w2=18; //molecular weight of water
+Dab=0.95*10^-9; //diffusivity of acetic water sol.in m^2/s
+ //av=d/m
+Mavg1=xa1*w1+(1-xa1)*w2; //average molecular wght of solution at pos 1
+d1=1012; // density of 10 % acid
+av1=d1/Mavg1; //value of (d/m) of copper solution
+
+ //for position 2
+d2=1003; //density of 4% acid
+Mavg2=xa2*w1+(1-xa2)*w2; //average molecular wght of solution at pos.2
+av2=d2/Mavg2; //value of (d/m) of water
+
+allavg=(av1+av2)/2; //average value of d/m
+
+ //assuming water to be non diffusing
+
+Na=Dab*(allavg)*log((1-xa2)/(1-xa1))/z; //diffusion rate of acetic acid aacross film of non diffusing water sol.
+printf("\n diffusion rate of acetic acid aacross film of non diffusing water sol. :%f *10^-7 kmol/m^2*s",Na/10^-7);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.14/example2_14.sce b/599/CH2/EX2.14/example2_14.sce new file mode 100755 index 000000000..f78cb66c7 --- /dev/null +++ b/599/CH2/EX2.14/example2_14.sce @@ -0,0 +1,22 @@ +
+clear;
+clc;
+printf("\t Example 2.14\n");
+ya1=0.2; //molefraction at pos.1
+ya2=0.1; //molefraction at pos.2
+T=(293); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=0.2*10^-2; //gas film thickness in m
+Dab=.206*10^-4; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+ //for ideal gases volume fraction =mole fraction
+//part (i)when N2 is non diffusing
+
+Na=Dab*pt*log((1-ya2)/(1-ya1))/(z*R*T); //diffusion flux in kmol/m^2*s
+printf("\n diffusion flux if N2 is non diffusing :%f *10^-5 kmol/m^2*s ",Na/10^-5);
+//part (ii) equimolar counter diffusion
+
+Na=Dab*pt*(ya1-ya2)/(z*R*T) //diffusion flux in kmol/m^2*s
+printf("\n diffusion flux of nitrogen during equimolar counter-diffusion :%f *10^-5 kmol/m^2*s ",Na/10^-5);
+
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.15/example2_15.sce b/599/CH2/EX2.15/example2_15.sce new file mode 100755 index 000000000..c41e04785 --- /dev/null +++ b/599/CH2/EX2.15/example2_15.sce @@ -0,0 +1,19 @@ +
+clear;
+clc;
+printf("\t Example 2.15\n");
+pa1=0.2*10^5; //partial pressure at pos.1
+pa2=0; //partial pressure at pos.2
+r=10/2; //radius of tank in which benzene is stored
+T=(298); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=10*10^-3; //gas film thickness in m
+Dab=.02/3600; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+ //benzene is stored in atank of dia 10m
+//part (i)when air is assumed to be stagnant
+
+Na=Dab*pt*log((pt-pa2)/(pt-pa1))/(z*R*T); //diffusion flux in kmol/m^2*s
+rate=Na*(3.14*r^2); //rate of loss of benzene if air is stagnant
+printf("\n diffusion rate of loss of benzene :%f *10^-4 kmol/s ",rate/10^-4);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.16/example2_16.sce b/599/CH2/EX2.16/example2_16.sce new file mode 100755 index 000000000..f4ff8ca2d --- /dev/null +++ b/599/CH2/EX2.16/example2_16.sce @@ -0,0 +1,26 @@ +
+clear;
+clc;
+printf("\t Example 2.16\n");
+ya2=0.1; //molefraction at pos.2
+ya1=0.8; //molefraction at pos.1
+T=(370); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=0.1*10^-3; //gas film thickness in m
+Dab=.15*10^-2; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+Area=10; //area of the film is 10m^2
+
+ //alcohol is being absorbed infrom amixture of alcohol vapour and water vapour by means of non-volatile solvent in which alcohol is soluble bt water is not
+ //for gase Dab=T^3/2
+ //Dab1/Dab2=(T1/T2)^3/2
+
+T2=370; //final temperature in kelvin
+T1=298; //initial temperature in kelvin
+Dab1=.15*10^-2; //initial diffusion coefficient
+Dab2=((T2/T1)^(3/2))*Dab1; //final diffusion coefficient
