103 files changed, 2308 insertions, 0 deletions
diff --git a/2762/CH1/EX1.4.1/1_4_1.sce b/2762/CH1/EX1.4.1/1_4_1.sce
new file mode 100755
index 000000000..c8952e7eb
--- /dev/null
+++ b/2762/CH1/EX1.4.1/1_4_1.sce
@@ -0,0 +1,20 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.4-1
+//Introduction to engineering principles and units
+//given data
+//calculation of gas constant R
+//Assuming standard conditions
+p=14.7; //atmospheric pressure in psia
+v=359;//volume in feet cube
+n=1;//number of moles in lb mol
+t=492;//temp in degree R
+r=(p*v)/(n*t);//gas constant unit: (feet*feet*feet*psia)/(lb mol*degree R)
+mprintf("the gas constant in given units %f (ft3.psia/lb mol deg R)",r);
+//calculation in SI units
+P=101325;//pressure in pascals
+V=22.414;//volume in meter cube
+N=1;//moles in kg mol
+T=273.15;//temperature in kelvin
+R=(P*V)/(N*T);//gas constant unit: 
+mprintf(" the gas constant in SI units %f (m3*Pa)/(kg mol K)",R);
diff --git a/2762/CH1/EX1.4.2/1_4_2.sce b/2762/CH1/EX1.4.2/1_4_2.sce
new file mode 100755
index 000000000..631027402
--- /dev/null
+++ b/2762/CH1/EX1.4.2/1_4_2.sce
@@ -0,0 +1,23 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.4-2
+//Introduction to engineering principles and units
+//given data
+//Pressure and composition calculation
+//A:carbon dioxide, B: carbon monoxide, C: Nitrogen gas, D: Oxygen gas
+pA=75;
+pB=50;
+pC=595; 
+pD=26;
+//above are the patial pressures of the given gases respectively in mm Hg
+P=pA+pB+pC+pD; //total pressure of the mixture
+mprintf("the total pressure in mm Hg is %d", P);//
+xA=pA/P;
+xB=pB/P;
+xC=pC/P;
+xD=pD/P;
+//above are the patial pressures of the given gases respectively
+mprintf(" the mole fractions of the carbon dioxide %f",xA);
+mprintf(" the mole fractions of the carbon monoxide %f",xB);
+mprintf(" the mole fractions of the nitrogen %f",xC);
+mprintf(" the mole fractions of the oxygen %f",xD);
diff --git a/2762/CH1/EX1.5.1/1_5_1.sce b/2762/CH1/EX1.5.1/1_5_1.sce
new file mode 100755
index 000000000..d4dbfa23d
--- /dev/null
+++ b/2762/CH1/EX1.5.1/1_5_1.sce
@@ -0,0 +1,10 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.5-1
+//Applying materal balance, input=output+accumulation => 1000= W+C
+//Applying material balance for solids, 1000*xf= W(0)+ C*xc  =>C= 1000*xf/xc
+C= (1000*7.08)/58; 
+W=1000-C;
+mprintf("The flow rate of water=%f kg/h",W);
+mprintf("the flow rate of concentrated juice=%f kg/h",C);
+//end\       
diff --git a/2762/CH1/EX1.5.2/1_5_2.sce b/2762/CH1/EX1.5.2/1_5_2.sce
new file mode 100755
index 000000000..aa35a11af
--- /dev/null
+++ b/2762/CH1/EX1.5.2/1_5_2.sce
@@ -0,0 +1,19 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.5-2
+//Introduction to engineering principles and units
+//given data
+//Basis: feed stream is 1000 kg/h
+//component balance for Pottasium Nitrate: 1000(20/100)=W(0)+P(96/100)
+P=1000*(0.2)*(100/96);//P is the outlet rate of KNO3 crystals
+
+//overall balance for a crystallizer: S-R=P
+//KNO3 balance on crystallizer: S(50/100)-R(37.5/100)=P(96/100)
+//solving 2 equations using matrix operations
+A=[1 -1;0.5 -0.375];
+B=[P;0.96*P];
+x=inv(A)*B;
+S=x(1,1);
+R=x(2,1);
+mprintf(" the recycle R %f kg/h",R)
+mprintf("the rate of crystals getting out of crystallizer %f kg/h",S)
diff --git a/2762/CH1/EX1.5.3/1_5_3.sce b/2762/CH1/EX1.5.3/1_5_3.sce
new file mode 100755
index 000000000..6dc819d35
--- /dev/null
+++ b/2762/CH1/EX1.5.3/1_5_3.sce
@@ -0,0 +1,36 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.5-3
+//Introduction to engineering principles and units
+//given data
+//Let A be the moles of air and f be the moles of flue gas
+//Basis 100 kgmol of flue gas
+//Reacttions: CO+0.5O2 -> CO2 ; H2 + 0.5O2 -> H2O
+//moles of O2 = 0.5*27.2(CO)+5.6(CO2)+(0.5)O2
+O=0.5*27.2+5.6+0.5;
+//for all H2 to be completely burnt we need: (3.1/2) moles of O2.Also for completely burning CO we need (27.2*0.5)
+//0.5 moles of O2 used up in fuel gas
+//calculating amount of O2 reqd theoretically
+Otheo=(3.1/2)+(27.2/2)-0.5;
+//for 20% excess we add:
+Oact=1.2*Otheo;
+//the amt of N2 added:
+N=(79/21)*Oact;
+//amt of unburnt CO (98% combustion)
+COun=0.02*27.2;
+//total carbon balance:
+C=27.2+5.6;
+//free CO2
+CO2=C-COun;
+//calculating inlet and outlet moles of o2
+Oin=Oact+O;
+//Oout=(3.1/2 in H2O)+(COun/2)(in CO)+CO2+ free O2(Ofree)
+Ofree=Oin-(3.1/2)-(COun/2)-CO2;
+//Nitrogen Balance in outlet(Nt): N in air + N in flue gas
+Nt=N+63.6;
+mprintf(" moles of H20=3.10 mol")
+mprintf(" moles of N2 %f mol",Nt)
+mprintf(" moles of CO %f mol",COun)
+mprintf(" moles of N2 %f mol",Nt)
+mprintf(" moles of CO2 %f mol",CO2)
+mprintf(" moles of free O2 %f mol",Ofree)
diff --git a/2762/CH1/EX1.6.1/1_6_1.sce b/2762/CH1/EX1.6.1/1_6_1.sce
new file mode 100755
index 000000000..566072dee
--- /dev/null
+++ b/2762/CH1/EX1.6.1/1_6_1.sce
@@ -0,0 +1,17 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.6-1
+//Introduction to engineering principles and units
+//given data
+//heat reqd= mCp)(delta T)
+//a) 298-673K
+m=3;//m is given as 3 g mol
+H1= m*29.68*(673-298);//for N2 at 673K Cp=29.68 J/g mol K
+//b) 298-1123K
+H2=m*31*(1123-298);//for N2 at 1123 K Cp=31.00 J/g mol K by linear interpolation
+//c) 673-1123K : As there id no mean Cp we subtract a) from b)
+H3=H2-H1;
+mprintf("the heat reqd in a) : %f joules",H1)
+mprintf(" the heat reqd in b): %f joules",H2)
+mprintf(" the heat reqd in c): %f joules",H3)
+
diff --git a/2762/CH1/EX1.6.2/1_6_2.sce b/2762/CH1/EX1.6.2/1_6_2.sce
new file mode 100755
index 000000000..b5a315da6
--- /dev/null
+++ b/2762/CH1/EX1.6.2/1_6_2.sce
@@ -0,0 +1,12 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.6-2
+//Introduction to engineering principles and units
+//given data
+//Avg Cp of cows milk is 3.85 kJ/kg K
+//heat reqd= mCp)(delta T)
+m=4536; //in kg/h
+delT=54.4-4.4;//Temp diff
+Cp=3.85;
+H=(m*Cp*delT)/3600;//heat reqd in kW
+mprintf("heat reqd for heating the milk is %f kW",H) 
diff --git a/2762/CH1/EX1.6.3/1_6_3.sce b/2762/CH1/EX1.6.3/1_6_3.sce
new file mode 100755
index 000000000..926a15b34
--- /dev/null
+++ b/2762/CH1/EX1.6.3/1_6_3.sce
@@ -0,0 +1,28 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.6-3
+//Introduction to engineering principles and units
+//given data
+//a)at 101.325 kPa
+
+H1=88.60;// enthalpy of water in kJ/kg at 21.11 degree celcius
+H2=251.13;// enthalpy of water kJ/kg at 60 degree celcius
+delH1=H2-H1;//change in enthalpy in SI units
+//enthalpy change in English units
+h1=38.09;//enthalpy of water in btu/lb at 70 degree F
+h2=107.96;//enthalpy of water in btu/lb at 140 degree F
+delh1=h2-h1;//change in enthalpy in English units
+mprintf("change in enthalpy in SI and English units :%f kJ/kg and %f btu/lb respectively",delH1,delh1);
+//b)Enthalpy change 172.2 kPa
+H3=2699.9;// enthalpy of water kJ/kg at 115.6 degree celcius
+delH2=H3-H1;
+h3=1160.7; //enthalpy of water in btu/lb
+delh2=h3-h1;//change in enthalpy in English units
+mprintf("change in enthalpy in SI and English units :%f kJ/kg and %f btu/lb respectively",delH2,delh2);
+//c)Enthalpy change at 172.2 kPa
+H4=484.9;//enthalpy of water in kJ/kg at 115.6 degree C
+delH3=H3-H4; //enthalpy change in SI units
+h4=208.4;//enthalpy of water in btu/lb at 240 degree F
+delh3=h3-h4;//change in enthalpy in English units
+mprintf("change in enthalpy in SI and English units :%f kJ/kg and %f btu/lb respectively",delH3,delh3);
+//end
diff --git a/2762/CH1/EX1.6.4/1_6_4.sce b/2762/CH1/EX1.6.4/1_6_4.sce
new file mode 100755
index 000000000..89ec7e124
--- /dev/null
+++ b/2762/CH1/EX1.6.4/1_6_4.sce
@@ -0,0 +1,13 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.6-4
+//Introduction to engineering principles and units
+//given data
+//heat of combustion for carbon to carbon dioxide is -393x10^3 kJ/kg mol or -94.0518 kCal/g mol
+//heat of combustion for carbon to carbon monoxide is -110x10^3 kJ/kg mol or -26.4157 kCal/g mol
+//basis: 10 g bol of carbon where 90% converts to carbon dioxide and rest to carbon monoxide
+HkJ=(90/100)*10*(-393.513)+(10/100)*10*(-110.523);//change in enthalpy= sum of heat of combustion of products(as reactant is a plain element)
+HkCal=(90/100)*10*(-94.0518)+(10/100)*10*(-26.4157);
+mprintf("change in enthalpy %f kJ",HkJ)
+mprintf("change in enthalpy  %f kCal",HkCal)
+//end
diff --git a/2762/CH1/EX1.6.5/1_6_5.sce b/2762/CH1/EX1.6.5/1_6_5.sce
new file mode 100755
index 000000000..5cea8b264
--- /dev/null
+++ b/2762/CH1/EX1.6.5/1_6_5.sce
@@ -0,0 +1,11 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.6-5
+//Introduction to engineering principles and units
+//given data
+//heat of formation at 298K for methane is -74.848x10^3 kJ/kg mol, water is 285.840x10^3 kJ/kg mol ,CO is -110.523x10^3 kJ/kg mol,H2 is 0x10^3 kJ/kg mol
+//basis: 1 kg mol of methane at 101.325 kPa and 298K: CH4 + H20 -> CO+ H2
+//std heat of reaction=(sum of heat of formation of pdts)-(sum of heat of formation of pdts)
+H=(-110.523*10^3-3*0)-(-74.848*10^3-285.840*10^3);
+mprintf("the std heat of reaction in  is %f kJ/kgmol",H)
+//end
diff --git a/2762/CH1/EX1.7.1/1_7_1.sce b/2762/CH1/EX1.7.1/1_7_1.sce
new file mode 100755
index 000000000..ff666cb28
--- /dev/null
+++ b/2762/CH1/EX1.7.1/1_7_1.sce
@@ -0,0 +1,23 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.7-1
+//Introduction to engineering principles and units
+//given data
+//Hf data at 298K 
+//Input items: sum of the enthalpies of two streams relative to 298K 
+//calculating H of liq
+Hil=2000*4.06*(30-25);//inlet mass flow rate of the liquid=2000 kg/h,Cp= 4.06 kJ/kg K, final temp-initial temp= 30 deg C - 25 deg C
+//Hiw(enthalpy at inlet of water)=W(4.21)(95-25) where W in kg/h Cp of water is 4.21 kJ/kg K, 95-25 is the temp diff
+//Output items 
+Hol=2000*4.06*(70-25);//outlet mass flow rate of liquid is 2000 kg/h, Cp= 4.06 kJ/kg K 70-25: temp diff
+//How= W(4.21)(85-25)
+//energy at inlet = energy at outlet
+//4.060*10^4 + 2.947*10^2 W= 3.654*10^5 + 2.526*10^2 W
+// solving these equations:
+W= ((4.060*10^4)-(3.654*10^5))/((2.526*10^2)-(2.947*10^2))
+mprintf("the outlet feed rate in kg/h is %f",W)
+//calculating enthalpy change of liquid: 
+delH= Hol-Hil;
+mprintf(" change in enthalpy in kw in kJ/h is %f",delH)
+//end
+//s=all the calculations performed are correct but there may be certein deviations.
