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+

+

+// Display mode

+mode(0);

+

+// Display warning for floating point exception

+ieee(1);

+

+clc;

+disp("Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 4 Example # 4.4 ")

+

+// Length of the crankcase in m is given as

+L = 0.6;

+// Width of the crankcase in m is given as

+b = 0.2;

+// Depth of the crankcase in m is given as

+d = 0.1;

+// Surface temperature in K is given as

+Ts = 350;

+// Air temperature in K is given as

+Tinfinity = 276;

+// Air velocity in m/sec is given as

+Uinfinity = 30;

+// It is stated that  boundary layer is turbulent over the entire surface

+

+//Average air temperature in degree K is

+T = (Ts+Tinfinity)/2;

+// At this average temperature, we get the following for air

+rho = 1.092;//density in kg/m^3

+mu = 0.000019123;//viscosity in SI units

+Pr = 0.71;//Prandtl number

+k = 0.0265;//Thermal conductivity in W/m-K

+

+// Reynold''s number is therefore given as

+ReL = ((rho*Uinfinity)*L)/mu;

+

+//From eq. 4.82, average nusselt number could be given as

+Nu = (0.036*(Pr^(1/3)))*(ReL^0.8);

+

+//We can write from the basic expression, Nu=hc*L/k, that

+//Heat transfer coefficient in W/m^2-K

+hc = (Nu*k)/L;

+

+// The surface area that dissipates heat is 0.28 m2

+disp("Total heat loss from the surface in W is therefore")

+//Heat loss from the surface in W

+q = (hc*0.28)*(Ts-Tinfinity)