initial commit / add all books

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+printf("\t example 13.5 \n");

+// for a Basis of one Hour

+printf("\t approximate values are mentioned in the book \n");

+c(1)=1544; // Flow rate of CO2, Lb/hr

+h(1)=4500; // Flow rate of H20, Lb/hr

+

+c(2)=35; //Flow rate of CO2, Mol/hr

+h(2)=250;//Flow rate of H20, Mol/hr

+

+t(1)=c(1)+h(1); //Total flow rate , Lb/hr

+t(2)=c(2)+h(2); //Total flow rate, Mol/hr

+

+Pt = (30+14.7)/(14.7); //Total Pressure in atm

+printf("\t Pt is %.2f\n",Pt);

+Pw = ( h(2)/t(2) )*Pt; //Partial pressure of Water in atm

+

+printf("\t Partial Pressure of Water: %.2f atm \n",Pw);

+

+Tw = 267; // from table 7 at 2.68atm

+Mm = (t(1)/t(2));

+

+printf("\t mean molecular weight : %.1f \n",Mm);

+// weighted temperature difference

+// overall balance

+//for Inlet

+Pv=2.68; // water vapour pressure, atm

+Pg=Pt-Pv; // Inert pressure

+//for Exit

+Pw1 = 0.1152 // Partial pressure of water at 120 F

+Pv1 = 0.115; // Water vapor pressure

+Pg1 = 2.935; // Inert pressure

+

+w1 = 250; //Pound mols steam inlet

+w2 = c(2)*(Pv1/Pg1);

+printf("\tPound mols steam exit:%.2f\n",w2);

+w3 = w1 - w2;

+printf("\tPound mols steam condessed:%.2f\n",w3);

+//Assume points at 267, 262, 255,225,150,120 deg F

+//For the interval from 267 to 262 F

+

+Pv2 = 2.49; // From table 7 at 262 F

+Pg2 = Pt - Pv2; //Inert pressure

+printf("\tPg is %.2f",Pg2);

+

+w4 = c(2) * (Pv2/Pg2); //Mol steam remaining

+w5 = h(2) - w4; //Mol steam condensed

+

+printf("\tMol steam remaining:%.0f\n",w4);

+printf("\tMol steam condensed:%.0f\n",w5);

+

+h1 = (w5*18*937.3) + (0.46*(267-262) * w5 * 18); //Heat of condensation

+h2 = (w4 * 18 * 0.46*(267-262)); //Heat from uncondensed steam

+h3 = c(1)*0.22*5.0; //Heat from noncondensable

+

+printf("\tHeat of condensation:%.2e\n",h1);

+printf("\tHeat from uncondensed steam:%.2e\n",h2);

+printf("\tHeat from noncondensable:%.1e\n",h3);

+

+ht = h1+h2+h3;//Total heat

+printf("\tTotal heat:%.0f\n",ht);

+

+//Similarily calculating the Heat balance for other intervals

+printf("\tInterval,F\tTotal Heat\n\t267-262\t1,598,000\n\t262-255\t1,104,000\n\t255-225\t1,172,000\n\t225-150\t751,000\n\t150-120\t177,000\n\tTotal\t4,802,000\n");

+

+w=4802000/(115-80); //Total water

+printf("\tTotal water: %.2e\n",w);

+//Water coefficient

+Nt = 246;

+at1 = 0.302;

+n = 4;

+

+at = Nt * (at1/(144*n)); // From eq 7.48

+printf("\tat is %.3f ft^2\n",at);

+Gt = w/at;

+printf("\tGt is %.2e lb/(hr)(ft^2)\n",Gt);

+ro = 62.5; 

+V = Gt/(3600*ro);

+printf("\tV is %.2f fps\n",V);

+hi = 1120; // From fig. 25

+ID = 0.62;

+OD = 0.75;

+hi0= hi *(ID/OD); //From eq 6.5

+printf("\thi0 is %.0f\n",hi0);

+//Mean properties at 267 F

+c = ((c(1)*0.22)+(h(1)*0.46))/t(1);  // Calculation mistake in Book

+printf("\tMean c:%.3f Btu/(lb)(F)\n",c);

+

+k = ((c(1)*0.0128)+(h(1)*0.015))/t(1); // Calculation mistake in Book

+printf("\tMean k:%.4f Btu/(hr)(ft^2)(F/ft)\n",k);

+

+mu = (((c(1)*0.019)+(h(1)*0.0136))/t(1))* 2.42; // Calculation mistake in Book

+printf("\tMean mu:%.4f lb/(hr)(ft)\n",mu);

+

+ID1 = 21.25;

+C = 0.25;

+B = 12;

+PT = 1.0;

+

+as = ID1 * C * (B/(144*PT)); //From eq 7.1

+printf("\tas is %.3f ft^2\n",as);

+Gs = t(1)/as //From eq 7.2

+printf("\tGs is %.3e lb/(hr)(ft^2)\n",Gs);

+Ds = 0.0792; // From Fig 28

+Res = Ds * (Gs/0.0363); // From eq 7.3

+printf("\tRes is %.2e\n",Res);

+jH = 102; // From Fig 28

+x = ((c*mu)/k)^(1/3);

+printf("\t(c.mu/k)^1/3 is %.0f\n",x);

+h0 = jH * 0.0146 * (x/Ds); //From eq 6.15b

+printf("\th0 is %.0f\n",h0);

+y = 0.62 // y = (mu/ro * kd)^(2/3)

+z = 1.01; // z = ((c*mu)/k)^(2/3)

+

+K = (h0*z)/(0.407*Mm*y); //KG = K/p0f

+printf("\tK is %.2f\n",K);

+//at point 1

+Tg = 244; // F

+tW = 115;

+delt=(Tg-tW);

+printf("\t delt is %.0f F \n",delt);

+