diff --git a/2762/CH1/EX1.7.2/1_7_2.sce b/2762/CH1/EX1.7.2/1_7_2.sce
new file mode 100755
index 000000000..553bb09a0
--- /dev/null
+++ b/2762/CH1/EX1.7.2/1_7_2.sce
@@ -0,0 +1,26 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.7-2
+//Introduction to engineering principles and units
+//given data: CO + 0.5O2 -> CO2
+//del H (298K)=-282.989*10^3 kJ/kg mol
+//inlet flow rate= 1 kg mol/h=i
+i=1;
+it=0.5*i;// moles of O2 theorettically reqd
+oa=it*(1.9);//moles of O2 actually added
+n=oa*(0.79/0.21);//moles of N2 added
+air=oa+n;
+oout=oa-it;
+// CO2 in outlet flue gas is 1 kg/h and N2 in outlet flue gas=n
+//calculating input enthalpy of diff components from formula H= m*Cp*(delta T) and Cp evaluated from data tables
+Hico=1*29.38*(473-298);
+Hiair=4.520*29.29*(373-298);
+Histd=-(-282.989*(10^3))*(1);
+//calculating output enthalpy of diff components from formula H= m*Cp*(delta T) and Cp evaluated from data tables
+HoCO2=1*49.91*(1273-298);
+HoO2=oout*33.25*(1273-298);
+HN2=n*31.43*(1273-298);
+//Energy in=Energy out ; Hico+Hiair+q(heat added)+Histd=HoCO2+HoO2+HN2
+q=HoCO2+HoO2+HN2-(Hico+Hiair+Histd);
+mprintf("the heat removed in  is %f kJ/h",q)
+//end
diff --git a/2762/CH1/EX1.7.3/1_7_3.sce b/2762/CH1/EX1.7.3/1_7_3.sce
new file mode 100755
index 000000000..f7bf09212
--- /dev/null
+++ b/2762/CH1/EX1.7.3/1_7_3.sce
@@ -0,0 +1,18 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 1
+//Example 1.7-3
+//Introduction to engineering principles and units
+//given data
+//datum temp= 25 deg C
+//input and output enthalpies are calculated: m*Cp*delT Cp obtained from data tables 
+delT=37-25;//temp diff
+Hil=342.3*1.20*delT ;
+HiO2=12*29.38*delT
+Hrxn=(-5648.8*10^3);//heat of reaction given
+//output items
+HoH2O=11*18.02*4.18*delT;
+HoCO2=12*37.45*delT;
+//Energy in= Energy out: Hil+HiO2-Hrxn=HoH2O+HoCO2-H310K
+H310K=HoH2O+HoCO2-(Hil+HiO2-Hrxn);
+mprintf("the heat reqd for complete oxidation  is %f J/mol",H310K)
+//end
diff --git a/2762/CH10/EX10.2.1/10_2_1.sce b/2762/CH10/EX10.2.1/10_2_1.sce
new file mode 100755
index 000000000..d50088d68
--- /dev/null
+++ b/2762/CH10/EX10.2.1/10_2_1.sce
@@ -0,0 +1,12 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 10
+//Example 10.2-1
+//Stage and continuous Gas-liquid Seperation Processes
+//given data
+pa=0.21;//atm
+H=4.38*10000;//henreys law constant
+xa=pa/H;
+mprintf("the amount of O2 dissolved is %f mol O2 in 1 mol water or",xa)
+xad=(xa/18)*100;
+mprintf(" %f per 100 parts",xad)
+ 
diff --git a/2762/CH10/EX10.3.1/10_3_1.sce b/2762/CH10/EX10.3.1/10_3_1.sce
new file mode 100755
index 000000000..d5fa7707e
--- /dev/null
+++ b/2762/CH10/EX10.3.1/10_3_1.sce
@@ -0,0 +1,22 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 10
+//Example 10.3-1
+//Stage and continuous Gas-liquid Seperation Processes
+//given data
+L0=300;//kg mol/h
+Ld=L0;
+V=100;//kg mol/h
+ya2=0.2;
+Vd=V*(1-ya2);
+//L0*(xa0/(1-xa0))+Vd*(ya2/(1-ya2))=Ld*(xa1/(1-xa1))+Vd*(ya1/(1-ya1))
+xa0=0;
+LHS=L0*(xa0/(1-xa0))+Vd*(ya2/(1-ya2));
+H=0.142*10000;//henrys law constant at 293 K (atm/mol frac)
+P=1;//atm
+Hd=H/P;
+xa1=1.41/10000
+ya1=Hd*xa1
+L1=Ld/(1-xa1);
+V1=Vd/(1-ya1);
+mprintf("the outlet liquid flow rate is %f kg/h",L1);
+mprintf("the outlet vapour flow rate is %f kg/h",V1);
diff --git a/2762/CH10/EX10.3.3/10_3_3.sce b/2762/CH10/EX10.3.3/10_3_3.sce
new file mode 100755
index 000000000..ceef2ac48
--- /dev/null
+++ b/2762/CH10/EX10.3.3/10_3_3.sce
@@ -0,0 +1,20 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 10
+//Example 10.3-3
+//Stage and continuous Gas-liquid Seperation Processes
+//given data
+V1=29.73;//kg mol/h
+ya1=0.00101;
+L0=90;
+xa0=0;
+m=2.53;
+A1=L0/(m*V1);//cross sectional area
+Vn1=30;
+yan1=0.01;
+Ln=90.27;
+xan=0.003;
+An=Ln/(m*Vn1);
+A=sqrt(A1*An);
+Np=log(((yan1-m*xa0)/(ya1-m*xa0))*(1-1/A)+(1/A))/log(A);//no. of plates
+mprintf("the no. of plates= %f",Np)
+//kremsor equations
diff --git a/2762/CH10/EX10.4.2/10_4_2.sce b/2762/CH10/EX10.4.2/10_4_2.sce
new file mode 100755
index 000000000..dd3551a3d
--- /dev/null
+++ b/2762/CH10/EX10.4.2/10_4_2.sce
@@ -0,0 +1,26 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 10
+//Example 10.4-2
+//Stage and continuous Gas-liquid Seperation Processes
+//given data from the graph mentioned in the example
+yastar=0.052;
+xal=0.1;//bulk concn of A in liquid phase
+yag=0.38;//bulk concn of A in gas phase
+yai=0.197;//concn of A at interface
+xai=0.247;//yai=f(xai)
+kdy=1.465/1000;//gas phase mass transfer coefficient
+kdx=1.967/1000;//liq phase mass transfer coefficient
+md=(yai-yastar)/(xai-xal);//graphical correlation
+yaim=((1-yai)-(1-yag))/((log((1-yai)/(1-yag)))/log(2.71828183))+1;//graphical correlation
+xaim=((1-xal)-(1-xai))/((log((1-xal)/(1-xai)))/log(2.71828183))+1;//graphical correlation
+yam=((1-yastar)-(1-yag))/((log((1-yastar)/(1-yag)))/log(2.71828183))+1;//graphical correlation
+//(1/(Kdy(1-yam)))=(1/(kdy/(1-yaim)))+(md/(kdx/(1-xaim)))
+A=(1/(kdy/(1-yaim)));
+B=(md/(kdx/(1-xaim)));
+Kdy=((A+B)^(-1))*(1-yam);
+R=(A/(A+B))*100;//
+Na=(Kdy/(1-yam))*(yastar-yag);
+mprintf("overall mass transfer coefficient= %f kg mol/s m2",Kdy)
+mprintf(" percentage resistace in gas film= %f percent",R)
+mprintf(" percentage resistace in liquid film= %f percent",(100-R))
+mprintf(" Flux= %f kg mol/s m2",Na)
diff --git a/2762/CH11/EX11.1.1/11_1_1.sce b/2762/CH11/EX11.1.1/11_1_1.sce
new file mode 100755
index 000000000..c37cc919d
--- /dev/null
+++ b/2762/CH11/EX11.1.1/11_1_1.sce
@@ -0,0 +1,78 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 11
+//Example 11.1-1
+//Vapour Liquid Seperation Processes
+//given data
+pa=[116.9 135.5 155.7 179.2 204.2 240]
+pb=[46 54 63.3 74.3 86 101.32]
+P=101.32
+//eq: paxa+pb(1-xa)=P
+xa=[];
+ya=[];
+for i=1:6
+    xa(i)=(P-pb(1,i))/(pa(1,i)-pb(1,i))
+    ya(i)=((pa(1,i))*xa(i))/P
+end
+xa(7)=1;
+ya(7)=1;
+m=linspace(0,1,10)
+n=linspace(0,1,10)
+plot(m,n)
+plot2d(xa,ya,rect=[0 0 1 1])
+xtitle("raolts law","xa","ya")
+mprintf("xa=%f",xa);
+mprintf("ya=%f",ya);
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+
+
diff --git a/2762/CH11/EX11.2.1/11_2_1.sce b/2762/CH11/EX11.2.1/11_2_1.sce
new file mode 100755
index 000000000..ae61d0ced
--- /dev/null
+++ b/2762/CH11/EX11.2.1/11_2_1.sce
@@ -0,0 +1,40 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 11
+//Example 11.3-1
+//Vapour Liquid Seperation Processes
+//given data
+//suffix d means values are theoretical. other values withoud d are graphical values
+ya2d=0.4;//mole fraction of benzene
+xa0d=0.3;
+V2=100;//100 kg mol mixture of benzene and toluene 
+L0=110;//mixed with 110 kg mol of benzene and toluene
+L1=L0;//as the 2 streams are in equilibrium with each other
+V1=V2;
+//material balance on A
+//L0xa0+V2ya2=L1xa1+V1ya1
+//assuming xa1=0.2
+xa=[];
+ya=[];
+xa1d=0.2;
+ya1d=(L0*xa0d+V2*ya2d-L1*xa1d)/V1
+xa1dd=0.4;
+ya1dd=(L0*xa0d+V2*ya2d-L1*xa1dd)/V1
+xa1ddd=0.3;
+ya1ddd=(L0*xa0d+V2*ya2d-L1*xa1ddd)/V1
+xad=[xa1d xa1dd xa1ddd];
+yad=[ya1d ya1dd ya1ddd];
+plot(xad,yad,rec=[0,0,1,1])
+pa=[116.9 135.5 155.7 179.2 204.2 240]
+pb=[46 54 63.3 74.3 86 101.32]
+P=101.32;//dew point pressure
+//eq: paxa+pb(1-xa)=P
+m=linspace(0,1,10)
+n=linspace(0,1,10)
+for i=1:6
+    xa(i)=(P-pb(1,i))/(pa(1,i)-pb(1,i))
+    ya(i)=((pa(1,i))*xa(i))/P
+end
+plot(m,n)
+plot2d(xa,ya,rect=[0 0 1 1])
+xtitle("equilibrium diagram for benzene-toluene system","xa","ya")
+disp("from the graph intersection we can say ya1=0.455 and xa1=0.25")
diff --git a/2762/CH11/EX11.3.1/11_3_1.sce b/2762/CH11/EX11.3.1/11_3_1.sce
new file mode 100755
index 000000000..49d52f990
--- /dev/null
+++ b/2762/CH11/EX11.3.1/11_3_1.sce
@@ -0,0 +1,17 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 11
+//Example 11.3-1
+//Vapour Liquid Seperation Processes
+//given data
+//85 deg C
+Pa1=116.9;
+Pb1=46;
+alpha1=Pa1/Pb1;
+// 105 deg C
+Pa2=204.2;
+Pb2=86;
+alpha2=Pa2/Pb2;
+mprintf("alpha at 85 deg C = %f",alpha1);
+mprintf(" alpha at 105 deg C = %f",alpha2);
+p=((alpha1-alpha2)/alpha2)*100;
+mprintf(" percentage variation=%f",p)
diff --git a/2762/CH12/EX12.3.1/12_3_1.sce b/2762/CH12/EX12.3.1/12_3_1.sce
new file mode 100755
index 000000000..ff5ac9bae
--- /dev/null
+++ b/2762/CH12/EX12.3.1/12_3_1.sce
@@ -0,0 +1,33 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 12
+//Example 12.1-1
+//Liquid Liquid and Fluid Solid Seperation Process
+//given data
+t=[0 3 3.5 4 4.5 5 5.5 6 6.2 6.5 6.8];
+cbyc0=[0 0 0.002 0.030 0.155 0.396 0.658 0.903 0.933 0.975 0.993];
+ncbyc0=1-cbyc0
+plot2d(t,cbyc0)
+xtitle("Breakthrough curve","time,t h","c/c0")
+v = inttrap(t,ncbyc0)
+mprintf("the break point time as seen from the graph is 3.65 h");
+tb=3.65;
+cbyc01=[0 0.002 0.030 0.1];
+t1=[0 3 3.5 3.65];
+ncbyc01=1-cbyc01
+v1 = inttrap(t1,ncbyc01)
+mprintf(" the fraction capacity broken to the used point is %f",tb/v)
+//total bed length: Hb=(tu/tt)*Ht
+Ht=14;
+Hunb=(1-v1/v)*Ht;//ht of unused bed
+Hb=(v1/v)*Ht;
+mprintf(" the total bed length is %f ",Hunb)
+//b) tb=6, hence new bed height Hbb
+tbn=6;
+Hbb=(tbn/tb)*Hb;
+HT=Hbb+Hunb;
+AFR=754*3600*0.00115;//air flow rate=754 cm3/s, 1h=3600s, density=0.00115 g/cm3
+c0=600;//inlet stream concn of 600 ppm
+TAA=(c0/(10^6))*AFR*v;//total alcohol absorbed
+m=79.2;//mass of carbon present in grams
+SC=TAA/m;//saturation capacity
+mprintf("fraction of new bed used up to the break point is %f",Hbb/HT)
diff --git a/2762/CH12/EX12.5.2/12_5_2.sce b/2762/CH12/EX12.5.2/12_5_2.sce
new file mode 100755
index 000000000..09f4d0881
--- /dev/null
+++ b/2762/CH12/EX12.5.2/12_5_2.sce
@@ -0,0 +1,23 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 12
+//Example 12.5-2
+//Liquid Liquid and Fluid Solid Seperation Process
+//given data
+ya=0.04;//concn of A in upper layer
+yb=0.02;//concn of B in upper layer
+yc=0.94;//concn of C in upper layer
+xa=0.12;//concn of A in lower layer
+xb=0.86;//concn of B in lower layer
+xc=0.02;//concn of C in lower layer
+M=100;//mass of mixture in kg
+xam=0.1;
+//Material Balance Equations
+//eq 1: V+L=M
+//eq 2:ya*V+xa*L=xam*M
+A=[1 1;ya xa];
+B=[M; M*xam];
+X=inv(A)*B
+V=X(1,1);
+L=X(2,1);
+mprintf("amt of vapour phase=%f kg ",V)
+mprintf("amt of liquid phase=%f kg ",L)
diff --git a/2762/CH12/EX12.8.1/12_8_1.sce b/2762/CH12/EX12.8.1/12_8_1.sce
new file mode 100755
index 000000000..f6d0a5109
--- /dev/null
+++ b/2762/CH12/EX12.8.1/12_8_1.sce
@@ -0,0 +1,13 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 12
+//Example 12.8-1
+//Liquid Liquid and Fluid Solid Seperation Process
+//given data
+//for 80% extraction, unextracted fraction, Es=0.2
+Es=0.2;
+//Dab*(t/a*a) is constant
+t1=3.11;//time reqd to leach 80% of solution
+a2=1.5/2;//radius of reduced particale size
+a1=2/2;//radius od the particle size
+t2=(t1*a2*a2)/(a1*a1);
+mprintf("the time of leaching= %f h",t2)
diff --git a/2762/CH13/EX13.2.1/13_2_1.sce b/2762/CH13/EX13.2.1/13_2_1.sce
new file mode 100755
index 000000000..43d3db873
--- /dev/null
+++ b/2762/CH13/EX13.2.1/13_2_1.sce
