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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 /858 | |
| 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 '858')
176 files changed, 3370 insertions, 0 deletions
diff --git a/858/CH1/EX1.1/example_1.sce b/858/CH1/EX1.1/example_1.sce new file mode 100755 index 000000000..1478d98cc --- /dev/null +++ b/858/CH1/EX1.1/example_1.sce @@ -0,0 +1,26 @@ +clc
+clear
+printf("example 1.1 page number 19\n\n")
+//to find composition of air by weight
+y_oxygen = 0.21 //mole fraction of oxygen
+y_nitrogen = 0.79 //mole fraction of nitrogen
+molar_mass_oxygen = 32
+molar_mass_nitrogen = 28
+
+molar_mass_air = y_oxygen*molar_mass_oxygen+y_nitrogen*molar_mass_nitrogen;
+mass_fraction_oxygen =y_oxygen*molar_mass_oxygen/molar_mass_air;
+mass_fraction_nitrogen = y_nitrogen*molar_mass_nitrogen/molar_mass_air;
+
+printf("mass fraction of oxygen = %f \n\n",mass_fraction_oxygen)
+printf("mass fraction of nitrogen = %f \n\n",mass_fraction_nitrogen)
+
+V1 = 22.4 //in liters
+P1 = 760 //in mm Hg
+P2= 735.56 //in mm Hg
+T1= 273 //in K
+T2 = 298 //in K
+
+V2= (P1*T2*V1)/(P2*T1);
+density = molar_mass_air/V2;
+
+printf("density = %f gm/l",density)
diff --git a/858/CH1/EX1.10/example_10.sce b/858/CH1/EX1.10/example_10.sce new file mode 100755 index 000000000..65a6dd993 --- /dev/null +++ b/858/CH1/EX1.10/example_10.sce @@ -0,0 +1,19 @@ +clc
+clear
+
+printf('example 1.10 page number 33\n')
+
+//to find relation between friction factor and reynold's number
+
+//log f=y, log Re=x, log a=c
+sigma_x=23.393;
+sigma_y=-12.437;
+sigma_x2=91.456
+sigma_xy=-48.554;
+m=((6*sigma_xy)-(sigma_x*sigma_y))/(6*sigma_x2-(sigma_x)^2);
+printf("m = %f \n",m)
+
+c=((sigma_x2*sigma_y)-(sigma_xy*sigma_x))/(6*sigma_x2-(sigma_x)^2);
+printf("c = %f \n",c)
+
+printf("f=0.084*Re^-0.256")
diff --git a/858/CH1/EX1.11/example_11.sce b/858/CH1/EX1.11/example_11.sce new file mode 100755 index 000000000..997b50935 --- /dev/null +++ b/858/CH1/EX1.11/example_11.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 1.11 page number 35\n\n")
+
+//to find the average velocity
+
+u = [2;1.92;1.68;1.28;0.72;0];
+r = [0;1;2;3;4;5];
+
+z = u.*r;
+plot(r,z)
+title("variation of ur with r")
+xlabel("r")
+ylabel("ur")
+
+//by graphical integration, we get
+u_avg = (2/25)*12.4
+printf("average velocity = %f cm/s\n",u_avg)
diff --git a/858/CH1/EX1.12/example_12.sce b/858/CH1/EX1.12/example_12.sce new file mode 100755 index 000000000..3365cffef --- /dev/null +++ b/858/CH1/EX1.12/example_12.sce @@ -0,0 +1,25 @@ +clc
+clear
+
+printf('example 1.12 page number 37\n')
+
+//to find the average velocity
+printf('using trapezoid rule\n')
+
+n = 6;
+h = (3 - 0)/n;
+I = (h/2)*(0+2*0.97+2*1.78+2*2.25+2*2.22+2*1.52+0);
+u_avg = (2/3^2)*I;
+
+printf("average velocity = %f cm/s\n",u_avg)
+
+disp('Simpsons rule')
+
+n = 6;
+h = 3/n;
+I = (h/3)*(0+4*(0.97+2.25+1.52)+2*(1.78+2.22)+0);
+u_avg = (2/3^2)*I;
+
+printf("average velocity = %f cm/s\n",u_avg)
+
+
diff --git a/858/CH1/EX1.13/example_13.sce b/858/CH1/EX1.13/example_13.sce new file mode 100755 index 000000000..5122ae9b0 --- /dev/null +++ b/858/CH1/EX1.13/example_13.sce @@ -0,0 +1,37 @@ +clc
+clear
+
+printf('example 1.13 page number 38\n\n')
+
+//to find the settling velocity as a function of time
+z0 = 30.84;
+z1 = 29.89;
+z2 = 29.10;
+h = 4;
+
+u1_t0 = (-3*z0+4*z1-z2)/(2*h);
+u1_t4 = (-z0+z2)/(2*h);
+u1_t8 = (z0-4*z1+3*z2)/(2*h);
+
+//considering data set for t = 4,8,12 min
+z0 = 29.89;
+z1 = 29.10;
+z2 = 28.30;
+u2_t4 = (-3*z0+4*z1-z2)/(2*h);
+u2_t8 = (-z0+z2)/(2*h);
+u2_t12 = (z0-4*z1+3*z2)/(2*h);
+
+//considering data set for t = 8,12,16 min
+z0 = 29.10;
+z1 = 28.30;
+z2 = 27.50;
+u3_t8 = (-3*z0+4*z1-z2)/(2*h);
+u3_t12 = (-z0+z2)/(2*h);
+u3_t16 = (z0-4*z1+3*z2)/(2*h);
+
+//taking average
+u_t4 = (u1_t4+u2_t4)/2;
+u_t8 = (u1_t8+u2_t8+u3_t8)/3;
+u_t12 = (u2_t12+u3_t12)/2;
+
+printf("u_t0 = %f cm/min\n u_t4 = %f cm/min\n u_t8 = %f cm/min \n u_t12 = %f/n cm/min\n u_t16 =%f/n cm/min ",u1_t0,u_t4,u_t8,u_t12,u3_t16)
diff --git a/858/CH1/EX1.14/example_14.sce b/858/CH1/EX1.14/example_14.sce new file mode 100755 index 000000000..c6359ec01 --- /dev/null +++ b/858/CH1/EX1.14/example_14.sce @@ -0,0 +1,2 @@ +printf('example 1.14 page number 45')
+disp ("this is a theoritical question, book shall be referred for solution")
diff --git a/858/CH1/EX1.15/example_15.sce b/858/CH1/EX1.15/example_15.sce new file mode 100755 index 000000000..e7c7c47c2 --- /dev/null +++ b/858/CH1/EX1.15/example_15.sce @@ -0,0 +1,2 @@ +printf('example 1.15 page number 46')
+disp ("this is a theoritical question, book shall be referred for solution")
diff --git a/858/CH1/EX1.16/example_16.sce b/858/CH1/EX1.16/example_16.sce new file mode 100755 index 000000000..ff29eab7c --- /dev/null +++ b/858/CH1/EX1.16/example_16.sce @@ -0,0 +1,24 @@ +clc
+clear
+printf('example 1.16 page number 49\n')
+
+//to find the flow rate and pressure drop
+density_water=988 //in kg/m3
+viscosity_water=55*10^-5 //in Ns/m2
+density_air=1.21 //in kg/m3
+viscosity_air=1.83*10^-5 //in Ns/m2
+L=1 //length in m
+
+L1=10*L //length in m
+Q=0.0133;
+
+Q1=((Q*density_water*viscosity_air*L)/(L1*viscosity_water*density_air))
+
+printf("flow rate = %f cubic meter/s\n",Q1)
+
+//equating euler number
+
+p=9.8067*10^4; //pressure in pascal
+p1=(p*density_water*Q^2*L^4)/(density_air*Q1^2*L1^4);
+
+printf("pressure drop corresponding to 1kp/square cm = %f kP/square cm",p1/p)
diff --git a/858/CH1/EX1.17/example_17.sce b/858/CH1/EX1.17/example_17.sce new file mode 100755 index 000000000..7f0cb4be6 --- /dev/null +++ b/858/CH1/EX1.17/example_17.sce @@ -0,0 +1,14 @@ +clc
+clear
+printf('example 1.17 page number 50\n')
+
+//to find the specific gravity of plasstic
+
+L=1 //length of prototype in m
+L1=10*L //length of model in m
+density_prototype=2.65 //gm/cc
+density_water=1 //gm/cc
+
+density_model=(L^3*(density_prototype-density_water))/(L1^3)+1;
+
+printf("specific gravity of plastic = %f",density_model)
diff --git a/858/CH1/EX1.18/example_18.sce b/858/CH1/EX1.18/example_18.sce new file mode 100755 index 000000000..03b5e4003 --- /dev/null +++ b/858/CH1/EX1.18/example_18.sce @@ -0,0 +1,27 @@ +clc
+clear
+printf('example 1.18 page number 53\n\n')
+
+//to find error in actual data and nomographic chat value
+
+//for my
+ly = 8 //in cm
+my = ly/((1/0.25) - (1/0.5));
+lz = 10.15 //in cm
+mz = lz/((1/2.85) - (1/6.76));
+mx = (my*mz)/(my+mz);
+printf("mx = %f cm\n",mx)
+err = ((1-0.9945)/0.9945)*100;
+printf("\nerror = %f \n",err)
+
+x = 2
+y = 0.5:0.5:2.5;
+
+plot(x,y)
+title("nomograph")
+xlabel("x")
+ylabel("y")
+
+x = 3
+y = 0.4:0.2:2;
+plot(x,y)
diff --git a/858/CH1/EX1.19/example_19.sce b/858/CH1/EX1.19/example_19.sce new file mode 100755 index 000000000..bf812c496 --- /dev/null +++ b/858/CH1/EX1.19/example_19.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf('example 1.19 page number 54\n')
+
+//to find the economic pipe diameter from nomograph
+//from the nomograph,we get the values of w and density
+
+w=450 //in kg/hr
+density=1000 //in kg/m3
+d=16 //in mm
+
+u=(w/density)/(3.14*d^2/4);
+Re=u*density*d/0.001;
+
+if Re>2100 then printf("flow is turbulent and d= %f mm",d)
+else disp ("flow is laminar and this nomograph is not valid")
+end
diff --git a/858/CH1/EX1.2/example_2.sce b/858/CH1/EX1.2/example_2.sce new file mode 100755 index 000000000..39a696339 --- /dev/null +++ b/858/CH1/EX1.2/example_2.sce @@ -0,0 +1,11 @@ +clc
+clear
+printf("example 1.2 page number 20\n\n")
+//find the volume occupied by propane
+
+mass_propane=14.2 //in kg
+molar_mass=44 //in kg
+moles=(mass_propane*1000)/molar_mass;
+volume=22.4*moles; //in liters
+
+printf("volume = %d liters\n\n",volume)
diff --git a/858/CH1/EX1.3/example_3.sce b/858/CH1/EX1.3/example_3.sce new file mode 100755 index 000000000..7edee4da4 --- /dev/null +++ b/858/CH1/EX1.3/example_3.sce @@ -0,0 +1,40 @@ +clc
+clear
+printf("example 1.3 page number 20\n\n")
+//to find the average weight, weight composition, gas volume in absence of SO2
+y_CO2 = 0.25;
+y_CO = 0.002;
+y_SO2 = 0.012;
+y_N2 = 0.680;
+y_O2 = 0.056;
+
+Mm = y_CO2*44+y_CO*28+y_SO2*64+y_N2*28+y_O2*32;
+printf ("\n molar mass = %d \n",Mm)
+
+printf("\n finding weight composition \n")
+w_CO2 = y_CO2*44*100/Mm;
+printf ("\n weight_CO2 = %f \n\n",w_CO2)
+w_CO = y_CO*28*100/Mm;
+printf ("weight_CO = %f \n\n",w_CO)
+w_SO2 = y_SO2*64*100/Mm;
+printf ("weight_SO2 = %f \n\n", w_SO2)
+w_N2 = y_N2*28*100/Mm;
+printf ("weight_N2 = %f \n\n", w_N2)
+w_O2 = y_O2*32*100/Mm;
+printf ("weight_O2 = %f \n\n", w_O2)
+
+printf("if SO2 is removed \n\n")
+v_CO2 = 25;
+v_CO = 0.2;
+v_N2 = 68.0;
+v_O2 = 5.6;
+v = v_CO2+v_CO+v_N2+v_O2;
+v1_CO2 = (v_CO2*100/98.8);
+
+printf ("volume_CO2 = %f \n\n", v1_CO2)
+v1_CO = (v_CO*100/98.8);
+printf ("volume_CO = %f \n\n",v1_CO)
+v1_N2 = (v_N2*100/98.8);
+printf ("volume_N2 = %f \n\n",v1_N2)
+v1_O2 = (v_O2*100/98.8);
+printf ("volume_O2 = %f \n\n",v1_O2 )
diff --git a/858/CH1/EX1.4/example_4.sce b/858/CH1/EX1.4/example_4.sce new file mode 100755 index 000000000..af15455ac --- /dev/null +++ b/858/CH1/EX1.4/example_4.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 1.4 page number 24\n\n")
+//to find volume of NH3 dissolvable in water
+
+p=1 //atm
+H=2.7 //atm
+x=p/H;
+
+mole_ratio = (x)/(1-x);
+moles_of_water=(100*1000)/18;
+moles_of_NH3=mole_ratio*moles_of_water;
+
+printf("moles of NH3 dissolved = %f\n\n",moles_of_NH3)
+
+volume_NH3=(moles_of_NH3*22.4*293)/273;
+printf("volume of NH3 dissolved = %f liters",volume_NH3)
diff --git a/858/CH1/EX1.5/example_5.sce b/858/CH1/EX1.5/example_5.sce new file mode 100755 index 000000000..68ae9dcb7 --- /dev/null +++ b/858/CH1/EX1.5/example_5.sce @@ -0,0 +1,16 @@ +clc
+clear
+
+printf("example 1.5 page number 24\n\n")
+
+//to calculate amount of CO2 released by water
+p=746 //in mm Hg
+H=1.08*10^6 //in mm Hg, Henry's constant
+
+x= p/H; //mole fraction of CO2
+X=x*(44/18); //mass ratio of CO2 in water
+
+initial_CO2 = 0.005; //kg CO2/kg H20
+G=1000*(initial_CO2-X);
+
+printf("CO2 given up by 1 cubic meter of water = %f kg CO2/cubic meter H20",G)
diff --git a/858/CH1/EX1.6/example_6.sce b/858/CH1/EX1.6/example_6.sce new file mode 100755 index 000000000..9b978ffd4 --- /dev/null +++ b/858/CH1/EX1.6/example_6.sce @@ -0,0 +1,20 @@ +clc
+clear
+
+printf('example 1.6 page number 27 \n\n')
+//to find vapor pressre of ethyl alchohal
+
+pa1 = 23.6; //VP of ethyl alchohal at 10 degree C
+pa3=760 //VP of ethyl alchohal at 78.3 degree C in mm Hg
+pb1 = 9.2 //VP of ethyl water at 10 degree C in mm Hg
+pb3=332 //VP of ethyl water at 78.3 degree C in mm Hg
+
+C=(log10(pa1/pa3)/(log10(pb1/pb3)));
+
+pb2=149 //VP of water at 60 degree C in mm Hg
+
+pas=(pb3/pb2);
+pa=C*log10(pas);
+pa2=pa3/(10^pa);
+
+printf("vapor pressure of ethyl alcholoh at 60 degree C = %f mm Hg",pa2)
diff --git a/858/CH1/EX1.7/example_7.sce b/858/CH1/EX1.7/example_7.sce new file mode 100755 index 000000000..1a062897c --- /dev/null +++ b/858/CH1/EX1.7/example_7.sce @@ -0,0 +1,16 @@ +clc
+clear
+
+printf('example 1.7 page number 28 \n\n')
+
+//to find vapor pressure using duhring plot
+
+t1 = 41 //in degree C
+t2=59 //in degree C
+theta_1 =83 //in degree C
+theta_2=100 //in degree C
+
+K = (t1-t2)/(theta_1-theta_2);
+t=59+(K*(104.2-100));
+
+printf ("boiling point of SCl2 at 880 Torr = %f degree celcius",t)
diff --git a/858/CH1/EX1.8/example_8.sce b/858/CH1/EX1.8/example_8.sce new file mode 100755 index 000000000..36833818f --- /dev/null +++ b/858/CH1/EX1.8/example_8.sce @@ -0,0 +1,13 @@ +clc
+clear
+printf('example 1.8 page number 29\n\n')
+//to find the amount of steam released
+
+vp_C6H6 = 520 //in torr
+vp_H2O = 225 //in torr
+mass_water=18
+mass_benzene=78
+
+amount_of_steam = (vp_H2O/vp_C6H6)/(mass_benzene/mass_water);
+
+printf("amount of steam = %f", amount_of_steam)
diff --git a/858/CH1/EX1.9/example_9.sce b/858/CH1/EX1.9/example_9.sce new file mode 100755 index 000000000..548af9bf0 --- /dev/null +++ b/858/CH1/EX1.9/example_9.sce @@ -0,0 +1,24 @@ +clc
+clear
+printf('example 1.9 page number 30\n\n')
+
+//to find equilibrium vapor liquid composition
+p0b = 385 //vapor pressue of benzene at 60 degree C in torr
+p0t=140 //vapor pressue of toluene at 60 degree C in torr
+xb=0.4;
+xt=0.6;
+
+pb=p0b*xb;
+pt=p0t*xt;
+P=pb+pt;
+
+printf("total pressure = %f torr\n\n",P)
+
+yb=pb/P;
+yt=pt/P;
+printf("vapor composition of benzene = %f \n vapor composition of toluene = %f\n\n",yb,yt)
+
+//for liquid boiling at 90 degree C and 760 torr, liquid phase composition
+//x=(760-408)/(1013-408);
+(1013*x)+(408*(1-x))==760;
+printf("mole fraction of benzene in liquid mixture = %f \n mole fraction of toluene in liquid mixture= %f",x,1-x)
diff --git a/858/CH2/EX2.1/example_1.sce b/858/CH2/EX2.1/example_1.sce new file mode 100755 index 000000000..ac461f69c --- /dev/null +++ b/858/CH2/EX2.1/example_1.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 2.1 page number 71\n\n")
+
+//to find the volume of oxygen that can be obtained
+
+p1=15 //in bar
+p2=1.013 //in bar
+t1=283 //in K
+t2=273 //in K
+v1=10 //in l
+
+v2=p1*v1*t2/(t1*p2);
+
+printf("volume of oxygen = %f liters",v2)
diff --git a/858/CH2/EX2.10/example_10.sce b/858/CH2/EX2.10/example_10.sce new file mode 100755 index 000000000..89a458150 --- /dev/null +++ b/858/CH2/EX2.10/example_10.sce @@ -0,0 +1,14 @@ +clc
+clear
+printf("example 2.10 page number 74\n\n")
+
+//to find amount of KClO3 precipitated
+
+solubility_70=30.2 //in gm/100gm
+w_solute=solubility_70*350/130.2; //in gm
+
+w_water=350-w_solute;
+solubility_30=10.1 //in gm/100gm
+precipitate=(solubility_70-solubility_30)*w_water/100
+
+printf("amount precipitated = %f gm",precipitate)
diff --git a/858/CH2/EX2.11/example_11.sce b/858/CH2/EX2.11/example_11.sce new file mode 100755 index 000000000..70c5af3e9 --- /dev/null +++ b/858/CH2/EX2.11/example_11.sce @@ -0,0 +1,12 @@ +clc
+clear
+printf("example 2.11 page number 74\n\n")
+
+//to find the pressure for solubility of CO2
+
+absorbtion_coefficient=1.71 //in liters
