initial commit / add all books

5 files changed, 131 insertions, 0 deletions

diff --git a/172/CH14/EX14.1/ex1.sce b/172/CH14/EX14.1/ex1.sce
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+//ques1

+//to determine the sublimation pressure of water

+clear

+clc

+//from table in appendix B.1.5

+T1=213.2;//K, Temperature at state 1

+P2=0.0129;//kPa, pressure at state 2

+T2=233.2;//K, Temperature at state 2

+hig=2838.9;//kJ/kg, enthalpy of sublimation 

+R=.46152;//Gas constant 

+//using relation log(P2/P1)=(hig/R)*(1/T1-1/T2) 

+P1=P2*exp(-hig/R*(1/T1-1/T2));

+printf('Sublimation Pressure = %.5f kPa \n',P1);
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diff --git a/172/CH14/EX14.4/ex4.sce b/172/CH14/EX14.4/ex4.sce
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+//ques4

+//Volume expansivity, Isothermal and Adiabatic compressibility

+clear

+clc

+//known data

+ap=5*10^-5;//K^-1 Volume expansivity

+bt=8.6*10^-12;//m^2/N, Isothermal compressibility

+v=0.000114;//m^3/kg, specific volume

+P2=100*10^6;//pressure at state 2 in kPa

+P1=100;//pressure at state 1 in kPa

+w=-v*bt*(P2^2-P1^2)/2;//work done in J/kg

+//q=T*ds and ds=-v*ap*(P2-P1)

+//so q=-T*v*ap*(P2-P1)

+T=288.2;//Temperature in K

+q=-T*v*ap*(P2-P1);//heat in J/kg

+du=q-w;//change in internal energy in J/kg

+printf('Change in internal energy = %.1f J/kg ',du);

+ 
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diff --git a/172/CH14/EX14.5/ex5.sce b/172/CH14/EX14.5/ex5.sce
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+//ques5

+//adiabatic steady state processes

+clear

+clc

+//from table A.2

+P1=20;//pressure at state 1 in MPa

+P2=2;//pressure at state 2 in MPa

+T1=203.2;//Temperature at state 1 in K

+Pr1=P1/3.39;//Reduced pressure at state 1

+Pr2=P2/3.39;//Reduced pressure at state 2

+Tr1=T1/126.2;//Reduced temperature

+//from compressibility chart h1*-h1=2.1*R*Tc

+//from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)

+//h2*-h2=0.5*R*Tc

+//this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.5*R*Tc

+R=0.2968;//gas constant for given substance

+Tc=126.2;//K, Constant temperature

+Cp=1.0416;//heat enthalpy at constant pressure in kJ/kg

+T2=146;//temperature at state 2

+dh=-1.6*R*Tc+Cp*(T1-T2);//

+printf('Enthalpy change = %.0f kJ/kg \n',dh);

+printf(' Since Enthalpy change is %.0f kJ/kg so Temperature = %.1f K',dh,T2); 
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diff --git a/172/CH14/EX14.6/ex6.sce b/172/CH14/EX14.6/ex6.sce
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+//ques6

+//isothermal steady state processes

+clear

+clc

+//from table A.2

+P1=8;//pressure at state 1 in MPa

+P2=0.5;//pressure at state 2 in MPa

+T1=150;//Temperature at state 1 in K

+Pr1=P1/3.39;//Reduced pressure at state 1

+Pr2=P2/3.39;//Reduced pressure at state 2

+Tr1=T1/126.2;//Reduced temperature

+T2=125;//temperature at state 2

+//from compressibility chart h1*-h1=2.1*R*Tc

+//from zero pressure specific heat data h1*-h2*=Cp*(T1a-T2a)

+//h2*-h2=0.5*R*Tc

+//this gives dh=h1-h2=-2.1*R*Tc+Cp*(T1a-T2a)+0.15*R*Tc

+R=0.2968;//gas constant for given substance

+Tc=126.2;//K, Constant temperature

+Cp=1.0416;//heat enthalpy at constant pressure in kJ/kg

+dh=(2.35)*R*Tc+Cp*(T2-T1);//

+printf('Enthalpy change = %.2f kJ/kg \n',dh);

+//change in entropy 

+//ds= -(s2*-s2)+(s2*-s1*)+(s1*-s1)

+//s1*-s1=1.6*R

+//s2*-s2=0.1*R

+//s2*-s1*=Cp*log(T2/T1)-R*log(P2/P1)

+//so

+ds=1.6*R-0.1*R+Cp*log(T2/T1)-R*log(P2/P1);

+printf(' Entropy Change = %.4f kJ/kg.K ',ds);
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diff --git a/172/CH14/EX14.7/ex7.sce b/172/CH14/EX14.7/ex7.sce
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+//ques7

+//percent deviation using specific volume calculated by kays rule and vander waals rule

+clear

+clc

+//a-denotes C02

+//b-denotes CH4

+T=310.94;//Temperature of mixture K

+P=86.19;//Pressure of mixture in MPa

+//Tc- critical Temperature

+//Pc-critical pressure

+Tca=304.1;//K

+Tcb=190.4;//K

+Pca=7.38;//MPa

+Pcb=4.60;//MPa

+Ra=0.1889;//gas constant for a in kJ/kg.K

+Rb=0.5183;//gas constant for b in kJ/kg.K

+xa=0.8;//fraction of CO2

+xb=0.2;//fraction of CH4

+Rm=xa*Ra+xb*Rb;//mean gas constant in kJ/kg.K

+Ma=44.01;//molecular mass of a

+Mb=16.043;//molecular mass of b

+//1.Kay's rule

+ya=xa/Ma/(xa/Ma+xb/Mb);//mole fraction of a

+yb=xb/Mb/(xa/Ma+xb/Mb);//mole fraction of b

+Tcm=ya*Tca+yb*Tcb;//mean critical temp in K

+Pcm=ya*Pca+yb*Tcb;//mean critical pressure n MPa

+//therefore pseudo reduced property of mixture

+Trm=T/Tcm;

+Prm=P/Pcm;

+Zm=0.7;//Compressiblity from generalised compressibility chart

+vc=Zm*Rm*T/P/1000;//specific volume calculated in m^3/kg

+ve=0.0006757;//experimental specific volume in m^3/kg

+pd1=(ve-vc)/ve*100;//percent deviation

+printf('Percentage deviation in specific volume using Kays rule = %.1f percent \n',pd1);   

+

+//2. using vander waals equation

+//values of vander waals constant

+Aa=27*Ra^2*Tca^2/(64*Pca*1000);

+Ba=Ra*Tca/(8*Pca*1000);

+Ab=27*Rb^2*Tcb^2/(64*Pcb*1000);

+Bb=Rb*Tcb/(8*Pcb*1000);

+//mean vander waals constant

+Am=(xa*sqrt(Aa)+xb*sqrt(Ab))^2;

+Bm=(xa*Ba+xb*Bb);

+//using vander waals equation we get cubic equation 

+//solving we get

+vc=0.0006326;//calculated specific volume in m^3/kg

+pd2=(ve-vc)/ve*100;

+printf(' Percentage deviation in specific volume using vander waals eqn = %.1f percent \n',pd2); 
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