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

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 839/CH6/EX6.1/Example_6_1.sce | 61 +++++++++++++++++++++++++++++++++++++++++++
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diff --git a/839/CH6/EX6.1/Example_6_1.sce b/839/CH6/EX6.1/Example_6_1.sce
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+//clear//
+clear;
+clc;
+
+//Example 6.1
+//Given
+gama = 1.4;
+M = 29;
+R = 82.0568*10^-3; //[atm-m^3/Kg mol-K]
+Nma = 0.8;
+gc = 1; //[ft-lb/lbf-s^2]
+//At Entrance 
+p0 = 20; //[atm]
+T0 = 555.6; //[K]
+
+//(a)
+// Using Eq.(6.28)
+//Pressure at throat
+pt  = (1/(1+((gama-1)/2)*Nma^2)^(1/(1-1/gama)))*p0 //[atm]
+//From Eq.(6.10)
+rho0 = (p0*M)/(R*T0); //[kg/m^3]
+// Using Eq.(6.10) and Eq.(6.26), the velocity in the throat
+ut = sqrt((2*gama*gc*R*T0)/(M*(gama-1))*(1-(pt/p0)^(1-1/gama))); // [m^3-am/kg]^0.5
+//In terms of [m/s], Using Appendix 2, 1 atm = 1.01325*10^ [N/m^2]
+ut = ut*sqrt(1.01325*10^5) //[m/s]
+//Using Eq.(6.23), density at throat 
+rho_t  = rho0*(pt/p0)^(1/gama) //[kg/m^3]
+//The mass velocity at the throat,
+Gt = ut*rho_t //[kg/m^2-s]
+//Using Eq.(6.24), The temperature at throat 
+Tt = T0*(pt/p0)^(1-1/gama) // [K]
+
+//(b)
+// From Eq.(6.29)
+pstar = ((2/(gama+1))^(1/(1-1/gama)))*p0 //[atm]
+//From Eq.(6.24) and (6.29)
+Tstar = T0*(pstar/p0)^(1-1/gama) //[K]
+//From Eq.(6.23)
+rho_star = rho0*(pstar/p0)^(1/gama) //[Kg/m^3]
+//From Eq.(6.30)
+G_star = sqrt(2*gama*gc*rho0*p0*101.325*10^3/(gama-1))*(pstar/p0)^(1/gama)*sqrt(1-(pstar/p0)^(1-1/gama)) //[Kg-m^2/s]
+u_star = G_star/rho_star //[m/s]
+
+//(c) 
+// By continuity, G inversely proportional to S, the mass velocity at dischage is
+G_r  = G_star/2 // [Kg/m^3-s]
+//Using Eq.(6.30)
+// Let x = pr/p0
+err = 1;
+eps = 10^-3;
+x = rand(1,1);
+
+while(err>eps)  
+  xnew  = ((0.1294)/sqrt(1-x^(1-1/1.4)))^1.4;
+  err = x-xnew;
+  x=xnew;
+end
+
+//Using Eq.(6.27)
+//The Mach Number at dischage is
+Nmr = sqrt((2/(gama-1))*(1/x^(1-1/gama)-1))
-- 
cgit