+molar_mass=44;
+solubility=absorbtion_coefficient*molar_mass/22.4; //in gm
+pressure=8/solubility*101.3;
+
+printf("pressure required = %f kPa",pressure)
diff --git a/858/CH2/EX2.12/example_12.sce b/858/CH2/EX2.12/example_12.sce new file mode 100755 index 000000000..0108a5adc --- /dev/null +++ b/858/CH2/EX2.12/example_12.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 2.12 page number 74\n\n")
+
+//to find the vapor pressure of water
+
+w_water=540 //in gm
+w_glucose=36 //in gm
+m_water=18; //molar mass of water
+m_glucose=180; //molar mass of glucose
+
+x=(w_water/m_water)/(w_water/m_water+w_glucose/m_glucose);
+p=8.2*x;
+depression=8.2-p;
+
+printf("depression in vapor pressure = %f Pa",depression*1000)
diff --git a/858/CH2/EX2.13/example_13.sce b/858/CH2/EX2.13/example_13.sce new file mode 100755 index 000000000..83470bdc8 --- /dev/null +++ b/858/CH2/EX2.13/example_13.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 2.13 page number 75\n\n")
+
+//to find the boiling point of solution
+
+w_glucose=9 //in gm
+w_water=100 //in gm
+E=0.52;
+m=90/180; //moles/1000gm water
+
+delta_t=E*m;
+boiling_point=100+delta_t;
+
+printf("boiling_point of water = %f degreeC",boiling_point)
diff --git a/858/CH2/EX2.14/example_14.sce b/858/CH2/EX2.14/example_14.sce new file mode 100755 index 000000000..87413d016 --- /dev/null +++ b/858/CH2/EX2.14/example_14.sce @@ -0,0 +1,20 @@ +clc
+clear
+printf("example 2.14 page number 75\n\n")
+
+//to find the molar mass and osmotic pressure
+
+K=1.86;
+c=15 //concentration of alcohol
+delta_t=10.26;
+
+m=delta_t/K; //molality
+M=c/(m*85); //molar mass
+printf("molar mass = %f gm\n\n",M*1000)
+
+density=0.97 //g/ml
+cm=c*density/(M*100);
+printf("molar concentration of alcohol = %f moles/l\n\n",cm)
+
+p=cm*8.314*293 //osmotic pressure
+printf("osmotic pressure = %f Mpa\n\n",p/1000)
diff --git a/858/CH2/EX2.15/example_15.sce b/858/CH2/EX2.15/example_15.sce new file mode 100755 index 000000000..668dd839c --- /dev/null +++ b/858/CH2/EX2.15/example_15.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 2.15 page number 75\n\n")
+
+//to find u_in, M_v, k'
+
+u_in = 0.575 //from the graph
+u_s = 0.295 //in mPa-s
+
+M_v = (u_in/(5.80*10^-5))^(1/0.72);
+u_red = 0.628; //in dl/g
+
+c = 0.40 //in g/dl
+k = (u_red-u_in)/((u_in^2)*c);
+
+printf("k = %f \nMv = %f\nu_in = %f dl/gm",k,M_v,u_in)
diff --git a/858/CH2/EX2.16/example_16.sce b/858/CH2/EX2.16/example_16.sce new file mode 100755 index 000000000..18944e881 --- /dev/null +++ b/858/CH2/EX2.16/example_16.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 2.16 page number 76\n\n")
+
+//to find the molecular formula
+
+C=54.5 //% of carbon
+H2=9.1 //% of hydrogen
+O2=36.4 //% of oxygen
+x=C/12; //number of carbon molecules
+y=O2/16; //number of oxygen molecules
+z=H2/2 //number of hydrogen molecules
+molar_mass=88;
+density=44;
+
+ratio=molar_mass/density;
+x=ratio*2;
+y=ratio*1;
+z=ratio*4;
+
+printf("x = %f, y = %f, z = %f",x,y,z)
+printf("\n\nformula of butyric acid is = C4H8O2")
diff --git a/858/CH2/EX2.17/example_17.sce b/858/CH2/EX2.17/example_17.sce new file mode 100755 index 000000000..9a7ffc6a8 --- /dev/null +++ b/858/CH2/EX2.17/example_17.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 2.17 page number 77\n\n")
+
+//to find molecular foemula
+C=93.75 //% of carbon
+H2=6.25 //% of hydrogen
+x=C/12 //number of carbon atoms
+y=H2/2 //number of hydrogen atoms
+molar_mass=64
+density=4.41*29;
+
+ratio=density/molar_mass;
+
+x=ratio*5;
+y=ratio*4;
+
+
+printf("x = %f, y = %f",x,y)
+printf("\n\nformula of butyric acid is = C10H8")
+
diff --git a/858/CH2/EX2.18/example_18.sce b/858/CH2/EX2.18/example_18.sce new file mode 100755 index 000000000..b4fda753a --- /dev/null +++ b/858/CH2/EX2.18/example_18.sce @@ -0,0 +1,31 @@ +clc
+clear
+printf("example 2.18 page number 77\n\n")
+
+//to find molecular formula
+C=50.69 //% of carbon
+H2=4.23 //% of hydrogen
+O2=45.08 //% of oxygen
+a=C/12; //number of carbon molecules
+c=O2/16; //number of oxygen molecules
+b=H2/2; //number of hydrogen molecules
+molar_mass=71;
+
+function M=f(m)
+ M=(2.09*1000)/(60*m);
+
+endfunction
+
+M=f((1.25/5.1));
+
+printf("actual molecular mass = %f\n\n",M)
+
+ratio=M/molar_mass;
+a=ratio*3;
+b=ratio*3;
+c=ratio*2;
+
+
+printf("a = %f, b = %f, c = %f",a,b,c)
+printf("\n\nformula of butyric acid is = C6H6O4")
+
diff --git a/858/CH2/EX2.19/example_19.sce b/858/CH2/EX2.19/example_19.sce new file mode 100755 index 000000000..74672fa19 --- /dev/null +++ b/858/CH2/EX2.19/example_19.sce @@ -0,0 +1,24 @@ +clc
+clear
+printf("example 2.19 page number 78\n\n")
+
+//to find the molecular formula
+C=64.6 //% of carbon
+H2=5.2 //% of hydrogen
+O2=12.6 //% of oxygen
+N2=8.8 //% of nitrogen
+Fe=8.8 //% of iron
+
+a=C/12; //number of carbon molecules
+c=8.8/14; //number of nitrogen molecules
+b=H2/2; //number of hydrogen molecules
+d=O2/16; //number of oxygen molecules
+e=Fe/56 //number of iron atoms
+
+cm=243.4/(8.31*293) //concentration
+
+ molar_mass=63.3/cm;
+
+ printf("a = %f, b = %f, c = %f, d = %f, e = %f",a*6.5,b*6.5,c*6.5,d*6.5,e*6.5)
+printf("\n\nformula of butyric acid is = C34H33N4O5Fe")
+
diff --git a/858/CH2/EX2.2/example_2.sce b/858/CH2/EX2.2/example_2.sce new file mode 100755 index 000000000..0c2176d66 --- /dev/null +++ b/858/CH2/EX2.2/example_2.sce @@ -0,0 +1,25 @@ +clc
+clear
+printf("example 2.2 page number 71\n\n")
+
+//to find volumetric composition,partial pressue of each gas and total pressure of mixture
+
+nCO2=2/44; //moles of CO2
+nO2=4/32; //moles of O2
+nCH4=1.5/16; //moles of CH4
+
+total_moles=nCO2+nO2+nCH4;
+yCO2=nCO2/total_moles;
+yO2=nO2/total_moles;
+yCH4=nCH4/total_moles;
+
+printf (" Composition of mixture = \nCH4 = %f \nO2 = %f \n CO2 = %f \n\n",yCH4,yO2,yCO2)
+
+pCO2=nCO2*8.314*273/(6*10^-3);
+pO2=nO2*8.314*273/(6*10^-3);
+pCH4=nCH4*8.314*273/(6*10^-3);
+
+printf ("pressure of CH4 = %f kPa \npressure of O2 = %f kPa\n pressure of CO2 =%f kPa\n\n",pCH4*10^-3,pO2*10^-3,pCO2*10^-3)
+
+total_pressure=pCO2+pCH4+pO2;
+printf ("total pressure = %f Kpa",total_pressure*10^-3)
diff --git a/858/CH2/EX2.20/example_20.sce b/858/CH2/EX2.20/example_20.sce new file mode 100755 index 000000000..aaf4f41d0 --- /dev/null +++ b/858/CH2/EX2.20/example_20.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 2.20 page number 78\n\n")
+
+//to find sequence of deposition
+E1=-0.25;
+E2=0.80;
+E3=0.34;
+
+a=[E1;E2;E3];
+b=gsort(a);
+
+printf("sorted potential in volts =")
+disp (b)
+disp ("E2>E3>E1")
+disp ("silver>copper>nickel")
+
+
diff --git a/858/CH2/EX2.21/example_21.sce b/858/CH2/EX2.21/example_21.sce new file mode 100755 index 000000000..d7da98cf4 --- /dev/null +++ b/858/CH2/EX2.21/example_21.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 2.21 page number 79\n\n")
+
+//to find the emf of cell
+
+E0_Zn=-0.76;
+E0_Pb=-0.13;
+c_Zn=0.1;
+c_Pb=0.02;
+
+E_Zn=E0_Zn+(0.059/2)*log10(c_Zn);
+E_Pb=E0_Pb+(0.059/2)*log10(c_Pb);
+E=E_Pb-E_Zn;
+
+printf("emf of cell = %f V",E)
+printf("\n\nSince potential of lead is greater than that of zinc thus reduction will occur at lead electrode and oxidation will occur at zinc electrode")
diff --git a/858/CH2/EX2.22/example_22.sce b/858/CH2/EX2.22/example_22.sce new file mode 100755 index 000000000..4e0ac0209 --- /dev/null +++ b/858/CH2/EX2.22/example_22.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 2.22 page number 79\n\n")
+
+//to find the emf of cell
+E0_Ag=0.80;
+E0_AgNO3=0.80;
+c_Ag=0.001;
+c_AgNO3=0.1;
+
+E_Ag=E0_Ag+(0.059)*log10(c_Ag);
+E_AgNO3=E0_AgNO3+(0.059)*log10(c_AgNO3);
+E=E_AgNO3-E_Ag;
+
+printf("emf of cell = %f V" ,E)
+printf("\n\nsince E is positive, the left hand electrode will be anode and the electron will travel in the external circuit from the left hand to the right hand electrode")
diff --git a/858/CH2/EX2.23/example_23.sce b/858/CH2/EX2.23/example_23.sce new file mode 100755 index 000000000..c3ea8b288 --- /dev/null +++ b/858/CH2/EX2.23/example_23.sce @@ -0,0 +1,10 @@ +clc
+clear
+printf("example 2.23 page number 79\n\n")
+
+//to find emf of cell
+pH=12; //pH of solution
+E_H2=0;
+E2=-0.059*pH;
+E=E_H2-E2;
+printf("EMF of cell = %f V",E)
diff --git a/858/CH2/EX2.24/example_24.sce b/858/CH2/EX2.24/example_24.sce new file mode 100755 index 000000000..d279c1848 --- /dev/null +++ b/858/CH2/EX2.24/example_24.sce @@ -0,0 +1,12 @@ +clc
+clear
+printf("example 2.24 page number 80\n\n")
+
+//to find amount of silver deposited
+I=3 //in Ampere
+t=900 //in s
+m_eq=107.9 //in gm/mol
+F=96500;
+
+m=(I*t*m_eq)/F;
+printf("mass = %f gm",m)
diff --git a/858/CH2/EX2.25/example_25.sce b/858/CH2/EX2.25/example_25.sce new file mode 100755 index 000000000..508af8f6c --- /dev/null +++ b/858/CH2/EX2.25/example_25.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 2.25 page number 80\n\n")
+
+//to find the time for electroplating
+volume=10*10*0.005; //in cm3
+mass=volume*8.9;
+F=96500;
+atomic_mass=58.7 //in amu
+current=2.5 //in Ampere
+
+charge=(8.9*F*2)/atomic_mass;
+yield=0.95;
+actual_charge=charge/(yield*3600);
+t=actual_charge/current;
+
+printf("time required = %f hours",t)
diff --git a/858/CH2/EX2.26/example_26.sce b/858/CH2/EX2.26/example_26.sce new file mode 100755 index 000000000..f763adc9c --- /dev/null +++ b/858/CH2/EX2.26/example_26.sce @@ -0,0 +1,12 @@ +clc
+clear
+printf("example 2.26 page number 80\n\n")
+
+//to find hardness of water
+m_MgSO4=90 //in ppm
+MgSO4_parts=120;
+CaCO3_parts=100;
+
+hardness=(CaCO3_parts/MgSO4_parts)*m_MgSO4;
+
+printf("hardness of water = %f mg/l",hardness)
diff --git a/858/CH2/EX2.27/example_27.sce b/858/CH2/EX2.27/example_27.sce new file mode 100755 index 000000000..ff5328e3c --- /dev/null +++ b/858/CH2/EX2.27/example_27.sce @@ -0,0 +1,30 @@ +clc
+clear
+printf("example 2.26 page number 80\n\n")
+
+m1 = 162 //mass of calcium bi carbonate in mg
+m2 = 73 //mass of magnesium bi carbonate in mg
+m3 = 136 // mass of calsium sulfate in mg
+m4 = 95 // mass of magnesium cloride
+m5 = 500 //mass of sodium cloride in mg
+m6 = 50 // mass of potassium cloride in mg
+
+content_1 = m1*100/m1; //content of calcium bi carbonate in mg
+content_2 = m2*100/(2*m2); //content of magnesium bi carbonate in mg
+content_3 = m3*100/m3; // content of calsium sufate in mg
+content_4 = m4*100/m4; // content of magnesium cloride
+
+//part_1
+
+temp_hardness = content_1 + content_2; //depends on bicarbonate only
+total_hardness = content_1+content_2+content_3+content_4;
+printf("total hardness = %f\n temporary hardness = %f \n",temp_hardness,total_hardness)
+
+//part 2
+wt_lime = (74/100)*(content_1+2*content_2+content_4);
+actual_lime = wt_lime/0.85;
+printf("amount of lime required = %f \n",actual_lime)
+
+soda_required = (106/100)*(content_1+content_4);
+actual_soda = soda_required/0.98;
+printf("amount of soda required = %f \n",actual_soda)
diff --git a/858/CH2/EX2.28/example_28.sce b/858/CH2/EX2.28/example_28.sce new file mode 100755 index 000000000..95249c066 --- /dev/null +++ b/858/CH2/EX2.28/example_28.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 2.28 page number 82\n\n")
+
+//to find hardness of water
+
+volume_NaCl=50 //in l
+c_NaCl=5000 //in mg/l
+
+m=volume_NaCl*c_NaCl;
+equivalent_NaCl=50/58.5;
+
+hardness=equivalent_NaCl*m;
+
+printf("hardness of water = %f mg/l",hardness/1000)
diff --git a/858/CH2/EX2.29/example_29.sce b/858/CH2/EX2.29/example_29.sce new file mode 100755 index 000000000..8eb909d7a --- /dev/null +++ b/858/CH2/EX2.29/example_29.sce @@ -0,0 +1,27 @@ +clc
+clear
+printf("example 2.29 page number 82\n\n")
+
+//to find the total vapor pressure and molar compositions
+
+m_benzene = 55 //in kg
+m_toluene = 28 //in kg
+m_xylene = 17 // in kg
+
+mole_benzene = m_benzene/78;
+mole_toluene = m_toluene/92;
+mole_xylene = m_xylene/106;
+
+mole_total = mole_benzene+mole_toluene+mole_xylene;
+x_benzene = mole_benzene/mole_total;
+x_toluene = mole_toluene/mole_total;
+x_xylene = mole_xylene/mole_total;
+
+P = x_benzene*178.6+x_toluene*74.6+x_xylene*28;
+printf("total pressure = %f kPa\n",P)
+
+benzene = (x_benzene*178.6*100)/P;
+toluene = (x_toluene*74.6*100)/P;
+xylene = (x_xylene*28*100)/P;
+
+printf("xylene = %f \n toluene = %f \n benzene = %f",xylene,toluene,benzene)
diff --git a/858/CH2/EX2.3/example_3.sce b/858/CH2/EX2.3/example_3.sce new file mode 100755 index 000000000..d4eaa19c6 --- /dev/null +++ b/858/CH2/EX2.3/example_3.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 2.3 page number 72\n\n")
+
+//to find equivalent mass of metal
+
+P=104.3 //total pressure in KPa
+pH2O=2.3 //in KPa
+pH2=P-pH2O; //in KPa
+
+VH2=209*pH2*273/(293*101.3)
+
+printf("volume of hydrogen obtained = %f ml\n\n",VH2)
+
+//calculating amount of metal having 11.2l of hydrogen
+
+m=350/196.08*11.2 //mass of metal in grams
+printf("mass of metal equivalent to 11.2 litre/mol of hydrogen = %f gm",m)
diff --git a/858/CH2/EX2.30/example_30.sce b/858/CH2/EX2.30/example_30.sce new file mode 100755 index 000000000..8f659237c --- /dev/null +++ b/858/CH2/EX2.30/example_30.sce @@ -0,0 +1,36 @@ +clc
+clear
+printf("example 2.30 page number 83\n\n")
+
+//to find the mixture composition
+
+vapor_pressure=8 //in kPa
+pressure=100 //in kPa
+
+//part 1
+volume=1 //in m3
+volume_ethanol=volume*(vapor_pressure/pressure);
+volume_air=1-volume_ethanol;
+printf("volumetric composition:- \nair composition = %f\n ethanol compostion = %f",volume_air*100,volume_ethanol*100)
+
+//part 2
+molar_mass_ethanol=46;
+molar_mass_air=28.9;
+mass_ethanol=0.08*molar_mass_ethanol; //in kg
+mass_air=0.92*molar_mass_air; //in kg
+fraction_ethanol=(mass_ethanol*100)/(mass_air+mass_ethanol);
+fraction_air=(mass_air*100)/(mass_air+mass_ethanol);
+printf("\n\ncomposition by weight:-\nAir = %f Ethanol vapor = %f",fraction_air,fraction_ethanol)
+
+//part 3
+mixture_volume=22.3*(101.3/100)*(299/273); //in m3
+weight_ethanol=mass_ethanol/mixture_volume;
+printf("\n\nweight of ethanol/cubic meter = %f Kg",weight_ethanol)
+
+//part 4
+w_ethanol=mass_ethanol/mass_air;
+printf("\n\nweight of ethanol/kg vapor free air = %f Kg",w_ethanol)
