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diff --git a/530/CH5/EX5.7.iii/example_5_7iii.sce b/530/CH5/EX5.7.iii/example_5_7iii.sce
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+clear;
+clc;
+
+// A Textbook on HEAT TRANSFER by S P SUKHATME
+// Chapter 5
+// Heat Transfer by Forced Convection
+
+
+// Example 5.7(iii)
+// Page 241
+printf("Example 5.7(iii), Page 241 \n \n");
+
+D = 0.0125 ; // [m]
+ST = 1.5*D ;
+SL = 1.5*D ;
+V_inf = 2 ; // [m/s]
+N = 5;
+Tw = 70; // [degree C]
+Tmi = 30; // [degree C]
+L = 1; // [m]
+
+rho = 1.165 ; // [kg/m^3]
+v = 16.00 *10^-6 ; // [m^2/s]
+Cp = 1.005*1000 ; // [J/kg K]
+k = 0.0267 ; // [W/m K]
+Pr = 0.701;
+
+// From eqn 5.10.2
+Vmax = ST/(SL-D)*V_inf ; // [m/s]
+Re = Vmax*D/v ;
+
+// From fig 5.15
+f = 0.37/4;
+// Also, tube arrangement is square
+X = 1;
+// From eqn 5.10.6
+delta_P = 4*f*N*X*(rho*Vmax^2)/2 ; // [N/m^2]
+
+// At 70 degree C
+Pr1 = 0.694 ; 
+// From table 5.4 and 5.5
+
+C1 = 0.27;
+m = 0.63;
+C2 = 0.93;
+
+// Substituting in Eqn 5.10.5
+Nu = C1*C2*(Re^m)*(Pr^0.36)*(Pr/Pr1)^(1/4);
+h = Nu*k/D; // [W/m^2 K]
+
+// For 1 m long tube
+m_dot = rho*(10*1.5*D*1)*2; // [kg/s]
+
+// Substituting m_dot in 5.3.4 and solving, we get
+function[f]=temp(Tmo)
+    f(1) = h*(%pi*D*L)*50*[(Tw-Tmi)-(Tw-Tmo(1))]/[log((Tw-Tmi)/(Tw-Tmo(1)))]-m_dot*Cp*(Tmo(1)-Tmi) ;
+    // h*(%pi*D*L)*N*((Tw-Tmi)-(Tw-Tmo))/log[(Tw-Tmi)/(Tw-Tmo)] - m_dot*Cp*(Tmo - Tmi);
+    funcprot(0);
+endfunction
+
+Tmo = 40; // Initial assumed value for fsolve function
+y = fsolve(Tmo,temp);
+
+// Heat transfer rate q
+q = h*(%pi*D*L)*50*((Tw-Tmi)-(Tw-y))/(log((Tw-Tmi)/(Tw-y)));
+
+printf("(iii) Heat transfer rate per unit length to air = %f W",q);
+\ No newline at end of file