+Na=Dab2*pt*log((1-ya2)/(1-ya1))/(z*R*T); //diffusion flux in kmol/m^2*s
+rate=Na*3600*46*Area; //rate of diffusion of alcohol-water vapour in kg/hour
+printf("\n rate of diffusion of alcohol-water vapour :%f *10^6 kg/hour ",rate/10^6);
+
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.17/example2_17.sce b/599/CH2/EX2.17/example2_17.sce new file mode 100755 index 000000000..7a3c7fe8f --- /dev/null +++ b/599/CH2/EX2.17/example2_17.sce @@ -0,0 +1,36 @@ +
+clear;
+clc;
+printf("\t Example 2.17\n");
+ya2=0; //molefraction at pos.2
+ya1=0.1; //molefraction at pos.1
+T=(273); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=2*10^-3; //gas film thickness in m
+Dab=.198*10^-4; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+ //ammonia is diffusing through an inert film 2mm thick
+
+ //for gase Dab=T^3/2
+ //Dab1/Dab2=(T1/T2)^3/2
+T2=293; //final temperature in kelvin
+T1=273; //initial temperature in kelvin
+Dab1=0.198*10^-4; //initial diffusion coefficient
+Dab2=((T2/T1)^(3/2))*Dab1; //final diffusion coefficient
+Na=Dab2*pt*log((1-ya2)/(1-ya1))/(z*R*T2); //diffusion flux in kmol/m^2*s
+printf("\n flux of diffusion of ammonia through inert film :%f *10^-5 kmol/m^2*s ",Na/10^-5);
+
+//if pressure is also incresed from 1 to 5 atm
+ //for gases Dab=(T^3/2)/pt;
+ //Dab1/Dab2=(T1/T2)^3/2*(p2/p1)
+T2=293; //final temperature in kelvin
+T1=273; //initial temperature in kelvin
+pa2=5; //final pressure in atm
+pa1=1; //initial pressure in atm
+p=pa2*1.013*10^5;
+Dab1=.198*10^-4; //initial diffusion coefficient
+Dab2=((T2/T1)^(3/2))*Dab1*(pa1/pa2); //final diffusion coefficient
+Na=Dab2*p*log((1-ya2)/(1-ya1))/(z*R*T2); //diffusion flux in kmol/m^2*s
+printf("\n flux of diffusion of ammonia if temp. is 20 and pressure is 5 atm :%f*10^-5 kmol/m^2*s ",Na/10^-5);
+printf("\n \n so there is no change in flux when pressure is changed");
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.18/example2_18.sce b/599/CH2/EX2.18/example2_18.sce new file mode 100755 index 000000000..5c27a5ff0 --- /dev/null +++ b/599/CH2/EX2.18/example2_18.sce @@ -0,0 +1,19 @@ +
+clear;
+clc;
+printf("\t Example 2.18\n");
+pa1=0.418*10^5; //partial pressure initially
+pa2=0; //partial pressure of pure air
+r=10/2; //radius of tank in which benzene is stored
+T=(350); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=2*10^-3; //gas film thickness in m
+Dab=.2*10^-4; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+r=0.2/2; //radius of open bowl is 0.2
+//when air layer is assumed to be stagnant of thickness 2mm
+
+Na=Dab*pt*log((pt-pa2)/(pt-pa1))/(z*R*T); //diffusion flux in kmol/m^2*s
+rate=Na*(3.14*r^2)*18; //rate of loss of evaporation
+printf("\n diffusion rate loss of evaporation :%f *10^-4 kg/s ",rate/10^-4);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.19/example2_19.sce b/599/CH2/EX2.19/example2_19.sce new file mode 100755 index 000000000..12096e8da --- /dev/null +++ b/599/CH2/EX2.19/example2_19.sce @@ -0,0 +1,27 @@ +
+clear;
+clc;
+printf("\t Example 2.19\n");
+ //stefan tube experiment
+
+Ml=92; //molecular weight of toluene
+T=(312.4); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+R=8314; //universal gas constant
+t=275*3600; //after 275 hours the level dropped to 80mm from the top
+zo=20*10^-3; //intially liquid toluene is at 20mm from top
+zt=80*10^-3; //finally liquid toluene is at 80mm from top
+ //air is assumed to be satgnant
+d=850; //density in kg/m^3
+pa=7.64*10^3; //vapour pressure of toluene in at 39.4degree celcius
+cal=d/Ml; //conc. at length at disxtance l
+ca=pt/(R*T); //total conc.