+
+//part 5
+moles_ethanol=0.08/0.92;
+printf("\n\nkmol of ethanol per kmol of vapor free air = %f",moles_ethanol)
diff --git a/858/CH2/EX2.31/example_31.sce b/858/CH2/EX2.31/example_31.sce new file mode 100755 index 000000000..93b25e8b9 --- /dev/null +++ b/858/CH2/EX2.31/example_31.sce @@ -0,0 +1,27 @@ +clc
+clear
+printf("example 2.31 page number 84\n\n")
+
+//to find relative saturation and dew point
+
+vapor_pressure=8 //in kPa
+volume_ethanol=0.05;
+
+//basis 1kmol of mixture
+
+partial_pressure=volume_ethanol*100;
+relative_saturation=partial_pressure/vapor_pressure;
+mole_ratio=volume_ethanol/(1-volume_ethanol);
+printf("mole ratio = %f \nrelative saturation = %f",mole_ratio,relative_saturation*100)
+
+//basis 1kmol saturated gas mixture at 100kPa
+volume_vapor=(8/100)*100;
+ethanol_vapor=volume_vapor/100;
+air_vapor=1-ethanol_vapor;
+saturation_ratio=ethanol_vapor/air_vapor;
+percentage_saturation=mole_ratio/saturation_ratio;
+
+printf("\n\npercentage saturation = %f",percentage_saturation)
+
+//dew point
+printf("\n\ncorresponding to partial pressure of 5kPa we get a dew point of 17.3 degree celcius")
diff --git a/858/CH2/EX2.32/example_32.sce b/858/CH2/EX2.32/example_32.sce new file mode 100755 index 000000000..459b9b9a4 --- /dev/null +++ b/858/CH2/EX2.32/example_32.sce @@ -0,0 +1,31 @@ +clc
+clear
+printf("example 2.32 page number 84\n\n")
+
+//to find the properties of humid air
+
+p = 4.24 //in kPa
+H_rel = 0.8;
+p_partial = p*H_rel;
+molal_H = p_partial/(100-p_partial);
+printf("initial molal humidity = %f\n\n",molal_H)
+
+//part 2
+P = 200 //in kPa
+p_partial = 1.70 //in kPa
+final_H = p_partial/(P-p_partial);
+printf("final molal humidity = %f\n\n",final_H)
+
+//part 3
+p_dryair = 100 - 3.39;
+v = 100*(p_dryair/101.3)*(273/303);
+moles_dryair = v/22.4;
+vapor_initial = molal_H*moles_dryair;
+vapor_final = final_H*moles_dryair;
+water_condensed = (vapor_initial-vapor_final)*18;
+printf("amount of water condensed = %f \n\n",water_condensed)
+
+//part 4
+total_air = moles_dryair+vapor_final;
+final_v = 22.4*(101.3/200)*(288/273)*total_air;
+printf("final volume of wety air = %f \n\n",final_v)
diff --git a/858/CH2/EX2.4/example_4.sce b/858/CH2/EX2.4/example_4.sce new file mode 100755 index 000000000..c4a8d2f30 --- /dev/null +++ b/858/CH2/EX2.4/example_4.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 2.4 page number 72\n\n")
+
+//to find NaCl content in NaOH solution
+
+w=2 //in gm
+m=0.287 //in gm
+
+//precipitate from 58.5gm of NaCl=143.4gm
+
+mNaCl=58.5/143.4*m;
+
+printf("mass of NaCl = %f gm\n",mNaCl )
+
+percentage_NaCl=mNaCl/w*100;
+printf("amount of NaCl = %f",percentage_NaCl)
diff --git a/858/CH2/EX2.5/example_5.sce b/858/CH2/EX2.5/example_5.sce new file mode 100755 index 000000000..b4989d530 --- /dev/null +++ b/858/CH2/EX2.5/example_5.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 2.5 page number 72\n\n")
+
+//to find the carbon content in sample
+
+w=4.73 //in gm5
+VCO2=5.30 //in liters
+
+weight_CO2=44/22.4*VCO2;
+carbon_content=12/44*weight_CO2;
+
+percentage_content=(carbon_content/w)*100;
+
+printf("percentage amount of carbon in sample = %f",percentage_content)
diff --git a/858/CH2/EX2.6/example_6.sce b/858/CH2/EX2.6/example_6.sce new file mode 100755 index 000000000..a9d65e5dd --- /dev/null +++ b/858/CH2/EX2.6/example_6.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 2.6 page number 73\n\n")
+//to find the volume of air
+
+volume_H2=0.5 //in m3
+volume_CH4=0.35 //in m3
+volume_CO=0.08 //in m3
+volume_C2H4=0.02 //in m3
+volume_oxygen=0.21 //in m3 in air
+
+//required oxygen for various gases
+H2=0.5*volume_H2;
+CH4=2*volume_CH4;
+CO=0.5*volume_CO;
+C2H4=3*volume_C2H4;
+
+total_O2=H2+CH4+CO+C2H4;
+oxygen_required=total_O2/volume_oxygen;
+
+printf("amount of oxygen required = %f cubic meter",oxygen_required)
diff --git a/858/CH2/EX2.7/example_7.sce b/858/CH2/EX2.7/example_7.sce new file mode 100755 index 000000000..11020ae1e --- /dev/null +++ b/858/CH2/EX2.7/example_7.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 2.7 page number 73\n\n")
+
+//to find the volume of sulphuric acid and mass of water consumed
+
+density_H2SO4 = 1.10 //in g/ml
+mass_1 = 100*density_H2SO4; //mass of 100ml of 15% solution
+mass_H2SO4 = 0.15*mass_1;
+density_std = 1.84 //density of 96% sulphuric acid
+mass_std = 0.96*density_std; //mass of H2SO4 in 1ml 96% H2SO4
+
+volume_std = mass_H2SO4/mass_std; //volume of 96%H2SO4
+mass_water = mass_1 - mass_H2SO4;
+
+printf("volume of 0.96 H2SO4 required = %f ml",volume_std)
+printf("\nmass of water required = %f g",mass_water)
diff --git a/858/CH2/EX2.8/example_8.sce b/858/CH2/EX2.8/example_8.sce new file mode 100755 index 000000000..08e4c91ff --- /dev/null +++ b/858/CH2/EX2.8/example_8.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 2.8 page number 73\n\n")
+
+//to find molarity,molality and normality
+
+w_H2SO4=0.15 //in gm/1gm solution
+density=1.10 //in gm/ml
+m=density*1000; //mass per liter
+weight=m*w_H2SO4; //H2SO4 per liter solution
+molar_mass=98;
+
+Molarity=weight/molar_mass;
+printf("Molarity = %f mol/l\n\n",Molarity)
+
+equivalent_mass=49;
+normality=weight/equivalent_mass;
+printf("Normality = %f N\n\n",normality)
+
+molality=176.5/molar_mass;
+printf("Molality = %f",molality)
diff --git a/858/CH2/EX2.9/example_9.sce b/858/CH2/EX2.9/example_9.sce new file mode 100755 index 000000000..025b0e232 --- /dev/null +++ b/858/CH2/EX2.9/example_9.sce @@ -0,0 +1,9 @@ +clc
+clear
+printf("example 2.9 page number 74\n\n")
+
+molar_mass_BaCl2=208.3; //in gm
+equivalent_H2SO4=0.144;
+normality=equivalent_H2SO4*1000/28.8;
+
+printf("Normality = %f N",normality)
diff --git a/858/CH3/EX3.1/example_1.sce b/858/CH3/EX3.1/example_1.sce new file mode 100755 index 000000000..df337ce67 --- /dev/null +++ b/858/CH3/EX3.1/example_1.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("example 3.1 page number 90\n\n")
+
+//to find the coal consumption
+w_C = 0.6; //amount of carbon in coal
+N2_content = 40 //in m3 per 100m3 air
+
+air_consumed = N2_content/0.79;
+weight_air = air_consumed*(28.8/22.4);
+O2_content = air_consumed*32*(0.21/22.4); //in kg
+
+H2_content = 20 //in m3
+
+steam_consumed = H2_content*(18/22.4);
+
+C_consumption1 = (12/18)*steam_consumed; //in reaction 1
+C_consumption2 = (24/32)*O2_content; //in reaction 2
+
+total_consumption = C_consumption1+C_consumption2;
+coal_consumption = total_consumption/w_C;
+
+printf("coal consumption = %f kg",coal_consumption)
diff --git a/858/CH3/EX3.10/example_10.sce b/858/CH3/EX3.10/example_10.sce new file mode 100755 index 000000000..06a1e7a48 --- /dev/null +++ b/858/CH3/EX3.10/example_10.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 3.10 page number 98\n\n")
+
+//to find the amount of water evaporated
+xf = 0.15;
+xl = (114.7)/(114.7+1000);
+xc = 1;
+
+K2Cr2O7_feed = 1000*0.15; //in kg
+
+n = 0.8;
+C = n*K2Cr2O7_feed;
+V = (K2Cr2O7_feed-120 - 880*0.103)/(-0.103);
+
+printf("amount of water evaporated = %f kg",V)
diff --git a/858/CH3/EX3.11/example_11.sce b/858/CH3/EX3.11/example_11.sce new file mode 100755 index 000000000..bbd396719 --- /dev/null +++ b/858/CH3/EX3.11/example_11.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 3.10 page number 98\n\n")
+
+//to find the yield of crystals
+
+xc = 106/286;
+xf = 0.25;
+xl = 27.5/127.5;
+
+water_present = 100*(1-xf); //in kg
+V = 0.15*75; //in kg
+C = (100*xf - 88.7*xl)/(xc-xl);
+Na2CO3_feed = 25/xc;
+
+yield = (C/Na2CO3_feed)*100;
+
+printf("yield = %f ",yield)
diff --git a/858/CH3/EX3.12/example_12.sce b/858/CH3/EX3.12/example_12.sce new file mode 100755 index 000000000..acb84650c --- /dev/null +++ b/858/CH3/EX3.12/example_12.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 3.12 page number 99\n\n")
+
+//to find the fraction of air recirculated
+
+r = 50 //weight of dry air passing through drier
+w1 = 1.60 //in kg per kg dry solid
+w2 = 0.1 //in kg/kg dry solid
+H0 = 0.016 //in kg water vapor/kg dry air
+H2 = 0.055 //in kg water vapor/kg dry air
+
+y = 1 - (w1-w2)/(r*(H2-H0));
+printf("fraction of air recirculated = %f",y)
+
+H1 = H2 - (w1-w2)/r;
+printf("\n\nhumidity of air entering the drier = %f kg water vapor/kg kg dry air",H1)
+
+//check
+H11 = H2*y+H0*(1-y);
+if H1 == H11 then printf("\n\nfraction of air recirculated = %f \n verified",y)
+end
diff --git a/858/CH3/EX3.13/example_13.sce b/858/CH3/EX3.13/example_13.sce new file mode 100755 index 000000000..6b2c194a0 --- /dev/null +++ b/858/CH3/EX3.13/example_13.sce @@ -0,0 +1,28 @@ +clc
+clear
+printf("example 3.13 page number 100\n\n")
+
+//to find the volumetric flow rate and fraction of air passing through the cooler
+
+//basis 60m3/h of conditioned air at 25 degree C and 60% RH
+
+Hf = 0.012;
+Hi = 0.033;
+H1 = 0.0075;
+
+water_vapor = Hf/18; //in kmol of water vapor
+dry_air = 1/28.9; //in kmol
+total_mass = water_vapor+dry_air;
+
+volume = 22.4*(298/273)*total_mass;
+weight = 60/volume;
+printf("weight of dry air handled per hr = %f kg",weight)
+
+//part 1
+inlet_watervapor = 0.033/18; //in kmol of water vapor
+volume_inlet = 22.4*(308/273)*(inlet_watervapor+dry_air);
+printf("\n\nvolumetric flow rate of inlet air = %f cubic meter",volume_inlet*weight)
+
+//part 2
+y = (Hf - Hi)/(H1 - Hi);
+printf("\n\nfraction of inlet air passing through cooler = %f",y)
diff --git a/858/CH3/EX3.14/example_14.sce b/858/CH3/EX3.14/example_14.sce new file mode 100755 index 000000000..439a75167 --- /dev/null +++ b/858/CH3/EX3.14/example_14.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 3.14 page number 102\n\n")
+
+//to find the fraction of purged recycle and total yield
+
+//x- moles of N2 and H2 recycled; y - moles of N2 H2 purged
+
+Ar_freshfeed = 0.2;
+//argon in fresh feed is equal to argon in purge
+
+y = 0.2/0.0633; //argon in purge = 0.0633y
+x = (0.79*100 - y)/(1-0.79);
+printf("y = %f kmol\nx = %f kmol",y,x)
+
+//part 1
+fraction = y/x;
+printf("\n\nfration of recycle that is purged = %f",fraction)
+
+//part 2
+yield = 0.105*(100+x);
+printf("\n\noverall yield of ammonia = %f kmol",yield)
diff --git a/858/CH3/EX3.15/example_15.sce b/858/CH3/EX3.15/example_15.sce new file mode 100755 index 000000000..c83ccebf4 --- /dev/null +++ b/858/CH3/EX3.15/example_15.sce @@ -0,0 +1,11 @@ +clc
+clear
+printf("example 3.15 page number 107\n\n")
+
+//to find change in enthalpy
+H0_CH4 = -74.9 //in kJ
+H0_CO2 = -393.5 //in kJ
+H0_H2O = -241.8 //in kJ
+
+delta_H0 = H0_CO2+2*H0_H2O-H0_CH4;
+printf("change in enthalpy = %f kJ",delta_H0)
diff --git a/858/CH3/EX3.16/example_16.sce b/858/CH3/EX3.16/example_16.sce new file mode 100755 index 000000000..372f5cac7 --- /dev/null +++ b/858/CH3/EX3.16/example_16.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 3.16 page number 107\n\n")
+
+//to compare the enthalpy change in two reactions
+
+H0_glucose = -1273 //in kJ
+H0_ethanol = -277.6 //in kJ
+H0_CO2 = -393.5 //in kJ
+H0_H2O = -285.8 //in kJ
+
+//for reaction 1
+delta_H1 = 2*H0_ethanol+2*H0_CO2-H0_glucose;
+printf("enthalpy change in reaction 1 = %f KJ",delta_H1)
+
+//for reaction 2
+delta_H2 = 6*H0_H2O+6*H0_CO2-H0_glucose;
+printf("\n\nenthalpy change in reaction 2 = %f kJ",delta_H2)
+
+if delta_H1>delta_H2 then disp ("reaction 2 supplies more energy")
+ else disp ("reaction 1 supplies more energy")
+end
diff --git a/858/CH3/EX3.17/example_17.sce b/858/CH3/EX3.17/example_17.sce new file mode 100755 index 000000000..18eb0c538 --- /dev/null +++ b/858/CH3/EX3.17/example_17.sce @@ -0,0 +1,14 @@ +clc
+clear
+printf("example 3.17 page number 108\n\n")
+
+//to find enthalpy of formation of CuSO4.5H2O
+
+delta_H2 = 11.7 //in kJ/mol
+m_CuSO4 = 16 //in gm
+m_H2O = 384 //in gm
+
+delta_H3 = -((m_CuSO4+m_H2O)*4.18*3.95*159.6)/(16*10^3)
+delta_H1 = delta_H3 - delta_H2;
+
+printf("enthalpy of formation = %f kJ/mol",delta_H1)
diff --git a/858/CH3/EX3.18/example_18.sce b/858/CH3/EX3.18/example_18.sce new file mode 100755 index 000000000..37db2eb07 --- /dev/null +++ b/858/CH3/EX3.18/example_18.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 3.18 page number 108\n\n")
+
+//to find the temperature of combustion
+
+H_combustion = 1560000 //in kJ/kmol
+H0_CO2 = 54.56 //in kJ/kmol
+H0_O2 = 35.2 //in kJ/kmol
+H0_steam = 43.38 //in kJ/kmol
+H0_N2 = 33.32 //in kJ/kmol
+
+t = H_combustion/(2*H0_CO2+3*H0_steam+0.875*H0_O2+16.46*H0_N2);
+
+printf("theoritical temperature of combustion = %f degree C",t)
diff --git a/858/CH3/EX3.19/example_19.sce b/858/CH3/EX3.19/example_19.sce new file mode 100755 index 000000000..59cde39aa --- /dev/null +++ b/858/CH3/EX3.19/example_19.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 3.19 page number 109\n\n")
+
+//to find the heat of reaction and consumption of coke
+
+H_NaCl = 410.9 //in MJ/kmol
+H_H2SO4 = 811.3 //in MJ/kmol
+H_Na2SO4 = 1384 //in MJ/kmol
+H_HCl = 92.3 //in MJ/kmol
+
+Q = H_Na2SO4 + 2*H_HCl -2*H_NaCl-H_H2SO4;
+printf("heat of reaction = %f MJ\n\n",Q)
+
+heat_required = 64.5*(500/73);
+coke_consumption = heat_required/19
+printf("amount of coke oven gas consumed = %f cubic meter",coke_consumption)
diff --git a/858/CH3/EX3.2/example_2.sce b/858/CH3/EX3.2/example_2.sce new file mode 100755 index 000000000..11d7be1d1 --- /dev/null +++ b/858/CH3/EX3.2/example_2.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 3.2 page number 91\n\n")
+
+//to find amount of ammonia and air consumed
+
+NH3_required = (17/63)*1000; //NH3 required for 1 ton of nitric acid
+NO_consumption = 0.96;
+HNO3_consumption = 0.92;
+NH3_consumed = NH3_required/(NO_consumption*HNO3_consumption);
+volume_NH3 = NH3_consumed*(22.4/17);
+printf("volume of ammonia consumed= %f cubic metre/h",volume_NH3)
+
+NH3_content = 11 //% by volume
+air_consumption = volume_NH3*((100-11)/11);
+printf("\n\nvolume of air consumed = %f cubic metre/h",air_consumption)
diff --git a/858/CH3/EX3.20/example_20.sce b/858/CH3/EX3.20/example_20.sce new file mode 100755 index 000000000..804cafb66 --- /dev/null +++ b/858/CH3/EX3.20/example_20.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 3.20 page number 109\n\n")