@@ -0,0 +1,23 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 13
+//Example 13.2-1
+//Membrane Seperation Processes
+//given data
+//i) is to find the eqation
+disp('the original equation is Na= (c1-c2)/((1/kc1)+(1/pm)+(1/kc2))')
+disp(' but 1/kc2=0(as it is given that kc2 is large or almost infinite) hence it is Na=(c1-c2)/((1/kc1)+(1/pm))')
+//(b)
+Dab=7e-11;//membrane diffusivity in m2/s
+Kd=1.5;//distribution coefficint
+L=3e-5;//membrane thickness in m
+pm=(Dab*Kd)/L;
+c1=3e-2;//concn of the dilute soln containing A
+c2=0.5e-2;//concn on the other side
+kc2=2.02e-5;//mass transfer coefficient
+Na=(c1-c2)/((1/kc2)+(1/pm));//to interface concn C2i, Na=kc2*(C2i-c2) hence,
+C2i=(Na/kc2)+c2;
+//also Kd=c2is/C2i
+c2is=Kd*C2i;
+mprintf("flux= %f kg mol/s m2",Na);
+mprintf("interface concentration: %f kg mol/m3",C2i)
+mprintf("interface concentration: %f kg mol/m3",c2is)
diff --git a/2762/CH13/EX13.2.2/13_2_2.sce b/2762/CH13/EX13.2.2/13_2_2.sce
new file mode 100755
index 000000000..0b2dcd1ec
--- /dev/null
+++ b/2762/CH13/EX13.2.2/13_2_2.sce
@@ -0,0 +1,16 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 13
+//Example 13.2-2
+//Membrane Seperation Processes
+//given data
+//nomenclature similar to previous problem
+c1=200;//urea in g/m3
+c2=0;
+kc1=1.25e-5;
+pm=8.73e-6;
+kc2=3.33e-5;
+Na= (c1-c2)/((1/kc1)+(1/pm)+(1/kc2));
+A=2;//area in m2
+R=Na*3600*A;//rate of removal
+mprintf("flux=%f g/s m2",Na)
+mprintf("rate of removal= %f g urea/h",R)
diff --git a/2762/CH13/EX13.4.1/13_4_1.sce b/2762/CH13/EX13.4.1/13_4_1.sce
new file mode 100755
index 000000000..46b923ee2
--- /dev/null
+++ b/2762/CH13/EX13.4.1/13_4_1.sce
@@ -0,0 +1,23 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 13
+//Example 13.4-1
+//Membrane Seperation Processes
+//given data
+pda=50e-10;
+pdb=5e-10;
+alphastar=pda/pdb;
+a=1-alphastar;
+ph=80;
+p1=20;
+x0=0.25;
+b=(ph/p1)*(1-x0)-1+(alphastar*(ph/p1)*x0)+alphastar;
+c=-alphastar*(ph/p1)*x0;
+yp=(-b+sqrt(b*b-4*a*c))/(2*a);//permeate composition
+xf=0.5;
+theta=(xf-x0)/(yp-x0);//fraction permeated
+qf=10000;
+t=2.54e-3;
+Am=(theta*qf*yp)/((pda/t)*(ph*x0-p1*yp));//membrane area
+mprintf("1. permeate composition= %f",yp)
+mprintf(" 2. fraction permeated= %f",theta)
+mprintf(" 3. membrane area=%f m2",Am)
diff --git a/2762/CH13/EX13.4.2/13_4_2.sce b/2762/CH13/EX13.4.2/13_4_2.sce
new file mode 100755
index 000000000..853ed9d00
--- /dev/null
+++ b/2762/CH13/EX13.4.2/13_4_2.sce
@@ -0,0 +1,22 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 13
+//Example 13.4-2
+//Membrane Seperation Processes
+//given data
+theta=0.2;//fraction cut
+p1=19;//pressure selected for use
+ph=190;//pressure selectred for use
+alphas=10
+a1=theta+(p1/ph)-(theta*(p1/ph))-(alphas*theta)-(alphas*(p1/ph))+(alphas*(p1/ph)*theta);
+xf=0.209;//feed composition
+b1=1-theta-xf-(p1/ph)+(theta*(p1/ph))+(alphas*theta)+(alphas*(p1/ph))-(alphas*(p1/ph)*theta)+alphas*xf;
+c1= -alphas*xf;
+yp=(-b1+sqrt(b1*b1-4*a1*c1))/(2*a1);//permeate composition
+x0=(xf-theta*yp)/(1-theta);//outlet reject composition
+qf=1000000;//feed rate
+Pad=500e-10;
+t=2.54e-3;
+Am=(theta*qf*yp)/((Pad/t)*(ph*x0-p1*yp));//area of membrane
+mprintf("the permeate composition= %f ",yp)
+mprintf("outlet reject composition= %f ",x0)
+mprintf("area of membrane= %f cm2",Am)
diff --git a/2762/CH13/EX13.4.3/13_4_3.sce b/2762/CH13/EX13.4.3/13_4_3.sce
new file mode 100755
index 000000000..c49dc499b
--- /dev/null
+++ b/2762/CH13/EX13.4.3/13_4_3.sce
@@ -0,0 +1,16 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 13
+//Example 13.4-3
+//Membrane Seperation Processes
+//given data
+//nomenclature similar to previous problem
+xf=0.5;
+theta=0.2;//fraction cut
+p1=20;//pressure selected for use
+ph=80;//pressure selectred for use
+alphas=10;
+xom=(xf*(1+(alphas-1)*(p1/ph)*(1-xf)))/(alphas*(1-xf)+xf);
+xf1=0.65;
+xom1=(xf1*(1+(alphas-1)*(p1/ph)*(1-xf1)))/(alphas*(1-xf1)+xf1);
+mprintf("minimum reject concentration at xf=0.5= %f",xom);
+mprintf(" minimum reject concentration at xf=0.65= %f",xom1);
diff --git a/2762/CH13/EX13.9.1/13_9_1.sce b/2762/CH13/EX13.9.1/13_9_1.sce
new file mode 100755
index 000000000..1cd6a3a04
--- /dev/null
+++ b/2762/CH13/EX13.9.1/13_9_1.sce
@@ -0,0 +1,13 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 13
+//Example 13.9-1
+//Membrane Seperation Processes
+//given data
+rhow=997;//density of water in kg/m3
+m=0.1/1000;
+n=2*m;//NaCl gives 2 ions n in kg mol
+Vm=1/rhow;//specific volume 
+R=82.057/1000;//gas constant
+T=298.15;//25 deg C to K
+pi=(n/Vm)*(R*T)
+mprintf("osmotic pressure= %f atm" ,pi)
diff --git a/2762/CH14/EX14.2.1/14_2_1.sce b/2762/CH14/EX14.2.1/14_2_1.sce
new file mode 100755
index 000000000..dd52d333a
--- /dev/null
+++ b/2762/CH14/EX14.2.1/14_2_1.sce
@@ -0,0 +1,30 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.2-1
+//Mechanical-Physical Seperation Processes
+//given data
+t=[4.4 9.5 16.3 24.6 34.7 46.1 59 73.6 89.4 107.3];
+V=(1/1000)*[0.498 1 1.501 2 2.498 3.002 3.506 4.004 4.502 5.009]
+tbyVd=[];
+for(i=1:10)
+    tbyVd(i)=(t(1,i)/V(1,i))
+end
+tbyV=(1/1000)*tbyVd
+plot2d((V*1000),tbyV);
+xtitle("Determination of the constants","Vx1000","(t/V)x(10^-3)")
+//as seen from the graph,
+B=6400;//intercept in s/m3
+//as it is seen the given graph resembles a straight line we can find Kp/2
+Kpby2=[tbyV(6)-tbyV(5)]*1000/(V(6)-V(5));//slope of st line in s/m6
+Kp=Kpby2*2;
+mu=8.937e-4;
+A=0.0439;
+delP=338*1000;
+//Now Kp=mu*alpha*cs/(A*A*delP)
+Cs=23.47;
+alpha=Kp*(A*A*delP)/(mu*Cs);
+//also B=(mu*Rm)/(A*A*(-delP))
+Rm=B*(A*(delP))/mu
+mprintf("alpha= %f m/kg",alpha)
+mprintf(" Rm= %f m-1",Rm)
+//As the slope has been calculated in numerical value deviations have been found.Kpby2=6*10^6 according to the example but the value calculated here is 5.815*10^6
diff --git a/2762/CH14/EX14.2.2/14_2_2.sce b/2762/CH14/EX14.2.2/14_2_2.sce
new file mode 100755
index 000000000..f7e02995e
--- /dev/null
+++ b/2762/CH14/EX14.2.2/14_2_2.sce
@@ -0,0 +1,27 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.2-2
+//Mechanical-Physical Seperation Processes
+//given data
+t=[4.4 9.5 16.3 24.6 34.7 46.1 59 73.6 89.4 107.3];
+V=(1/1000)*[0.498 1 1.501 2 2.498 3.002 3.506 4.004 4.502 5.009]
+tbyVd=[];
+for(i=1:10)
+    tbyVd(i)=(t(1,i)/V(1,i))
+end
+tbyV=(1/1000)*tbyVd
+plot2d((V*1000),tbyV);
+xtitle("Determination of the constants","Vx1000","(t/V)x(10^-3)")
+//as seen from the graph,
+B=6400;//intercept in s/m3
+//as it is seen the given graph resembles a straight line we can find Kp/2
+Kpby2=[tbyV(6)-tbyV(5)]*1000/(V(1,6)-V(1,5));//slope of st line in s/m6
+Kp=Kpby2*2;
+A=20*0.837;//total area= (no. of plates)*(area of one plate)
+Aold=0.0439;
+Kpn=Kp*((Aold/A)^2);
+Bn=B*((Aold/A));
+V=3.37;//amt6 of filtrate to be recovered inm3
+t=((Kpn/2)*V*V)+(Bn*V);
+mprintf("the time taken to recover the filtrate= %f s",t)
+//As the slope has been calculated in numerical value deviations have been found.Kpby2=6*10^6 according to the example but the value calculated here is 5.815*10^6
diff --git a/2762/CH14/EX14.3.2/14_3_2.sce b/2762/CH14/EX14.3.2/14_3_2.sce
new file mode 100755
index 000000000..614b2d6dd
--- /dev/null
+++ b/2762/CH14/EX14.3.2/14_3_2.sce
@@ -0,0 +1,24 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.3-2
+//Mechanical-Physical Seperation Processes
+//given data
+rhowater=998;
+muwater=1.005e-3;
+rhop=2467;
+Dp=1.554e-4;
+solid=60;//solid weight percent
+liquid=100-solid;//liquid weight percent
+E=(liquid/rhowater)/((liquid/rhowater)+(solid/rhop));//volume fraction
+rhom=E*rhowater+(1-E)*rhop;
+ship=(1/(10^(1.82*(1-E))));
+g=9.807;//gravity accelaration
+vt=((g*Dp*Dp)*(rhop-rhowater)*(E*E*ship))/(18*muwater);
+Re=(Dp*vt*rhom)/((muwater/ship)*E)
+if (Re<2100)
+    disp(Re)
+    disp('Settling is in laminar range')
+else
+    disp('Settling is not in laminar range')
+end
+//end
diff --git a/2762/CH14/EX14.4.3/14_4_3.sce b/2762/CH14/EX14.4.3/14_4_3.sce
new file mode 100755
index 000000000..7574e312c
--- /dev/null
+++ b/2762/CH14/EX14.4.3/14_4_3.sce
@@ -0,0 +1,12 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.4-3
+//Mechanical-Physical Seperation Processes
+//given data
+rhol=919.5;//density of oil in kg/m3
+rhoh=980.3;//density of the aqueous phase
+rin=10.16;
+rout=10.414;
+r2=sqrt((rhoh*(rout^2)-rhol*(rin^2))/(rhoh-rhol));
+mprintf("the location of centrifuge,r= %f mm",r2)
+//end
diff --git a/2762/CH14/EX14.5.1/14_5_1.sce b/2762/CH14/EX14.5.1/14_5_1.sce
new file mode 100755
index 000000000..297001bab
--- /dev/null
+++ b/2762/CH14/EX14.5.1/14_5_1.sce
@@ -0,0 +1,12 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 14
+//Example 14.5-1
+//Mechanical-Physical Seperation Processes
+//given data in english units
+Df=3/12;//feed size in feet
+Dp=(1/8)/12;//productr size in ft
+T=10/60;//feed rate in ton/min
+//Bonds Equation(English units):(P/T)=1.46E1((1/Dp^0.5)-(1/Df^0.5))
+E1=12.68;
+P=T*(1.46*E1*((1/sqrt(Dp))-(1/sqrt(Df))));
+mprintf("gross power reqd= %f hp",P)
diff --git a/2762/CH2/EX2.10.1/2_10_1.sce b/2762/CH2/EX2.10.1/2_10_1.sce
new file mode 100755
index 000000000..ad2c15e39
--- /dev/null
+++ b/2762/CH2/EX2.10.1/2_10_1.sce
@@ -0,0 +1,25 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.10-1
+//Principles of Momentum Transfer and Overall Balances
+//given data
+h=0.0655;//pressure drop reading in m
+rhow=996;//density of water in kg/m3
+g=9.80665;//gravity force
+dP=h*rhow*g;//pressure drop
+//dP=(32*mu*v*(l2-l1/(D*D))) this implies---> v= dP*(D*D)/(32*mu*(l2-l1)
+D=2.22/1000;//capillary internal diameter
+mu=1.13/1000;//viscosity of liquid
+dl=0.317;//l2-l1= length of capillary being used in m
+v= dP*(D*D)/(32*mu*dl);//velocity in m/s
+V=(v*3.14*D*D)/4;//volumetric flow rate D=2.22/1000;//capillary internal diameter
+mu=1.13/1000;//viscosity of liquid
+rhol=875;//density of liquid in kg/m3
+Re=(D*v*rhol)/mu;
+if(Re<2100) then
+    disp("the flow is laminar")
+else
+    disp("the flow is turbulent")
+end
+mprintf("volumetric flow rate= %f m3/s",V)
+//end
diff --git a/2762/CH2/EX2.10.2/2_10_2.sce b/2762/CH2/EX2.10.2/2_10_2.sce
new file mode 100755
index 000000000..5bde53067
--- /dev/null
+++ b/2762/CH2/EX2.10.2/2_10_2.sce
@@ -0,0 +1,21 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.10-2
+//Principles of Momentum Transfer and Overall Balances
+//given data
+D=2.22/1000;//capillary internal diameter
+mu=1.13/1000;//viscosity of liquid
+D=2.22/1000;//capillary internal diameter
+rhol=875;//density of liquid in kg/m3
+dl=0.317;//l2-l1= length of capillary being used in m
+v=0.275;//velcity of liq in m/s
+Re=(D*v*rhol)/mu;//reynolds number
+if(Re<2100) then
+    disp("the flow is laminar")
+    f=16/Re;//fannings friction factor
+else
+    disp("the flow is turbulent")
+end
+dP=(4*f*rhol*dl*v*v)/(2*D);//pressure drop in Pa (Hagen Poiseulle equation)
+mprintf("pressure drop= %f Pa",dP)
+//end
diff --git a/2762/CH2/EX2.10.3/2_10_3.sce b/2762/CH2/EX2.10.3/2_10_3.sce