+xa1=pa/pt; //mole fraction of toluene at pt1 i.e before evaporation
+xb1=1-xa1; //mole fraction of air before evaporation i.e at pt1
+xb2=1; //mole fraction of air after evaporation i.e at pt.2
+xa2=0; //mole fraction of toluene at point 2
+xbm=(xb2-xb1)/(log(xb2/xb1));
+ //t/(zt-zt0) = (xbm*cal*(zt+zo))/(2*c*(xa1-xa2)*t);
+Dab=(xbm*cal*(zt^2-zo^2))/(2*ca*t*(xa1-xa2));
+printf("\n the diffusivity of the mixture in stefan tube of toluene in air is :%f*10^-5 m^2/s",Dab/10^-5);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.2/example2_2.sce b/599/CH2/EX2.2/example2_2.sce new file mode 100755 index 000000000..a55167831 --- /dev/null +++ b/599/CH2/EX2.2/example2_2.sce @@ -0,0 +1,15 @@ +
+clear;
+clc;
+printf("\t Example 2.2\n");
+ //kopp's law is valid
+
+u=1.145*10^-3; //viscosity of water1.145cp
+v_a=5*.0148+12*.0037+1*.0074; //by kopp's law
+t=288; //temperature of water in kelvin
+MB=18; //molecular weight of water
+phi=2.26; //association parameter for solvent-water
+
+D_ab=(117.3*10^-18)*((phi*MB)^.5)*(t)/(u*(v_a)^.6);
+printf("\n the diffusivity of isoamyl alcohol is :%f *10^-9 m^2/s",D_ab/10^-9);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.20/example2_20.sce b/599/CH2/EX2.20/example2_20.sce new file mode 100755 index 000000000..b68b30242 --- /dev/null +++ b/599/CH2/EX2.20/example2_20.sce @@ -0,0 +1,20 @@ +
+clear;
+clc;
+printf("\t Example 2.20\n");
+ //this is the case of equimolar counter diffusion as the latent heat of vaporisation are very close to each other
+
+T=(360); //temperature in kelvin
+pt=372.4/760; //total pressure in atm
+R=82.06; //universal gas constant
+Dab=0.0506; //diffusion coefficient in cm^2/s
+z=0.254; //gas layer thickness in cm
+vp=368/760; //vapour pressure of toluene in atm
+xtol=.3; //mole fractoin of toluene in atm
+pb1=xtol*vp; //partial pressure of toluene
+//since pb1 is .045263 bt in book it is rounded to 0.145
+pb2=xtol*pt; //parial pressure of toluene in vapour phase
+Na=Dab*(pb1-pb2)/(z*R*T); //diffusion flux
+printf("\n the diffusion flux of a mixture of benzene and toluene %f*10^-8 gmol/cm^2*s\n",Na/10^-8);
+printf("\nthe negative sign indicates that the toluene is getting transferred from gas phase to liquid phase(hence the transfer of benzene is from liquid to gas phase)")
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.21/example2_21.sce b/599/CH2/EX2.21/example2_21.sce new file mode 100755 index 000000000..0d578ec8b --- /dev/null +++ b/599/CH2/EX2.21/example2_21.sce @@ -0,0 +1,27 @@ +
+clear;
+clc;
+printf("\t Example 2.21\n");
+ //stefan tube experiment(pseudo steady state diffusion)
+
+Ml=92; //molecular weight of toluene
+T=(303); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+R=8314; //universal gas constant
+t=275*3600; //after 275 hours the level dropped to 80mm from the top
+zo=20*10^-3; //intially liquid toluene is at 20mm from top
+zt=77.5*10^-3; //finally liquid toluene is at 80mm from top
+ //air is assumed to be satgnant
+d=820; //density in kg/m^3
+pa=(57/760)*1.0135*10^5; //vapour pressure of toluene in at 39.4degree celcius
+cal=d/Ml; //conc. at length at disxtance l
+ca=pt/(R*T); //total conc.
