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+clc;
+warning("off");
+printf("\n\n example10.8 - pg439");
+// given
+mu=6.72*10^-4;  //[lb/ft*sec] - viscosity
+p=62.4;  //[lb/ft^3] - density
+S=0.03322;  //[ft^2] - flow area
+d=0.206;  //[ft]
+e=1.5*10^-4;  // absolute roughness for steel pipe
+ebyd=e/d;
+Nre=10^5;
+// friction factor as read from fig in book for the given reynolds no. and relative roughness is-
+f=0.0053;
+U=(Nre*mu)/(p*d);
+Q=U*S;
+gc=32.174;
+// (a) equivalent length method
+deltapbyL=f*(4/d)*(p*(U^2))*(1/(2*gc))*(6.93*10^-3);
+// using L=Lpipe+Lfittings+Lloss;
+Lfittings=2342.1*d;
+kc=0.50;  //  due to contraction loss
+ke=1;  // due to enlargement loss
+Lloss=(kc+ke)*(1/(4*f))*d;
+Lpipe=137;
+L=Lpipe+Lfittings+Lloss;
+deltap=deltapbyL*L;
+patm=14.696;  //[psi] - atmospheric pressure
+p1=patm+deltap;
+printf("\n\n (a)The inlet pressure is\n p1=%f psi",p1);
+// (b) loss coefficient method
+// using the equation deltap/p=-(Fpipe+Ffittings+Floss)
+L=137;
+kfittings=52.39;
+sigmaF=((4*f*(L/d))+kc+ke+kfittings)*((U^2)/(2*gc));
+deltap=(p*sigmaF)/(144);
+p1=patm+deltap;
+printf("\n\n (b)The inlet pressure is \n p1=%f psi",p1);
+printf("\n\n Computation of the pressure drop by the loss coefficient method differs from the equivalent length method by less than 1 psi");
+
+
+