new file mode 100755
index 000000000..363a46c90
--- /dev/null
+++ b/2762/CH2/EX2.10.3/2_10_3.sce
@@ -0,0 +1,23 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.10-3
+//Principles of Momentum Transfer and Overall Balances
+//given data
+D=0.0525;//diameter of the pipe in m
+v=4.57;//vel of fluid in m/s
+rhol=801;//density of fluid in kg/m3
+mu=4.46/1000;//viscosity in kg/m.s
+Re=(D*v*rhol)/mu;//reynolds number
+E=4.6*(10^-5)
+if(Re<2100) then
+    disp("the flow is laminar")
+    f=16/Re;//fannings friction factor
+else
+    disp("the flow is turbulent")
+    k=E/D;
+end
+dl=36.6;//l2-l1= length of pipe being used in m
+f=0.0060;//for given Re
+Ff=(4*f*dl*v*v)/(2*D);//mechanical energy friction loss
+mprintf("mechanical energy friction loss= %f J/kg",Ff)
+//end
diff --git a/2762/CH2/EX2.10.5/2_10_5.sce b/2762/CH2/EX2.10.5/2_10_5.sce
new file mode 100755
index 000000000..0006e330f
--- /dev/null
+++ b/2762/CH2/EX2.10.5/2_10_5.sce
@@ -0,0 +1,18 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.10-5
+//Principles of Momentum Transfer and Overall Balances
+//given data
+D=0.01;//diameter of tube in m
+G=9;// rate of flow of N through tube in kg/s.m2
+mu=1.77/100000;//viscosity of gas
+Re=D*G/mu;//reynolds number
+p1=2.0265*100000;//entrance pressure
+f=0.009;//friction factor for given Re
+dl=200;//section of tube used in m
+R=8314.3;//gas constant
+T=298.15;//std temp given
+M=28.02;//molecular weight of N2
+p2=((p1^2)-((4*f*dl*G*G*R*T)/(M*D)))^0.5;//outlet pressure
+mprintf("outlet pressure= %f Pa",p2)
+//end
diff --git a/2762/CH2/EX2.2.1/2_2_1.sce b/2762/CH2/EX2.2.1/2_2_1.sce
new file mode 100755
index 000000000..6b3b19e62
--- /dev/null
+++ b/2762/CH2/EX2.2.1/2_2_1.sce
@@ -0,0 +1,15 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.2-1
+//Principles of Momentum Transfer and Overall Balances
+//given data
+//a)
+Fl=3*32.174*(1/32.174);// lbm*(ft/s^2)*(1/(ft/s^2))=m*g/gc
+//b)
+Fdyne=3*453.59*980.665;//lbm*(g/lbm)8(cm/s^2)
+//c)
+Fnewt=3*(1/2.2046)*9.80665;//(lbm*(kg/lbm)*(m/s^2))
+mprintf("the force in lb is %f lb force ",Fl)
+mprintf("the force in dynes is %f dyn ",Fdyne)
+mprintf("the force in Newtons is %f N ",Fnewt)
+//end
diff --git a/2762/CH2/EX2.2.2/2_2_2.sce b/2762/CH2/EX2.2.2/2_2_2.sce
new file mode 100755
index 000000000..c20cb5885
--- /dev/null
+++ b/2762/CH2/EX2.2.2/2_2_2.sce
@@ -0,0 +1,24 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.2-2
+//Principles of Momentum Transfer and Overall Balances
+//given data
+//P(total pressure)=h(height of the column)*rho(density of fluid)*g(gravity force)+P(absolute pressure)
+h1=10 ;//ht of oil layer in ft
+rhooil=917;//density of oil in kg/m3
+g=9.8;//gravity force in m/s2
+Patmsi=1.01325*10^5;//atm pressure in si units
+Patm=14.696;//lbf/in2
+Ptot1=h1*(rhooil*62.43/1000)*1*(1/144)+Patm;//ft*(lbm/ft3)*(1/(in2/ft2));
+Ptot1si=(h1*0.3048)*rhooil*g+Patmsi;//total pressure of oil in si units
+h2=2;//ht in ft
+rhowater=1000;//density of water in kg/m3
+Ptot2=h2*(rhowater*62.43/1000)*1*(1/144)+Ptot1;//ft*(lbm/ft3)*(1/(in2/ft2))
+Ptot2si=(h2*0.3048)*rhowater*g+Ptot1si;//total pressure of water in si units
+Pgage=Ptot2-Patm
+mprintf("the pressure on oil layer is %f psia",Ptot1)
+mprintf("the pressure on oil layer is %f pa",Ptot1si)
+mprintf("the pressure on bottom layer is %f psia",Ptot2)
+mprintf("the pressure on oil layer is %f pa",Ptot2si)
+mprintf("the gage pressure %f psia",Pgage)
+//end
diff --git a/2762/CH2/EX2.2.3/2_2_3.sce b/2762/CH2/EX2.2.3/2_2_3.sce
new file mode 100755
index 000000000..8e1f055c3
--- /dev/null
+++ b/2762/CH2/EX2.2.3/2_2_3.sce
@@ -0,0 +1,15 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.2-3
+//Principles of Momentum Transfer and Overall Balances
+//given data
+//a) si units
+rhow=1000;
+g=9.80665;
+P=101325;
+hw=P/(rhow*g);//water head 
+mprintf("a) head= %f m of water 4 deg C",hw)
+//b) for Hg
+rhom=13595.5;
+hm=(rhow/rhom)*hw;
+mprintf(" b) head= %f m of Mercury",hm)
diff --git a/2762/CH2/EX2.2.4/2_2_4.sce b/2762/CH2/EX2.2.4/2_2_4.sce
new file mode 100755
index 000000000..1f8acf7a6
--- /dev/null
+++ b/2762/CH2/EX2.2.4/2_2_4.sce
@@ -0,0 +1,12 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.2-4
+//Principles of Momentum Transfer and Overall Balances
+//given data: 
+R=32.7;//manometer reading in cm
+rhom=13.6;//density of mercury in g/cc
+rhow=1;//density of water in g/cc
+g=9.81//gravity force in m/s2
+Pdiff=(R/100)*(rhom-rhow)*1000*g;//Pressure diff in N/m2
+mprintf("the pressure diff is %f N/m2",Pdiff)
+//end
diff --git a/2762/CH2/EX2.3.1/2_3_1.sce b/2762/CH2/EX2.3.1/2_3_1.sce
new file mode 100755
index 000000000..5aa486b81
--- /dev/null
+++ b/2762/CH2/EX2.3.1/2_3_1.sce
@@ -0,0 +1,19 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.3-1
+//Principles of Momentum Transfer and Overall Balances
+//given data
+//a)
+del=0.013;//diffusivity
+T1=1.37e-2;//concn at pt1 amt of prop/m3
+T2=0.72e-2;//concn at pt2
+z2=0.4;
+z1=0;
+shi1=(del*(T1-T2))/(z2-z1);
+mprintf("%f amt of property/s m2",shi1)
+//b
+disp('T=T1+(shi1/del)*(z1-z)')
+//c)
+z=0.2;//point where concn is being found
+T=T1+(shi1/del)*(z1-z);//concentration at mid point
+mprintf("%f amt of property/s m3",T)
diff --git a/2762/CH2/EX2.4.1/2_4_1.sce b/2762/CH2/EX2.4.1/2_4_1.sce
new file mode 100755
index 000000000..74c3e2129
--- /dev/null
+++ b/2762/CH2/EX2.4.1/2_4_1.sce
@@ -0,0 +1,24 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.4-1
+//Principles of Momentum Transfer and Overall Balances
+//given data
+dely=0.5;//dist between 2 plates in cm
+delv=10;//vel diff along z in cm/s
+mu=0.0177//viscosity in CP
+//a) after integrating shear stress(tao)= mu*(delv)/dely
+tao=mu*(delv/dely);//g/(s2*cm)
+//shear rate= delv/dely as the vel change is linear with y
+SR=delv/dely;
+mprintf("shear stress in cgs = %f dyn/cm2",tao)
+mprintf(" shear rate in cgs = %f s-1",SR)
+//b) in lb force
+mulb=mu*(6.7197/100);//viscosity changes to lbm/(ft*s)
+taolb=(mulb*delv)/(dely*32.174);//lbf/ft2
+mprintf(" shear stress in english units = %f lbf/ft2",taolb)
+mprintf(" shear rate in english units = %f s-1",SR)
+//c)
+taosi=(mu*0.1*delv)/(dely);
+mprintf(" shear stress in si = %f N/m2",taosi)
+mprintf(" shear rate in si = %f s-1",SR)
+//end
diff --git a/2762/CH2/EX2.5.1/2_5_1.sce b/2762/CH2/EX2.5.1/2_5_1.sce
new file mode 100755
index 000000000..3d3b3a4fe
--- /dev/null
+++ b/2762/CH2/EX2.5.1/2_5_1.sce
@@ -0,0 +1,21 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.5-1
+//Principles of Momentum Transfer and Overall Balances
+//given data
+f=10*(1/7.481)*(1/60);//flow rate(ft3/s)=10(gal/min)*(1ft3/7.481 gal)(1min/60s)
+D=2.067/12;//diameter in ft
+A=(3.14*D*D)/4;//cross sectional area in ft2
+v=f/A;//velocity in ft/s
+rhoeng=0.996*62.43;//lbm/ft3
+mueng=0.8007*6.7197/10000;//viscosity in lbm/ft*s
+Nreeng= (D*v*rhoeng)/(mueng);//reynolds number in english units
+//in SI units
+rhosi=0.996*1000;//density in kg/m3
+Dsi=2.067*1/3.2808;//in*(ft/12in)*(m/ft)
+vsi=v*(1/3.2808)*(1/12);//velocity in m/s
+musi=0.8007/1000;//viscosity in SI units
+Nresi=(Dsi*rhosi*vsi)/musi;
+mprintf("the reynolds number in si units is %f",Nresi)
+mprintf("the reynolds number in english units is %f",Nreeng)
+//end
diff --git a/2762/CH2/EX2.6.1/2_6_1.sce b/2762/CH2/EX2.6.1/2_6_1.sce
new file mode 100755
index 000000000..49b62ebbd
--- /dev/null
+++ b/2762/CH2/EX2.6.1/2_6_1.sce
@@ -0,0 +1,20 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.6-1
+//Principles of Momentum Transfer and Overall Balances
+//given data
+//oil density=892 kg/m3 , volumetric flow rate= (1.388*10^-3) m3/s, schedule 40 pipes are being used
+//a) 
+A1=0.02330*0.0929;//cross sectional area in m2
+A3=0.01414*0.0929;//cross sectional area in m2
+rho=892;//oil density=892 kg/m3
+m1=(1.388/1000)*rho;//mass flow rate in kg/s in pipes 1 and 2
+m3=m1/2;//mas flow rate divides eqully in 3 pipes
+v1=m1/(rho*A1);//velocity at 1 pipe'
+v3=m3/(rho*A3);//
+G1=(v1)*rho;
+mprintf("a) the total mass flow rate in pipe 1 is %f kg/s",m1)
+mprintf(" b) the velocity in pipe 1 is %f m/s",v1)
+mprintf(" the velocity in pipe 3 is %f m/s",v3)
+mprintf(" c) the mass velocity in pipe 1 is %f kg/s m2",G1)
+//end
diff --git a/2762/CH2/EX2.7.1/2_7_1.sce b/2762/CH2/EX2.7.1/2_7_1.sce
new file mode 100755
index 000000000..80c44fd45
--- /dev/null
+++ b/2762/CH2/EX2.7.1/2_7_1.sce
@@ -0,0 +1,22 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.7-1
+//Principles of Momentum Transfer and Overall Balances
+//given data
+//Q=(z2-z1)*g+(v2^2-v1^2)/2 + (H2-H1)
+//for KE terms
+v1=1.52;//velociy of inlet stream
+v2=9.14;//velocity of exit stream
+KE=((v1^2)-(v2^2))/2;
+//for PE terms 
+z1=0 ,// (z2-z1)*g
+z2=15.2;//ht of exit stream
+g=9.80665;// g force
+PE=(z2-z1)*g;
+//enthalpy change
+H2=2771.4*1000;//exit ,enthalpy data from appendix
+H1=76.97*1000;//inlet enthalpy
+H=H2-H1;
+Q=KE+PE+H;
+mprintf("the amount of heat to be added %f J/kg",Q)
+//end
diff --git a/2762/CH2/EX2.7.2/2_7_2.sce b/2762/CH2/EX2.7.2/2_7_2.sce
new file mode 100755
index 000000000..1a3ade16b
--- /dev/null
+++ b/2762/CH2/EX2.7.2/2_7_2.sce
@@ -0,0 +1,22 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.7-2
+//Principles of Momentum Transfer and Overall Balances
+//given data
+m=0.567;//inlet vol flow rate 
+rho1=968.5;//density of fluid
+m1=m*rho1/60;// mass flow rate in kg/s
+m2=m1;//steady state
+E1=7.45*1000;//energy supplied by the pump in J/s
+Ws=-E1/m1;//work done by shaft
+E2=-1408*1000;//energy given up in J/s
+Q=E2/m2;
+g=9.80665;
+z2=20;
+z1=0;
+//H2-H1+(z2-z1)g+(del v^2)/2= Ws+Q
+H1=355.9*1000;//enthalpy from tables
+H2=Q-Ws-(z2-z1)*g+H1;
+mprintf("the enthalpy is %f J/kg",H2)
+mprintf(" the temp as ssen from steam tables is 48.41 deg C")
+//end
diff --git a/2762/CH2/EX2.7.3/2_7_3.sce b/2762/CH2/EX2.7.3/2_7_3.sce
new file mode 100755
index 000000000..aa17d2ca3
--- /dev/null
+++ b/2762/CH2/EX2.7.3/2_7_3.sce
@@ -0,0 +1,13 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.7-2
+//Principles of Momentum Transfer and Overall Balances
+//given data
+//si units
+KE=0;PE=0;Ws=0;//steady state
+Power=19.63*1000;
+m1=0.3964/60;//mass flow rate in si units
+Q=Power/m1;//heat added
+H1=0;
+H2=H1+Q;//exit enthalpy
+mprintf("exit enthalpy= %f kJ/kg",H2/1000)
diff --git a/2762/CH2/EX2.7.4/2_7_4.sce b/2762/CH2/EX2.7.4/2_7_4.sce
new file mode 100755
index 000000000..a3b74406e
--- /dev/null
+++ b/2762/CH2/EX2.7.4/2_7_4.sce
@@ -0,0 +1,16 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.7-4
+//Principles of Momentum Transfer and Overall Balances
+//given data
+Ws=-155.4;
+z1=0;
+z2=3.05;
+g=9.806;
+PE=(z1-z2)*g;
+KE=0;//const dia pipe'
+rho=998;
+p1=68.9*1000;
+p2=137.8*1000;
+sumF=-Ws+PE+KE+(p1-p2)/rho
+mprintf("frictional loss= %f J/kg",sumF)
diff --git a/2762/CH2/EX2.7.5/2_7_5.sce b/2762/CH2/EX2.7.5/2_7_5.sce
new file mode 100755
index 000000000..ea4cf47e0
--- /dev/null
+++ b/2762/CH2/EX2.7.5/2_7_5.sce
@@ -0,0 +1,32 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.7-5