+xa1=pa/pt; //mole fraction of toluene at pt1 i.e before evaporation
+xb1=1-xa1; //mole fraction of air before evaporation i.e at pt1
+xb2=1; //mole fraction of air after evaporation i.e at pt.2
+xa2=0; //mole fraction of toluene at point 2
+xbm=(xb2-xb1)/(log(xb2/xb1));
+ //t/(zt-zt0) = (xbm*cal*(zt+zo))/(2*c*(xa1-xa2)*t);
+Dab=(xbm*cal*(zt^2-zo^2))/(2*ca*t*(xa1-xa2));
+printf("\n the diffusivity of the mixture in stefan tube of toluene in air is :%f*10^-5 m^2/s",Dab/10^-5);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.22/example2_22.sce b/599/CH2/EX2.22/example2_22.sce new file mode 100755 index 000000000..c1f681148 --- /dev/null +++ b/599/CH2/EX2.22/example2_22.sce @@ -0,0 +1,36 @@ +
+clear;
+clc;
+printf("\t Example 2.22\n");
+ //variation in liquid level with respect to time is given below
+
+t=[26 185 456 1336 1958 2810 3829 4822 6385]
+// let Zt-Zo= x;
+x=[.25 1.29 2.32 4.39 5.47 6.70 7.38 9.03 10.48]
+i=1; //looping starts
+while(i<10)
+ y(i)=t(i)/x(i); //for calculating the t/Zt-Zo value
+i=i+1;
+end
+plot(x,y,"o-");
+xtitle(" Fig.2.2 Example 22 ","X--(zi-zo),cm --->","Y-- vs (t/(zi-zo))min/cm ---->");
+slope=51.4385*60 *10^4; //slope of the curve in 1/sec*m^2
+//slope = Cal *(xblm)/(2*Dab*C*(xa1-xa2))
+d=1540; //density in kg/m^3
+Ml=154; //molecular weight of toluene
+Cal=d/Ml ; //conc. at length at disxtance l in mol/m^3
+
+T=(321); //temperature in kelvin
+pt=1; //total pressure in atm
+R=82.06; //universal gas constant
+C=pt/(R*T) *10^3; //total conc. in kg mol/m^3
+
+pa=(282/760); //vapour pressure of toluene
+xa1=pa/pt; //mole fraction of toluene at pt1 i.e before evaporation
+xb1=1-xa1; //mole fraction of air before evaporation i.e at pt1
+xb2=1; //mole fraction of air after evaporation i.e at pt.2
+xa2=0; //mole fraction of toluene at point 2
+xblm=(xb2-xb1)/(log(xb2/xb1));//log mean temp. difference
+Dab = Cal *(xblm)/(2*slope*C*(xa1-xa2)); //diffusivity coefficient
+printf("\n the diffusivity of the mixture by winklemann method of toluene in air is :%f*10^-6 m^2/s",Dab/10^-6);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.23/example2_23.sce b/599/CH2/EX2.23/example2_23.sce new file mode 100755 index 000000000..b91e82b9c --- /dev/null +++ b/599/CH2/EX2.23/example2_23.sce @@ -0,0 +1,20 @@ +
+clear;
+clc;
+printf("\t Example 2.23\n");
+ //it is the case of equimolar conter diffusion as the tube is perfectly sealed to two bulbs at the end and the pressure throughout is constant
+d=0.001;
+area=3.14*(d/2)^2; //area of the bulb
+T=298; //temperature in kelvin
+p=1.013*10^5; //total pressure of both the bulbs
+R=8314; //universal gas constant
+c=p/(R*T); //total concentration
+Dab=.784*10^-4; //diffusion coefficient in m^2/s
+xa1=0.8; //molefraction of nitrogen gas at the 1 end
+xa2=0.25; //molefraction of nitrogen gas at the 2nd end
+z=.15; //distance between the bulbs
