initial commit / add all books

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diff --git a/536/CH5/EX5.2/Example_5_2.sce b/536/CH5/EX5.2/Example_5_2.sce
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+clc;

+clear;

+

+printf("\n Example 5.2\n");

+

+M_p_d=0.2e-3; // Mean particle diameter

+printf("\n Given:\n Mean particle diameter = %.1f mm",M_p_d*1e3);

+f_r_w=0.5; //Flow rate of water

+printf("\n Flow rate of water = %.1f kg/s",f_r_w);

+id=25e-3; //Diameter of pipe

+printf("\n Diameter of pipe = %d mm",id*1e3);

+l=100; //length of pipe

+printf("\n length of pipe = %d m",l);

+t_vel=0.0239; //Terminal velocity of falling sand particles

+printf("\n Terminal velocity of falling sand particles = %.4f m/s",t_vel);

+//Assuming the mean velocity of the suspension is equal to the water velocity, that is, neglecting slip, then:

+Um=f_r_w/(1000*%pi*id^2/4);

+printf("\n\n Calculations:\n Mean velocity of suspension = %.2f m/s",Um);

+Re=id*Um*1000/0.001;

+printf("\n Reynolds no. of water alone = %d",Re);

+//Assuming e/d = 0.008, then, from Figure 3.7:

+phi=0.0046;

+f=0.0092;

+//From, equation 3.20, the head loss is:

+hf=4*phi*l*Um^2/(9.81*id);

+printf("\n Head loss = %.1f m water",hf);

+iw=hf/l;

+printf("\n Hydraulic gradient = %.3f m water/m",iw);

+i=300*1000/(1000*9.81*100);

+// Substituting in equation 5.20:

+C=(iw/(i-iw)*(1100*9.81*id*(2.6-1)*t_vel)/(Um^2*Um))^-1;

+printf("\n C = %.2f",C);

+//If G kg/s is the mass flow of sand, then:

+G=poly([0],'G');

+p=2600^-1*G-0.30*(2600^-1*G+.0005);

+printf("\n Mass flow of sand = %.2f kg/s",roots(p));

+printf("")
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