+//Principles of Momentum Transfer and Overall Balances
+//given data
+//cgs units
+V=69.1;//volumetric flow rate in gallon/min
+fr1=V*(1/60)*(1/7.481);//converting flow rate in ft3/s
+A1=0.0233;//cross section area in ft2 of the pipe
+v2=fr1/A1;//vel in ft/s
+v1=0;//since the tank is very large
+p1=1;//atm pressure
+p2=p1;
+gc=32.174;
+z1=0;//datum
+g=32.174;
+z2=50;//length of discharge line in ft
+rho=114.8;//density of liq soln
+F1=10;//friction loss in ft-lb force/lb mass
+Ws=(z1*(g/gc))-(z2*(g/gc))+(v1*v1/(2*gc))-(v2*v2/(2*gc))+((p1-p2)/rho)-F1;//shaft work calculated by mechanical energy equation
+n=0.65;//efficiency
+Wp=(-Ws/n);
+m=fr1*rho;//mass flow rate in lbm/s
+P=m*Wp*(1/550);//pump horsepower
+A2=0.05134;//area of cross section
+v3=fr1/A2;
+v4=v2;
+F2=0;//friction losses in the second pipe is negligible
+Pdbyrho=(v3*v3/(2*gc))-(v4*v4/(2*gc))-Ws-F2;
+Pdiff=Pdbyrho*(rho/144);//pressure diff in lb force/in2
+mprintf("pressure developed by pump = %f psia",Pdiff)
+mprintf(" pump horsepower= %f hp",P)
diff --git a/2762/CH2/EX2.8.2/2_8_2.sce b/2762/CH2/EX2.8.2/2_8_2.sce
new file mode 100755
index 000000000..845bbc69f
--- /dev/null
+++ b/2762/CH2/EX2.8.2/2_8_2.sce
@@ -0,0 +1,21 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.8-2
+//Principles of Momentum Transfer and Overall Balances
+//given data
+V=0.03154;//vol flow rate in si units
+D1=0.0635;// upstream ID
+A1=(%pi/4)*D1*D1;//area of cross section
+D2=0.0286;// downstream ID
+A2=(%pi/4)*D2*D2;//area of cross section
+rho=1000;//density of water
+m=V*rho;//mass flow rate of water upstream
+m1=m;
+m2=m;
+v1=V/A1;//vel at pt 1
+v2=V/A2;//vel at pt 2
+p2=0;//gage pressure
+p1=(((v2*v2/2)-(v1*v1/2))+(p2/rho))*rho;
+//for x direction the momentum balance equation is used
+Rx=m*(v2-v1)-A1*p1;
+mprintf("the resultant force towards the negative x direction is %f N",-Rx)
diff --git a/2762/CH2/EX2.8.5/2_8_5.sce b/2762/CH2/EX2.8.5/2_8_5.sce
new file mode 100755
index 000000000..976265c45
--- /dev/null
+++ b/2762/CH2/EX2.8.5/2_8_5.sce
@@ -0,0 +1,17 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.8-5
+//Principles of Momentum Transfer and Overall Balances
+//given data
+v1=30.5;
+D1=2.54/100;
+alpha2=60;
+rho=1000;
+A1=(%pi*D1*D1)/4;
+m=v1*A1*rho;
+Rx=m*v1*(cos(%pi/3)-1);
+Ry=m*v1*sin(%pi/3);
+mprintf("the force on x direction =%f N",-Rx)
+mprintf("the force on y direction =%f N",-Ry)
+R=sqrt(Rx*Rx+Ry*Ry);
+mprintf("the resultant force =%f N",-R)
diff --git a/2762/CH2/EX2.9.1/2_9_1.sce b/2762/CH2/EX2.9.1/2_9_1.sce
new file mode 100755
index 000000000..818b38206
--- /dev/null
+++ b/2762/CH2/EX2.9.1/2_9_1.sce
@@ -0,0 +1,16 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.9-1
+//Principles of Momentum Transfer and Overall Balances
+//given data
+rho=820;//density in kg/m3
+del=1.7/1000;//film thikness in m
+g=9.806;//g force
+mu=0.2;//viscocity in Pa.s
+T=(rho^2)*(del^3)*g/(3*mu);//T= mass flow rate per unit width of wall
+Re=(4*T)/mu;//Reynolds Number
+v=(rho*g*(del^2))/(3*mu);//avg velocity
+mprintf("mass flow rate per unit width of wall= %f kg/(s.m)",T)
+mprintf(" Reynolds Number= %f",Re)
+mprintf(" avg velocity= %f m/s",v)
+//end
diff --git a/2762/CH3/EX3.1.1/3_1_1.sce b/2762/CH3/EX3.1.1/3_1_1.sce
new file mode 100755
index 000000000..80d14caec
--- /dev/null
+++ b/2762/CH3/EX3.1.1/3_1_1.sce
@@ -0,0 +1,16 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.1-1
+//Principles of Momentum Transfer and Applications
+//given data
+rho=1.137;
+mu=1.9e-5;
+Dp=0.042;
+v0=23;
+Re=(Dp*v0*rho)/mu
+//from the mentioned graph,
+Cd=0.47;//drag coefficient as seen from the graph
+Ap=(%pi*Dp*Dp)/4;//surface area of sphere
+Fd=Cd*(v0*v0/2)*rho*Ap;//drag force
+mprintf("drag coefficient= %f",Cd)
+mprintf("drag force= %f N",Fd)
diff --git a/2762/CH3/EX3.1.2/3_1_2.sce b/2762/CH3/EX3.1.2/3_1_2.sce
new file mode 100755
index 000000000..4ef052c2b
--- /dev/null
+++ b/2762/CH3/EX3.1.2/3_1_2.sce
@@ -0,0 +1,15 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.1-2
+//Principles of Momentum Transfer and Applications
+//given data
+rho=997.2;//density of water at 300K in kg/m3
+v=1;//velocity of air in m/s
+Dp=0.09;//diameter of a cylinder
+mu=0.9142/1000;//viscosity of water Pa.s
+Re=Dp*v*rho/mu;//Reynolds Number
+Cd=1.4;//drag coefficient; found from a graph by the american chemical society
+L=1;//length of the tube in m
+Ap=L*Dp;
+Fd=Cd*(v*v/2)*rho*Ap;//drag force in newtons
+mprintf("the drag force is %f N",Fd)
diff --git a/2762/CH3/EX3.1.3/3_1_3.sce b/2762/CH3/EX3.1.3/3_1_3.sce
new file mode 100755
index 000000000..3829dd2c6
--- /dev/null
+++ b/2762/CH3/EX3.1.3/3_1_3.sce
@@ -0,0 +1,19 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.1-3
+//Principles of Momentum Transfer and Applications
+//given data
+basis=1;//taking basis as 1 m3 of packed bed
+rho=962;//bulk density of packed bed
+m=rho*basis;//total mass
+rho2=1600;//density of solid cylinders 
+V=m/rho2;//volume of the cylinder
+E=(basis-V)/(basis);//void fraction
+mprintf("void fraction = %f",E);
+D=0.02;//diameter of cylinder
+Av=6/D;// Av= Sp/Vp where Sp is the surface area of the particle and D is the diameter of the particle
+Dp=6/Av;//effectice diameter
+mprintf(" ii) effectice diameter = %f m",Dp);
+a=(6/Dp)*(1-E);//value of a
+mprintf(" iii) value of a= %f m-1",a)
+//end
diff --git a/2762/CH3/EX3.1.4/3_1_4.sce b/2762/CH3/EX3.1.4/3_1_4.sce
new file mode 100755
index 000000000..87325f335
--- /dev/null
+++ b/2762/CH3/EX3.1.4/3_1_4.sce
@@ -0,0 +1,27 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.1-4
+//Principles of Momentum Transfer and Applications
+//given data
+D=0.61;//diameter of bed in m
+h=2.44;//height of bed
+A=(3.14*D*D)/4;//cross section area in m2
+mdot=0.358;//mass flow rate of air in kg/s
+E=0.38//void fraction
+G=mdot/A;
+Dp=0.0127;//diameter of spheres in m
+mu=1.9e-5;//viscosity of air Pa.s
+delL=2.44;
+Re= (Dp*G/((1-E)*mu));//Reynolds Number
+delP=0.05e+5;//assumed pressure difference in pascal
+p1=1.115*101325;//air entering at this pressure in Pa
+p2=p1-delP;
+avgP=(p1+p2)/2;//average pressure
+M=28.97;//molecular weight of air in SI units 
+R=8314.34;//gas constant
+T=311;//temp of air in K
+avgrho=(M/(R*T))*avgP;//Avg density 
+//(delP*rho/G*G)*(Dp/delL)*(E^3/(1-E))= (150/Re)+1.75: erguns equation in dimensionless groups
+delPn=((150/Re)+1.75)*((G*G*delL*(1-E))/(avgrho*E*E*E*Dp));//calculated pressure drop
+mprintf("calculated pressure drop= %f Pa",delPn)
+//end
diff --git a/2762/CH3/EX3.1.5/3_1_5.sce b/2762/CH3/EX3.1.5/3_1_5.sce
new file mode 100755
index 000000000..eba51209a
--- /dev/null
+++ b/2762/CH3/EX3.1.5/3_1_5.sce
@@ -0,0 +1,15 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.1-5
+//Principles of Momentum Transfer and Applications
+//given data
+p=[0.25 0.40 0.35];//percentages of diff particle sizes
+s=[25 50 75];//sizes of the particles in mm
+phi=0.68;//sphericity
+sump=0
+for i=1:3
+    term=p(1,i)/(phi*s(1,i))
+    sump=sump+term
+end
+Dpm=1/sump;
+mprintf("mean diameter= %f mm",Dpm)
diff --git a/2762/CH3/EX3.2.1/3_2_1.sce b/2762/CH3/EX3.2.1/3_2_1.sce
new file mode 100755
index 000000000..89c3c9e1e
--- /dev/null
+++ b/2762/CH3/EX3.2.1/3_2_1.sce
@@ -0,0 +1,26 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.2-1
+//Principles of Momentum Transfer and Applications
+//given data
+rho=1.043;//density of air at 328.5K in kg/m3
+v=23;//velocity of air in m/s
+D=0.6;//diameter of a cylinder
+mu=2.03/100000;//viscosity of air Pa.s
+delh=0.205;// 0.205m of water pitot tube reading
+rhow=1000;//density of water 
+delP=delh*(rhow-rho)*9.80665;//pressure diff and g=9.80655 m/s2
+patm=101325;//atm pressure in pascals
+p1=patm+0.02008*100000;//absolute pressure+ pressure diff
+rhoc=(p1/patm)*1.043;//corrected air density
+delH=10.7/1000;//manometer reading, m of water
+Cp=0.98;
+delP=delH*(rhow-rhoc)*9.80655;//pressure diff in Pa
+v=Cp*((2*delP)/rhoc)^0.5;//max vel at center
+Re=D*v*rhoc/mu;//Reynolds Number
+vr=0.85;//from the given graph the ratio of avg vel/max vel is 0.85
+vavg=vr*v;//the average velcity in m/s
+mprintf(" average velcity = %f m/s",vavg)
+A=(3.14/4)*(D*D);//cross sec area in m2
+V=A*vavg;//volumetric flow rate in m3/s
+mprintf("volumetric flow rate = %f m3/s",V)
diff --git a/2762/CH3/EX3.2.2/3_2_2.sce b/2762/CH3/EX3.2.2/3_2_2.sce
new file mode 100755
index 000000000..8f18271b8
--- /dev/null
+++ b/2762/CH3/EX3.2.2/3_2_2.sce
@@ -0,0 +1,16 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.2-2
+//Principles of Momentum Transfer and Applications
+//given data
+delP=9.32e+4;//pressure diff in N/m2
+D1=0.1541;//external diameter in m
+D0=0.0566;//internal diameter in m
+Dr=D0/D1;
+Co=0.61;
+rho=878; //oil density in kg/m3
+v0=(Co/(sqrt(1-(Dr^4))))*sqrt((2*delP)/rho);//velocity calculation in m/s
+A=(%pi/4)*D0*D0;//cross section area
+V=A*v0;//volumetric flow rate
+mprintf("the volumetric flow rate is %f m3/s",V);
+//end
diff --git a/2762/CH3/EX3.3.2/3_3_2.sce b/2762/CH3/EX3.3.2/3_3_2.sce
new file mode 100755
index 000000000..723c71c89
--- /dev/null
+++ b/2762/CH3/EX3.3.2/3_3_2.sce
@@ -0,0 +1,25 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.3-2
+//Principles of Momentum Transfer and Applications
+//given data
+Ps=741.7;//suction pressure in mm hg
+Pd=769.6;//discharge pressure in mm hg
+Patm=760;//atmospheric pressure in mm Hg
+rho1=28.97*(1/22.414)*(273.2/366.3)*(Ps/Patm);//air density at suction: mol wt= 28.97 kg air/kg mol for 22.414 m3/kg mol at 101.3 kPa and 273.2 K
+rho2=rho1*(Pd/Ps);
+rhoavg=(rho1+rho2)/2
+V=28.32;//volumetric flow rate in m3/s
+Ts=294.1;//temp at suction
+mdot=V*(1/60)*(1/22.414)**(273.2/Ts)*28.97;//mass flow rate of gas
+Patm=760;//atm pressure in mm Hg
+Hp=((Pd-Ps)/Patm)*(101325/rhoavg);//pressure head in J/kg
+v1=0;//air is stationary
+v2=45.7;//discharge velocity in m/s
+vd=(((v2^2)-(v1^2))/2);//developed velocity
+z1=0;
+sumF=0;
+Ws=Hp+vd;//substituting and solving for Ws by mechanical energy balance equation for a closed system in J/kg
+n=60/100;//efficiency given is 60%
+bkW= (Ws*mdot)/(n*1000);// brake kW
+mprintf(" brake kW= %f hP",bkW)
diff --git a/2762/CH3/EX3.3.3/3_3_3.sce b/2762/CH3/EX3.3.3/3_3_3.sce
new file mode 100755
index 000000000..6106624c6
--- /dev/null
+++ b/2762/CH3/EX3.3.3/3_3_3.sce
@@ -0,0 +1,20 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.3-3
+//Principles of Momentum Transfer and Applications
+//given data
+p1=137.9*1000;
+p2=551.6*1000;
+T1=26.7+273.2;
+mmol=7.56/1000;
+M=16;
+mdot=mmol*M;
+gam=1.31;
+R=8314.3;
+nWs1=(gam/(gam-1))*(R*T1/M)*((p2/p1)^((gam-1)/gam)-1);
+n=80/100;
+bkW1=(nWs1*mdot)/(n*1000);
+mprintf("i) brake power= %f kW",bkW1)
+nWs2=(R*T1/M)*log(p2/p1);
+bkW2=(nWs2*mdot)/(n*1000);
+mprintf("ii) brake power= %f kW",bkW2)
diff --git a/2762/CH3/EX3.4.3/3_4_3.sce b/2762/CH3/EX3.4.3/3_4_3.sce
new file mode 100755
index 000000000..77f1c3057
--- /dev/null
+++ b/2762/CH3/EX3.4.3/3_4_3.sce