+
+ //rate=area*Na;
+rate=area*Dab*c*(xa1-xa2)/z; //rate of transfer of hydrogen and hydrogen
+printf("\n the rate of transfer from 1 to 2 of nitrogen and 2 to 1of hydrogen is :%f *10^-11kmol/s",rate/10^-11);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.24/example2_24.sce b/599/CH2/EX2.24/example2_24.sce new file mode 100755 index 000000000..68203cc64 --- /dev/null +++ b/599/CH2/EX2.24/example2_24.sce @@ -0,0 +1,15 @@ +
+clear;
+clc;
+printf("\t Example 2.24\n");
+ //using wilke and chang empirical correlation
+//Dab=(117.3*10^-18)*(phi*Mb)^0.5*T/(u*va^0.6);
+
+T=288; //temperature in kelvin
+Mb=32; //molecular weight of methanol
+phi=1.9; //association factor for solvent
+va=(14.8+(4*24.6))*10^-3 //solute(CCl4) volume at normal BP in m^3/kmol
+u=.6*10^-3; //viscosity of solution in kg/m*s
+Dab=(117.3*10^-18)*(phi*Mb)^0.5*T/(u*va^0.6); //diffusion coefficient in m^2/s
+printf("\ndiffusivity of methanol in carbon tetrachloride is :%f*10^-9 m^2/s",Dab/10^-9);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.25/example2_25.sce b/599/CH2/EX2.25/example2_25.sce new file mode 100755 index 000000000..910d085ca --- /dev/null +++ b/599/CH2/EX2.25/example2_25.sce @@ -0,0 +1,15 @@ +
+clear;
+clc;
+printf("\t Example 2.25\n");
+ //using wilke and chang empirical correlation
+//Dab=(117.3*10^-18)*(phi*Mb)^0.5*T/(u*va^0.6);
+
+T=288; //temperature in kelvin
+Mb=18; //molecular weight of methanol
+phi=2.26; //association factor for solvent
+va=(2*14.8+(6*3.7)+7.4)*10^-3 //solute(water) volume at normal BP in m^3/kmol
+u=1*10^-3; //viscosity of solution in kg/m*s
+Dab=(117.3*10^-18)*(phi*Mb)^0.5*T/(u*va^0.6); //diffusion coefficient in m^2/s
+printf("\ndiffusivity of methanol in water is :%f*10^-9 m^2/s",Dab/10^-9);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.26/example2_26.sce b/599/CH2/EX2.26/example2_26.sce new file mode 100755 index 000000000..ab5663733 --- /dev/null +++ b/599/CH2/EX2.26/example2_26.sce @@ -0,0 +1,39 @@ +
+clear;
+clc;
+printf("\t Example 2.26\n");
+u=20*10^-6; //viscosity in Ns/m^2
+pt=2666; //total pressure in N/m^2
+pa1=pt; //pressure at 1
+pa2=0; //pressure at 2
+mw=32; //molecular weight of oxygen
+R=8314; //universal law constant
+T=373; //temp. in kelvin
+gc=1;
+l=(3.2*u/pt)*((R*T)/(2*3.14*gc*mw))^0.5;//mean free path
+d=.2*10^-6; //pore diameter
+s=d/l; //value of dia/l
+ //hence knudsen diffusion occurs
+Na=0.093*20*273/(760*373*22414*10^-1); //diffusion coefficient in kmol/m^2*s
+Dka=(d/3)*((8*gc*R*T)/(3.14*mw))^0.5;
+len=Dka*(pa1-pa2)/(R*T*Na); //length of the plate
+printf("\n the length of the plate is :%f m ",len);
+
+
+ //for diffusion with hydrogen
+u=8.5*10^-6; //viscosity in Ns/m^2
+pt=1333; //total pressure in N/m^2
+pa1=pt; //pressure at 1
+pa2=0; //pressure at 2
+mw=2; //molecular weight of oxygen
+R=8314; //universal law constant
