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+clear;

+clc;

+

+// Illustration 7.15

+// Page: 267

+

+printf('Illustration 7.15 - Page: 267\n\n');

+

+// solution

+

+//***Data***//

+w = 0.75;// [m]

+OD = 19.05/1000;// [m]

+l = 3.75;// [m]

+n = 20;

+t = 1.65/1000;// [m]

+Ws = 2.3;// [kg/s]

+Wal = 10;// [kg/s]

+Wt = 4;// [kg/s]

+Density = 800;// [kg/cubic m]

+viscocity = 0.005;// [kg/m.s]

+K = 0.1436;// [W/m.K]

+Ct = 2010;// [J/kg.K]

+Cal = 4187;// [J/kg.K]

+Y1_prime = 0.01;// [kg H2O/kg dry air]

+Y2_prime = 0.06;// [kg H2O/kg dry air]

+TempT = 95;// [OC]

+//*****//

+

+Free_area = (w-(n*OD))*l;// [square m]

+Gs_min = 2.3/Free_area;// [kg/square m.s]

+Yav_prime = (Y1_prime+Y2_prime)/2;// [kg H2O/kg dry air]

+// From Eqn. 7.86:

+ky = 0.0493*(Gs_min*(1+Yav_prime))^0.905;// [kg/square m.s.delta_Y_prime]

+// From Fig. 7.5:

+H1_prime = 56000;// [J/kg]

+Ao = 400*%pi*OD*l;// [square m]

+// Cooling water is distributed over 40 tubes & since tubes are staggered

+geta = Wal/(40*2*l);// [kg/m.s]

+geta_by_OD = geta/OD;// [kg/square m.s]

+// Assume:

+TempL = 28;// [OC]

+// From Eqn. 7.84:

+hL_prime = (982+(15.58*TempL))*(geta_by_OD^(1/3));// [W/square m.K]

+// From Eqn. 7.85:

+hL_dprime = 11360;// [W/square m.K]

+// From Fig. 7.5 (Pg 232)

+m = 5000;// [J/kg.K]

+Ky = 1/((1/ky)+(m/hL_dprime));

+ID = (OD-(2*t));// [m]

+Ai = %pi*(ID^2)/4;// [square m]

+Gt_prime = Wt/(n*Ai);// [kg/square m.s]

+Re = ID*Gt_prime/viscocity;

+Pr = Ct*viscocity/K;

+// From a standard correlation:

+hT = 364;// [W/square m.K]

+Dav = (ID+OD)/2;// [m]

+Zm = (OD-ID)/2;// [m]

+Km = 112.5;// [W/m.K]

+// From Eqn. 7.67:

+Uo = 1/((OD/(ID*hT))+((OD/Dav)*(Zm/Km))+(1/hL_prime));// [W/square m.K]

+// From Eqn. 7.75:

+alpha1 = -(((Uo*Ao)/(Wt*Ct))+((Uo*Ao)/(Wal*Cal)));

+alpha2 = m*Uo*Ao/(Wt*Ct);

+// From Eqn. 7.76:

+beeta1 = Ky*Ao/(Wal*Cal);

+beeta2 = -((m*Ky*Ao/(Wal*Cal))-(Ky*Ao/Ws));

+y = deff('[y] = f26(r)','y = (r^2)+((alpha1+beeta2)*r)+((alpha1*beeta2)-(alpha2*beeta1))');

+r1 = fsolve(10,f26);

+r2 = fsolve(0,f26);

+beeta2 = 1.402;

+// From Eqn. 7.83:

+// N1-(M1*(r1+alpha1)/beeta1) = 0............................................(1)

+// N2-(M2*(r2+alpha2)/beeta2) = 0............................................(2)

+// From Eqn. 7.77:

+// At the top:

+x1 = 1;

+// TempL2+(M1*exp(r1*x1))+(M2*exp(-(r2*x1))) = TempL.........................(3)

+// From Eqn. 7.78:

+// At the bottom:

+x2 = 0;

+// H1_star-N1-N2 = H1_prime..................................................(4)

+// From Eqn. 7.80:

+// ((M1/r1)*(exp(r1)-1))+((M2*r2)*(exp(r2)-1)) = (Tempt-TempL)...............(5)

+// From Eqn. 7.81:

+// ((N1/r1)*(exp(r1)-1))+((N2*r2)*(exp(r2)-1)) = (H1_star-H1_prime)..........(6)

+// From Eqn. 7.91 & Eqn. 7.92:

+// Uo*Ao*(TempT-TempL)=Ky*Ao*(H1_star-H1_prime)..............................(7)

+

+// Elimination of M's & N's by solving Eqn. (1) to (4) and (7) simultaneously:

+// and from Fig. 7.5 (Pg 232):

+TempL1=28;// [OC]

+H1_star=(Uo*Ao*(TempT-TempL)/(Ky*Ao))+H1_prime;// [J/kmol]

+// Solving (1) to (4) simultaneously:

+a = [1 -(r1+alpha1)/beeta1 0 0;0 0 1 -(r2+alpha1)/beeta1;0 exp(r1*x1) 0 exp(r2*x1);1 0 1 0];

+b = [0;0;TempT-TempL1;H1_star-H1_prime];

+soln = a\b;

+N1 = soln(1);

+M1 = soln(2);

+N2 = soln(3);

+M2 = soln(4);

+// By Eqn. 5

+delta_Temp = ((M1/r1)*(exp(r1)-1))+((M2*r2)*(exp(r2)-1));// [OC]

+Q = Uo*delta_Temp*Ao;

+TempT1 = TempT-(Q/(Wt*Ct));// [OC]

+H2_prime = Q/(Ws)+H1_prime;// [J/kg]

+printf("Temparature to which oil was cooled: %f OC\n",TempT1);
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