From 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:27:19 +0530
Subject: initial commit / add all books

---
 3511/CH9/EX9.6/Ex9_6.sce | 29 +++++++++++++++++++++++++++++
 1 file changed, 29 insertions(+)
 create mode 100644 3511/CH9/EX9.6/Ex9_6.sce

(limited to '3511/CH9/EX9.6/Ex9_6.sce')

diff --git a/3511/CH9/EX9.6/Ex9_6.sce b/3511/CH9/EX9.6/Ex9_6.sce
new file mode 100644
index 000000000..dfbb9325d
--- /dev/null
+++ b/3511/CH9/EX9.6/Ex9_6.sce
@@ -0,0 +1,29 @@
+clc;
+T1=290; // Temperature at inlet in kelvin
+n=10; // Number of stages
+rp=6.5; // Pressure ratio
+m=3; // mass flow rate in kg/s
+eff_C=0.9; // isentropic efficiency of the compression
+ca=110; // Axial velocity in m/s
+u=180; // Mean blade velocity in m/s
+Cp=1.005; // Specific heat in kJ/kg K
+r=1.4; // Specific heat ratio
+R=287; // Characteristic gas constant in J/kg K
+
+T_2=(rp)^((r-1)/r)*T1; // temperature after isentropic compression
+T2=((T_2-T1)/eff_C)+T1; // Temperature after actual compression
+P=m*Cp*(T2-T1); // Power given to the air
+Del_Tstage=(T2-T1)/n; // Temperature rise per stage
+Del_ct=Cp*10^3*Del_Tstage/u; // For work done per kg of air per second
+// To find blade angles let solve the following equations
+// Del_ct=ca(tan beta_1-tan beta_2) for symmetrical stages
+// u=ca(tan beta_1=tan beta_2) for degree of reaction = 0.5
+// Solving by matrix method
+A=[1,-1;1,1]; C=[Del_ct/ca;u/ca];
+B=A\C;
+// Blade angles at entry and exit
+beta_1=atand(B(1));
+beta_2=atand(B(2));
+
+disp ("kW   (roundoff error)",P,"Power given to the air = ");
+disp ("degree",beta_2,"Blade angle at exit = ","degree",beta_1,"Blade angle at inlet = ");
-- 
cgit