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diff --git a/1328/CH17/EX17.6/17_6.sce b/1328/CH17/EX17.6/17_6.sce
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+printf("\t example 17.6 \n");
+printf("\t approximate values are mentioned in the book \n");
+// basis 1ft^2 ground area
+//Assumption: 20 per cent of the initial vapor content of the gas enters the water body
+X1=(1.69/(14.7-1.69))*(18/29);
+printf("\t X1 : %.4f lb/lb \n",X1);
+G=1500;
+w1=G*X1;
+printf("\t total water in inlet gas : %.2f lb/hr \n",w1);
+// The inlet gas is at 300F and a 120F dew point. Use 0.25 Btu/(lb)(°F) for the specific heat of nitrogen
+H1=(0.0807*120)+(0.0807*1025.8)+(0.45*0.0807*(300-120))+(0.25*300); // eq 17.55
+printf("\t H1 : %.0f Btu/lb dry air \n",H1);
+X2=(w1*(1-.2)/G);
+printf("\t outlet gas humidity : %.5f lb/lb \n",X2);
+pw=(X2*29*14.7/18)/(1+(X2*29/18));
+printf("\t pw : %.3f psia \n",pw);
+Tw=112.9; // F, from table 7 for above pw
+// The outlet gas has a temperature of 200°F and a 112.9°F dew point
+H2=(X2*Tw)+(X2*1029.8)+(X2*0.45*(200-Tw))+(0.25*200); // eq 17.55
+printf("\t H2 : %.1f Btu/lb dry air \n",H2);
+q=G*(H1-H2);
+printf("\t total heat load : %.2e Btu/hr \n",q);
+w2=q/(120-85);
+printf("\t water loading : %.2e lb/hr \n",w2);
+printf("\t interval 1 \n");
+// (Kxa*delV/L)= 0 t0 0.05
+nd=0.05; // nd=Kxa*V/L
+Le=0.93; // fig 17.4 at 300F
+C=(0.25)+(0.45*X1);
+printf("\t C : %.3f Btu/(lb)*(F) \n",C);
+haV=(nd*w2*Le*C);
+printf("\t haV : %.1f Btu/(hr)*(F) \n",haV);
+qc=(haV*(300-120));
+printf("\t qc : %.2e Btu/hr \n",qc);
+delT=(qc/(C*G));
+printf("\t delT : %.1f F \n",delT);
+T1=(300-delT);
+printf("\t T(0.05) : %.1f F \n",T1);
+delt=(qc/w2);
+printf("\t delt : %.2f F \n",delt);
+t1=(120-delt);
+printf("\t t(0.05) : %.1f F \n",t1);
+printf("\t interval 2 \n");
+// (Kxa*delV/L)= 0.05 to 0.15
+nd1=0.1;
+haV1=(nd1*w2*Le*C);
+printf("\t haV1 : %.1f Btu/(hr)*(F) \n",haV1);
+qc1=(haV1*(T1-t1));
+printf("\t qc1 : %.1e Btu/hr \n",qc1);
+delT1=(qc1/(C*G));
+printf("\t delT1 : %.1f F \n",delT1);
+T2=(T1-delT1);
+printf("\t T(0.15) : %.2f F \n",T2);
+X3=0.0748; // at 117.6F
+w3=(nd1*w2*(0.0807-X3));
+printf("\t water diffused during interval : %.3f lb/hr \n",w3);
+w4=(w1-w3);
+printf("\t water remaining : %.2f lb/hr \n",w4);
+l1=1027; // Btu/lb, l1= lamda at 117.6F
+qd=(w3*l1);
+printf("\t qd : %.0f Btu/hr \n",qd);
+q1=(qd+qc1);
+printf("\t q1 : %.0f Btu/hr \n",q1);
+delt1=(q1/w2);
+printf("\t delt1 : %.2f F \n",delt1);
+t2=(t1-delt1);
+printf("\t t(0.15) : %.1f F \n",t2);
+X4=0.0640; // at 112.5
+X5=(w4/G);
+printf("\t X(112.5F) : %.4f lb/lb \n",X5);
+printf("\t interval 3 \n");
+// (Kxa*delV/L)= 0.15 to 0.25
+nd1=0.1;
+haV1=(nd1*w2*Le*C);
+printf("\t haV1 : %.1f Btu/(hr)*(F) \n",haV1);
+qc2=(haV1*(T2-t2));
+printf("\t qc2 : %.2e Btu/hr \n",qc2);
+delT2=(qc2/(C*G));
+printf("\t delT2 : %.1f F \n",delT2);
+T3=(T2-delT2);
+printf("\t T(0.25) : %.1f F \n",T3);
+w5=(nd1*w2*(X5-X4));
+printf("\t water diffused during interval : %.3f lb/hr \n",w5);
+w6=(w4-w5);
+printf("\t water remaining : %.2f lb/hr \n",w6);
+l2=1030; // Btu/lb, l1= lamda at 112.5F
+qd1=(w5*l2);
+printf("\t qd1 : %.2e Btu/hr \n",qd1);
+q2=(qd1+qc2);
+printf("\t q2 : %.3e Btu/hr \n",q2);
+delt2=(q2/w2);
+printf("\t delt2 : %.2f F \n",delt2);
+t3=(t2-delt2);
+printf("\t t(0.25) : %.1f F \n",t3);
+X6=0.0533; // at 106.5
+X7=(w6/G);
+printf("\t X(106.5F) : %.4f lb/lb \n",X7);
+// The calculations of the remaining intervals until a. gas temperature of 200°F is reached are shown in Fig. 17.17
+w7=21.92; // total water diffused from table in solution
+d=(w7/w1)*100;
+printf("\t calculated diffusion : %.0f \n",d);
+printf("\t Using some standard low-pressure-drop data \n");
+// For G = 1500, extrapolate to L = 2040 on logarithmic coordinates. Kxa = 510.
+ndt=.54; // from 1st table in solution
+Kxa=510; // from 2nd table in solution
+Z=(ndt*w2/Kxa);
+printf("\t tower height : %.2f ft \n",Z);
+A=(50000/G);
+printf("\t cross section : %.1f ft^2 \n",A);
+// end
+
+
+