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/CH8/EX8.6/Example_8_6.sce | 41 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 41 insertions(+)
 create mode 100755 536/CH8/EX8.6/Example_8_6.sce

(limited to '536/CH8/EX8.6/Example_8_6.sce')

diff --git a/536/CH8/EX8.6/Example_8_6.sce b/536/CH8/EX8.6/Example_8_6.sce
new file mode 100755
index 000000000..00425619e
--- /dev/null
+++ b/536/CH8/EX8.6/Example_8_6.sce
@@ -0,0 +1,41 @@
+clc;
+clear;
+printf("\n Example 8.6\n");
+
+P1=101.3e3;
+Q_watr=0.01;
+printf("\n Given:\n Flow rate of Water = %.2f m^3/s",Q_watr);
+depth=100;
+printf("\n Depth of well = %d m",depth);
+d=100e-3;
+printf("\n Diameter of pipe = %d mm",d*1e3);
+depth_watr=40;
+printf("\n Level of water below water = %d m",depth_watr);
+Q_air=0.1;
+printf("\n Flow rate of Air = %.2f m^3/s",Q_air);
+P2=800e3;
+Gamma=1.4;
+//V1=Q_air;
+G_watr=Q_watr*1000;//Mass flow of water
+W=G_watr*depth_watr*9.81;
+//The energy needed to compress 0. 1 m^3/s of air is given by:
+E=P1*Q_air*(1.4/0.4)*((P2/P1)^(0.4/1.4)-1);//     equation 8.37
+printf("\n\n Calculations:\n The power required for this compression is = %d W",E);
+effi=W/E*100;
+printf("\n Efficiency = %.1f per cent",effi);
+//the mean pressure
+P=345e3;
+printf("\n The mean pressure = %d kN/m^2",P);
+v1=8314*273/(29*P);
+printf("\n The specific volume v of air at 273 K and given pressure is = %.3f m^3/kg",v1);
+v2=8314*273/(29*P1);
+printf("\n The specific volume v of air at 273 K and 101.3 kN/m^2 is = %.3f m^3/kg",v2);
+G_air=Q_air/v2; //mass flowrate of the air is:
+Q_mean=G_air*v1;//Mean volumetric flowrate of air
+Q_tot=Q_watr+Q_mean;//Total volumetric flowrate
+A=%pi/4*d^2;//Area of pipe
+v_mean=Q_tot/A;
+printf("\n Mean velocity of the mixture = %.2f m/s",v_mean);
+
+
+
-- 
cgit