@@ -0,0 +1,45 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3
+//Example 3.4-3
+//Principles of Momentum Transfer and Applications
+//given data
+H1=1.83;
+DT1=H1;
+V1=(%pi*DT1*DT1*H1)/4;
+V2=3*V1;//given
+R=(V2/V1)^(1/3);
+DT2=R*DT1;
+Da1=0.61;
+Da2=R*Da1;
+W1=0.122;
+W2=R*W1;
+J1=0.15;
+J2=R*J1;
+N1=1.5;//no. of revs
+N2=N1*((1/R)^(2/3))
+rho=929;
+mu=0.01;
+Re=(Da2*Da2*N2*rho)/(mu)
+Np=5;
+P2=Np*rho*(N2^3)*(Da2^5);
+P1=Np*rho*(N1^3)*(Da1^5);
+//a)
+N2=N1*((1/R)^(2/3));
+sP1=P1/V1;
+sP2=P2/V2;
+
+mprintf("scaled up no. of revs %f rev/s",N2);
+mprintf("scaled up Power %f W",P2);
+mprintf(" power per unit volume= %f kW/m3",sP1/1000)
+if (sP1/1000)<0.8  then
+    disp(" Value of power is less than permissible condition(0.8 kW/m3 for mass transfer)")
+end
+mprintf(" scaled up Power %f m3",P2);
+mprintf(" power per unit volume %f W/m3",(P2/(V2*1000)));
+//b)
+N2b=N1*(1/R);
+mprintf(" scaled up revolutions %f rev/s",N2b);
+P2b=Np*rho*(N2b^3)*(Da2^5);
+mprintf(" scaled up Power %f kW",P2b);
+mprintf(" power per unit volume %f W/m3",(P2b/V2));
+
diff --git a/2762/CH3/EX3.5.1/3_5_1.sce b/2762/CH3/EX3.5.1/3_5_1.sce
new file mode 100755
index 000000000..fb7720c55
--- /dev/null
+++ b/2762/CH3/EX3.5.1/3_5_1.sce
@@ -0,0 +1,18 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 3//Example 3.5-1
+//Principles of Momentum Transfer and Applications
+//given data
+Kd=15.23;
+nd=0.4;
+D=0.0524;
+V=0.0728;
+L=14.9;
+rho=1041;
+delP=(Kd*4*L/D)*((8*V/D)^nd);//pressure drop
+Ff=delP/rho;//friction loss
+nd=0.4;
+g=8;
+Re=((D^nd)*(V^(2-nd))*rho)/(Kd*(g^(nd-1)));
+f=16/Re;//friction factor
+delP=4*f*rho*(L/D)*(V*V/2);
+mprintf("pressure drop= %f kN/m2",delP/1000)
diff --git a/2762/CH4/EX4.1.1/4_1_1.sce b/2762/CH4/EX4.1.1/4_1_1.sce
new file mode 100755
index 000000000..5e5178fa6
--- /dev/null
+++ b/2762/CH4/EX4.1.1/4_1_1.sce
@@ -0,0 +1,12 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.1-1
+//Principles of Steady State Heat Transfer
+//si units
+//given data
+k=0.048;//thermal conductivity
+delx=0.0254;//thickness
+T1=352.7;//temp of one surface
+T2=297.1;
+qbya=(k/delx)*(T1-T2);//heAT lost per m2
+mprintf("heat loss per m2= %f W/m2",qbya)
diff --git a/2762/CH4/EX4.2.1/4_2_1.sce b/2762/CH4/EX4.2.1/4_2_1.sce
new file mode 100755
index 000000000..8c63f39ab
--- /dev/null
+++ b/2762/CH4/EX4.2.1/4_2_1.sce
@@ -0,0 +1,25 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.2-1
+//Principles of Steady State Heat Transfer
+//given data
+//si units
+//nomenclature of unmentioned specifications similar to previous example
+k=0.151;
+T1=274.9;
+T2=297.1;
+r1=5/1000;
+r2=20/1000;
+L=1;
+A1=2*3.14*L*r1;//area of inner surface of cylinder
+A2=2*3.14*L*r2;
+Am=(A2-A1)/(log(A2/A1)/log(2.71828183));//log mean of inner surface area and outer surface area
+q=(k*Am*(T1-T2))/(r2-r1);//fouriers law
+if(q<0)
+    disp("HT is from r2 to r1")
+    else
+    disp("HT is from r1 to r2")
+end
+qd=-14.65;//amt of heat that needs to be dissipitated
+l=qd/q;//length reqd
+mprintf("length of tubing required= %f m",l)
diff --git a/2762/CH4/EX4.3.1/4_3_1.sce b/2762/CH4/EX4.3.1/4_3_1.sce
new file mode 100755
index 000000000..60eb7b7e9
--- /dev/null
+++ b/2762/CH4/EX4.3.1/4_3_1.sce
@@ -0,0 +1,21 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.3-2
+//Principles of Steady State Heat Transfer
+//given data
+//si units
+//nomenclature of unmentioned specifications similar to previous example
+ka=0.151;//thermal conductivity
+kb=0.0433;
+kc=0.762;
+T1=255.4;//temperatures
+T4=297.1;
+dx1=0.0127;//thickness
+dx2=0.1016;
+dx3=0.0762;
+Ra=dx1/(ka);//per unit area calculation
+Rb=dx2/(kb);
+Rc=dx3/kc;
+q=(T1-T4)/(Ra+Rb);
+T2=T1-(q*Ra);
+mprintf("the intermediate wall temp is %f K",T2)
diff --git a/2762/CH4/EX4.3.2/4_3_2.sce b/2762/CH4/EX4.3.2/4_3_2.sce
new file mode 100755
index 000000000..897e92923
--- /dev/null
+++ b/2762/CH4/EX4.3.2/4_3_2.sce
@@ -0,0 +1,25 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.3-2
+//Principles of Steady State Heat Transfer
+//given data
+//si units
+//nomenclature of unmentioned specifications similar to previous example
+ka=21.63;
+kb=0.2423;
+T1=811;
+T3=310.8;
+r1=0.0254/2;
+r2=0.0508/2;
+r3=0.0508;
+L=0.305;
+A1=2*3.14*L*r1;//area of inner surface of cylinder
+A2=2*3.14*L*r2;
+A3=2*3.14*L*r3;
+A1m=(A2-A1)/(log(A2/A1)/log(2.71828183));//log mean of inner surface area and outer surface area
+A2m=(A3-A2)/(log(A3/A2)/log(2.71828183))
+Ra=(r2-r1)/(ka*A1m);
+Rb=(r3-r2)/(kb*A2m);
+q=(T1-T3)/(Ra+Rb);
+T2=T1-(q*Ra);
+mprintf("the intermediate wall temp is %f K",T2)
diff --git a/2762/CH4/EX4.3.3/4_3_3.sce b/2762/CH4/EX4.3.3/4_3_3.sce
new file mode 100755
index 000000000..a812004a6
--- /dev/null
+++ b/2762/CH4/EX4.3.3/4_3_3.sce
@@ -0,0 +1,29 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.3-3
+//Principles of Steady State Heat Transfer
+//given data
+//nomenclature of unmentioned specifications similar to previous example
+ri=0.412/12;//inside radius of the steel pipe in ft
+r1=0.525/12;//outside radius of the steel pipe in ft
+ro=2.025/12;//lagging radius of the steel pipe in ft
+L=1;//unit length of the pipe
+Ai=2*%pi*ri*L;//Area of the resp surfaces
+A1=2*%pi*r1*L;
+Ao=2*%pi*ro*L;
+Aa1lm=(A1-Ai)/log(A1/Ai);
+Ab1lm=(Ao-A1)/log(Ao/A1);
+ka=26;
+kb=0.037;
+hi=1000;
+Ri=1/(hi*Ai);
+Ra=(r1-ri)/(ka*Aa1lm);
+Rb=(ro-r1)/(kb*Ab1lm);
+ho=2;
+Ro=1/(ho*Ao);
+Ti=267;To=80;
+q=(Ti-To)/(Ri+Ra+Rb+Ro);
+Ui=1/(Ai*(Ri+Ra+Rb+Ro));
+mprintf("the overall HTC= %f btu/h ft2 deg F",Ui)
+Q=Ui*Ai*(Ti-To);
+mprintf(" the heat ejected = %f btu/h",Q)
diff --git a/2762/CH4/EX4.3.4/4_3_4.sce b/2762/CH4/EX4.3.4/4_3_4.sce
new file mode 100755
index 000000000..f7bb43685
--- /dev/null
+++ b/2762/CH4/EX4.3.4/4_3_4.sce
@@ -0,0 +1,16 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.3-4
+//Principles of Steady State Heat Transfer
+//given data
+//nomenclature of unmentioned specifications similar to previous example
+I=200;//current in A
+R=0.126;// resistance in ohms
+P=I*I*R;//Power in watts
+Tw=422.1;//watt temp in K
+L=0.91;//length of wire
+r=0.001268;//radius of wire
+qdot=P/(%pi*L*r*r);
+k=22.5;
+T0=(qdot*r*r/(4*k))+Tw
+mprintf("centre temperature= %f K",T0)
diff --git a/2762/CH4/EX4.3.5/4_3_5.sce b/2762/CH4/EX4.3.5/4_3_5.sce
new file mode 100755
index 000000000..955c58904
--- /dev/null
+++ b/2762/CH4/EX4.3.5/4_3_5.sce
@@ -0,0 +1,23 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.3-5
+//Principles of Steady State Heat Transfer
+//given data in si units
+//nomenclature of unmentioned specifications similar to previous example
+k=0.4;
+h0=20
+rcr=k/h0;//critical radius
+mprintf("critical radius= %f mm",rcr*1000)
+L=1;
+D2=1.5;//diameter in m
+r2=D2/(2*1000);//radius in si units
+A=2*%pi*r2*L;
+t2=400;//wire surface temp
+T0=300;//temp of air
+q=h0*A*(t2-T0);
+mprintf(" heat loss per m of length without insulation %f W",q)
+r1i=r2;//with insulation
+x=2.5;//thickness of insulation in mm
+r2i=r2+(x/1000);
+qi=(2*%pi*L*(t2-T0))/(((log(r2i/r1i))/k)+(1/(r2i*h0)))
+mprintf(" heat loss per m of length with insulation %f W",qi)
diff --git a/2762/CH4/EX4.4.1/4_4_1.sce b/2762/CH4/EX4.4.1/4_4_1.sce
new file mode 100755
index 000000000..777004505
--- /dev/null
+++ b/2762/CH4/EX4.4.1/4_4_1.sce
@@ -0,0 +1,14 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.4-1
+//Principles of Steady State Heat Transfer
+//given data
+//si units
+N=4;
+M=9.25;
+L=5;
+k=0.9;
+T1=600;
+T2=400;
+q=4*((M/N)*k*L*(T1-T2));
+mprintf("heat transfer through the walls= %f W",q)
diff --git a/2762/CH4/EX4.5.1/4_5_1.sce b/2762/CH4/EX4.5.1/4_5_1.sce
new file mode 100755
index 000000000..b66ff96ac
--- /dev/null
+++ b/2762/CH4/EX4.5.1/4_5_1.sce
@@ -0,0 +1,27 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.5-1
+//Principles of Steady State Heat Transfer
+//given data
+//si units
+//nomenclature of unmentioned specifications similar to previous example
+mub=2.6e-5;//viscosity of air 
+Tw=488.7;//temp of water
+k=0.03894;
+Pr=0.686;//prandtl number
+muw=2.64e-5;//viscosity of water
+Mair=28.97;//mol wt of air
+P1=206.8;//inlet pressure of air
+Patm=101.33;//atmospheric pressure of air
+V=22.414;//mol vol of air
+T0=273.2;//STP temp
+T1=477.6;//temp of air
+rhoair=Mair*(1/V)*(P1/Patm)*(T0/T1);//density of air
+D=0.0254;//diameter of tube
+v=7.62;//vel of air
+Re=(D*v*rhoair)/mub;//reynolds number
+Nu=0.027*(Re^0.8)*(Pr^(1/3))*((mub/muw)^0.14);//nusselts number
+hl=Nu*k/D;//HT coeffiecient
+qbyA=hl*(Tw-T1);//flux
+mprintf("heat flux= %f W/m2",qbyA)
+mprintf(" HT coefficient= %f W/m2 K",hl)
diff --git a/2762/CH4/EX4.5.2/4_5_2.sce b/2762/CH4/EX4.5.2/4_5_2.sce
new file mode 100755
index 000000000..9cb80a11f
--- /dev/null
+++ b/2762/CH4/EX4.5.2/4_5_2.sce
@@ -0,0 +1,38 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.5-2
+//Principles of Steady State Heat Transfer
+//given data in si units
+//nomenclature of unmentioned specifications similar to previous example
+Di=0.0266;
+Do=0.0334;
+Pr=2.72;//Prandtl Number
+L=0.305;
+rho=0.98*1000;//density of water
+k=0.633;
+mu=4.32e-4;
+Tw1=80;//assumed for the first trial temp in deg C
+muw1=3.56e-4;
+v=2.44;//vel in m/s
+Re1=Di*v*rho/mu;
+k=0.663;
+hl=(k*0.027*(Re1^0.8)*(Pr^(1/3))*((mu/muw1)^0.14))/Di;
+mprintf("i) the convective HTC= %f W/m2 K",hl)
+Ai=%pi*Di*L;
+Alm=%pi*((Do+Di)/2)*L;
+Ao=%pi*Do*L;
+ksteel=45;
+Ri=1/(hl*Ai);
+Rm=((Do-Di)/2)/(ksteel*Alm);
+ho=10500;
+Ro=(1/(Ao*ho));
+sumR=Ri+Rm+Ro;
+Tdiff=107.8-65.6;
+Tdrop=(Ri/sumR)*Tdiff;
+Ui=1/(Ai*sumR);
+mprintf(" ii) overall HTC= %f W/m2 K",Ui);
+q=Ui*Ai*(Tdiff);
+mprintf(" ii) heat transfer rate= %f W",q)
+mprintf(" the calculations performed in the above will vary from the example as fraction exponents have been used")
+//the calculations performed in the above will vary from the example as fraction exponents have been used)
+//the calculations performed in the above will vary from the example as fraction exponents have been used
diff --git a/2762/CH4/EX4.5.3/4_5_3.sce b/2762/CH4/EX4.5.3/4_5_3.sce
new file mode 100755
index 000000000..69d2367e5
--- /dev/null
+++ b/2762/CH4/EX4.5.3/4_5_3.sce
@@ -0,0 +1,22 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.5-3
+//Principles of Steady State Heat Transfer
+//given data in si units
+//nomenclature of unmentioned specifications similar to previous example
+D=0.05;//diameter of the tube in m
+A=%pi*D*D/4;
+fr=4;//mass flow rate in kg/s
+G=fr/A;
+mu=7.1e-4;
+Re=(D*G)/mu;
+Cp=120;//Specific Heat in J/kg K
+k=13;
+Pr=(Cp*mu)/k
+hl=(k/D)*0.625*((Re*Pr)^0.4);
+dT=505-500;//when liq is heated from 500 to 505 K
+q=fr*Cp*dT;
+dTw=30;//temp diff b/w fluid and Tw
+Ad=q/(hl*dTw);
+L=Ad/(%pi*D);
+mprintf("the tube length= %f m",L)
diff --git a/2762/CH4/EX4.6.1/4_6_1.sce b/2762/CH4/EX4.6.1/4_6_1.sce
new file mode 100755
index 000000000..a37e6162f
--- /dev/null
+++ b/2762/CH4/EX4.6.1/4_6_1.sce
@@ -0,0 +1,24 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.6-1
+//Principles of Steady State Heat Transfer