+T=298; //temp. in kelvin
+gc=1;
+l=(3.2*u/pt)*((R*T)/(2*3.14*gc*mw))^0.5;//mean free path
+d=.2*10^-6; //pore diameter
+s=d/l; //value of dia/l
+ //hence knudsen diffusion occurs
+Dka=(d/3)*((8*gc*R*T)/(3.14*mw))^0.5;
+Na=Dka*(pa1-pa2)/(R*T*len); //diffusion coefficient in kmol/m^2*s
+printf("\n the diffusion coefficient is :%f *10^-4 kmol/m^2*s",Na/10^-6);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.3/example2_3.sce b/599/CH2/EX2.3/example2_3.sce new file mode 100755 index 000000000..39d493c33 --- /dev/null +++ b/599/CH2/EX2.3/example2_3.sce @@ -0,0 +1,22 @@ +
+clear;
+clc;
+printf("\t Example 2.3\n");
+
+pa1=(33/760)*1.013*10^5; //vapour pressure of ccl4 at 273 in pascal
+pa2=0;
+d=1.59; //density of liquid ccl4 in g/cm^3
+ //considering o2 to be non diffusing and with
+T=273; //temperature in kelvin
+pt=(755/780)*1.013*10^5; //total pressure in pascal
+z=.171; //thickness of film
+a=.82*10^-4; //cross-sectional area of cell in m^2
+v=.0208; //volume of ccl4 evaporated
+t=10; //time of evaporation
+MB=154; //molecular wght of ccl4
+rate=v*d/(MB*t); //.0208cc of ccl4 is evaporating in 10hrs
+Na=rate*10^-3/(3600*a); //flux in kmol/m^2*S
+
+D_ab=Na*z*8314*273/(pt*log((pt-pa2)/(pt-pa1))); //molecular diffusivity in m^2/s
+printf("\n the diffusivity of ccl4 through oxygen:%f *10^-6 m^2/s",D_ab/10^-6);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.4/example2_4.sce b/599/CH2/EX2.4/example2_4.sce new file mode 100755 index 000000000..a8c91fa20 --- /dev/null +++ b/599/CH2/EX2.4/example2_4.sce @@ -0,0 +1,23 @@ +
+clear;
+clc;
+printf("\t Example 2.4\n");
+z=.0305*10^-3; //wall thickness sorrounding the crystal
+x1=0.0229;
+w1=160; //molecular weight of copper sulphate
+w2=18; //molecular weight of water
+Dab=7.29*10^-10; //diffusivity of copper sulphatein m^2/s
+ //av=d/m
+Mavg=x1*w1+(1-x1)*w2; //average molecular wght of solution
+d1=1193; //density of copper sulphate solution
+av1=d1/Mavg; //value of (d/m) of copper solution
+
+ //for pure water
+d2=1000; //density of water
+m2=18; //molecular wght of water
+av2=d2/m2; //value of (d/m) of water
+allavg=(av1+av2)/2; //average value of d/m
+xa2=0;
+Na=Dab*(allavg)*log((1-xa2)/(1-x1))/z; //flux of cuso4 from crystal surface to bulk solution
+printf("\n the rate at which crystal dissolves :%f *10^-5 kmol/m^2*s",Na/10^-5);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.5.a/example2_5_a.sce b/599/CH2/EX2.5.a/example2_5_a.sce new file mode 100755 index 000000000..f9a67c3f4 --- /dev/null +++ b/599/CH2/EX2.5.a/example2_5_a.sce @@ -0,0 +1,23 @@ +
+clear;
+clc;
+printf("\t Example 2.5.a\n");
+//position 1 moles molefraction
+// air 80 0.8
+// water 20 0.2
+
+//position 2 moles molefraction
+// air 10 0.1
+// water 90 0.9
+ya1=0.8;
+ya2=0.1;
+T=(273+35); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=0.3*10^-3; //gas film thickness in m
