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/CH7/EX7.6/ex7_6.sce | 41 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 41 insertions(+)
 create mode 100755 914/CH7/EX7.6/ex7_6.sce

(limited to '914/CH7/EX7.6/ex7_6.sce')

diff --git a/914/CH7/EX7.6/ex7_6.sce b/914/CH7/EX7.6/ex7_6.sce
new file mode 100755
index 000000000..cd6b07980
--- /dev/null
+++ b/914/CH7/EX7.6/ex7_6.sce
@@ -0,0 +1,41 @@
+clc;
+warning("off");
+printf("\n\n example7.6 - pg283");
+// given
+id=6;  //[inch] - inlet diameter
+od=4;  //[inch] - outlet diameter
+Q=10;  //[ft^3/sec] - water flow rate
+alpha2=%pi/3;  //[radians] - angle of reduction of elbow
+alpha1=0;
+p1=100;  //[psi] - absolute inlet pressure
+p2=29;  //[psi] - absolute outlet pressure
+patm=14.7;  //[psi] - atmospheric pressure
+p1gauge=p1-patm;
+p2gauge=p2-patm;
+S1=(%pi*((id/12)^2))/4;
+S2=(%pi*((od/12)^2))/4;
+U1=Q/S1;
+U2=Q/S2;
+p=62.4;  //[lb/ft^3]
+b=1;
+w1=p*Q;  //[lb/sec] - mass flow rate
+w2=w1;
+gc=32.174;
+// using the equation Fpress=p1gauge*S1-p2gauge*S2*cos(alpha2);
+Fpressx=p1gauge*144*S1-p2gauge*144*S2*cos(alpha2);
+Fpressy=p1gauge*144*S1*sin(alpha1)-p2gauge*144*S2*sin(alpha2);
+wdeltaUx=(w1/gc)*((U2)*(cos(alpha2))-(U1)*(cos(alpha1)));
+wdeltaUy=(w1/gc)*((U2)*(sin(alpha2))-(U1)*(sin(alpha1)));
+Fextx=wdeltaUx-Fpressx;
+Fexty=wdeltaUy-Fpressy;
+Fext=((Fextx)^2+(Fexty)^2)^(1/2);
+alpha=180+(atan(Fexty/Fextx))*(180/%pi);
+printf("\n\n The force has a magnitude of %flb and a direction of %f from the positive x direction(in the second quadrant",Fext,alpha);
+printf("\n\n Also there is a force on the elbow in the z direction owing to the weight of the elbow plus the weight of the fluid inside");
+
+
+
+
+
+
+
-- 
cgit