+
+//to find the rate of heat flow
+
+cp_water = 146.5 //in kj/kg
+cp_steam = 3040 //in kJ/kg
+d = 0.102 //in m
+u = 1.5 //in m/s
+density = 1000 //in kg/m3
+
+m = (3.14/4)*d^2*u*density;
+Q = m*(cp_steam-cp_water);
+
+printf("rate of heat flow = %f kW",Q)
diff --git a/858/CH3/EX3.21/example_21.sce b/858/CH3/EX3.21/example_21.sce new file mode 100755 index 000000000..67fc2f755 --- /dev/null +++ b/858/CH3/EX3.21/example_21.sce @@ -0,0 +1,4 @@ +clc
+//EXAMPLE 3.21
+//To find the calorific value of coal
+disp('this is a theoritical problem.Refer the book for solution')
diff --git a/858/CH3/EX3.22/example_22.sce b/858/CH3/EX3.22/example_22.sce new file mode 100755 index 000000000..893cbe949 --- /dev/null +++ b/858/CH3/EX3.22/example_22.sce @@ -0,0 +1,48 @@ +clc
+clear
+printf("example 3.22 page number 110\n\n")
+
+//to find the amount of air required for combustion and composition of flue gas
+wt_C = 0.75 //in kg
+wt_H2 = 0.05 //in kg
+wt_O2 = 0.12 //in kg
+wt_N2 = 0.03 //in kg
+wt_S = 0.01 //in kg
+wt_ash = 0.04 //in kg
+
+O2_C = wt_C*(32/12); //in kg
+O2_H2 = wt_H2*(16/2); //in kg
+O2_S = wt_S*(32/32); //in kg
+O2_required = O2_C+O2_H2+O2_S;
+
+oxygen_supplied = O2_required - wt_O2;
+air_needed = oxygen_supplied/0.23;
+printf("amount of air required = %f kg",air_needed)
+
+volume = (22.4/28.8)*air_needed;
+printf("\n\nvolume of air needed = %f cubic meter",volume)
+
+air_supplied = 1.20*air_needed;
+N2_supplied = air_supplied*0.77;
+total_N2 = N2_supplied+wt_N2;
+
+O2_fluegas = air_supplied*0.23 - oxygen_supplied;
+
+wt_CO2 = wt_C+O2_C;
+wt_SO2 = wt_S+O2_S;
+
+moles_CO2 = wt_CO2/44;
+moles_SO2 = wt_SO2/64;
+moles_N2 = total_N2/28;
+moles_O2 = O2_fluegas/32;
+total_moles = moles_CO2+moles_SO2+moles_N2+moles_O2;
+
+x_CO2 = moles_CO2/total_moles;
+x_SO2 = moles_SO2/total_moles;
+x_N2 = moles_N2/total_moles;
+x_O2 = moles_O2/total_moles;
+
+printf("\n\nCO2 = %f",x_CO2*100)
+printf("\n\nSO2 = %f",x_SO2*100)
+printf("\n\nN2 = %f",x_N2*100)
+printf("\n\nO2 = %f",x_O2*100)
diff --git a/858/CH3/EX3.23/example_23.sce b/858/CH3/EX3.23/example_23.sce new file mode 100755 index 000000000..8dbd593fe --- /dev/null +++ b/858/CH3/EX3.23/example_23.sce @@ -0,0 +1,73 @@ +clc
+clear
+printf("example 3.23 page number 110\n\n")
+
+//to find the composition of flue gas
+
+C = 0.8 //in kg
+H2 = 0.05 //in kg
+S = 0.005 //in kg
+ash = 0.145 //in kg
+
+//required oxygen in kg
+C_O2 = C*(32/12);
+H2_O2 = H2*(16/2);
+S_O2 = S*(32/32);
+O2_supplied = C_O2+S_O2+H2_O2;
+printf("amount of O2 supplied = %f kg\n\n",O2_supplied)
+
+wt_air = O2_supplied*(100/23);
+wt_airsupplied = 1.25*wt_air;
+printf("amount of air supplied = %f kg\n\n",wt_airsupplied)
+
+//flue gas composition
+m_N2 = wt_airsupplied*0.77; //in kg
+mole_N2 = m_N2/28;
+
+m_O2 = (wt_airsupplied-wt_air)*0.23; //in kg
+mole_O2 = m_O2/32;
+
+m_CO2 = C*(44/12); //in kg
+mole_CO2 = m_CO2/44;
+
+m_H2O = H2*(18/2); //in kg
+mole_H2O = m_H2O/18;
+
+m_SO2 = S*(64/32); //in kg
+mole_SO2 = m_SO2/64;
+
+m = m_N2+m_O2+m_CO2+m_H2O+m_SO2
+
+//percent by weight
+w_N2 = m_N2/m;
+printf("percentage of N2 by weight = %f\n\n",w_N2*100)
+
+w_O2 = m_O2/m;
+printf("percentage of O2 by weight = %f\n\n",w_O2*100)
+
+w_CO2 = m_CO2/m;
+printf("percentage of CO2 by weight = %f\n\n",w_CO2*100)
+
+w_H2O = m_H2O/m;
+printf("percentage of H2O by weight = %f\n\n",w_H2O*100)
+
+w_SO2 = m_SO2/m;
+printf("percentage of SO2 by weight = %f\n\n",w_SO2*100)
+
+m1 = mole_N2+mole_O2+mole_CO2+mole_H2O+mole_SO2
+
+//percent by mole
+x_N2 = mole_N2/m1;
+printf("percentage of N2 by mole = %f\n\n",x_N2*100)
+
+x_O2 = mole_O2/m1;
+printf("percentage of O2 by mole = %f\n\n",x_O2*100)
+
+x_CO2 = mole_CO2/m1;
+printf("percentage of CO2 by mole = %f\n\n",x_CO2*100)
+
+x_H2O = mole_H2O/m1;
+printf("percentage of H2O by mole = %f\n\n",x_H2O*100)
+
+x_SO2 = mole_SO2/m1;
+printf("percentage of SO2 by mole = %f\n\n",x_SO2*100)
diff --git a/858/CH3/EX3.24/example_24.sce b/858/CH3/EX3.24/example_24.sce new file mode 100755 index 000000000..d336acfef --- /dev/null +++ b/858/CH3/EX3.24/example_24.sce @@ -0,0 +1,29 @@ +clc
+clear
+printf("example 3.24 page number 112\n\n")
+
+//to find volumetric composition of flue glass
+
+wt_H2 = 0.15;
+wt_C = 0.85;
+O2_H2 = wt_H2*(16/2);
+O2_C = wt_C*(32/12);
+
+total_O2 = O2_H2+O2_C;
+
+wt_air = total_O2/0.23;
+
+air_supplied = 1.15*(wt_air);
+N2_supplied = 0.77*air_supplied/28;
+O2_supplied = 0.23*(air_supplied-wt_air)/32;
+moles_CO2 = 0.85/12;
+
+printf("moles of CO2 = %f kmol\n\n",moles_CO2)
+printf("moles of N2 = %f kmol \n\n",N2_supplied)
+printf("moles of O2 = %f kmol\n\n",O2_supplied)
+
+total_moles = N2_supplied+O2_supplied+moles_CO2;
+
+printf("percentage of CO2 = %f\n\n",(moles_CO2/total_moles)*100)
+printf("percentage of N2 = %f\n\n",(N2_supplied/total_moles)*100)
+printf("percentage of O2 = %f",(O2_supplied/total_moles)*100)
diff --git a/858/CH3/EX3.25/example_25.sce b/858/CH3/EX3.25/example_25.sce new file mode 100755 index 000000000..4882ba72b --- /dev/null +++ b/858/CH3/EX3.25/example_25.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 3.25 page number 113\n\n")
+
+//to find the excess air supplied
+
+N2 = 80.5 //in m3
+air_supplied = N2/0.79 //in m3
+volume_O2 = air_supplied*0.21; //in m3
+O2_fluegas = 6.1 //in m3
+
+O2_used = volume_O2 - O2_fluegas;
+excess_air_supplied = (O2_fluegas/O2_used)*100;
+
+printf("percentage of excess air supplied = %f",excess_air_supplied)
diff --git a/858/CH3/EX3.26/example_26.sce b/858/CH3/EX3.26/example_26.sce new file mode 100755 index 000000000..f5fdcf056 --- /dev/null +++ b/858/CH3/EX3.26/example_26.sce @@ -0,0 +1,25 @@ +clc
+clear
+printf("example 3.26 page number 114\n\n")
+
+//to find the outlet temperature of water
+
+q_NTP = 10*(200/101.3)*(273/313);
+m_CO2 = 44*(q_NTP/22.4);
+s_CO2 = 0.85 //in kJ/kg K
+
+Q = m_CO2*s_CO2*(40-20) //Q = ms*delta_T
+
+d0 = 0.023 //in mm
+A0 = (3.14/4)*d0^2;
+di = 0.035 //in mm
+Ai = (3.14/4)*di^2;
+
+A_annular = Ai-A0;
+u = 0.15 //in m/s
+m_water = A_annular*(u*3600)*1000 //in kg/hr
+
+s_water = 4.19 //in kJ/kg K
+t = 15+(Q/(m_water*s_water));
+
+printf("exit water temperature = %f degree C",t)
diff --git a/858/CH3/EX3.27/example_27.sce b/858/CH3/EX3.27/example_27.sce new file mode 100755 index 000000000..e99d03ca8 --- /dev/null +++ b/858/CH3/EX3.27/example_27.sce @@ -0,0 +1,38 @@ +clc
+clear
+printf("example 3.27 page number 114\n\n")
+
+//to find the area of heating surface
+F = 1000 //in kg
+xF = 0.01
+
+solid_feed = F*xF;
+water_feed = F - solid_feed;
+
+tF = 40 //in degree C
+hF = 167.5 //in kJ/kg
+xL = 0.02;
+
+solid_liquor = 10 //in kg
+L = solid_liquor/xL;
+tL = 100 //in degree C
+hL = 418.6 //in kJ/kg
+
+V = F -L;
+
+tv = 100 //in degree C
+Hv = 2675 //in kJ/kg
+ts = 108.4 //in degree C
+Hs = 2690 //in kJ/kg
+tc = 108.4 //in degree C
+hc = 454 //in kJ/kg
+
+//applying heat balance
+S = (F*hF-V*Hv-L*hL)/(hc-Hs);
+printf("weight of steam required = %f kg/hr",S)
+
+Q = S*(Hs-hc);
+U = 1.4 //in kW/m2K
+delta_t = ts-tL;
+A = 383.2/(U*delta_t);
+printf("\n\narea of heating surface = %f square meter",A)
diff --git a/858/CH3/EX3.28/example_28.sce b/858/CH3/EX3.28/example_28.sce new file mode 100755 index 000000000..26b253515 --- /dev/null +++ b/858/CH3/EX3.28/example_28.sce @@ -0,0 +1,32 @@ +clc
+clear
+printf("example 3.28 page number 115\n\n")
+
+//to find the top and bottom product,condenser duty,heat input to rebpoiler
+hF = 171 //in kJ/kg
+hD = 67 //in kJ/kg
+hL = hD;
+
+hW = 200 //in kJ/kg
+H = 540 //in kJ/kg
+
+disp('part 1')
+F = 1000 //in kg/h
+xF = 0.40
+xW = 0.02;
+xD = 0.97;
+D = F*(xF-xW)/(xD-xW);
+W = F-D;
+
+printf("bottom product = %f kg/hr",W)
+printf("\ntop product = %f kg/hr\n\n",D)
+
+disp('part 2')
+L = 3.5*D;
+V = L+D;
+Qc = V*H-L*hL-D*hD;
+printf("condenser duty = %f KJ/hr\n\n",Qc)
+
+disp('part 3')
+Qr = Qc - 24200;
+printf("rate of heat input to reboiler = %f kJ/hr",Qr)
diff --git a/858/CH3/EX3.29/example_29.sce b/858/CH3/EX3.29/example_29.sce new file mode 100755 index 000000000..9315a25b5 --- /dev/null +++ b/858/CH3/EX3.29/example_29.sce @@ -0,0 +1,27 @@ +clc
+clear
+printf("example 3.29 page number 117\n\n")
+
+//to find the rate of crystal formation, cooling water rate, required area
+
+F = 1000; //in kg
+V = 0.05*F; //in kg
+xF = 0.48;
+xL = 75/(100+75);
+xC = 1;
+C = (F*xF-950*xL)/(1-0.429);
+printf ("rate of crystal formation = %f kg",C)
+
+L = F-C-V;
+
+//cooling water
+W = (F*2.97*(85-35)+126.9*75.2-V*2414)/(4.19*11);
+printf("\n\nrate of cooling water = %f kg",W)
+
+delta_T1 = 56;
+delta_T2 = 17;
+delta_Tm = (delta_T1-delta_T2)/(log(delta_T1/delta_T2))
+U = 125;
+
+A=(F*2.97*(85-35)+126.9*75.2-V*2414)/(U*delta_Tm*3.6);
+printf("\n\narea = %f square meter",A)
diff --git a/858/CH3/EX3.3/example_3.sce b/858/CH3/EX3.3/example_3.sce new file mode 100755 index 000000000..780428bac --- /dev/null +++ b/858/CH3/EX3.3/example_3.sce @@ -0,0 +1,20 @@ +clc
+clear
+printf("example 3.3 page number 91\n\n")
+
+//to find the consumption of NaCl and H2SO4 in HCl consumption
+
+HCl_production = 500 //required to be produced in kg
+NaCl_required = (117/73)*HCl_production;
+yield = 0.92;
+purity_NaCl= 0.96;
+
+actual_NaCl = NaCl_required/(purity_NaCl*yield);
+printf("amount of NaCl required = %f kg",actual_NaCl)
+
+purity_H2SO4 = 0.93;
+H2SO4_consumption = (98/73)*(HCl_production/(yield*purity_H2SO4));
+printf("\n\namount of H2SO4 consumed = %f kg",H2SO4_consumption)
+
+Na2SO4_produced = (142/73)*HCl_production;
+printf("\n\namount of Na2SO4 produced = %f kg",Na2SO4_produced)
diff --git a/858/CH3/EX3.30/example_30.sce b/858/CH3/EX3.30/example_30.sce new file mode 100755 index 000000000..5ae626691 --- /dev/null +++ b/858/CH3/EX3.30/example_30.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 3.30 page number 118\n\n")
+
+//to find the heat of combustion
+
+delta_n = 10-12; //mole per mole napthanlene
+
+//basis 1g
+moles_napthalene = (1/128);
+
+disp('part 1')
+Qv = 40.28 //in kJ
+Qp = Qv-(delta_n*moles_napthalene*8.3144*298/1000);
+printf("heat of combustion = %f kJ\n\n",Qp)
+
+disp('part 2')
+delta_H = 44.05 //in kJ/gmol
+water_formed = 4/128; //in g mol
+Qp1 = Qp - (delta_H*water_formed);
+printf("heat of combustion = %f kJ",Qp1)
diff --git a/858/CH3/EX3.4/example_4.sce b/858/CH3/EX3.4/example_4.sce new file mode 100755 index 000000000..1a0fb8293 --- /dev/null +++ b/858/CH3/EX3.4/example_4.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 3.4 page number 92\n\n")
+
+//to find the period of service
+
+C2H2_produced = (1/64)*0.86; //in kmol
+volume_C2H2 = C2H2_produced*22.4*1000; //in l
+
+//assuming ideal behaviour,
+volume = (100/101.3)*(273/(273+30));
+time = (volume_C2H2/volume)*(1/60);
+printf("time of service = %f hr",time)
+
+
+
+
diff --git a/858/CH3/EX3.5/example_5.sce b/858/CH3/EX3.5/example_5.sce new file mode 100755 index 000000000..6f284f6a0 --- /dev/null +++ b/858/CH3/EX3.5/example_5.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("example 3.5 page number 92\n\n")
+
+//to find the screen effectiveness
+
+xv = 0.88;
+xf = 0.46;
+xl = 0.32;
+F= 100 //in kg
+
+L = (F*(xf-xv))/(xl-xv);
+V = F-L;
+printf("L = %f Kg \nV = %f Kg",L,V)
+Eo = (V*xv)/(F*xf);
+
+printf(" \n\neffectiveness based on oversized partices = %f \n\n",Eo)
+Eu = (L*(1-xl))/(F*(1-xf));
+
+printf("effectiveness based on undersized partices = %f",Eu)
+E=Eu*Eo;
+
+printf("\n\noverall effectiveness = %f",E)
diff --git a/858/CH3/EX3.6/example_6.sce b/858/CH3/EX3.6/example_6.sce new file mode 100755 index 000000000..d6085e1ef --- /dev/null +++ b/858/CH3/EX3.6/example_6.sce @@ -0,0 +1,41 @@ +clc
+clear
+printf("example 3.6 page number 94\n\n")
+
+//to find the flow rate and concentration
+
+G1 = 3600 //in m3/h
+P = 106.6 //in kPa
+T = 40 //in degree C
+q = G1*(P/101.3)*(273/((273+T))); //in m3/s
+m = q/22.4; //in kmol/h
+y1 = 0.02;
+Y1 = y1/(1-y1);
+
+printf("mole ratio of benzene = %f kmol benzene/kmol dry gas",Y1)
+
+Gs = m*(1-y1);
+printf("\n\nmoles of benzene free gas = %f kmol drygas/h",Gs)
+
+//for 95% removal
+Y2 = Y1*(1-0.95);
+printf("\n\nfinal mole ratio of benzene = %f kmol benzene/kmol dry gas",Y2)
+
+x2 = 0.002
+X2 = 0.002/(1-0.002);
+
+//at equilibrium y* = 0.2406X
+//part 1
+//for oil rate to be minimum the wash oil leaving the absorber must be in equilibrium with the entering gas
+
+y1 = 0.02;
+x1 = y1/(0.2406);
+X1 = x1/(1-x1);
+min_Ls = Gs*((Y1-Y2)/(X1-X2));
+printf("\n\nminimum Ls required = %f kg/h",min_Ls*260)
+
+//for 1.5 times of the minimum
+Ls = 1.5*min_Ls;
+printf("\n\nflow rate of wash oil = %f kg/h",Ls*260)
+X1 = X2 + (Gs*((Y1-Y2)/Ls));
+printf("\n\nconcentration of benzene in wash oil = %f kmol benzene/kmol wash oil",X1)
diff --git a/858/CH3/EX3.7/example_7.sce b/858/CH3/EX3.7/example_7.sce new file mode 100755 index 000000000..2438d9d76 --- /dev/null +++ b/858/CH3/EX3.7/example_7.sce @@ -0,0 +1,49 @@ +clc
+clear
+printf("example 3.7 page number 95\n\n")
+
+//to find the extraction of nicotine
+xf = 0.01
+Xf = xf/(1-xf);
+Feed = 100 //feed in kg
+c_nicotine = Feed*Xf; //nicotine conc in feed
+c_water = Feed*(1-Xf) //water conc in feed
+
+//part 1
+function[f] = F1(x)
+ funcprot(0)
+ f = (x/150)-0.9*((1-x)/99);
+endfunction
+
+//initial guess
+x = 10;
+y = fsolve(x,F1);
+printf("amount of nicotine removed N = %f kg",y)
+//part 2
+function[f] = F1(x)
+ f = (x/50)-0.9*((1-x)/99);
+endfunction
+
+//initial guess
+x = 10;
+N1 = fsolve(x,F1);
+printf("\n\namount of nicotine removed in stage 1, N1 = %f kg",N1)
+function[f] = F1(x,N1)
+ f = (x/50)-0.9*((1-x-N1)/99);
+endfunction
+
+//initial guess
+x = 10;
+N2 = fsolve(x,F1);
+printf("\n\namount of nicotine removed in stage 2, N2 = %f kg",N2)
+function[f] = F1(x,N1,N2)
+ f = (x/50)-0.9*((1-x-N2-N1)/99);
+endfunction
+
+//initial guess
+x = 10;
+N3 = fsolve(x,F1);
+
+printf("\n\namount of nicotine removed in stage 3, N3 = %f kg",N3)
+N = N1+N2+N3;
+printf("\n\ntotal amount of nicotine removed = %f kg",N)
