initial commit / add all books

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diff --git a/2762/CH1/EX1.4.1/1_4_1.sce b/2762/CH1/EX1.4.1/1_4_1.sce
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+//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
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+//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
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+//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
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+//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
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+//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
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+//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
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+//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
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+//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
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+//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
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+//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
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+//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
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+//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
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+//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