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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+clc;
+clear;
+vis=0.4;//Ns/(m^2)
+d=900;//kg/(m^3)
+D=0.02;//m
+Q=2.0*(10^-5);//(m^3)/s
+x1=0;
+x2=10;//m
+p1=200;//kPa
+x3=5;//m
+V=Q/(%pi*(D^2)/4);//m/s
+Re=d*V*D/vis;
+disp("Hence the flow is laminar.",Re,"a) Reynolds number =")
+pdiff=128*vis*(x2-x1)*Q/(%pi*(D^4)*1000);
+//for part b0 p1=p2; Q=%pi*(pdiff-(sw*l*sin(ang)))*(D^4)/(128*vis*l)
+ang=(asin(-128*vis*Q/(%pi*d*9.81*(D^4))))*180/%pi;
+//since sin(ang) doesn= not depend on pdiff, the the pressure is constant all along the pipe
+//hence for c)
+p3=p1;//kPa
+disp("kPa.",pdiff,"The pressure drop required if the pipe is horizontal=")
+disp("degrees.",ang,"b) The angle of the hill the pipe must be on if the oil is to flow at the same rate as a) but with (p1=p2) =")
+disp("kPa",p3,"c) For conditions of part b), the pressure at x3=5 m = ") \ No newline at end of file