diff --git a/858/CH3/EX3.8/example_8.sce b/858/CH3/EX3.8/example_8.sce new file mode 100755 index 000000000..f6948ef03 --- /dev/null +++ b/858/CH3/EX3.8/example_8.sce @@ -0,0 +1,19 @@ +clc
+clear
+printf("example 3.8 page number 96\n\n")
+
+//to find the amount of water in residue
+
+vp_water = 31.06 //in kPa
+vp_benzene = 72.92 //in kPa
+
+P = vp_water +vp_benzene;
+x_benzene = vp_benzene/P;
+x_water = vp_water/P;
+
+initial_water = 50/18; //in kmol of water
+initial_benzene = 50/78 //in kmol of benzene
+water_evaporated = initial_benzene*(x_water/x_benzene);
+water_left = (initial_water - water_evaporated);
+
+printf("amount of water left in residue = %f kg",water_left*18)
diff --git a/858/CH3/EX3.9/example_9.sce b/858/CH3/EX3.9/example_9.sce new file mode 100755 index 000000000..4df8514ed --- /dev/null +++ b/858/CH3/EX3.9/example_9.sce @@ -0,0 +1,29 @@ +clc
+clear
+printf("example 3.9 page number 97\n\n")
+
+//to find the vapor content of dimethylanaline
+po_D = 4.93 //in kPa
+po_W = 96.3 //in kPa
+n = 0.75 //vaporization efficiency
+
+P = n*po_D+po_W;
+printf("P = %f kPa",P)
+
+x_water = 96.3/100;
+x_dimethylanaline = 1-x_water;
+wt_dimethylanaline = (x_dimethylanaline*121)/(x_dimethylanaline*121+x_water*18);
+printf("\n\nweight of dimethylanaline in water = %f",wt_dimethylanaline*100)
+
+//part 1
+n = 0.8;
+po_D = 32 //in kPa
+actual_vp = n*po_D;
+p_water = 100 - actual_vp;
+steam_required = (p_water*18)/(actual_vp*121);
+printf("\n\namount of steam required = %f kg steam/kg dimethylanaline",steam_required)
+
+//part 2
+x_water = p_water/100;
+wt_water = x_water*18/(x_water*18+(1-x_water)*121);
+printf("\n\nweight of water vapor = %f \nweight of dimethylanaline =%f",wt_water*100,100*(1-wt_water))
diff --git a/858/CH4/EX4.1/example_1.sce b/858/CH4/EX4.1/example_1.sce new file mode 100755 index 000000000..d87134ff3 --- /dev/null +++ b/858/CH4/EX4.1/example_1.sce @@ -0,0 +1,9 @@ +clc
+clear
+printf("example 4.1 page number 125\n\n")
+
+//to find water compressibility
+delta_p=70; //in bar
+Et=20680 //in bar
+compressibility = delta_p/Et;
+printf("compressibilty of water = %f",compressibility)
diff --git a/858/CH4/EX4.10/example_10.sce b/858/CH4/EX4.10/example_10.sce new file mode 100755 index 000000000..ed8b4f6e7 --- /dev/null +++ b/858/CH4/EX4.10/example_10.sce @@ -0,0 +1,13 @@ +clc
+clear
+printf("example 4.10 page number 139\n\n")
+
+//to find the temperature increase
+
+Q=0.001*10^5 //in J/s
+w=0.001*1000 //in kg/s
+density=1000 //in kg/m3
+cp=4.19*10^3 //in J/kg K
+
+delta_T=Q/(w*cp);
+printf("Temperature increase = %f degree celcius",delta_T)
diff --git a/858/CH4/EX4.11/example_11.sce b/858/CH4/EX4.11/example_11.sce new file mode 100755 index 000000000..a10120041 --- /dev/null +++ b/858/CH4/EX4.11/example_11.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("example 4.11 page number 142\n\n")
+
+//to find the pressure
+
+u1=0; //in m/s
+ws=0;
+P1=0.7*10^5 //in Pa
+P3=0
+density=1000 //in kg/m3
+
+u3=((2*(P1-P3))/density)^0.5;
+printf("u3 = %f m/s",u3)
+
+ratio_area=0.5;
+u2=u3/ratio_area;
+printf("\n\nu2 = %f m/s",u2)
+
+//applying bernoulli's equation
+P2=1.7*10^5-((density*u2^2)/2)
+printf("\n\nP2 = %f Pa",P2)
+printf("\nthis flow is physically unreal")
diff --git a/858/CH4/EX4.12/example_12.sce b/858/CH4/EX4.12/example_12.sce new file mode 100755 index 000000000..0d524f6f8 --- /dev/null +++ b/858/CH4/EX4.12/example_12.sce @@ -0,0 +1,19 @@ +clc
+clear
+printf("example 4.12 page number 143\n\n")
+
+//to find the power requirements
+
+Q=3800/(24*3600) //in m3/s
+d=0.202 //in m
+
+u=Q/((3.14/4)*d^2); //in m/s
+delta_P=5.3*10^6 //in Pa
+density=897 //in kg/m3
+F=delta_P/density; //in J/kg
+ws=9.8*30+F;
+mass_flow_rate= Q*density;
+power=(ws*mass_flow_rate)/0.6;
+
+printf("power required = %f kW",power/1000)
+
diff --git a/858/CH4/EX4.13/example_13.sce b/858/CH4/EX4.13/example_13.sce new file mode 100755 index 000000000..6d0973283 --- /dev/null +++ b/858/CH4/EX4.13/example_13.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 4.13 page number 146\n\n")
+
+//to find the tube length
+density=1000 //in kg/m3
+viscosity=1*10^-3 //in Pa s
+P=100*1000 //in Pa
+
+vdP=P/density;
+
+Q=2.5*10^-3/(24*3600)
+A=3.14*(0.0005)^2/4;
+u=Q/A;
+printf("u = %f m/s",u)
+
+Re=density*u*0.0005/viscosity;
+printf("\n\nRe = %f",Re)
+
+//F=18.86*L
+L=(-u^2+vdP)/18.86;
+printf("\n\nL = %f m",L)
diff --git a/858/CH4/EX4.14/example_14.sce b/858/CH4/EX4.14/example_14.sce new file mode 100755 index 000000000..4cf05a660 --- /dev/null +++ b/858/CH4/EX4.14/example_14.sce @@ -0,0 +1,20 @@ +clc
+clear
+printf("example 4.14 page number 151\n\n")
+
+//to find the discharge pressure
+d=0.025 //in m
+u=3 //in m/s
+density=894 //in kg/m3
+viscosity=6.2*10^4 //in Pa-s
+
+Re=(u*d*density)/viscosity;
+f=0.0045;
+L=50;
+
+delta_P=2*f*density*u^2*(L/d)
+printf("frictional head loss = %f kPa",delta_P/1000)
+
+required_P=25*density*9.8;
+total_head=delta_P+required_P;
+printf("\n\ntotal pressure head = %f bar",total_head/10^5)
diff --git a/858/CH4/EX4.15/example_15.sce b/858/CH4/EX4.15/example_15.sce new file mode 100755 index 000000000..2aa7c554e --- /dev/null +++ b/858/CH4/EX4.15/example_15.sce @@ -0,0 +1,20 @@ +clc
+clear
+printf("example 4.15 page number 152\n\n")
+
+//to find the level difference
+
+Q=0.8*10^-3; //in m3/s
+d=0.026 //in m
+A=(3.14*(d^2))/4 //in m2
+
+u=Q/A; //in m/s
+density=800 //in kg/m3
+viscosity=0.0005 //in Pa-s
+
+Re=(u*density*d)/viscosity;
+f=0.079*(Re)^-0.25;
+L=60
+h_f=2*f*((u^2)/9.8)*(L/d);
+
+printf("level difference = %f m",h_f)
diff --git a/858/CH4/EX4.16/example_16.sce b/858/CH4/EX4.16/example_16.sce new file mode 100755 index 000000000..ad8351654 --- /dev/null +++ b/858/CH4/EX4.16/example_16.sce @@ -0,0 +1,24 @@ +clc
+clear
+printf("example 4.16 page number 153\n\n")
+
+//to find the engery cost
+delta_z=50; //in m
+L=290.36 //in m
+d=0.18 //in m
+Q=0.05 //in m3/s
+
+A=(3.14*d^2)/4; //in m2
+u=Q/A; //in m/s
+density=1180; //in kg/m3
+viscosity=0.0012 //in Pa-s
+Re=u*density*d/viscosity;
+
+f=0.004;
+sigma_F=2*f*u^2*L/d;
+ws=((9.8*50)+sigma_F)/0.6;
+mass_flow_rate=Q*density; //in Kg/s
+power=mass_flow_rate*ws/1000; //in KW
+energy_cost=power*24*0.8;
+
+printf("Energy cost = Rs %f",energy_cost)
diff --git a/858/CH4/EX4.17/example_17.sce b/858/CH4/EX4.17/example_17.sce new file mode 100755 index 000000000..2656cc805 --- /dev/null +++ b/858/CH4/EX4.17/example_17.sce @@ -0,0 +1,20 @@ +clc
+clear
+printf("example 4.17 page number 154\n\n")
+
+//to find the pressure loss
+density=998 //in kg/m3
+viscosity=0.0008 //in Pa-s
+d=0.03 //in m
+u=1.2 //in m/s
+
+Re=density*d*u/viscosity;
+
+f=0.0088;
+D=1 //in m
+N=10
+L=3.14*D*N;
+delta_P=(2*f*u^2*L)/d; //in Pa
+delta_P_coil=delta_P*(1+(3.54*(d/D)));
+
+printf("frictional pressure drop = %f kPa",delta_P_coil)
diff --git a/858/CH4/EX4.18/example_18.sce b/858/CH4/EX4.18/example_18.sce new file mode 100755 index 000000000..5f7d40a9f --- /dev/null +++ b/858/CH4/EX4.18/example_18.sce @@ -0,0 +1,20 @@ +clc
+clear
+printf("example 4.18 page number 154\n\n")
+
+//to find pressure drop per unit length
+
+b=0.050 //in m
+a=0.025 //in m
+d_eq=b-a //in m
+density=1000 //in kg/m3
+u=3 //in m/s
+viscosity = 0.001
+
+Re=d_eq*u*density/viscosity;
+
+e=40*10^6 //in m
+f=0.0062;
+P_perunit_length=2*f*density*u^2/d_eq; //in Pa/m
+
+printf("pressure per unit length = %f Pa/m",P_perunit_length)
diff --git a/858/CH4/EX4.19/example_19.sce b/858/CH4/EX4.19/example_19.sce new file mode 100755 index 000000000..eb57ad25e --- /dev/null +++ b/858/CH4/EX4.19/example_19.sce @@ -0,0 +1,10 @@ +clc
+clear
+printf("example 4.19 page number 155\n\n")
+
+//to find the flow rate
+d = 0.3 //in m
+u = 17.63 //avg velocity in m/s
+
+q = (3.14/4)*d^2*u;
+printf("volumetric flow rate = %f cubic meter per second",q)
diff --git a/858/CH4/EX4.2/example_2.sce b/858/CH4/EX4.2/example_2.sce new file mode 100755 index 000000000..faf242b5a --- /dev/null +++ b/858/CH4/EX4.2/example_2.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 4.2 page number 125\n\n")
+
+disp("this is a theoritical problem,book shall be referred for solution")
diff --git a/858/CH4/EX4.20/example_20.sce b/858/CH4/EX4.20/example_20.sce new file mode 100755 index 000000000..2ecddb63f --- /dev/null +++ b/858/CH4/EX4.20/example_20.sce @@ -0,0 +1,10 @@ +clc
+clear
+printf("example 4.20 page number 156\n\n")
+
+//to find the size of pipe required
+d = 0.15 //in m
+u = (0.0191/0.15^2); //in m/s
+
+q = (3.14/4)*d^2*u;
+printf("volumetric flow rate = %f cubic meter/s",q)
diff --git a/858/CH4/EX4.21/example_21.sce b/858/CH4/EX4.21/example_21.sce new file mode 100755 index 000000000..f00ef1cb9 --- /dev/null +++ b/858/CH4/EX4.21/example_21.sce @@ -0,0 +1,26 @@ +clc
+clear
+printf("example 4.21 page number 160\n\n")
+
+//to find the pressure gradient
+
+Q=0.0003 //in m3/s
+d=0.05 //in m
+A=(3.14*d^2)/4;
+
+u=Q/A;
+
+density=1000; //in kg/m3
+viscosity=0.001; //in Pa-s
+e=0.3;
+dp=0.00125; //particle diameter in m
+
+Re=(dp*u*density)/(viscosity*(1-e));
+fm=(150/Re)+1.75;
+L=0.5 //in m
+delta_Pf=fm*((density*L*u^2)/dp)*((1-e)/e^3); //in Pa
+
+//applying bernoulli's equation, we get
+delta_P=delta_Pf-(density*9.8*L);
+pressure_gradient=delta_P/(L*1000); //in kPa/m
+printf("required pressure gradient = %f kPa/m of packed height",pressure_gradient)
diff --git a/858/CH4/EX4.22/example_22.sce b/858/CH4/EX4.22/example_22.sce new file mode 100755 index 000000000..8508ea21f --- /dev/null +++ b/858/CH4/EX4.22/example_22.sce @@ -0,0 +1,29 @@ +clc
+clear
+printf("example 4.22 page number 163\n\n")
+
+//to find minimum fluidization velocity
+
+d=120*10^-6 //in m
+density=2500 //particle density in kg/m3
+e_min=0.45;
+density_water=1000 //in kg/m3
+viscosity=0.9*10^-3; //in Pa-s
+umf=(d^2*(density-density_water)*9.8*e_min^3)/(150*viscosity*(1-e_min));
+printf("minimum fludization velocity = %f m/s",umf)
+
+Re_mf=(d*umf*density_water)/(viscosity*(1-e_min));
+
+
+//given that uo/umf=10
+function[f] = F(e)
+ f = e^3+1.657*e-1.675;
+endfunction
+
+//initial guess
+x = 10;
+e = fsolve(x,F);
+
+printf("\n\ne = %f",e)
+length_ratio=(1-e_min)/(1-e);
+printf("\n\nratio of heights = %f",length_ratio)
diff --git a/858/CH4/EX4.23/example_23.sce b/858/CH4/EX4.23/example_23.sce new file mode 100755 index 000000000..a8c7ddc34 --- /dev/null +++ b/858/CH4/EX4.23/example_23.sce @@ -0,0 +1,17 @@ +clc
+clear
+printf("example 4.23 page number 167\n\n")
+
+//to find the power requirements
+
+P=9807 //in Pa
+density=1000 //in kg/m3
+Q=250/(60*density)
+head=25 //in m
+
+w= head*Q*P; //in kW
+power_delivered=w/0.65;
+power_taken=power_delivered/0.9;
+
+printf("power_delivered = %f kW",power_delivered/1000)
+printf("\n\npower taken by motor = %f kW",power_taken/1000)
diff --git a/858/CH4/EX4.3/example_3.sce b/858/CH4/EX4.3/example_3.sce new file mode 100755 index 000000000..89b3cddc0 --- /dev/null +++ b/858/CH4/EX4.3/example_3.sce @@ -0,0 +1,14 @@ +clc
+clear
+printf("example 4.3 page number 128\n\n")
+
+//to find the viscosity of oil
+
+F=0.5*9.8; //in N
+A=3.14*0.05*0.15; //in m2
+shear_stress=F/A; //in Pa
+printf("shear_stress = %f Pa",shear_stress)
+
+velocity_distribution =0.1/(0.05*10^-3);
+viscosity=shear_stress/velocity_distribution;
+printf("\n\nviscosity = %f Pa-s",viscosity)
diff --git a/858/CH4/EX4.4/example_4.sce b/858/CH4/EX4.4/example_4.sce new file mode 100755 index 000000000..22ffeefa5 --- /dev/null +++ b/858/CH4/EX4.4/example_4.sce @@ -0,0 +1,4 @@ +clc
+clear
+printf("example 4.4 page number 130\n\n")
+printf("this is a theoritical problem,book shall be referred for solution")
diff --git a/858/CH4/EX4.5/example_5.sce b/858/CH4/EX4.5/example_5.sce new file mode 100755 index 000000000..718f55335 --- /dev/null +++ b/858/CH4/EX4.5/example_5.sce @@ -0,0 +1,9 @@ +clc
+clear
+printf("example 4.5 page number 133\n\n")
+
+//to find variation of losses with velocity
+loss_ratio=3.6; //delta_P2/delta_P1=3.6
+velocity_ratio=2; //u2/u1=2
+n=log2(loss_ratio); //delta_P2/delta_P1=(u2/u1)^n
+printf("power constant = %f flow is turbulent",n)
diff --git a/858/CH4/EX4.6/example_6.sce b/858/CH4/EX4.6/example_6.sce new file mode 100755 index 000000000..77bf07c1e --- /dev/null +++ b/858/CH4/EX4.6/example_6.sce @@ -0,0 +1,4 @@ +clc
+clear
+printf("example 4.6 page number 133\n\n")
+printf("this is a theoritical problem,book shall be referred for solution")
diff --git a/858/CH4/EX4.7/example_7.sce b/858/CH4/EX4.7/example_7.sce new file mode 100755 index 000000000..1d0b4af71 --- /dev/null +++ b/858/CH4/EX4.7/example_7.sce @@ -0,0 +1,4 @@ +clc
+clear
+printf("example 4.7 page number 134")
+disp("this is a theoritical problem,book shall be referred for solution")
diff --git a/858/CH4/EX4.8/example_8.sce b/858/CH4/EX4.8/example_8.sce new file mode 100755 index 000000000..f0d9f2fd7 --- /dev/null +++ b/858/CH4/EX4.8/example_8.sce @@ -0,0 +1,31 @@ +clc
+clear
+printf("example 4.8 page number 137\n\n")
+
+//to find the boundary layer properties
+
+disp('part 1')
+x=0.05 //in m
+density=1000 //in kg/m3
+viscosity=1*10^-3 //in Pa-s
+u=1 //in m/s
+Re=(density*u*x)/viscosity;
+
+printf("Reynolds Number = %f",Re)
+
+thickness=4.65*x*(Re)^-0.5;
+printf("\nboundary layer thickness = %f m\n",thickness)
+
+disp('part 2')
+Re_x=3.2*10^5;
+x_cr=(Re_x*viscosity)/(density*u);
+printf("transition takes place at x = %f m\n",x_cr)
+
+disp('part 3')
+x=0.5 //in m
+Re=(density*u*x)/viscosity;
+thickness=0.367*x*(Re)^-0.2;
+printf("boundary layer thickness= %f m",thickness)
+
+t_sublayer=71.5*x*(Re)^-0.9;
+printf("\nsub layer thickness= %f m",t_sublayer)
diff --git a/858/CH4/EX4.9/example_9.sce b/858/CH4/EX4.9/example_9.sce new file mode 100755 index 000000000..3ccdea2eb --- /dev/null +++ b/858/CH4/EX4.9/example_9.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 4.9 page number 138\n\n")
+
+//to find the flow properties
+d1=0.05 //in m
+A1=(3.14*d1^2)/4;
+density_1=2.1 //in kg/m3
+u1=15 //in m/s
+P1=1.8; //in bar
+P2=1.3; //in bar
+
+w=density_1*A1*u1;
+density_2=density_1*(P2/P1);
+printf("density at section 2 = %f kg/cubic meter",density_2)
+