+//given data
+//si units
+Tw=82.2;//temp of the fin
+Tb=15.6;//temp of cooling air
+Tf=(Tw+Tb)/2;
+k=0.028;//thermal conductivity
+rho=1.097;//density of water
+mu=1.95e-5;//viscosity
+Pr=0.704;//Prandtl No
+L=0.051;//fin thickness
+v=12.2;
+Re=(L*v*rho)/mu;//reynolds number
+Nu=0.664*(Re^0.5)*(Pr^(1/3));
+h=Nu*k/L;
+mprintf("a) heat transfer coefficient= %f W/m2 K",h);
+//part b
+Reb=4*100000//reynolds number
+Nub=0.0366*(Re^0.8)*(Pr^(1/3));
+hb=Nub*(k/L);
+mprintf("b) heat transfer coefficient= %f W/m2 K",hb);
diff --git a/2762/CH4/EX4.7.3/4_7_3.sce b/2762/CH4/EX4.7.3/4_7_3.sce
new file mode 100755
index 000000000..35d374e9f
--- /dev/null
+++ b/2762/CH4/EX4.7.3/4_7_3.sce
@@ -0,0 +1,23 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.7-3
+//Principles of Steady State Heat Transfer
+//given data
+//si units
+T1=394.3;
+T2=366.5;
+Tf=(T1+T2)/2;
+del=30/1000;
+rho=0.9295;
+mu=2.21e-5;
+k=0.03219;
+Pr=0.693;
+betaa=1/Tf;
+L=0.6;
+g=9.806;
+Gr=(del^3)*(rho^2)*g*betaa*(T1-T2)/(mu*mu)
+h=((k/del)*(0.20)*((Gr*Pr)^0.25))/((L/del)^(1/9));
+A=0.6*0.4;//area= length*breadth
+q=h*A*(T1-T2);
+mprintf("heat transfer rate= %f W",q)
+//calculation deviations may occur
diff --git a/2762/CH4/EX4.8.2/4_8_2.sce b/2762/CH4/EX4.8.2/4_8_2.sce
new file mode 100755
index 000000000..110821cfc
--- /dev/null
+++ b/2762/CH4/EX4.8.2/4_8_2.sce
@@ -0,0 +1,32 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 4
+//Example 4.8-2
+//Principles of Steady State Heat Transfer
+//given data
+//si units
+Tsat=89.44;
+Tw=87.8;
+Tf=(Tsat+Tw)/2;
+D=0.0254;
+hf=2657.8-374.6;//latent heat using steam tables
+rho1=60.3*16.018;//density of condensed steam
+rhov=0.391;//density of ateam at 10 psia
+mu1=3.24e-4;
+kt=0.675;
+L=0.305;
+Tsat=193;//in K
+delT=3.33;
+g=9.806;
+//assuming laminar film
+Nu=1.13*(((rho1^2)*g*hf*1000*(L^3)/(mu1*kt*(delT)))^(0.25));
+h=Nu*(kt/L);
+A=%pi*D*L;
+q=h*A*(Tsat-Tw);
+m=q/hf;
+kteng=0.390;//
+Leng=1;//in english units
+heng=Nu*(kteng/Leng);
+Deng=1/12;
+Aeng=%pi*Deng*Leng;
+mprintf("avg HT coefficient= %f W/m2 K in si units",h)
+mprintf("avg HT coefficient= %f btu/h ft2 F in english units",heng)
diff --git a/2762/CH5/EX5.2.1/5_2_1.sce b/2762/CH5/EX5.2.1/5_2_1.sce
new file mode 100755
index 000000000..e59ff9048
--- /dev/null
+++ b/2762/CH5/EX5.2.1/5_2_1.sce
@@ -0,0 +1,44 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 5
+//Example 5.2-1
+//Principles of Unsteady State Heat Transfer
+//given data
+//english units  
+r=1/12;//radius
+x1=r/3;//ratio of volume to area
+h=2;//convective coefficient
+k=25;//thermal conductivity
+Bi=(h*x1)/k;//biot number
+Cp=0.11;//specific heat
+rho=490;//density
+
+if(Bi<0.1)
+    M=h/(Cp*rho*x1);
+    Tinf=250;//constant temp
+    T0=800;//initial temp
+    t=1;
+    T=((T0-Tinf)*((2.718)^(-M*t)))+Tinf;//solving arhenius equation
+    mprintf("the temp in english units %f deg F",T)
+else 
+    mprintf("some other method must be employed")
+end
+//si units
+rsi=25.4/1000;
+x1si=rsi/3;
+hsi=11.36;
+ksi=43.3;
+Bisi=(hsi*x1si)/ksi;
+Cpsi=0.4606*1000;
+rhosi=7849;
+if(Bisi<0.1)
+    Msi=hsi/(Cpsi*rhosi*x1si);
+    Tinfsi=394.3;
+    T0si=699.9;
+    tsi=3600;
+    Tsi=((T0si-Tinfsi)*((2.718)^(-Msi*tsi)))+Tinfsi;//solving arhenius equation
+    mprintf(" the temp in si units %f deg K",Tsi)
+else 
+    mprintf("some other method must be employed")
+end
+    
+
diff --git a/2762/CH5/EX5.2.2/5_2_2.sce b/2762/CH5/EX5.2.2/5_2_2.sce
new file mode 100755
index 000000000..290829f87
--- /dev/null
+++ b/2762/CH5/EX5.2.2/5_2_2.sce
@@ -0,0 +1,20 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 5
+//Example 5.2-2
+//Principles of Unsteady State Heat Transfer
+//given data
+//si units
+//nomenclature similar to previous question
+r=25.4/1000;
+x1=r/3;
+h=11.36;
+k=43.3;;
+Cp=0.4606*1000;
+rho=7849;
+M=h/(Cp*rho*x1);
+V=(4/3)*(%pi)*(r^3);
+T0=699.9;
+Tinf=394.3;
+t=3600;//time
+Q=Cp*V*rho*(T0-Tinf)*(1-((%e)^(-M*t)));
+mprintf("heat removed= %f J",Q)
diff --git a/2762/CH5/EX5.3.3/5_3_3.sce b/2762/CH5/EX5.3.3/5_3_3.sce
new file mode 100755
index 000000000..6cea7e14f
--- /dev/null
+++ b/2762/CH5/EX5.3.3/5_3_3.sce
@@ -0,0 +1,23 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 5
+//Example 5.3-3
+//Principles of Unsteady State Heat Transfer
+//given data
+//si units
+//nomenclature similar to previous questions;
+//center is at x=0
+x=0;
+x1=0.0681/2;//radius of can
+n=x/x1;
+k=0.830;
+h=4540;
+m=(k/(h*x1));
+alpha=2.007e-7;
+t=0.75*3600;
+X=(alpha*t)/(x1*x1);
+Y=0.13;//heislers figure or graph
+//Y=(T1-T)/(T1-T0)
+T1=115.6;
+T0=29.4;
+T=T1-(Y*(T1-T0));
+mprintf("temp at the center of the can= %f deg C",T)
diff --git a/2762/CH5/EX5.3.4/5_3_4.sce b/2762/CH5/EX5.3.4/5_3_4.sce
new file mode 100755
index 000000000..f4ff8d831
--- /dev/null
+++ b/2762/CH5/EX5.3.4/5_3_4.sce
@@ -0,0 +1,31 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 5
+//Example 5.3-4
+//Principles of Unsteady State Heat Transfer
+//given data
+//si units
+//nomenclature similar to previous questions
+//conduction in x dirn
+x1=0.03405;
+y1=0.1016/2;
+k=0.83;
+alpha=2.007e-7;
+t=0.75*3600;//time in s
+r=0.75*3600;//hours to mins
+x=0;
+n=x/x1;
+h=4540;
+m=k/(h*x1);
+X=(alpha*t)/(x1*x1);
+//conduction in y direction
+y=0;
+y1=0.0508;
+my=k/(h*y1);
+Y=(alpha*t)/(y1*y1);
+Yx=0.13;//taken from the figure
+Yy=0.8;
+Yxy=Yx*Yy;
+T1=115.6;
+T0=29.4;
+Txy=T1-(Yxy*(T1-T0));
+mprintf("temp at the centre of the can= %f deg C",Txy)
diff --git a/2762/CH5/EX5.5.1/5_5_1.sce b/2762/CH5/EX5.5.1/5_5_1.sce
new file mode 100755
index 000000000..a3f81d89f
--- /dev/null
+++ b/2762/CH5/EX5.5.1/5_5_1.sce
@@ -0,0 +1,24 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 5
+//Example 5.5-1
+//Principles of Unsteady State Heat Transfer
+//given data
+//english units
+h=39.7;
+k=0.498;//thermal conductivity
+Cp=3.48*1000;//specific heat
+rho=1073;//density
+alpha=k/(rho*Cp)//temp coefficient 
+w=0.203;
+x1=w/2;
+x=0;
+n=x/x1;
+m=k/(h*x1);
+T1=1.7+273.2;
+T0=37.8+273.2;
+T=10+273.2;
+Y=(T1-T)/(T1-T0);
+X=0.9;//from the given graph, X=(alpha*t)/(x1^2)
+t=(X*x1*x1)/(alpha);
+mprintf("the time taken is %f s = %f h",t,t/3600)
+//end
diff --git a/2762/CH5/EX5.5.2/5_5_2.sce b/2762/CH5/EX5.5.2/5_5_2.sce
new file mode 100755
index 000000000..c7dd2d24a
--- /dev/null
+++ b/2762/CH5/EX5.5.2/5_5_2.sce
@@ -0,0 +1,16 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 5
+//Example 5.5-2
+//Principles of Unsteady State Heat Transfer
+//given data
+m=75/100;//percentage of moisture
+H=335*1000;//latent heat of fusion
+lemda=m*H;
+a=0.0635;//meat slab thickness
+Tf=270.4;
+T1=244.3;
+rho=1057;
+h=17;
+k=1.038;
+t=(lemda*rho/(Tf-T1))*((a/(2*h))+((a*a)/(8*k)));
+mprintf("the time taken is %f h",t/3600)
diff --git a/2762/CH6/EX6.1.1/6_1_1.sce b/2762/CH6/EX6.1.1/6_1_1.sce
new file mode 100755
index 000000000..c7677a2ab
--- /dev/null
+++ b/2762/CH6/EX6.1.1/6_1_1.sce
@@ -0,0 +1,15 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 6
+//Example 6.1-1
+//Principles of Mass Transfer
+//given data
+pa1=0.6*101325;//partial pressure of He at one end
+pa2=0.2*101325;//partial pressure of He at other end
+Dab=0.687/10000;//diffusivity of He-N2 mixture
+R=8314;//gas constant
+T=298;//temp of mixture
+z2=0.2;//z2-z1= dist b/w 2 ends
+z1=0;
+Jaz=(Dab*(pa1-pa2))/(R*T*(z2-z1));//ficks law of diffusion
+mprintf("the flux at steady state is %f kg mol He/s m2",Jaz)
+//end
diff --git a/2762/CH6/EX6.2.1/6_2_1.sce b/2762/CH6/EX6.2.1/6_2_1.sce
new file mode 100755
index 000000000..245c1ea8b
--- /dev/null
+++ b/2762/CH6/EX6.2.1/6_2_1.sce
@@ -0,0 +1,17 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 6
+//Example 6.2-1
+//Principles of Mass Transfer
+//given data
+pa2=1.013*10000;//partial pressure of NH4 at one end
+pa1=00.507*10000;//partial pressure of NH4 at other end
+Dab=0.23/10000;//diffusivity of He-N2 mixture
+R=8314;//gas constant
+T=298;//temp of mixture
+z2=0.1;//z2-z1= dist b/w 2 ends
+z1=0;
+Ja=(Dab*(pa1-pa2))/(R*T*(z2-z1));//ficks law of diffusion
+P=1.013*100000
+Jb=(Dab*((P-pa1)-(P-pa2))/(R*T*(z2-z1)));
+mprintf("the flux at steady state for A is %f kg mol He/s m2",Ja);
+mprintf("the flux at steady state for B is %f kg mol He/s m2",Jb);
diff --git a/2762/CH6/EX6.2.2/6_2_2.sce b/2762/CH6/EX6.2.2/6_2_2.sce
new file mode 100755
index 000000000..921d7c44b
--- /dev/null
+++ b/2762/CH6/EX6.2.2/6_2_2.sce
@@ -0,0 +1,33 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 6
+//Example 6.2-2
+//Principles of Mass Transfer
+//given data
+//english units
+pa1=(17.54/760);//partial pressure of water at one end
+pa2=0;//partial pressure of water at other end
+Dab=(0.25/10000)*(3.875*10000);//diffusivity of H20-air mixture
+P=1;
+pb1=P-pa1;
+pb2=P-pa2;
+pbm=(pb2-pb1)/(log(pb2/pb1)/log(%e));
+R=0.73;//gas constant
+T=528;//temp of mixture
+z2=0.5;//z2-z1= dist b/w 2 ends
+z1=0;
+Na=(Dab*P*(pa1-pa2))/(R*T*(z2-z1)*pbm);//ficks law of diffusion
+mprintf("the flux at steady state for A is %f lb mol /h m2",Na)
+//si units
+pa1si=(17.54/760)*101325;//partial pressure of water at one end
+pa2si=0;//partial pressure of water at other end
+Dabsi=(0.25/10000);//diffusivity of H20-air mixture
+Psi=101325;
+pb1si=Psi-pa1si;
+pb2si=Psi-pa2si;
+pbmsi=(pb2si-pb1si)/(log(pb2si/pb1si)/log(%e));
+Rsi=8314//gas constant
+Tsi=293;//temp of mixture
+z2si=0.5*0.3048;//z2-z1= dist b/w 2 ends
+z1si=0;
+Nasi=(Dabsi*Psi*(pa1si-pa2si))/(Rsi*Tsi*(z2si-z1si)*pbmsi);//ficks law of diffusion
+mprintf(" the flux at steady state for A is %f kg mol /s m2",Nasi)
diff --git a/2762/CH6/EX6.2.4/6_2_4.sce b/2762/CH6/EX6.2.4/6_2_4.sce
new file mode 100755
index 000000000..a34d97224
--- /dev/null
+++ b/2762/CH6/EX6.2.4/6_2_4.sce
@@ -0,0 +1,20 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 6
+//Example 6.2-4
+//Principles of Mass Transfer
+//given data
+//si units
+pa1=(0.555/760)*101325;//partial pressure of napthalene at one end
+pa2=0;//partial pressure of water at other end
+r1=2/1000;
+Dab=(6.92/1000000);//diffusivity of H20-air mixture
+P=101325;
+pb1=P-pa1;
+pb2=P-pa2;
+pbm=(pb2+pb1)/2;
+R=8314;//gas constant
+T=318;//temp of mixture
+z2=0.5;//z2-z1= dist b/w 2 ends
+z1=0;
+Na=(Dab*P*(pa1-pa2))/(R*T*r1*pbm);//ficks law of diffusion
+mprintf("the flux at steady state for A is %f kg mol A/s m2",Na)
diff --git a/2762/CH6/EX6.2.5/6_2_5.sce b/2762/CH6/EX6.2.5/6_2_5.sce
new file mode 100755
index 000000000..adf71bbc0
--- /dev/null
+++ b/2762/CH6/EX6.2.5/6_2_5.sce
@@ -0,0 +1,25 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 6
+//Example 6.2-5
+//Principles of Mass Transfer
+//given data
+//si units
+P1=1;
+T1=0+273;
+Mb=74.1;//
+Ma=29;
+k=0
+X=[4 10 1;16.5 1.98 5.48];
+for i=1:3
+    s=X(1,i)*X(2,i);
+    k=k+s;
+end
+sumvb=20.1;
+Dab1=((1/(10^7))*(T1^1.75)*(((1/Ma)+(1/Mb))^0.5))/(P1*((k^(1/3))+(sumvb^(1/3)))^2);
+mprintf("i) Diffusivity= %f m2/s",Dab1)
+T2=25.9+273;
+Dab2=((1/(10^7))*(T2^1.75)*(((1/Ma)+(1/Mb))^0.5))/(P1*((k^(1/3))+(sumvb^(1/3)))^2);
+mprintf("ii) Diffusivity= %f m2/s",Dab2);