+Dab=.18*10^-4; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+Na=Dab*pt*(ya1-ya2)/(z*R*T) //diffusion flux in kmol/m^2*s
+rate=Na*100*10^-4*3600*46; //since molecular weight of mixture is 46
+printf("\n rate of diffusion of alcohol-water vapour :%f kg/hr ",rate);
+
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.5.b/example2_5_b.sce b/599/CH2/EX2.5.b/example2_5_b.sce new file mode 100755 index 000000000..28239bbbe --- /dev/null +++ b/599/CH2/EX2.5.b/example2_5_b.sce @@ -0,0 +1,17 @@ +
+clear;
+clc;
+printf("\t Example 2_5_b\n");
+ya1=0.8;
+ya2=0.1;
+T=(273+35); //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+z=0.3*10^-3; //gas film thickness in m
+Dab=.18*10^-4; //diffusion coefficient in m^2/s
+R=8314; //universal gas constant
+
+//diffusion through stagnant film
+Na=Dab*pt*log((1-ya2)/(1-ya1))/(z*R*T); //diffusion flux in kmol/m^2*s
+rate=Na*100*10^-4*3600*46; //since molecular weight of mixture is 46
+printf("\n rate of diffusion if water layer is stagnant :%f *10^-3 kg/s ",rate/(3600*10^-3));
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.6/example2_6.sce b/599/CH2/EX2.6/example2_6.sce new file mode 100755 index 000000000..c2a6c168d --- /dev/null +++ b/599/CH2/EX2.6/example2_6.sce @@ -0,0 +1,19 @@ +
+clear;
+clc;
+printf("\t Example 2.6\n");
+T=298; //temperature in kelvin
+pt=1*1.013*10^5; //total pressure in pascal
+ID=25*10^-3; //internal diameter in m of unvulcanised rubber in m
+OD=50*10^-3; //internal diameter in m of unvulcanised rubber in m
+Ca1=2.37*10^-3; //conc. of hydrogen at the inner surface of the pipe in kmol/m^3
+Ca2=0; //conc. of hydrogen at 2
+Dab=1.8*10^-10; //diffusion coefficient in cm^2/s
+l=2; //length of pipe in m
+// Va=Da*Sa*(pa1-pa2)/z;
+z=(50-25)*10^-3/2; //wall thickness in m
+Va=Dab*(Ca1-Ca2)/z; //diffusion through a flat slab of thickness z
+Sa=2*3.14*l*(OD-ID)/(2*log(OD/ID)); //average mass transfer area of
+rate=Va*Sa; //rate of loss of hydrogen by diffusion
+printf("\n rate of loss hydrogen by diffusion through a pipe of 2m length :%f*10^-12kmol/s",rate/10^-12);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.7/example2_7.sce b/599/CH2/EX2.7/example2_7.sce new file mode 100755 index 000000000..f24144612 --- /dev/null +++ b/599/CH2/EX2.7/example2_7.sce @@ -0,0 +1,16 @@ +
+clear;
+clc;
+printf("\t Example 2.7\n");
+
+pa1=(1.5)*10^4; //vapour pressure of ammonia at pt.1
+pa2=(0.5)*10^4; //vapour pressure of ammonia at pt.2
+Dab=2.3*10^-5 //molecular diffusivity in m^2/s
+z=0.15; //diffusion path in m
+R=8314; //universal gas constant
+ //ammonia diffuses through nitrogen under equimolar counter diffusion
+T=298; //temperature in kelvin
+pt=1.013*10^5; //total pressure in pascal
+Na=Dab*(pa1-pa2)/(z*R*T); //flux in kmol/m^2*S