+u2=u1*(density_1/density_2)*(0.05/0.075)^2;
+printf("\n\nvelocity at section 2 = %f m/s",u2)
diff --git a/858/CH5/EX5.1/example_1.sce b/858/CH5/EX5.1/example_1.sce new file mode 100755 index 000000000..9015d394d --- /dev/null +++ b/858/CH5/EX5.1/example_1.sce @@ -0,0 +1,16 @@ +clc
+clear
+printf("example 5.1 page number 171\n\n")
+
+//to find the rate of heat loss
+A=5*4 //in m2
+T1=100; //in K
+T2=30; //in K
+
+delta_T=T1-T2;
+
+x=0.25 //in m
+k=0.70 //in W/mK
+Q=k*A*(delta_T/x);
+
+printf("rate of heat loss = %f W",Q)
diff --git a/858/CH5/EX5.10/example_10.sce b/858/CH5/EX5.10/example_10.sce new file mode 100755 index 000000000..449febfe1 --- /dev/null +++ b/858/CH5/EX5.10/example_10.sce @@ -0,0 +1,10 @@ +clc
+clear
+printf("example 5.10 page number 191\n\n")
+
+//to find temperature of earth
+R=7*10^10; //in cm
+Ts=6000; //in K
+l=1.5*10^13; //in m
+To=((R^2/(4*l^2))^0.25)*Ts;
+printf("temperature of earth = %f K",To)
diff --git a/858/CH5/EX5.11/example_11.sce b/858/CH5/EX5.11/example_11.sce new file mode 100755 index 000000000..b185c64f9 --- /dev/null +++ b/858/CH5/EX5.11/example_11.sce @@ -0,0 +1,10 @@ +clc
+clear
+printf("example 5.11 page number 192\n\n")
+
+//to find the equilibrium temperature
+R=6.92*10^5 //in km
+l=14.97*10^7 //in km
+Ts=6200; //in K
+To=(R^2/l^2)^0.25*Ts;
+printf("Equilibrium temperature = %f K",To)
diff --git a/858/CH5/EX5.12/example_12.sce b/858/CH5/EX5.12/example_12.sce new file mode 100755 index 000000000..9704fc14b --- /dev/null +++ b/858/CH5/EX5.12/example_12.sce @@ -0,0 +1,11 @@ +clc
+clear
+printf("example 5.12 page number 192\n\n")
+
+//to find the equilibrium temperature
+view_factor=0.5;
+R=6.92*10^5 //in km
+l=14.97*10^7 //in km
+Ts=6200; //in K
+To=(view_factor*(R^2/l^2))^0.25*Ts;
+printf("Equilibrium temperature = %f K",To)
diff --git a/858/CH5/EX5.13/example_13.sce b/858/CH5/EX5.13/example_13.sce new file mode 100755 index 000000000..ebc44c531 --- /dev/null +++ b/858/CH5/EX5.13/example_13.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("example 5.13 page number 193\n\n")
+
+//to find the surface temperature
+view_factor=0.25;
+R=7.1*10^10 //in cm
+l=1.5*10^13 //in cm
+Ts=5973; //in K
+alpha=0.2;
+epsilon=0.1;
+
+ratio=alpha/epsilon;
+To=(ratio*view_factor*(R^2/l^2))^0.25*Ts;
+printf("Equilibrium temperature = %f K",To)
diff --git a/858/CH5/EX5.14/example_14.sce b/858/CH5/EX5.14/example_14.sce new file mode 100755 index 000000000..d44c28f74 --- /dev/null +++ b/858/CH5/EX5.14/example_14.sce @@ -0,0 +1,12 @@ +clc
+clear
+printf("example 5.14 page number 193\n\n")
+
+//to find the solar constant
+R=7*10^10; //in cm
+l=1.5*10^13; //in cm
+sigma=5.3*10^-5; //in erd/s(cm2)(K)4
+T=6000; //in K
+
+S=(R/l)^2*(sigma)*(T^4)*60;
+printf("solar constant = %f J/sq cm min",S/10^7)
diff --git a/858/CH5/EX5.15/example_15.sce b/858/CH5/EX5.15/example_15.sce new file mode 100755 index 000000000..edb638d86 --- /dev/null +++ b/858/CH5/EX5.15/example_15.sce @@ -0,0 +1,30 @@ +clc
+clear
+printf("example 5.15 page number 207\n\n")
+
+//to find the amount of vapor and liquid and amount of heat transfer
+
+F = 5000 //in kg/hr
+xF = 0.01
+xL = 0.02;
+
+L = F*xF/xL;
+V = F-L;
+printf("L = %f Kg/hr\n V = %f kg/hr",L,V)
+
+TF= 303 //in K
+hF = 125.9 //in KJ/kg
+T1 = 373.2 //in K
+Hv = 2676.1 //in kJ/kg
+hL = 419.04; //in kJ/kg
+Ts = 383.2 //in K
+Hs = 2691.5 //in kJ/kg
+hs = 461.30 //in kJ/kg
+
+S = (F*hF-L*hL-V*Hv)/(hs-Hs);
+printf("\n\namount of steam = %f kg steam/h",S)
+
+q = S*(Hs - hs);
+q = q*1000/3600 //conversion to Watt
+U = q/(69.9*10);
+printf("\n\nheat reansfer coefficient = %f W/sq m K",U)
diff --git a/858/CH5/EX5.16/example_16.sce b/858/CH5/EX5.16/example_16.sce new file mode 100755 index 000000000..1ff974f78 --- /dev/null +++ b/858/CH5/EX5.16/example_16.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 5.16 page number 208\n\n")
+
+//to find the amount of liquid and vapor leaving and outlet concentration
+//we have two linear equations in L and V so we will write them in form of a matrix and then solve using principles of linear algebra
+
+b1 = 6000*125.79+3187.56*2691.5-3187.56*461.30; //data from previous problem
+b2 = 6000;
+A = [419.04 2676.1;1 1];
+
+b = [b1;b2];
+x = A\b;
+L = x(1);
+V = x(2);
+
+printf("L = %f kg/hr\nV = %f kg/hr",L,V)
+
+F = 6000 //in kg/hr
+xF = 0.01;
+xL = F*xF/L;
+printf("\n\npercentage increase in outlet concentration = %f",xL*100)
diff --git a/858/CH5/EX5.17/example_17.sce b/858/CH5/EX5.17/example_17.sce new file mode 100755 index 000000000..25b4bb51b --- /dev/null +++ b/858/CH5/EX5.17/example_17.sce @@ -0,0 +1,19 @@ +clc
+clear
+printf("example 5.17 page number 209\n\n")
+
+//to find the change in heat trnasfer area
+
+Hv=2635.3 //kJ/kg
+hL=313.93 //in kJ/kg
+S=(2500*313.93+2500*2635.3-5000*125.79)/(2691.5-461.30);
+printf("steam flow rate = %f kg steam/hr",S)
+
+q = S*(2691.5 - 461.30);
+q = q*1000/3600 //in W
+U = 2833.13; //in W/m2 K
+delta_T = 383.2-348.2; //in K
+A = q/(U*delta_T);
+
+printf("\n\nArea = %f sq meter",A)
+printf("\n\nin this case a condensor and vaccum pump should be used")
diff --git a/858/CH5/EX5.2/example_2.sce b/858/CH5/EX5.2/example_2.sce new file mode 100755 index 000000000..37e2ddb7b --- /dev/null +++ b/858/CH5/EX5.2/example_2.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("example 5.1 page number 171\n\n")
+
+//to find the heat loss
+
+d1=0.15 //in m
+d2=0.16 //in m
+l=1 //in m
+
+A1=3.14*d1*l;
+A2=3.14*d2*l
+Am=(A1-A2)/log (A1/A2);
+
+T1=120; //in K
+T2=119.8; //in K
+
+delta_T=T1-T2;
+x=(d2-d1)/2;
+k=50 //in W/mK
+Q=k*Am*(delta_T/x);
+
+printf("rate of heat loss per unit length = %f W/m",Q)
diff --git a/858/CH5/EX5.3/example_3.sce b/858/CH5/EX5.3/example_3.sce new file mode 100755 index 000000000..6feeeb471 --- /dev/null +++ b/858/CH5/EX5.3/example_3.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 5.3 page number 172\n\n")
+
+//to find the rate of heat loss
+
+ri=0.5 //in m
+ro=0.6; //in m
+A1=4*3.14*ri^2;
+A2=4*3.14*ro^2;
+
+Am=(A1*A2)^0.5;
+
+Ti=140; //in K
+To=50; //in K
+delta_T=Ti-To;
+x=0.1 //in m
+k=0.12 //in W/mK
+
+Q=k*Am*(delta_T/x);
+printf("Heat loss through sphere = %f W",Q)
diff --git a/858/CH5/EX5.4/example_4.sce b/858/CH5/EX5.4/example_4.sce new file mode 100755 index 000000000..577cc0d85 --- /dev/null +++ b/858/CH5/EX5.4/example_4.sce @@ -0,0 +1,25 @@ +clc
+clear
+printf("example 5.4 page number 173\n\n")
+
+//to find the heat loss from composite wall
+//for the red brick layer
+
+x1=0.250; //in m
+k1=0.7; //in W/mK
+A1=1; //in m2
+R1=x1/(k1*A1); //in K/W
+
+//for the felt layer
+x2=0.020; //in m
+k2=0.046; //in W/mK
+A2=1; //in m2
+R2=x2/(k2*A2); //in K/W
+R=R1+R2;
+printf("Total resistance = %f K/W",R)
+
+T1=110; //in K
+T2=25 //in K
+delta_T=T1-T2;
+Q=delta_T/R;
+printf("\n\nheat loss through wall = %f W/square m",Q)
diff --git a/858/CH5/EX5.5/example_5.sce b/858/CH5/EX5.5/example_5.sce new file mode 100755 index 000000000..d676a28bb --- /dev/null +++ b/858/CH5/EX5.5/example_5.sce @@ -0,0 +1,37 @@ +clc
+clear
+printf("example 5.5 page number 173\n\n")
+
+//to find the rate of heat loss through pipeline
+//resistance by pipeline
+
+d1=0.15 //in m
+d2=0.16 //in m
+l=1 //in m
+A1=3.14*d1*l;
+A2=3.14*d2*l
+Am1=(A2-A1)/log (A2/A1);
+x1=(d2-d1)/2;
+k1=50 //in W/mK
+R1=x1/(k1*Am1);
+
+//resistance by insulation
+d2=0.16 //in m
+d3=0.26 //in m
+l=1 //in m
+A2=3.14*d2*l;
+A3=3.14*d3*l
+Am2=(A3-A2)/log (A3/A2);
+x2=(d3-d2)/2;
+k2=0.08 //in W/mK
+R2=x2/(k2*Am2);
+R=R1+R2;
+
+printf("total resistance = %f K/W",R)
+
+T1=120; //in K
+T2=40; //in K
+delta_T=T1-T2;
+Q=delta_T/R;
+
+printf("\n\nheat loss = %f W/m",Q)
diff --git a/858/CH5/EX5.6/example_6.sce b/858/CH5/EX5.6/example_6.sce new file mode 100755 index 000000000..c51e4fd28 --- /dev/null +++ b/858/CH5/EX5.6/example_6.sce @@ -0,0 +1,33 @@ +clc
+clear
+printf("example 5.6 page number 174\n\n")
+
+//to find the increase in heat transfer rate
+
+x1=0.1; //in m
+x2= 0.25; //in m
+k_rb=0.93; //in W/mK
+k_ib=0.116 //in W/mK
+k_al=203.6 //in W/mK
+A=0.1 //in m2
+
+//to find resistance without rivets
+R=(1/A)*((x1/k_rb)+(x2/k_ib));
+T1=225 //in K
+T2=37 //in K
+delta_T=T1-T2;
+Q=delta_T/R;
+printf("heat transfer rate = %f W",Q)
+
+//to find resistance with rivet
+d=0.03 //in m
+rivet_area= (3.14/4)*d^2;
+R_r=(x1+x2)/(k_al*rivet_area);
+area_norivet=A-rivet_area;
+R_cl=(A/area_norivet)*R;
+R_eq=1/(1/R_r+1/R_cl);
+Q_new=delta_T/R_eq;
+
+printf("\n\nRate of heat transfer with rivet = %f W",Q_new)
+increase=((Q_new-Q)/Q)*100;
+printf("\n\npercentage increase in heat transfer rate = %f",increase)
diff --git a/858/CH5/EX5.7/example_6.sce b/858/CH5/EX5.7/example_6.sce new file mode 100755 index 000000000..c51e4fd28 --- /dev/null +++ b/858/CH5/EX5.7/example_6.sce @@ -0,0 +1,33 @@ +clc
+clear
+printf("example 5.6 page number 174\n\n")
+
+//to find the increase in heat transfer rate
+
+x1=0.1; //in m
+x2= 0.25; //in m
+k_rb=0.93; //in W/mK
+k_ib=0.116 //in W/mK
+k_al=203.6 //in W/mK
+A=0.1 //in m2
+
+//to find resistance without rivets
+R=(1/A)*((x1/k_rb)+(x2/k_ib));
+T1=225 //in K
+T2=37 //in K
+delta_T=T1-T2;
+Q=delta_T/R;
+printf("heat transfer rate = %f W",Q)
+
+//to find resistance with rivet
+d=0.03 //in m
+rivet_area= (3.14/4)*d^2;
+R_r=(x1+x2)/(k_al*rivet_area);
+area_norivet=A-rivet_area;
+R_cl=(A/area_norivet)*R;
+R_eq=1/(1/R_r+1/R_cl);
+Q_new=delta_T/R_eq;
+
+printf("\n\nRate of heat transfer with rivet = %f W",Q_new)
+increase=((Q_new-Q)/Q)*100;
+printf("\n\npercentage increase in heat transfer rate = %f",increase)
diff --git a/858/CH5/EX5.8/example_8.sce b/858/CH5/EX5.8/example_8.sce new file mode 100755 index 000000000..0687be856 --- /dev/null +++ b/858/CH5/EX5.8/example_8.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("example 5.8 page number 188\n\n")
+
+//to find the heat transfer coefficient
+density=984.1 //in kg/cubic meter
+v=3 //in m/s
+viscosity=485*10^-6; //in Pa-s
+k=0.657 //in W/mK
+cp=4178 //in J/kg K
+d=0.016 //in m
+
+Re=(density*v*d)/viscosity;
+Pr=(cp*viscosity)/k;
+
+//dittus boelter equation
+h=0.023*Re^0.8*Pr^0.3*(k/d);
+printf("heat transfer coefficient = %f W/sq meter K",h)
+
+//Sieder Tate equation
+viscosity_w=920*10^-6
+h1=0.023*Re^0.8*Pr^(1/3)*(k/d)*(viscosity/viscosity_w)^0.14;
+printf("\n\nheat transfer coefficient = %f W/sq meter K",h1)
diff --git a/858/CH5/EX5.9/example_9.sce b/858/CH5/EX5.9/example_9.sce new file mode 100755 index 000000000..235df2404 --- /dev/null +++ b/858/CH5/EX5.9/example_9.sce @@ -0,0 +1,11 @@ +clc
+clear
+printf("example 5.9 page number 191\n\n")
+
+//to find the surface temperature of earth
+T_sun = 5973 //in degree C
+d = 1.5*10^13 //in cm
+R = 7.1*10^10; //in cm
+
+T_earth = ((R/(2*d))^0.5)*T_sun;
+printf("Temperature of earth = %f C",T_earth-273)
diff --git a/858/CH6/EX6.1/example_1.sce b/858/CH6/EX6.1/example_1.sce new file mode 100755 index 000000000..3ba5e5909 --- /dev/null +++ b/858/CH6/EX6.1/example_1.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 6.1 page number 213\n\n")
+
+printf("This is a theoritical problem and book shall be referred for solution")
diff --git a/858/CH6/EX6.10/example_10.sce b/858/CH6/EX6.10/example_10.sce new file mode 100755 index 000000000..fd89f499a --- /dev/null +++ b/858/CH6/EX6.10/example_10.sce @@ -0,0 +1,14 @@ +clc
+clear
+printf("example 6.10 page number 231\n\n")
+
+// to find the equilibrium composition
+P_M = 53.32 //kPa
+P_W = 12.33 //in kpA
+P = 40 //IN K pA
+x = (P - P_W)/(P_M-P_W);
+
+printf("liquid phase composition = %f",x)
+
+y = P_M*x/P;
+printf("\n\nvapor phase composition = %f",y)
diff --git a/858/CH6/EX6.11/example_11.sce b/858/CH6/EX6.11/example_11.sce new file mode 100755 index 000000000..f35e89057 --- /dev/null +++ b/858/CH6/EX6.11/example_11.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 6.11 page number 232\n\n")
+
+printf("this is a theoritical question, book shall be referred for solution")
diff --git a/858/CH6/EX6.12/example_12.sce b/858/CH6/EX6.12/example_12.sce new file mode 100755 index 000000000..c6f2f2dbb --- /dev/null +++ b/858/CH6/EX6.12/example_12.sce @@ -0,0 +1,24 @@ +clc
+clear
+printf("example 6.12 page number 231\n\n")
+
+//to find the top and bottom composition
+x = [1;0.69;0.40;0.192;0.045;0];
+y = [1;0.932;0.78;0.538;0.1775;0];
+plot(x,y)
+xlabel("x")
+ylabel("y")
+title("distillation curve")
+x = 0:0.1:1;
+y = 0:0.1:1;
+plot(x,y)
+x = [0.5,0.31];
+y = [0.5,0.7];
+plot (x,y)
+Z=0.5;
+y_D=0.69;
+x_W=0.31;
+
+
+printf("composition of top product = %f mole percent of hexane",y_D*100)
+printf("\n\ncomposition of bottom product = %f mole percent of hexane",x_W*100)
diff --git a/858/CH6/EX6.13/example_13.sce b/858/CH6/EX6.13/example_13.sce new file mode 100755 index 000000000..716f3a0af --- /dev/null +++ b/858/CH6/EX6.13/example_13.sce @@ -0,0 +1,30 @@ +clc
+clear
+printf("example 6.13 page number 237\n\n")
+
+//to find the composite distillate and residue
+
+F = 100 //moles
+xf = 0.4;
+D = 60 //moles
+W = 40 //moles
+
+x = 0.2:0.05:0.45;
+for i =1:((0.45-0.2)/0.05)+1
+
+y(i) = 2.16*x(i)/(1+1.16*x(i));
+
+
+z(i) = (y(i)-x(i))^-1;
+
+end
+
+plot (x,z'/10)
+title('Batch Distillation Curve')
+xlabel('x')
+ylabel('y')
+xw = 0.22; //from the graph
+yd = (F*xf-W*xw)/D;
+
+printf("composition of distillate = %f",yd)
+printf("\n\ncomposition of residue = %f",xw)
diff --git a/858/CH6/EX6.14/example_14.sce b/858/CH6/EX6.14/example_14.sce new file mode 100755 index 000000000..38116a78e --- /dev/null +++ b/858/CH6/EX6.14/example_14.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 6.14 page number 238\n\n")
+
+printf('this is a theoritical question and solution can be referred from the book')
diff --git a/858/CH6/EX6.15/example_15.sce b/858/CH6/EX6.15/example_15.sce new file mode 100755 index 000000000..fd6210ab8 --- /dev/null +++ b/858/CH6/EX6.15/example_15.sce @@ -0,0 +1,26 @@ +clc
+clear
+printf("example 6.15 page number 249\n\n")
+
+//to find the top and bottom product composition
+
+//part 1
+x=0.4;
+y=0.8;
+x_D=y;
+x_W=0.135; //bottom concentration
+D=(100*x-100*x_W)/(y-x_W); //distillate amount
