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+// Example 8_4
+clc;funcprot(0);
+// Given data
+lambda=4;// W/mK
+// From example 6.4
+mu=6.7*10^-5;// Pa s
+V=18.85;// m/s
+h=1*10^-4;// m
+
+// Calculation
+// (a)
+q_w=-(mu)*((V^2)/h);// The heat flux to the wall (y =0) for the bearing in W/m^2
+// (b)
+deltaT=(mu/lambda)*((V^2)/(2*h));// The temperature difference T_h-T_o across the oil gap in K
+printf("\n(a)The heat flux to the wall (y =0) for the bearing,q_w=%1.3e W/m^2 \n(b)The temperature difference T_h-T_o across the oil gap is %2.2f K",q_w,deltaT);