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

---
 914/CH12/EX12.12/ex12_12.sce | 20 ++++++++++++++++++++
 1 file changed, 20 insertions(+)
 create mode 100755 914/CH12/EX12.12/ex12_12.sce

(limited to '914/CH12/EX12.12')

diff --git a/914/CH12/EX12.12/ex12_12.sce b/914/CH12/EX12.12/ex12_12.sce
new file mode 100755
index 000000000..32ec2ccae
--- /dev/null
+++ b/914/CH12/EX12.12/ex12_12.sce
@@ -0,0 +1,20 @@
+clc;
+warning("off");
+printf("\n\n example12.12 - pg 594");
+// given
+pp=1.13*10^4;  //[kg/m^3] - density of lead particle
+p=1.22;  //[kg/m^3] - density of air
+g=9.80;  //[m/sec^2] - acceleration due to gravity
+d=2*10^-3;  //[m] - diameter of particle
+mu=1.81*10^-5;  //[kg/m*sec] - viscosity of air
+// let us assume
+Cd=0.44;
+Ut=((4*d*g*(pp-p))/(3*p*Cd))^(1/2);
+disp(Ut)
+Nre=(Ut*d*p)/(mu);
+// from fig 12,16 value of Cd is
+Cd=0.4;
+Ut=((4*d*g*(pp-p))/(3*p*Cd))^(1/2);
+Nre=(Ut*d*p)/(mu);
+// Within the readibility of the chart Cd is unchanged and therefore the above obtained Cd is the final answer
+printf("\n\n The terminal velocity is \n Ut = %f m/sec",Ut);
-- 
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