From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 536/CH5/EX5.2/Example_5_2.sce | 37 +++++++++++++++++++++++++++++++++++++
 1 file changed, 37 insertions(+)
 create mode 100755 536/CH5/EX5.2/Example_5_2.sce

(limited to '536/CH5/EX5.2')

diff --git a/536/CH5/EX5.2/Example_5_2.sce b/536/CH5/EX5.2/Example_5_2.sce
new file mode 100755
index 000000000..1a8c072ba
--- /dev/null
+++ b/536/CH5/EX5.2/Example_5_2.sce
@@ -0,0 +1,37 @@
+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("")
\ No newline at end of file
-- 
cgit