+printf("amount of distillate =%f moles/h",D)
+
+//part 2
+alpha=6; //relative volatility
+x_R=y/(y+(alpha*(1-y))); //liquid leaving partial condensor
+printf("\n\nliquid leaving partial condenser = %f",x_R)
+
+y1=(1/3)*y+(2/3)*x;
+x1=y1/(y1+(alpha*(1-y1)));
+y_W = (1/3)*x_D+(2/3)*x1;
+x_W=y_W/(y_W+(alpha*(1-y_W)));
+D=(100*(x-x_W))/(y-x_W);
+
+printf("\n\namount of distillate = %f moles/h",D)
diff --git a/858/CH6/EX6.16/example_16.sce b/858/CH6/EX6.16/example_16.sce new file mode 100755 index 000000000..00bcd6581 --- /dev/null +++ b/858/CH6/EX6.16/example_16.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 6.16 page number 264\n\n")
+
+//to find the percentage extraction of nicotine
+x=0.01; //% of nicotine
+X0 = x/(1-x);
+w=150 //weight of nicotine water solution
+A0=w*(1-X0);
+B0=250; //kg keroscene
+X1 = A0*X0/(A0+B0*0.798);
+printf("final concentration of nicotine = %f",X1)
+
+c=A0*(X0-X1);
+printf("\n\namount of nicotine removed = %f kg",c)
+
+percentage = (c*100)/(A0*x);
+printf("\n\npercentage recovery = %f percent",percentage)
diff --git a/858/CH6/EX6.17/example_17.sce b/858/CH6/EX6.17/example_17.sce new file mode 100755 index 000000000..8e0eaafae --- /dev/null +++ b/858/CH6/EX6.17/example_17.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 6.17 page number 264\n\n")
+
+//to find the number of stages
+x=0.01 //mole fraction of nicotine
+yN = 0.0006; //mole fraction in solvent
+xN = 0.001; //final mole fraction in water
+
+X0=x/(1-x); //in kg nicotine/kg water
+YN =yN/(1-yN); //in kg nicotine/kg keroscene
+XN = xN/(1-xN);
+A0=100*(1-X0); //kgwater/h
+B0=150*(1-YN); //in kg kerosene/h
+
+Y1=((A0*(X0-XN))/B0)+YN; //in kg nicotine/kg kerosene
+printf("Y1 = %f kg nicotine/kg kerosene",Y1)
+
+//for graph refer to the book
+number_of_stages = 8.4;
+printf("\n\nnumnber of stages = %f",number_of_stages)
diff --git a/858/CH6/EX6.18/example_18.sce b/858/CH6/EX6.18/example_18.sce new file mode 100755 index 000000000..8baa8ed1a --- /dev/null +++ b/858/CH6/EX6.18/example_18.sce @@ -0,0 +1,19 @@ +clc
+clear
+printf("example 6.18 page number 274\n\n")
+
+//to calculate the humidity
+P = 101.3 //in kPa
+pA = 3.74 //in kPa
+p_AS = 7.415 //in kPa
+H = (18.02/28.97)*(pA/(P-pA));
+printf("humidity = %f kg H2O/kg air",H)
+
+Hs = (18.02/28.97)*(p_AS/(P-p_AS));
+printf("\n\nSaturated humidity = %f kg H2O/kg air",Hs)
+
+%_humidity = 100*(H/Hs);
+printf("\n\npercentage humidity = %f percent",%_humidity)
+
+relative_humidity = 100*(pA/p_AS);
+printf("\n\npercentage relative humidity = %f percent",relative_humidity)
diff --git a/858/CH6/EX6.19/example_19.sce b/858/CH6/EX6.19/example_19.sce new file mode 100755 index 000000000..b95725bbf --- /dev/null +++ b/858/CH6/EX6.19/example_19.sce @@ -0,0 +1,34 @@ +clc
+clear
+printf("example 6.17 page number 264\n\n")
+
+//to find the air flow rate and outlet humidity
+S=425.6 //in kg/h
+X1 = 0.035 //in kgwater/kg dry solid
+t_s1=25 //in degree C
+X2 = 0.017 //in kg H2O/kg dry air
+t_s2=60 //in degree C
+H2 = 0.0175 //in kg H2O/kg dry air
+t_G2 = 84.2 //in degree C
+t_G1= 32.8 //in degree C
+C_pS = 1.465 //in kJ/kg dry solid
+C_pA = 4.187 //in kg/ kg H2O K
+
+H_G2=(1.005+1.88*H2)*(t_G2-0)+H2*2501;
+H_S1 = C_pS*(t_s1-0)+X1*C_pA*(t_s1-0); //in kJ/kg
+H_S2 = C_pS*(t_s2-0)+X2*C_pA*(t_s2-0); //in kJ/kg
+Q=9300; //in kJ/h
+
+printf("Latent heat of water at 0C, HG2 = %f kJ/kg dryair",H_G2)
+printf("\n\nEnthalpy of entering solid, HS1 = %f kJ/kg dryair",H_S1)
+printf("\n\nEnthalpy of exit solid, HS2 = %f kJ/kg dryair",H_S2)
+
+//applying GHg2 + SHs1 = GHg1 +SHs2 +Q, we get two linear equations
+//0.0175G+14.17248 = GH1 and 98.194G-29745.398 = 2562.664GH1
+A = [0.0175 -1;98.194 -2562.664];
+b = [-14.17248;29745.398];
+x = A\b;
+G = x(1);
+H1 = x(2)/G;
+printf("\n\nAir flow rate, G = %f kg dryair/hr",G)
+printf("\n\nHumidity, H1 = %f kg dryair/hr",H1)
diff --git a/858/CH6/EX6.2/example_2.sce b/858/CH6/EX6.2/example_2.sce new file mode 100755 index 000000000..894f5fd47 --- /dev/null +++ b/858/CH6/EX6.2/example_2.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 6.2 page number 214\n\n")
+
+printf("This is a theoritical problem and book shall be referred for solution")
diff --git a/858/CH6/EX6.20/example_20.sce b/858/CH6/EX6.20/example_20.sce new file mode 100755 index 000000000..98bf3e702 --- /dev/null +++ b/858/CH6/EX6.20/example_20.sce @@ -0,0 +1,24 @@ +clc
+clear
+printf("example 6.20 page number 291\n\n")
+
+//to find the crystal yield
+
+M_Na2CO3 = 106
+M_10H2O = 180.2
+M_Na2CO3_10H2O = 286.2;
+w_Na2CO3 = 5000 //in kg
+%_water = 0.05 //% of water evaporated
+
+W = %_water*w_Na2CO3;
+//solving material balance, we have two equations
+//equation 1 -> 0.8230L +0.6296C = 3500
+//equation 2 -> 0.1769L + 0.3703C = 1250
+
+A=[0.8230 0.6296;0.1769 0.3703]
+b = [3500;1250]
+x = A\b;
+L = x(1);
+C = x(2);
+printf("L = %f kg solution",L)
+printf("\n\nC = %f kg of Na2CO3.10H2O crystals",C)
diff --git a/858/CH6/EX6.21/example_21.sce b/858/CH6/EX6.21/example_21.sce new file mode 100755 index 000000000..ca4c12731 --- /dev/null +++ b/858/CH6/EX6.21/example_21.sce @@ -0,0 +1,28 @@ +clc
+clear
+printf("example 6.21 page number 291\n\n")
+
+//to find the crystal yield
+//from material balance, we have two linear equations
+//1400 = 0.7380L+0.5117C and 600 = 0.2619L+0.4882C
+
+A=[0.7380 0.5117;0.2619 0.4882]
+b = [1400;600]
+x = A\b;
+L = x(1);
+C = x(2);
+printf("L = %f kg solution",L)
+printf("\n\nC = %f kg of MgSO4.7H2O crystals",C)
+
+F = 2000 //in kg/h
+cv = 2.93 //in kJ/kg K
+H1 = F*cv*(330-293);
+printf("\n\nenthalpy of feed = %f kJ",H1)
+
+wt = 246.49 //molar mass MgSO4.7H2O
+heat_soln = -13.31*10^3; //in kJ/kg mol
+heat = heat_soln/wt;
+heat_crystallization = abs(heat);
+H2 = heat_crystallization*C; //total heat
+q = -H1-H2;
+printf("\n\nheat absorbed = %f kJ\nthus heat shall be removed",q)
diff --git a/858/CH6/EX6.3/example_3.sce b/858/CH6/EX6.3/example_3.sce new file mode 100755 index 000000000..9619ba897 --- /dev/null +++ b/858/CH6/EX6.3/example_3.sce @@ -0,0 +1,20 @@ +clc
+clear
+printf("example 6.3 page number 215\n\n")
+
+//to find the flux and pressure difference
+
+D_AB=6.75*10^-5 //in m2/s
+Z=0.03 //in m
+R=8314
+p_A1=5.5*10^4 //in Pa
+p_A2=1.5*10^4 //in Pa
+T=298 //in K
+
+N_A=D_AB*(p_A1-p_A2)/(R*T*Z);
+printf("flux = %f kmol/sq m s",N_A)
+
+//for partial pressure
+Z=0.02; //in m
+p_A2=p_A1-((N_A*R*T*Z)/D_AB);
+printf("\n\npressure = %f Pa",p_A2)
diff --git a/858/CH6/EX6.4/example_4.sce b/858/CH6/EX6.4/example_4.sce new file mode 100755 index 000000000..c5feadb0a --- /dev/null +++ b/858/CH6/EX6.4/example_4.sce @@ -0,0 +1,26 @@ +clc
+clear
+printf("example 6.4 page number 216\n\n")
+
+//to find the flux of NH3 and equimolar counter diffusion flux
+
+Z=0.15 //in m
+P=1.103*10^5 //in Pa
+p_A1=1.5*10^4 //in Pa
+p_A2=5*10^3 //in Pa
+
+p_B1=P-p_A1;
+p_B2=P-p_A2;
+
+D_AB=2.30*10^-5 //in m2/s
+R=8314
+T=298 //in K
+
+//for non diffusing N2
+p_BM=(p_B2-p_B1)/log (p_B2/p_B1);
+N_A=D_AB*(p_A1-p_A2)*P/(R*T*Z*p_BM);
+printf("flux = %f kmol/sq m s",N_A)
+
+//for diffusing N2
+N_A=D_AB*(p_A1-p_A2)/(R*T*Z);
+printf("\n\nflux = %f kmol/sq m s",N_A)
diff --git a/858/CH6/EX6.5/example_5.sce b/858/CH6/EX6.5/example_5.sce new file mode 100755 index 000000000..1921cec57 --- /dev/null +++ b/858/CH6/EX6.5/example_5.sce @@ -0,0 +1,4 @@ +clc
+clear
+printf("example 6.5 page number 216\n\n")
+printf("This is a theoritical problem and book shall be referred for solution")
diff --git a/858/CH6/EX6.6/example_6.sce b/858/CH6/EX6.6/example_6.sce new file mode 100755 index 000000000..7473e7c95 --- /dev/null +++ b/858/CH6/EX6.6/example_6.sce @@ -0,0 +1,31 @@ +clc
+clear
+printf("example 6.6 page number 218\n\n")
+
+M_A=36.5 //molar mass of HCl
+M_B=18 //molar masss of water
+w_A1=12; //weight % of HCL
+w_A2=4 //weight % of HCL
+x_A1=(w_A1/M_A)/((w_A1/M_A)+((100-w_A1)/M_B));
+printf('x_A1 =%f',x_A1)
+
+x_B1=1-x_A1;
+M1=100/((w_A1/M_A)+((100-w_A1)/M_B));
+printf("\n\nmolar mass at point 1 = %f kg/kmol",M1)
+
+//at point 2
+x_A2=(w_A2/M_A)/((w_A2/M_A)+((100-w_A2)/M_B));
+x_B2=1-x_A2;
+M2=100/((w_A2/M_A)+((100-w_A2)/M_B)); //avg molecular weight at point 2
+printf("\n\nmolar mass at point 2 = %f Kg/kmol",M2)
+
+density_1=1060.7; //in kg/m3
+density_2=1020.15; //in kg/m3
+C_av=((density_1/M1)+(density_2/M2))/2;
+printf("\n\nC_av = %f kmol/cubic m",C_av)
+
+x_BM=(x_B2-x_B1)/(log (x_B2/x_B1));
+Z=0.004 //in m
+D_AB=2.5*10^-9;
+N_A=(D_AB*C_av*(x_A1-x_A2))/(x_BM*Z);
+printf("\n\nflux = %f kmol/sq m-s",N_A)
diff --git a/858/CH6/EX6.7/example_7.sce b/858/CH6/EX6.7/example_7.sce new file mode 100755 index 000000000..062ce8a37 --- /dev/null +++ b/858/CH6/EX6.7/example_7.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 6.7 page number 220\n\n")
+
+printf("This is a theoritical problem and book shall be referred for solution")
diff --git a/858/CH6/EX6.8/example_8.sce b/858/CH6/EX6.8/example_8.sce new file mode 100755 index 000000000..8d0ce2a7a --- /dev/null +++ b/858/CH6/EX6.8/example_8.sce @@ -0,0 +1,25 @@ +clc
+clear
+printf("example 6.8 page number 229\n\n")
+
+//to find the mean driving force and mass transfer area
+
+Gs=700/22.4 //in kmol of dry air/hr
+Ls=1500/18 //in kmol of dry air/hr
+y1=0.05
+Y1=y1/(1-y1);
+Y2=0.02*Y1;
+X2=0
+X1=(Gs/Ls)*(Y1-Y2);
+m=Gs*(Y1-Y2);
+
+//driving force
+delta_Y1=Y1-1.68*X1;
+delta_Y2=Y2-1.68*X2;
+delta_Y=(delta_Y1-delta_Y2)/(log (delta_Y1/delta_Y2));
+printf("driving force = %f kmol acetone/kmol dry air",delta_Y)
+
+//mass transfer area
+K_G=0.4 //in kmol acetone/kmol dry air
+A=m/(K_G*delta_Y);
+printf("\n\narea = %f sq m",A)
diff --git a/858/CH6/EX6.9/example_9.sce b/858/CH6/EX6.9/example_9.sce new file mode 100755 index 000000000..f2f3f11be --- /dev/null +++ b/858/CH6/EX6.9/example_9.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 6.9 page number 229\n\n")
+
+//to calculate minimum oil circulation rate
+
+G1=(855/22.4)*(106.6/101.3)*(273/299.7);
+y1=0.02;
+Y1=y1/(1-y1);
+Gs=G1*(1-y1);
+
+//for 95% removal
+Y2=0.05*Y1;
+x2=0.005;
+X2=x2/(1-x2);
+Y=0.204;
+X1=0.176; //in kmol bgenzene/kmol benzene free oil
+
+Ls_molar=(Gs*(Y1-Y2))/(X1-X2);
+Ls=Ls_molar*260;
+
+printf("minimum oil circulation rate = %f kg/hr",Ls)
diff --git a/858/CH7/EX7.1/example_1.sce b/858/CH7/EX7.1/example_1.sce new file mode 100755 index 000000000..4d88fbbc7 --- /dev/null +++ b/858/CH7/EX7.1/example_1.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 7.1 page number 305\n\n")
+
+printf("it is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH7/EX7.10/example_10.sce b/858/CH7/EX7.10/example_10.sce new file mode 100755 index 000000000..4942d297d --- /dev/null +++ b/858/CH7/EX7.10/example_10.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 7.10 page number 316\n\n")
+
+//to find the % transformation
+moles_A = 3;
+moles_B = 5;
+K = 1;
+
+function[f] = F(x)
+ f = 15-8*x;
+endfunction
+
+//initial guess
+x = 10;
+y = fsolve(x,F);
+printf("amount of A transformed = %f percent",y*(100/3))
+
diff --git a/858/CH7/EX7.11/example_11.sce b/858/CH7/EX7.11/example_11.sce new file mode 100755 index 000000000..cb3e0db5c --- /dev/null +++ b/858/CH7/EX7.11/example_11.sce @@ -0,0 +1,6 @@ +clc
+clear
+printf("example 7.11 page number 316\n\n")
+
+//to find the product concentration
+printf("this is a theoritical question, book shall be referred for solution")
diff --git a/858/CH7/EX7.12/example_12.sce b/858/CH7/EX7.12/example_12.sce new file mode 100755 index 000000000..1cc918120 --- /dev/null +++ b/858/CH7/EX7.12/example_12.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 7.11 page number 316\n\n")
+
+//to find the initial conc of A and B
+Cp = 0.02;
+Cq = 0.02;
+K = 4*10^-2;
+Cb = 0.05;
+Cb_i = Cb+Cp;
+a = (Cp*Cq)/(K*Cb);
+funcprot(0)
+function[f] = F(x,a)
+ f = x-0.02-a;
+endfunction
+
+//initial guess
+x = 10;
+y = fsolve(x,F);
+printf("conc of A= %f mol/l",y)
+printf("\n\nconc of B= %f mol/l",Cb_i)
diff --git a/858/CH7/EX7.13/example_13.sce b/858/CH7/EX7.13/example_13.sce new file mode 100755 index 000000000..c529c3055 --- /dev/null +++ b/858/CH7/EX7.13/example_13.sce @@ -0,0 +1,30 @@ +clc
+clear
+printf("example 7.11 page number 316\n\n")
+
+//to find the % transformation
+
+moles_A = 0.02;
+K = 1;
+
+//part 1
+moles_B = 0.02;
+function[f] = F(x)
+ f = moles_A*moles_B-(moles_A+moles_B)*x;
+endfunction
+
+//initial guess
+x = 10;
+y = fsolve(x,F);
+printf("amount of A transformed = %f percent",y*(100/0.02))
+
+//part 2
+moles_B = 0.1;
+y = fsolve(x,F);
+printf("\n\namount of A transformed = %f percent",y*(100/0.02))
+
+//part 1
+moles_B = 0.2;
+y = fsolve(x,F);
+printf("\n\namount of A transformed = %f percent',y*(100/0.02))
+
diff --git a/858/CH7/EX7.14/example_14.sce b/858/CH7/EX7.14/example_14.sce new file mode 100755 index 000000000..ddd653e53 --- /dev/null +++ b/858/CH7/EX7.14/example_14.sce @@ -0,0 +1,21 @@ +clc
+clear
+printf("example 7.14 page number 317\n\n")
+
+//to find the initial concentration and shift in equilibrium
+
+Ce_N2 = 3; //equilibrium conc of N2
+Ce_H2 = 9; //equilibrium conc of H2
+Ce_NH3 = 4; //equilibrium conc oh NH3
+C_N2 = Ce_N2 + 0.5*Ce_NH3;
+C_H2 = Ce_H2 + 1.5*Ce_NH3;
+
+printf("concentration of N2 = %f mol/l\nconcentration of H2 = %f mol/l",C_N2,C_H2)
+printf("\n\nsecond part is theoritical, book shall be referred for solution")
+
+n_H2 = 3; //stotiometric coefficient
+n_N2 = 1; //stotiometric coefficient
+n_NH3= 2; //stotiometric coefficient
+delta_n = n_H2+n_N2-n_NH3;
+if delta_n > 0 then printf ("\n\ndelta_n =%f\nsince delta_n is greater than 0,equilibrium will shift to right with increase in volume",delta_n)
+end
diff --git a/858/CH7/EX7.15/example_15.sce b/858/CH7/EX7.15/example_15.sce new file mode 100755 index 000000000..270a27e25 --- /dev/null +++ b/858/CH7/EX7.15/example_15.sce @@ -0,0 +1,34 @@ +clc
+//example 7.15
+//to find the rate equation
+t = [0;5;10;15;20;25]
+C_A = [25;18.2;13.2;9.6;7;5.1]
+
+//integral method of rate determination
+s = 0;
+for i = 2:6
+ k(i) = (1/t(i))*log(25/C_A(i))
+ //disp (k(i),"k values for various conc.")