+T3=0+273;P3=2;
+Dab3=((1/(10^7))*(T3^1.75)*(((1/Ma)+(1/Mb))^0.5))/(P3*((k^(1/3))+(sumvb^(1/3)))^2);
+mprintf("iii) Diffusivity= %f m2/s",Dab3)
diff --git a/2762/CH6/EX6.3.1/6_3_1.sce b/2762/CH6/EX6.3.1/6_3_1.sce
new file mode 100755
index 000000000..f72cecf47
--- /dev/null
+++ b/2762/CH6/EX6.3.1/6_3_1.sce
@@ -0,0 +1,26 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 6
+//Example 6.3-1
+//Principles of Mass Transfer
+//given data
+//si units
+Dab=(0.74/(10^9));//diffusivity of H20-air mixture
+Mb=18.02;//
+Ma=46.05;
+na2=6.8;
+nb2=100-na2;
+xa2=(na2/Ma)/((na2/Ma)+(nb2/Mb));//partial pressure of water at one end
+xb2=1-xa2;//partial pressure of water at other end
+na1=16.8;
+nb1=100-na1;
+xa1=(na1/Ma)/((na1/Ma)+(nb1/Mb))
+M2=100/((na2/Ma)+(nb2/Mb));
+xb1=1-xa1;
+M1=100/((na1/Ma)+(nb1/Mb));
+xbm=(xb1+xb2)/2;
+rho1=972.8;
+rho2=988.1;
+Cav=((rho1/M1)+(rho2/M2))/2;
+z=2/1000;
+Na=(Dab*Cav*(xa1-xa2))/(z*xbm);//ficks law of diffusion
+mprintf("the flux at steady state for A is %f kg mol /s m2",Na);
diff --git a/2762/CH7/EX7.1.1/7_1_1.sce b/2762/CH7/EX7.1.1/7_1_1.sce
new file mode 100755
index 000000000..2ab5e67df
--- /dev/null
+++ b/2762/CH7/EX7.1.1/7_1_1.sce
@@ -0,0 +1,33 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 7
+//Example 7.1-1
+//Principles of Unsteady state and convective mass transfer
+//given data
+//si units
+c0=0.1;//initial concentration
+c1=0;//final concentration
+K=1;//assumed as 1 
+Dab=4.72/(10^10);//diffusivity
+t1=10*3600;//time take
+x1=0;
+xc=10.6/(2*1000);
+n1=x1/xc;
+X1=(Dab*t1/(xc^2));
+Y1=0.275;//from graph
+ci=c1-(Y1*((c1/K)-c0));//graph intercept
+m=0;
+mprintf("the concentration at i) the centre is %f kg mol/m3",ci)
+x2=2.54/1000;
+n2=x2/xc;
+X=(Dab*t1/(xc^2));
+Y2=0.172;//from graph
+cii=c1-(Y2*((c1/K)-c0));
+m=0;
+mprintf(" the concentration at i) midpoint of the slab is %f kg mol/m3",cii)
+x3=2.54/1000;
+n2=x1/xc;
+X3=X1/(0.5*0.5);
+Y3=0.002;//from graph
+ciii=c1-(Y3*((c1/K)-c0));
+m=0;
+mprintf(" the concentration at iii) haLVING thickness is %f kg mol/m3",ciii)
diff --git a/2762/CH7/EX7.2.1/7_2_1.sce b/2762/CH7/EX7.2.1/7_2_1.sce
new file mode 100755
index 000000000..7e951829f
--- /dev/null
+++ b/2762/CH7/EX7.2.1/7_2_1.sce
@@ -0,0 +1,20 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 7
+//Example 7.2-1
+//Principles of Unsteady state and convective mass transfer
+//given data
+pa1=0.2*101325;//Pa
+pa2=0;
+P=2*101325;
+ya1=pa1/P;
+ya2=pa2/P;
+yb1=1-ya1;
+yb2=1-ya2;
+ybm=(yb2-yb1)/((log(yb2/yb1))/log(2.71828183));
+kdy=6.78/(10^5);
+ky=kdy/ybm;
+kg=ky/P;
+Na=ky*(ya1-ya2);
+mprintf("i) flux=%f kg mol/s m2",Na)
+mprintf(" ii) ky=%f kg mol/s m2 mol frac",ky)
+mprintf(" iii) kg=%f kg mol/s m2 Pa",kg)
diff --git a/2762/CH7/EX7.3.1/7_3_1.sce b/2762/CH7/EX7.3.1/7_3_1.sce
new file mode 100755
index 000000000..d3f275b61
--- /dev/null
+++ b/2762/CH7/EX7.3.1/7_3_1.sce
@@ -0,0 +1,27 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 7
+//Example 7.3-1
+//Principles of Unsteady state and convective mass transfer
+//given data
+//si units
+Dab=6.92*(10^-6);//m2/s
+pa1=74;//Pa
+R=8314.3;
+T=318;
+L=1.1;
+ca1=pa1/(R*T);
+mu=1.932*(10^-5);
+rho=1.114;
+Sc=mu/(rho*Dab);
+D=0.020;
+v=0.8;
+Re=(D*v*rho)/mu;
+if Re<2100 then
+y=Re*Sc*(D/L)*(3.14/4);
+end
+//using a graph
+X=0.55;//(Ca-Ca0)/(Cai-Ca0)
+Ca0=0;//initial concentration
+Cai=2.799*(10^-5);
+Ca=X*(Cai-Ca0)+Ca0;
+mprintf("the concentration is given %f kg mol/m3",Ca)
diff --git a/2762/CH7/EX7.4.1/7_4_1.sce b/2762/CH7/EX7.4.1/7_4_1.sce
new file mode 100755
index 000000000..47779b9be
--- /dev/null
+++ b/2762/CH7/EX7.4.1/7_4_1.sce
@@ -0,0 +1,19 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 7
+//Example 7.4-1
+//Principles of Unsteady state and convective mass transfer
+//given data
+Dp=100/(10^6);
+Dab=3.25/(10^9);
+muc=6.947/(10^4);
+rhoc=994;
+rhop=1.13;
+Sc=(muc/(rhoc*Dab));
+delrho=rhoc-rhop;
+g=9.806;//gravity force
+kdl=(2*Dab/Dp)+(0.31*(Sc^(-2/3))*(((delrho*muc*g)/(994*994))^(1/3)));
+kl=kdl;//for dilute solutions
+ca1=2.26/10000;
+ca2=0;
+Na=kl*(ca1-ca2);
+mprintf("the rate of absorption is %f kg mol O2/s m2",Na)
diff --git a/2762/CH7/EX7.5.3/7_5_3.sce b/2762/CH7/EX7.5.3/7_5_3.sce
new file mode 100755
index 000000000..f07885c56
--- /dev/null
+++ b/2762/CH7/EX7.5.3/7_5_3.sce
@@ -0,0 +1,14 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 7
+//Example 7.5-3
+//Principles of Unsteady state and convective mass transfer
+//given data
+//si units
+k=35;//rate constant first order
+t=0.01;
+sol=2.961/(10^7);
+P=101.32*1000;
+Dab=1.5/(10^9)
+ca0=sol*P;//initial concn
+Q=ca0*sqrt(Dab/k)*(k*t+0.5)*erf(sqrt(k*t)+sqrt(k*t/(3.14*(2.71828183^(k*t)))));//rate of absorption
+mprintf("the rate of absorption is %f kg mol CO2/m2",Q)
diff --git a/2762/CH7/EX7.6.1/7_6_1.sce b/2762/CH7/EX7.6.1/7_6_1.sce
new file mode 100755
index 000000000..4cab4d058
--- /dev/null
+++ b/2762/CH7/EX7.6.1/7_6_1.sce
@@ -0,0 +1,11 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 7
+//Example 7.6-1
+//Principles of Unsteady state and convective mass transfer
+//given data
+rb=6/(10^9);//pore radius
+Ma=2.016;//mol wt of hydrogen
+T=373;//temp in K
+Dka=97*rb*sqrt(T/Ma);
+mprintf("the knudsen diffusivity is %f m2/s",Dka)
+//end
diff --git a/2762/CH8/EX8.4.1/8_4_1.sce b/2762/CH8/EX8.4.1/8_4_1.sce
new file mode 100755
index 000000000..ab5853420
--- /dev/null
+++ b/2762/CH8/EX8.4.1/8_4_1.sce
@@ -0,0 +1,31 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 8
+//Example 8.4-1
+//Evaporation
+//given data
+Tf=37.8+273.2;//temp in K
+xf=1/100;//wt % of salt in the feed
+xl=1.5/100;//wt % of salt in the extract
+//2 eqns to be solved
+F=9072;//feel flow rate
+xv=0;
+A=[1 1;xl xv];
+B=[F;(F*xf)];
+LV=inv(A)*B;
+L=LV(1,1);//liquid flow rate
+V=LV(2,1);//vapour flow rate
+Cpf=4.14;//heat capacity of the feed
+T1=273.2+100;//bp of water at 101.32 Kpa(as given)
+hf=Cpf*(Tf-T1);//enthalpy of feed
+hl=0;//enthalpy of extract as it is at datum
+hv=2257;//enthalpy of vapour
+lemda=2230;//enthalpy of steam;
+S=(L*hl+V*hv-F*hf)/lemda;//flow rate of steam
+q=S*lemda*(1000/3600);//heat transfered through the heating surface area
+U=1704;//heat transfer coefficient 
+Ts=383.2;
+A=q/(U*(Ts-T1));//heat transfer area
+mprintf("i) extract flow rate= %f kg/h",L)
+mprintf("ii) vapour flow rate= %f kg/h",V)
+mprintf("iii) heat transfer area= %f m2",A)
+//end
diff --git a/2762/CH8/EX8.4.3/8_4_3.sce b/2762/CH8/EX8.4.3/8_4_3.sce
new file mode 100755
index 000000000..1a240b913
--- /dev/null
+++ b/2762/CH8/EX8.4.3/8_4_3.sce
@@ -0,0 +1,36 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 7
+//Example 8.4-3
+//Evaporation
+//given data
+F=4536;//feed flow rate in kg/h
+xf=0.2;//wt fraction of NaOH in feed
+Tf=60;
+P1=11.7;// In kPa
+Ps=172.4;//steam pressure in kPa
+xl=0.5;//wt fraction of NaOH in liq
+xv=0;
+A=[1 1;xl xv];
+B=[F;(F*xf)];
+LV=inv(A)*B;
+L=LV(1,1);//liquid flow rate
+V=LV(2,1);//vapour flow rate
+Tbp1=48.9;//boiling pt of water at P1 and 50% concn
+Tbp2=89.5;//boiling pt of soln at Tbp1 and 50% concn from durhing chart
+bpr1=Tbp2-Tbp1;//boiling point rise
+hf=214;//(20% sol) kJ/kg from enthalpy concn chart
+hl=505;//(50% sol)kJ/kg from enthalpy concn chart
+Hv=2667;//steam tables for superheated steam
+Cps=1.884;//enthalpy of superheated steam
+
+lemda=2214;//latent heat of steam
+S=(L*hl+V*Hv-F*hf)/lemda;//flow rate of steam
+q=S*lemda*(1000/3600);
+U=1560;
+T1=115.6;
+A=q/(U*(T1-Tbp2));//heat transfer area
+SE=V/S;//steam economy
+mprintf("i) steam flow rate= %f kg/h",S)
+mprintf("ii) steam economy %f",SE)
+mprintf("iii) heat transfer area= %f m2",A)
+//deviations may occur due apprroximated values taken in the textbook
diff --git a/2762/CH9/EX9.3.1/9_3_1.sce b/2762/CH9/EX9.3.1/9_3_1.sce
new file mode 100755
index 000000000..2abba7d1d
--- /dev/null
+++ b/2762/CH9/EX9.3.1/9_3_1.sce
@@ -0,0 +1,16 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 9
+//Example 9.3-1
+//Drying of Process Materials
+//given data
+P=101.32;
+pa=2.76;
+pas=3.5;
+H=(18.02/28.97)*(pa/(P-pa));//Humidity=(mol wt of water/mol wt of air)*(pa/(P-pa))
+Hs=(18.02/28.97)*(pas/(P-pas));//saturation humidity
+Hp=100*(H/Hs);//percentage humidity
+Hr=100*(pa/pas);//relative humidity
+mprintf("1) Humidity= %f kg H20/kg air",H);
+mprintf("2)saturation Humidity= %f kg H20/kg air",Hs);
+mprintf("3) percentage Humidity= %f percent",Hp);
+mprintf("4) relative Humidity= %f percent",Hr);
diff --git a/2762/CH9/EX9.6.3/9_6_3.sce b/2762/CH9/EX9.6.3/9_6_3.sce
new file mode 100755
index 000000000..e66f5c7cc
--- /dev/null
+++ b/2762/CH9/EX9.6.3/9_6_3.sce
@@ -0,0 +1,33 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 9
+//Example 9.6-3
+//Drying of Process Materials
+//given data
+//SI units
+H=0.010;
+Tdry=65.6+273;//dry bulb tempereature
+Twet=28.9+273;//wet bulb temp obtained from the humidity chart
+Hw=0.26;
+Vh=(22.41/273)*Tdry*((1/28.7)+(1/18.02)*H);//((humid volume= molar vol of ideal gas at STP)/(Temp at STP)*(22.41/273)*T*((1/28.7)+(1/18.02)*H))
+rho=(1+H)/Vh;//density of dry air + moiture
+v=6.1;//velocity of air
+G=v*3600*rho;
+h=0.0204*(G^0.8);
+lemdaw=2433*1000;
+Rc=(h/lemdaw)*(Tdry-Twet)*3600;
+A=0.457^2
+rate=A*Rc;
+//end933100355
+H=0.010;
+Tdryeng=150;//dry bulb tempereature
+Tweteng=84;//wet bulb temp obtained from the humidity chart
+rhoeng=0.0647;
+veng=20;//velocity of air
+Geng=veng*3600*rhoeng;
+heng=0.0128*(Geng^0.8);
+lemdaweng=1046;
+Rceng=(heng/lemdaweng)*(Tdryeng-Tweteng);
+Aeng=1.5*1.5;
+rateeng=Aeng*Rceng;
+mprintf("rate of drying using si units= %f kg H20/h",rate)
+mprintf(" rate of drying using english units= %f lbm H20/h",rateeng)
diff --git a/2762/CH9/EX9.7.1/9_7_1.sce b/2762/CH9/EX9.7.1/9_7_1.sce
new file mode 100755
index 000000000..afd8a0495
--- /dev/null
+++ b/2762/CH9/EX9.7.1/9_7_1.sce
@@ -0,0 +1,25 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 9
+//Example 9.7-1
+//Drying of Process Materials
+//given data
+X=(1/1000)*[195 150 100 65 50 40]
+R=0.01*[151 121 90 71 37 27]
+Rinv=[]
+for i=1:6
+    Rinv(i)=1/R(1,i);
+    
+end
+v=inttrap(X,Rinv)
+
+X1=0.38;
+X2=0.195;
+XC=X2;
+Rc=1.51;
+A=18.58;
+Ls=399;
+t1=(Ls/A)*(-v);//as the graph is a decreasing function
+t=(X1-X2)*(Ls/(A*Rc))
+plot(X,Rinv,rec=[0,0,0.3,5])
+xtitle("Graphical Representation of Falling Rate Period","X","1/R")
+mprintf("the time for drying= %f h",t+t1)
diff --git a/2762/CH9/EX9.7.2/9_7_2.sce b/2762/CH9/EX9.7.2/9_7_2.sce
new file mode 100755
index 000000000..8e9aa0a8f
--- /dev/null
+++ b/2762/CH9/EX9.7.2/9_7_2.sce
@@ -0,0 +1,22 @@
+//Transport Processes and Seperation Process Principles
+//Chapter 9
+//Example 9.7-2
+//Drying of Process Materials
+//given data
+X=(1/1000)*[195 150 100 65 50 40]
+R=0.01*[151 121 90 71 37 27]
+Rinv=[]
+for i=1:6
+    Rinv(i)=1/R(1,i);
+    
+end
+v=inttrap(X,Rinv)
+
+X1=0.38;
+XC=0.195;
+X2=0.040;
+Rc=1.51;
+A=18.58;
+Ls=399;
+t=(Ls*XC/(A*Rc))*(log(XC/X2)/log(%e));
+mprintf("Drying time= %f h",t)