initial commit / add all books

8 files changed, 197 insertions, 0 deletions

diff --git a/555/CH7/EX7.1/1.sce b/555/CH7/EX7.1/1.sce
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+// Implementation of example 7.1

+// Basic and Applied Thermodynamics by P.K.Nag

+

+clc

+clear

+

+// T is temperature,dS is entropy change

+T1=35 // degree celsius

+T2=37 // degree celsius

+t1=T1+273;

+t2=T2+273;

+// change in entropy is given by dS=mCvlog(t2/t1)

+m=1 // kg

+Cv=4.187

+dS=m*Cv*log(t2/t1);

+printf("change in entropy = %.4f kJ/K",dS);

+// end
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diff --git a/555/CH7/EX7.10/10.sce b/555/CH7/EX7.10/10.sce
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+// Implementation of example 7.10

+// Basic and Applied Thermodynamics by P.K.Nag

+

+clc

+clear

+

+// T is temperature,dS is change in entropy,S is entropy,Q is heat transfer

+l=5 // length in m

+b=7 // breadth in m

+th=0.32 // thickness in m

+k=0.71 // W/m*K

+t1=6 //degree celsius

+t2=21 //degree celsius

+T1=t1+273; // K

+T2=t2+273; // K

+Tr=27 //degree celsius

+Ts=2 //degree celsius

+tr=Tr+273;

+ts=Ts+273;

+Q=k*l*b*(T2-T1)/th;

+Sgen=(Q/T1)-(Q/T2);

+Sgent=(Q/ts)-(Q/tr);

+printf("rate of heat transfer through wall = %.2f W \n rate of entropy generation in wall = %.3f W/K \n rate of total entropy generation = %.3f W/K",Q,Sgen,Sgent);

+// end
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diff --git a/555/CH7/EX7.2/2.sce b/555/CH7/EX7.2/2.sce
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+// Implementation of example 7.2

+// Basic and Applied Thermodynamics by P.K.Nag

+

+clc

+clear

+

+// T for temperature,m for mass,S for entropy,dS is change in entropy

+t1=273 // K

+t2=373 // K

+m=1 // kg

+c=4.187

+// (a)

+dSw=m*c*log(t2/t1);

+// reservoir's temperature remains constant so dS=Q/T

+Q=m*c*(t2-t1);

+dSr=-(Q/t2);

+dSu=dSw+dSr;

+printf("entropy change of universe = %.3f kJ/K \n",dSu);

+// (b)

+// now water is heated in stages from two reservoirs..

+t3=323 // K

+dSw=m*c*log(t3/t1)+m*c*log(t2/t3);

+dSr1=-[m*c*(t3-t1)/t3];

+dSr2=-[m*c*(t2-t3)/t2];

+dSu2=dSw+dSr1+dSr2;

+printf("entropy change of universe in 2nd case = %.3f kJ/K \n",dSu2);

+// the entropy change of  universe would be less & less if water is heated in more & more stages...it will be zero if water is heated reversibly...

+// end
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diff --git a/555/CH7/EX7.3/3.sce b/555/CH7/EX7.3/3.sce
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+// Implementation of example 7.3

+// Basic and Applied Thermodynamics by P.K.Nag

+

+clc

+clear

+

+m = 1; //mass of ice in kg

+Lf = 333.3; //latent heat of fusion of ice in kJ/kg

+T2 = 0; //degree C

+T2 = T2 + 273; //K

+T1 = -5; //degree C

+T1 = T1 + 273; //K 

+Ta = 20; //degree C

+Ta = Ta + 273; //K

+Cp_ice = 2.093; //specific heat for ice in kJ/kg K 

+Cp_water = 4.187; //specific heat for water in kJ/kg K

+

+//(a)

+Q = m*Cp_ice*(T2-T1) + m*Lf + m*Cp_water*(Ta-T2); //kJ

+dS_atm = -Q/Ta; //kJ/K

+//change in entropy of system when temperature changes from -5 to 0 degree C 

+dS1_sys = m*Cp_ice*log(T2/T1); //kJ/K

+//change in entropy of system when ice melts at 0 degree C

+dS2_sys = m*Lf/T2;

+//change in entropy of when temperature of water changes from 0 to 20 degree C

+dS3_sys = m*Cp_water*log(Ta/T2); //kJ/K

+dS_sys = dS1_sys + dS2_sys + dS3_sys;

