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+clc;
+clear;
+printf("\t\t\tChapter9_example7\n\n\n");
+// (a)  Determine the  UA product for the  exchanger.  (b) Calculate  the  exit temperatures for the exchanger, assuming that only the inlet temperatures are known
+// properties of engine oil at (190 + 158)/2 = 174°F = 176 degree  F from appendix table C4
+rou_1= 0.852*62.4; // density in lbm/ft^3 
+cp_1=0.509; // specific heat BTU/(lbm-degree Rankine) 
+v_1= 0.404e-3; // viscosity in ft^2/s 
+kf_1 = 0.08; // thermal conductivity in BTU/(hr.ft.degree Rankine) 
+a_1 = 2.98e-3; // diffusivity in ft^2/hr 
+Pr_1 = 490; // Prandtl Number 
+m_1=39.8; // mass flow rate in lbm/min
+// temperatures in degree F
+T1=190;
+T2=158;
+// properties of air at (126 + 166)/2 = 146°F = 606 degree R from appendix table D1
+rou_2= 0.0653; // density in lbm/ft^3 
+cp_2=0.241; // specific heat BTU/(lbm-degree Rankine) 
+v_2= 20.98e-5; // viscosity in ft^2/s 
+kf_2 = 0.01677 ; // thermal conductivity in BTU/(hr.ft.degree Rankine) 
+a_2 = 1.066; // diffusivity in ft^2/hr 
+Pr_2 = 0.706; // Prandtl Number 
+m_2=67; // mass flow rate in lbm/min
+// temperatures in degree F
+t1=126;
+t2=166;
+// Heat Balance
+q_air=m_2*cp_2*60*(t2-t1);
+q_oil=m_1*cp_1*60*(T1-T2);
+printf("\nThe heat gained by air is %.2e BTU/hr",q_air);
+printf("\nThe heat lost by oil is %.2e BTU/hr",q_oil);
+// for counterflow
+LMTD=((T1-t2)-(T2-t1))/(log((T1-t2)/(T2-t1)));
+printf("\nThe LMTD for counter flow configuration is %.1f degree F",LMTD);
+// Frontal Areas for Each Fluid Stream
+Area_air=(9.82*8)/144;
+Area_oil=(3.25*9.82)/144;
+printf("\nThe Core frontal area on the air side is %.3f sq.ft\nThe Core frontal area on the oil side is %.3f sq.ft ",Area_air,Area_oil);
+// Correction Factors (parameters calculated first)
+S=(t2-t1)/(T1-t1);
+R=(T1-T2)/(t2-t1);
+F=0.87; //value of correction factor from figure 9.21a corresponding to above calculated values of S and R
+// Overall Coefficient (q = U*A*F*LMTD)
+UA=q_air/(F*LMTD);
+printf("\nThe Overall Coefficient is %.2e BTU/(hr. degree R)",UA);
+// determining the capacitances
+mcp_air=m_2*cp_2*60;
+mcp_oil=m_1*cp_1*60;
+printf("\nThe capacitance value of air is %d BTU/(hr. degree R)",mcp_air);
+printf("\nThe capacitance value of engine oil is %d BTU/(hr. degree R)",mcp_oil);
+if mcp_air>mcp_oil then
+    mcp_max=mcp_air;
+    mcp_min=mcp_oil;
+    printf("\nEngine Oil has minimum capacitance");
+    else mcp_max=mcp_oil;
+    mcp_min=mcp_air;
+    printf("\nAir has minimum capacitance");
+end
+// determination of parameters for determining effectiveness
+mcp_min_max=mcp_min/mcp_max;
+NTU=(UA/mcp_min);
+printf("\nThe required parameters are mcp_min/mcp_max=%.3f and (UoAo/mcp_min)=%.2f",mcp_min_max,NTU);
+effectiveness=0.62; //value of effectiveness from figure 9.21b corresponding to the above calculated values of capacitance ratio and (UoAo/mcp_min):');
+t2_c=(T1-t1)*effectiveness+t1;
+T2_c=T1-(mcp_min_max)*(t2_c-t1);
+printf("\n\t\t\tSummary of Requested Information\n");
+printf("\n(a) UA = %.2e BTU/(hr. degree R)",UA);
+printf("\n(b) The Outlet temperatures (degree F)");
+printf("\n\tCalculated\tGiven in Problem Statement");
+printf("\nAir\t\t%d\t%d",t2_c,t2);
+printf("\nEngine Oil\t%d\t%d",T2_c,T2);
+