+printf("\n the ammonia diffusion through nitrogen under equimolar counter diffusion:%f *10^-7 kmol/m^2*s",Na/10^-7);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.8/example2_8.sce b/599/CH2/EX2.8/example2_8.sce new file mode 100755 index 000000000..cdc1cfa88 --- /dev/null +++ b/599/CH2/EX2.8/example2_8.sce @@ -0,0 +1,31 @@ +
+clear;
+clc;
+printf("\t Example 2.8\n");
+//position 1 moles molefraction weight
+// ethanol 0.1478 0.02775 6.80
+// water 5.18 0.9722 93.20
+
+//position 2 moles molefraction weight
+// ethanol 0.235 0.0453 10.8
+// water 4.96 0.9547 89.20
+z=0.4*10^-2; //film thickness sorrounding the crystal
+xa1=0.0453; //mole fraction of ethanol at pos.2
+xa2=0.02775; //mole fraction of ethanol at pos.1
+w1=46; //molecular weight of ethanol
+w2=18; //molecular weight of water
+Dab=74*10^-5*10^-4; //diffusivity of ethanol water sol.in m^2/s
+ //av=d/m
+Mavg1=xa2*w1+(1-xa2)*w2; //average molecular wght of solution at pos 1
+d1=0.9881*10^3; // density of 6.8 wt% solution
+av1=d1/Mavg1; //value of (d/m) of copper solution
+
+ //for pure water
+d2=972.8; // density of 10.8 wt% solution
+Mavg2=xa1*w1+(1-xa1)*w2; //average molecular wght of solution at pos.2
+av2=d2/Mavg2; //value of (d/m) of water
+
+allavg=(av1+av2)/2; //average value of d/m
+Na=Dab*(allavg)*log((1-xa2)/(1-xa1))/z; //steady state flux in kmol/m^2*s of ethanol water sol.
+printf("\n the rate at which crystal dissolves :%f *10^-5 kmol/m^2*s",Na/10^-5);
+//end
\ No newline at end of file diff --git a/599/CH2/EX2.9/example2_9.sce b/599/CH2/EX2.9/example2_9.sce new file mode 100755 index 000000000..07d07e3ef --- /dev/null +++ b/599/CH2/EX2.9/example2_9.sce @@ -0,0 +1,34 @@ +
+clear;
+clc;
+printf("\t Example 2.9\n");
+//position 1 kmoles molefraction weight
+// acetic acid 0.167 0.0323 10
+// water 5 0.9677 90
+
+//position 2 kmoles molefraction weight
+// aceitic acid 0.067 0.0124 4
+// water 5.33 0.9876 96
+
+ //basis : 100kg of mixture
+z=2*10^-3; //film thickness sorrounding the water
+xa1=0.0323; //mole fraction of ethanol at pos.2
+xa2=0.0124; //mole fraction of ethanol at pos.1
+w1=60; //molecular weight of acetic acid
+w2=18; //molecular weight of water
+Dab=0.000095; //diffusivity of acetic water sol.in m^2/s
+ //av=d/m
+Mavg1=xa1*w1+(1-xa1)*w2; //average molecular wght of solution at pos 1
+d1=1013; // density of 10 % acid
+av1=d1/Mavg1; //value of (d/m) of copper solution
+
+ //for pure water
+d2=1004; //density of 4% acid
+Mavg2=xa2*w1+(1-xa2)*w2; //average molecular wght of solution at pos.2
+av2=d2/Mavg2; //value of (d/m) of water
+
+allavg=(av1+av2)/2; //average value of d/m
+ //assuming water to be non diffusing
+Na=Dab*(allavg)*log((1-xa2)/(1-xa1))/z; //diffusion rate of acetic acid aacross film of non diffusing water sol.
+printf("\n diffusion rate of acetic acid aacross film of non diffusing water sol. :%f kmol/m^2*s",Na);
+//end
\ No newline at end of file |