+dS_univ = dS_atm + dS_sys;

+printf("Entropy increase of the universe = %f kJ/K\n\n",dS_univ);

+

+

+//(b)

+Wmin = dS_sys*Ta - Q;

+printf("Minimum amount of work necessary to convert water back into ice, Wmin = %0.2f kJ",Wmin);

+// end
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diff --git a/555/CH7/EX7.6/6.sce b/555/CH7/EX7.6/6.sce
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+// Implementation of example 7.6

+// Basic and Applied Thermodynamics by P.K.Nag

+

+clc

+clear

+

+T1 = 200; //K

+T2 = 100; //K

+function Cv = f(T)

+  Cv = 0.042*T^2;

+endfunction

+

+Q1 = intg(T1,T2,f);

+

+function S = g(T)

+  S = f(T)/T;

+endfunction

+

+dS_sys = intg(T1,T2,g);

+Wmax = dS_sys*T2 + abs(Q1);

+printf("Maximum amount of work that can be recovered as system is cooled down to temperature of reservoir, Wmax = %d J",Wmax); 

+//end
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diff --git a/555/CH7/EX7.7/7.sce b/555/CH7/EX7.7/7.sce
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+// Implementation of example 7.5

+// Basic and Applied Thermodynamics by P.K.Nag

+

+clc

+clear

+

+// T for temperature,dS is change in entropy,p for pressure,V is volume

+n=1.3

+p1=500 // kPa

+V1=0.2 // m^3

+V2=0.05 // m^3

+// the fluid is undergoing reversible adiabatic compression according to the law p*(V^1.3)=constant

+p2=p1*(V1/V2)^1.3;

+dH=[n*(p2*V2-p1*V1)]/(n-1);

+dU=dH-(p2*V2-p1*V1);

+dS=0;

+Q12=0;

+W12=-dU;

+printf("change in enthalpy = %.2f kJ \n change in entropy = %.2f  \n change in internal energy = %.2f kJ \n heat transfer = %.2f \n work transfer = %.2f kJ",dH,dS,dU,Q12,W12);

+// end
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diff --git a/555/CH7/EX7.8/8.sce b/555/CH7/EX7.8/8.sce
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+// Implementation of example 7.8

+// Basic and Applied Thermodynamics by P.K.Nag

+

+clc

+clear

+

+Pa = 130; //kPa

+Pb = 100; //kPa

+Ta = 50; //degree C

+Ta = Ta + 273; //K

+Tb = 13; //degree C

+Tb = Tb + 273; //K

+Cp = 1.005; //kJ/kg K

+dS_sys = Cp * log(Tb/Ta) - 0.287 * log(Pb/Pa);

+dS_surr = 0;

+dS_univ = dS_sys + dS_surr;

+printf("dS_univ = %f kJ/kg K\n\n",dS_univ);

+

+if dS_univ<0 then

+  printf("Flow must be from B to A since entropy cannot be negative\n");

+elseif dS_univ>0 then

+  printf("Flow must be from A to B as entropy change is positive\n");

+else 

+    printf("Flow will not occur\n");

+end
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diff --git a/555/CH7/EX7.9/9.sce b/555/CH7/EX7.9/9.sce
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+// Implementation of example 7.9

+// Basic and Applied Thermodynamics by P.K.Nag

+

+clc

+clear

+

+//Mass flow rate of air entering the device m1, Pressure p1, Temperature T1

+//Mass flow rate of air exiting through stream1 m2 and stream2 m3

+

+m1 = 2;//kg/s

+p1 = 4;//bar

+T1 = 300;//K

+p2 = 1;//bar

+p3 = 1;//bar

+T2 = 330;//K

+T3 = 270;//K

+cp = 1.005;//kJ/kg K

+R = 0.287;//KJ/Kg K

+m2 = m1/2;

+m3 = m2;

+//s21 = s2 - s1

+s21 = cp*log(T2/T1)-R*log(p2/p1);

+s31 = cp*log(T3/T1)-R*log(p3/p1);

+Sgen = m2*s21 + m3*s31;

+printf('Sgen = %0.3f kW/K \nSince Sgen > 0, the device is possible',Sgen);

+//end
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