+ s = s+k(i)
+end
+
+printf("average value of k = %f",s/5)
+disp ("ra =- 0.06367*CA","since its a first order reaction,")
+
+subplot(221)
+plot(t,C_A)
+xlabel("time")
+ylabel("concentration")
+title("integral method")
+
+//differential method of rate determination
+ra = [-1.16;-0.83;-0.60;-0.43];
+C_A = [18.2;13.2;9.6;7];
+
+subplot(222)
+plot(ra,C_A)
+xlabel("Concentration")
+ylabel("-ra")
+title("differential method")
+
+printf("\n\nrate from differential method = -0.064*CA")
diff --git a/858/CH7/EX7.16/example_16.sce b/858/CH7/EX7.16/example_16.sce new file mode 100755 index 000000000..cf1148c2e --- /dev/null +++ b/858/CH7/EX7.16/example_16.sce @@ -0,0 +1,11 @@ +clc
+clear
+//example 7.16
+//to find the rate of reaction
+E = 75200 //in J/mol
+E1 = 50100 //in J/mol
+R = 8.314 //in J/mol K
+T = 298 //in K
+ratio = exp((E1-E)/(R*T));
+rate_increase = ratio^-1
+disp ("times",rate_increase,"increase in rate of reaction =")
diff --git a/858/CH7/EX7.2/example_2.sce b/858/CH7/EX7.2/example_2.sce new file mode 100755 index 000000000..2ed6eba4e --- /dev/null +++ b/858/CH7/EX7.2/example_2.sce @@ -0,0 +1,4 @@ +clc
+clear
+printf("example 7.2 page number 306\n\n")
+printf("it is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH7/EX7.3/example_3.sce b/858/CH7/EX7.3/example_3.sce new file mode 100755 index 000000000..2bdb3166a --- /dev/null +++ b/858/CH7/EX7.3/example_3.sce @@ -0,0 +1,18 @@ +clc
+clear
+printf("example 7.3 page number 305\n\n")
+
+//to find the change on rate of reaction
+//part 1
+//rate equation r = kC_NO^2*C_O2
+//if pressure increases 3 times
+
+r = 3^2*3; //according to the rate reaction
+printf("reaction reate will be increased by with 3 times increase in pressure = %f times",r)
+
+//part 2
+r = 3^2*3; //according to the rate reaction
+printf("\n\nreaction reate will be increased by with 3 times decrease in volume = %f times",r)
+
+r = 3^2; //according to the rate reaction
+printf("\n\nreaction reate will be increased by with 3 times increase in conc of NO = %f times",r)
diff --git a/858/CH7/EX7.4/example_4.sce b/858/CH7/EX7.4/example_4.sce new file mode 100755 index 000000000..b1db14a33 --- /dev/null +++ b/858/CH7/EX7.4/example_4.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 7.4 page number 308\n\n")
+
+printf("it is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH7/EX7.5/example_5.sce b/858/CH7/EX7.5/example_5.sce new file mode 100755 index 000000000..254f6940a --- /dev/null +++ b/858/CH7/EX7.5/example_5.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 7.5 page number 308\n\n")
+
+printf("it is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH7/EX7.6/example_6.sce b/858/CH7/EX7.6/example_6.sce new file mode 100755 index 000000000..c8b96654b --- /dev/null +++ b/858/CH7/EX7.6/example_6.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 7.6 page number 308\n\n")
+
+printf("it is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH7/EX7.7/example_7.sce b/858/CH7/EX7.7/example_7.sce new file mode 100755 index 000000000..ad3f35c99 --- /dev/null +++ b/858/CH7/EX7.7/example_7.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 7.7 page number 309\n\n")
+
+printf("it is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH7/EX7.8/example_8.sce b/858/CH7/EX7.8/example_8.sce new file mode 100755 index 000000000..b4f412a78 --- /dev/null +++ b/858/CH7/EX7.8/example_8.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 7.8 page number 312\n\n")
+
+printf("it is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH7/EX7.9/example_9.sce b/858/CH7/EX7.9/example_9.sce new file mode 100755 index 000000000..15ce39129 --- /dev/null +++ b/858/CH7/EX7.9/example_9.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 7.9 page number 312\n\n")
+
+printf("this is a theoretical question, book shall be referred for solution")
diff --git a/858/CH8/EX8.1/example_1.sce b/858/CH8/EX8.1/example_1.sce new file mode 100755 index 000000000..56bef86fa --- /dev/null +++ b/858/CH8/EX8.1/example_1.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 8.1 page number 334\n\n")
+
+printf("this is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH8/EX8.10/example_10.sce b/858/CH8/EX8.10/example_10.sce new file mode 100755 index 000000000..214357886 --- /dev/null +++ b/858/CH8/EX8.10/example_10.sce @@ -0,0 +1,26 @@ +clc
+clear
+printf("example 8.10 page number 370\n\n")
+
+//to find the rate of oil flow in l/s
+
+density_oil=900; //in kg/m3
+viscosity_oil=38.8*10^-3; //in Pa-s
+density_water = 1000; //in kg/m3
+diameter=0.102 //in m
+manometer_reading=0.9; //m of water
+delta_H=manometer_reading*(density_water-density_oil)/density_oil;
+printf("manometer reading as m of oil = %f m",delta_H)
+
+maximum_velocity=(2*9.8*delta_H)^0.5;
+printf("\n\nmaximum_velocity(Vmax) = %f m/s",maximum_velocity)
+
+Re=diameter*maximum_velocity*density_oil/viscosity_oil;
+printf("\n\nif Re<4000 then v=0.5*Vmax Re = %f",Re)
+if Re<4000 then velocity=maximum_velocity*0.5;
+end
+
+printf("\n\nvelocity = %f m/s",velocity)
+
+flow_rate=(3.14/4)*diameter^2*velocity*1000;
+printf("\n\nflow rate =%f litre/s",flow_rate)
diff --git a/858/CH8/EX8.11/example_11.sce b/858/CH8/EX8.11/example_11.sce new file mode 100755 index 000000000..4ccc8f228 --- /dev/null +++ b/858/CH8/EX8.11/example_11.sce @@ -0,0 +1,11 @@ +clc
+clear
+printf("example 8.11 page number 372\n\n")
+
+//to find the maximum capacity of keroscene
+flow_rate_steel=1.2; //l/s
+density_steel=7.92;
+density_kerosene=0.82;
+density_water=1;
+flow_rate_kerosene =(((density_steel-density_kerosene)/density_kerosene)/((density_steel-density_water)/density_water))^0.5*flow_rate_steel
+printf("maximum_flow rate of kerosene = %f litre/s",flow_rate_kerosene)
diff --git a/858/CH8/EX8.12/example_12.sce b/858/CH8/EX8.12/example_12.sce new file mode 100755 index 000000000..a4b9512fb --- /dev/null +++ b/858/CH8/EX8.12/example_12.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("example 8.12 page number 373\n\n")
+
+//to find the rate of flow of flue gas
+
+initial_CO2 = 0.02; //weight fraction
+flow_rate_CO2 = 22.5; //gm/s
+final_CO2=0.031; //weight fraction
+
+//flow rate of flue gas =x
+//amount of CO2 entering = 0.02*x
+//amount of CO2 leaving = 0.02x+0.0225
+//amount of gas leaving = x+0.0225
+//amount of CO2 leaving = 0.031*(x+0.0225)
+
+deff('y=f(x)','y=initial_CO2*x+0.0225 - 0.031*(x+0.0225)');
+
+flow_rate_flue_gas=fsolve(0,f)
+
+printf("flow rate of flue gas = %f kg/s",flow_rate_flue_gas)
+
diff --git a/858/CH8/EX8.2/example_2.sce b/858/CH8/EX8.2/example_2.sce new file mode 100755 index 000000000..c7ef7e59f --- /dev/null +++ b/858/CH8/EX8.2/example_2.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 8.2 page number 335\n\n")
+
+printf("this is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH8/EX8.3/example_3.sce b/858/CH8/EX8.3/example_3.sce new file mode 100755 index 000000000..02e7a6cf6 --- /dev/null +++ b/858/CH8/EX8.3/example_3.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 8.3 page number 335\n\n")
+
+printf("this is a theoritical problem, book shall be referred for solution")
diff --git a/858/CH8/EX8.4/example_4.sce b/858/CH8/EX8.4/example_4.sce new file mode 100755 index 000000000..3b63450cf --- /dev/null +++ b/858/CH8/EX8.4/example_4.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("example 8.4 page number 336\n\n")
+
+//Chapter 8 : Measuring Devices
+//Given: Balance Height=4m
+//side 1-air, side 2:- N2-H2 mixture
+
+pressure_difference = 3.4 //in mm water
+pressure = 1.0133*10^5 //in pa
+temperatue = 293 //in K
+mass_of_air = 29 //in Kg
+density_air = pressure/(temperatue*8314)*mass_of_air //in kg/m3
+printf("Density of air = %f kg/cu m",density_air)
+
+delta_p = pressure_difference*9.8 //in pascal, acceleration due to gravity, g=9.8
+Height=4
+density_difference = delta_p/(9.8*Height);
+printf("\n\nDensity difference = %f kg/cu m",density_difference)
+
+density_mixture= density_air-density_difference; //in kg/m3
+printf("\n\nDensity of mixture = %f kg/cu m",density_mixture)
+
diff --git a/858/CH8/EX8.5/example_5.sce b/858/CH8/EX8.5/example_5.sce new file mode 100755 index 000000000..eb168e7f9 --- /dev/null +++ b/858/CH8/EX8.5/example_5.sce @@ -0,0 +1,13 @@ +clc
+clear
+printf("example 8.5 page number 341\n\n")
+
+//to find viscosity of oil
+diameter=0.6; //in m
+disk_distance=1.25*10^-3; //in m
+speed=5; //revolutions/min
+torque=11.5; //in Joules
+
+//we know that torque= pi*omega*viscosity*radius^4/2*disc_distance
+viscosity=(2*disk_distance*torque)/(3.14*(10*3.14)*(diameter/2)^4);
+printf("viscosity = %f Pa-s",viscosity)
diff --git a/858/CH8/EX8.6/example_6.sce b/858/CH8/EX8.6/example_6.sce new file mode 100755 index 000000000..6eb51c96a --- /dev/null +++ b/858/CH8/EX8.6/example_6.sce @@ -0,0 +1,19 @@ +clc
+clear
+printf("example 8.6 page number 342\n\n")
+
+//to find the viscosity of solution using given parameters
+
+diameter =10; //in mm
+density_of_solution = 1750; //in kg/m3
+density_of_air = 1.2; //in kg/m3
+velocity = 0.9; //in mm/s
+viscosity = (density_of_solution-density_of_air)*9.8*(diameter*10^-3)^2/(18*velocity*10^-3); //expression for finding viscosity
+
+printf("viscosity of solution = %f Pa-s",viscosity)
+
+
+//checking stoke's region validity
+v=(0.2*viscosity)/(density_of_solution*diameter*10^-3);
+if v>0.9 then printf("\n\nsystem follows stokes law")
+end
diff --git a/858/CH8/EX8.7/example_7.sce b/858/CH8/EX8.7/example_7.sce new file mode 100755 index 000000000..d62873e68 --- /dev/null +++ b/858/CH8/EX8.7/example_7.sce @@ -0,0 +1,29 @@ +clc
+clear
+printf("example 8.7 page number 367\n\n")
+
+//to find the flow rate in an orifice
+density_of_water = 1000; //in kg/m3
+viscosity = 1*10^-3; //in Pa-s
+pipe_diameter = 250; //in mm
+orifice_diameter = 50; // in mm
+density_of_mercury = 13600; // in mm
+manometer_height = 242; //in mm
+
+//calculation
+height_water_equivalent = (density_of_mercury-density_of_water)*(manometer_height*10^-3)/(density_of_water) //in m
+
+//assuming Re>30000
+Co = 0.61;
+velocity = Co*(2*9.8*height_water_equivalent/(1-(orifice_diameter/pipe_diameter)^4))^0.5; //in m/s
+
+//checking Reynold's number
+Re = (orifice_diameter*10^-3*velocity*density_of_water)/viscosity;
+printf("reynolds number = %f\nwhich is greater than 30000",Re)
+
+if Re>30000 then printf("\n\nvelocity of water = %f m/s",velocity)
+
+end
+
+rate_of_flow = (3.14*(orifice_diameter*10^-3)^2/4)*velocity*density_of_water;
+printf("\n\nrate of flow = %f litre/s",rate_of_flow)
diff --git a/858/CH8/EX8.8/example_8.sce b/858/CH8/EX8.8/example_8.sce new file mode 100755 index 000000000..02cae8bc4 --- /dev/null +++ b/858/CH8/EX8.8/example_8.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("example 8.8 page number 368\n\n")
+
+//to find the coefficient of discharge for converging cone
+
+pipe_diameter=0.15; //in m
+venturi_diameter=0.05; //in m
+pressure_drop=0.12; //m of water
+flow_rate=3; //in kg/s
+density = 1000; //in kg/m3
+viscosity = 0.001 //in Pa-s
+
+velocity = ((4/3.14)*flow_rate)/(venturi_diameter^2*density);
+printf("velociy = %f m/s",velocity)
+
+//calculating coefficient of discharge
+Cv=velocity*((1-(venturi_diameter/pipe_diameter)^4)/(2*9.8*pressure_drop))^0.5;
+printf("\n\ncoefficient of discharge = %f",Cv)
+
+//calculating reynold's number
+Re = velocity*(venturi_diameter/pipe_diameter)^2*pipe_diameter*density/viscosity;
+printf("\n\nreynolds No = %f",Re)
diff --git a/858/CH8/EX8.9/example_9.sce b/858/CH8/EX8.9/example_9.sce new file mode 100755 index 000000000..650fd76a7 --- /dev/null +++ b/858/CH8/EX8.9/example_9.sce @@ -0,0 +1,24 @@ +clc
+clear
+printf("example 8.9 page number 369\n\n")
+
+//to find pA and pB
+//part 1
+
+h1=0.66; //in m
+h2=0.203; //in m
+h3=0.305 //in m
+density=1000; //in kg/m3
+pB=68900; //in Pa
+s1=0.83;
+s2=13.6;
+disp("part 1")
+pA=pB+(h2*s2-(h1-h3)*s1)*density*9.81; //in Pa
+printf("\npressure at A = %f Pa\n",pA)
+
+disp("part 2")
+pA1=137800 //in Pa
+pressure=735 //mm Hg
+pB1=pA1-(h2*s2-(h1-h3)*s1)*density*9.81;
+pressure_B=(pB1-pressure*133.3)/9810; //m of water
+printf("\npressure at B = %f m of water",pressure_B)
diff --git a/858/CH9/EX9.1/example_1.sce b/858/CH9/EX9.1/example_1.sce new file mode 100755 index 000000000..426bced1c --- /dev/null +++ b/858/CH9/EX9.1/example_1.sce @@ -0,0 +1,30 @@ +clc
+clear
+printf("example 9.1 page number 384\n\n")
+
+//to find the pressure drop in the coil
+
+D = 38*10^-3; //in m
+U = 1 //in m/s
+density = 998 //in kg/cubic m
+viscosity = 8*10^-4 //in Pa-s
+DC = 1 //in m
+N = 10
+e = 4*10^-6; //in m
+
+Re = (density*U*D)/viscosity;
+printf ("Reynolds number = %f",Re)
+
+f = (4*log10((e/D)/3.7+(6.81/Re)^0.9))^-2;
+printf("\n\nfriction factor = %f",f);
+
+L = 3.14*DC*N;
+
+delta_Pstr = (2*f*U*density*L)/D;
+printf("\n\npressure drop through straight pipe = %f Pa",delta_Pstr)
+
+S = 1+3.54*(D/DC);
+printf("\n\ncorrection factor = %f",S)
+
+delta_P = S*delta_Pstr
+printf("\n\npressure drop of coil = %f Pa",delta_P)
diff --git a/858/CH9/EX9.2/example_2.sce b/858/CH9/EX9.2/example_2.sce new file mode 100755 index 000000000..44c396fc5 --- /dev/null +++ b/858/CH9/EX9.2/example_2.sce @@ -0,0 +1,36 @@ +clc
+clear
+printf("example 9.2 page number 384\n\n")
+
+//to find the shell side pressure drop in heat exchanger
+
+U = 0.5 //in m/s
+N = 19;
+DT = 0.026 //in m
+L = 2.7 //in m
+DS = 0.2 //in m
+e = 0.0002 //in m
+density = 836 //in kg/cu m
+viscosity = 0.00032 //in Pa s
+Pr = 6.5;
+Prw = 7.6;
+
+
+HYDIA = (DS^2-N*DT^2)/(DS+N*DT);
+
+Re = HYDIA*U*density/viscosity;
+printf ("Reynolds number = %f",Re)
+
+f = (4*log10((e/HYDIA)/3.7+(6.81/Re)^0.9))^-2;
+printf("\n\nfriction factor = %f",f);
+
+L = 3.14*DT*N;
+
+delta_Pstr = (2*f*U*density*L)/HYDIA;
+printf("\n\npressure drop through straight pipe = %f Pa",delta_Pstr)
+
+S = (Prw/Pr)^0.33;
+printf("\n\ncorrection factor = %f",S)
+
+delta_P = S*delta_Pstr
+printf("\n\npressure drop of coil = %f Pa",delta_P)
diff --git a/858/CH9/EX9.3/example_3.sce b/858/CH9/EX9.3/example_3.sce new file mode 100755 index 000000000..666e30261 --- /dev/null +++ b/858/CH9/EX9.3/example_3.sce @@ -0,0 +1,38 @@ +clc
+clear
+printf("example 9.3 page number 385\n\n")
+
+MH = 10 //in kg/s
+MC = 12.5 //in kg/s
+CPH = 4.2 //in kJ/kg
+CPC = 4.2 //in kJ/kg
+THI = 353 //in K
+THO = 333 //in K
+TCI = 300 //in K
+U = 1.8 //in kW/sq m K
+
+Q = MH*CPH*(THI-THO);
+printf("heat load = %f J",Q)
+
+TCO = Q/(MC*CPC)+TCI;
+printf("\n\ncold fluid outlet temperature = %f K",TCO)
+
+//for co current flow
+
+DT1 = THI-TCO;
+DT2 = THO-TCO;
+
+LMTD = (DT1-DT2)/log(DT1/DT2);
+
+A = Q/(U*LMTD);
+printf("\n\nfor co current flow, area = %f sq m",A);
+
+//for counter current flow
+
+DT1 = THI-TCO;
+DT2 = THO-TCI;
+
+LMTD = (DT1-DT2)/log(DT1/DT2);
+
+A = Q/(U*LMTD);
+printf("\n\nfor counter current flow, area = %f sq m",A);
diff --git a/858/CH9/EX9.4/example_4.sce b/858/CH9/EX9.4/example_4.sce new file mode 100755 index 000000000..4158067fb --- /dev/null +++ b/858/CH9/EX9.4/example_4.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 9.4 page number 387\n\n")
+
+printf("this is a theoretical question, book shall be referred for solution")
diff --git a/858/CH9/EX9.5/example_5.sce b/858/CH9/EX9.5/example_5.sce new file mode 100755 index 000000000..e525db72d --- /dev/null +++ b/858/CH9/EX9.5/example_5.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 9.5 page number 392\n\n")
+
+printf("this is a theoretical question, book shall be referred for solution")
diff --git a/858/CH9/EX9.6/example_6.sce b/858/CH9/EX9.6/example_6.sce new file mode 100755 index 000000000..4cfdf559b --- /dev/null +++ b/858/CH9/EX9.6/example_6.sce @@ -0,0 +1,5 @@ +clc
+clear
+printf("example 9.6 page number 395\n\n")
+
+printf("this is a theoritical problem, book shall be referred for solution")
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