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diff --git a/599/CH5/EX5.10/example5_10.sce b/599/CH5/EX5.10/example5_10.sce new file mode 100755 index 000000000..b29ec98dc --- /dev/null +++ b/599/CH5/EX5.10/example5_10.sce @@ -0,0 +1,85 @@ +
+clear;
+clc;
+printf("\t Example 5.10\n");
+L=2000; //flow rate of water to be cooled in kg/min
+T1=50; //temperature of inlet water
+T2=30; //temp. of outlet water
+H1=.016; //humidity of incoming air
+cp=4.18; //specific heat of water
+cpair=1.005; //specific heat capcity of air
+cpwater=1.884; //specific heat capcity of water
+tg=20; //temperature in degree
+to=0;
+ybar=0.016; //saturated humidity at 20 degree
+d=2502; //latent heat
+Ky_a=2500; //value of masstransfer coefficient in kg/hr*m^3*dybar
+E=cpair*(tg-to)+(cpwater*(tg-to)+d)*ybar; //enthalpy
+ //similarly for other temperatures
+T=[20 30 40 50 55] //differnt temperature for different enthalpy calculation
+i=1;
+while(i<6) //looping for different enthalpy calculation of operating line
+E(i)=cpair*(T(i)-to)+(cpwater*(T(i)-to)+d)*ybar;
+printf("\n the enhalpy at :%f is :%f",T(i),E(i));
+i=i+1;
+end //end of lop
+ES=[60.735 101.79 166.49 278.72 354.92] //enthalpy of eqll condition
+
+plot(T,E,"o--");
+plot(T,ES,"+-");
+title("Fig.5.10(b),Temperature-Enthalpy plot");
+xlabel("X-- Temperature, degree celcius");
+ylabel("Y-- Enthalpy ,kj/kg");
+legend("operating line","Enthalpy at saturated cond")
+
+//locate (30,71.09) the operating conditions at the bottom of the tower and draw the tangent to the curve
+Hg1=71.09; //point on the oper. line(incoming air)
+Hg2=253; //point after drawing the tangent
+slope=(Hg2-Hg1)/(T1-T2); //we gt slope of the tangent
+ //slope = (L*Cl/G)_min
+Cl=4.18;
+G_min=L*60*Cl/slope; //tangent gives minimum value of the gas flow rate
+G_actual=G_min*1.3; //since actual flow rate is 1.3 times the minimum
+slope2=L*Cl*60/G_actual; //slope of operating line
+Hg2_actual=slope2*(T1-T2)+Hg1; //actual humidityat pt 2
+Ggas=10000; //minimum gas rate in kg/hr*m^2
+Area1=G_actual/Ggas; //maximum area of the tower(based on gas)
+Gliq=12000; //minimum liquid rate in kg/hr*m^2
+Area2=60*L/Gliq; //maximum area of the tower(based on liquid)
+printf("\n \n the maximum area of the tower(based on gas) is :%f m^2",Area1);
+printf("\n the maximum area of the tower(based on liquid) is :%f m^2",Area2);
+dia=(Area1*4/3.14)^0.5; //diameter of the tower in m
+
+//let us assume the resistance to mass transfer lies basically in gas phase. hence the,interfacial conditions and the eqlb cond. are same.vertical line drawn between oper. and equl. line we get conditions of gas and equl. values are tabulated below as follows
+
+
+//table
+T=[20 30 40 50 55] //differnt temperature for different enthalpy calculation
+//enthaly
+H_bar=[101.79 133.0 166.49 210.0 278.72] //H_bar i.e. at equl.
+Hg=[71.09 103.00 140.00 173.00 211.09] //Hg i.e. of operating line
+i=1;
+while(i<6) //looping for different enthalpy calculation of operating line
+y(i)=1/(H_bar(i)-Hg(i));
+printf("\n the enhalpy at :%f is :%f",T(i),y(i));
+i=i+1;
+end //end of lop
+xset('window',1);
+plot(Hg,y,"o-");
+xtitle(" Fig.5.10(c) Example 10 (1/(Hf-Hg)) vs Hg","X-- Hg --->","Y-- 1/(Hf-Hg) ---->");
+
+//area under this curve gives Ntog =4.26
+Ntog=4.26; //no. of transfer unit
+Gs=10000; //gas flow rate
+Htog=Gs/Ky_a; // height of transfer unit
+height=Ntog*Htog; //height of the tower
+printf("\n \nthe tower height is :%f m",height);
+
+
+//make up water is based onevaporation loss(E),blow down loss(B),windage loss(W) M = E + B + W
+W=.2/100 *L*60; //windage loss(W)
+B=0; //blow down loss neglected
+E=G_actual*(.064-.016); //assuming air leaves fully saturated
+M = E + B + W; //make up water is based onevaporation loss(E),blow down loss(B),windage loss(W)
+printf("\n make up water is based onevaporation loss(E),blow down loss(B),windage loss(W) is :%f kg /hr",M);
+//end
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