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diff --git a/534/CH11/EX11.1/11_1_Counterflow_tube_HeatX.sce b/534/CH11/EX11.1/11_1_Counterflow_tube_HeatX.sce
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+++ b/534/CH11/EX11.1/11_1_Counterflow_tube_HeatX.sce
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+clear;
+clc;
+printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 11.1   Page 680 \n'); //Example 11.1
+// Tube Length to achieve a desired hot fluid temperature
+
+//Operating Conditions
+Tho = 60+273    ;//[K] Hot Fluid outlet Temperature
+Thi = 100+273   ; //[K] Hot Fluid intlet Temperature
+Tci = 30+273    ;//[K] Cold Fluid intlet Temperature
+mh = .1          ;//[kg/s] Hot Fluid flow rate
+mc = .2          ;//[kg/s] Cold Fluid flow rate
+Do = .045        ;//[m] Outer annulus
+Di = .025        ;//[m] Inner tube
+
+//Table A.5 Engine Oil Properties T = 353 K
+cph = 2131              ;//[J/kg.K] Specific Heat
+kh = .138               ; //[W/m.K] Conductivity
+uh = 3.25*10^-2           ; //[N.s/m^2] Viscosity
+//Table A.6 Saturated water Liquid Properties Tc = 308 K
+cpc = 4178              ;//[J/kg.K] Specific Heat
+kc = 0.625               ; //[W/m.K] Conductivity
+uc = 725*10^-6           ; //[N.s/m^2] Viscosity
+Pr = 4.85                ;//Prandtl Number
+
+q = mh*cph*(Thi-Tho);
+
+Tco = q/(mc*cpc)+Tci;
+
+T1 = Thi-Tco;
+T2 = Tho-Tci;
+Tlm = (T1-T2)/(2.30*log10(T1/T2));
+
+//Through Tube
+Ret = 4*mc/(%pi*Di*uc);
+printf("\n Flow through Tube has Reynolds Number as %i. Thus the flow is Turbulent", Ret);
+//Equation 8.60
+Nut = .023*Ret^.8*Pr^.4;
+hi = Nut*kc/Di;
+
+//Through Shell
+Reo = 4*mh*(Do-Di)/(%pi*uh*(Do^2-Di^2));
+printf("\n Flow through Tube has Reynolds Number as %i. Thus the flow is Laminar", Reo);
+//Table 8.2
+Nuo = 5.63;
+ho = Nuo*kh/(Do-Di);
+
+U = 1/[1/hi+1/ho];
+L = q/(U*%pi*Di*Tlm);
+
+printf("\n Tube Length to achieve a desired hot fluid temperature is %.1f m",L);
+//END
+\ No newline at end of file