From f35ea80659b6a49d1bb2ce1d7d002583f3f40947 Mon Sep 17 00:00:00 2001
From: prashantsinalkar
Date: Tue, 10 Oct 2017 12:38:01 +0530
Subject: updated the code

---
 839/CH6/EX6.2/Example_6_2.sce | 114 +++++++++++++++++++++---------------------
 1 file changed, 57 insertions(+), 57 deletions(-)

(limited to '839/CH6/EX6.2/Example_6_2.sce')

diff --git a/839/CH6/EX6.2/Example_6_2.sce b/839/CH6/EX6.2/Example_6_2.sce
index 4abb1d09a..ed2213d65 100755
--- a/839/CH6/EX6.2/Example_6_2.sce
+++ b/839/CH6/EX6.2/Example_6_2.sce
@@ -1,57 +1,57 @@
-//clear//
-clear;
-clc;
-
-//Example 6.2
-//Given
-Tr = 1000; //[R] 
-pr = 20; //[atm]
-Ma_a = 0.05;
-gama = 1.4;
-gc = 32.174; //[ft-lb/lbf-s^2]
-M = 29;
-R = 1545;
-//(a)
-//Using Eq.(6.45)
-A = 2*(1+((gama-1)/2)*Ma_a^2)/((gama+1)*Ma_a^2);
-fLmax_rh = (1/Ma_a^2-1-(gama+1)*log(A)/2)/gama  
-
-//(b)
-//Using Eq.(6.28), the pressure at the end of the isentropic nozzle pa
-A = (1+(gama-1)*(Ma_a^2)/2); 
-pa = pr/(A^(gama/(gama-1)))  // [atm]
-//From Example 6.1, the density of air at 20atm and 1000R is 0.795 lb/ft^3
-//Using Eq.(6.17), the acoustic velocity 
-Aa  = sqrt(gc*gama*Tr*R/M) //[m/s]
-//The velocity at the entrance of the pipe 
-ua  = Ma_a*Aa //[m/s]
-//When L_b = L_max, the gas leaves the pipe at the asterisk conditions, where 
-Ma_b = 1;
-// Using Eq.(6.43)
-A = (gama-1)/2;
-Tstar = Tr *(1+A*Ma_a^2)/(1+A*Ma_b^2) // [K]
-// Using Eq.(6.44)
-rho_star = 0.795*Ma_a/sqrt(2*(1+(gama-1)*Ma_a^2/2)/(2.4)) //[lb/ft^3]
-//Using Eq.(6.39)
-pstar = p0*Ma_a/sqrt(1.2) // [atm]
-//Mass velocity through the entire pipe
-G = 0.795*ua //[lb/ft^2-s]
-ustar = G/rho_star //[ft/s]
-
-//(c)
-//Using Eq.(6.45) with f_Lmax_rh = 400
-
-err = 1;
-eps = 10^-3;
-Ma_ac = rand(1,1);
-i =1;
-while((err>eps))  
-  A = 2*(1+((gama-1)/2)*Ma_ac^2)/((gama+1)*Ma_ac^2);
-  B = gama*400+1+(gama+1)*log(A)/2;
-  Ma_anew  = sqrt(1/B);
-  err = Ma_ac-Ma_anew;
-  Ma_ac = Ma_anew;
-end
-Ma_ac;
-uac  = Ma_ac*ua/Ma_a //[ft/s]
-Gc = uac*0.795 //[lb/ft^2-s]
+//clear//
+clear;
+clc;
+
+//Example 6.2
+//Given
+Tr = 1000; //[R] 
+pr = 20; //[atm]
+Ma_a = 0.05;
+gama = 1.4;
+gc = 32.174; //[ft-lb/lbf-s^2]
+M = 29;
+R = 1545;
+//(a)
+//Using Eq.(6.45)
+A = 2*(1+((gama-1)/2)*Ma_a^2)/((gama+1)*Ma_a^2);
+fLmax_rh = (1/Ma_a^2-1-(gama+1)*log(A)/2)/gama  
+
+//(b)
+//Using Eq.(6.28), the pressure at the end of the isentropic nozzle pa
+A = (1+(gama-1)*(Ma_a^2)/2); 
+pa = pr/(A^(gama/(gama-1)))  // [atm]
+//From Example 6.1, the density of air at 20atm and 1000R is 0.795 lb/ft^3
+//Using Eq.(6.17), the acoustic velocity 
+Aa  = sqrt(gc*gama*Tr*R/M) //[m/s]
+//The velocity at the entrance of the pipe 
+ua  = Ma_a*Aa //[m/s]
+//When L_b = L_max, the gas leaves the pipe at the asterisk conditions, where 
+Ma_b = 1;
+// Using Eq.(6.43)
+A = (gama-1)/2;
+Tstar = Tr *(1+A*Ma_a^2)/(1+A*Ma_b^2) // [K]
+// Using Eq.(6.44)
+rho_star = 0.795*Ma_a/sqrt(2*(1+(gama-1)*Ma_a^2/2)/(2.4)) //[lb/ft^3]
+//Using Eq.(6.39)
+pstar = pa*Ma_a/sqrt(1.2) // [atm]
+//Mass velocity through the entire pipe
+G = 0.795*ua //[lb/ft^2-s]
+ustar = G/rho_star //[ft/s]
+
+//(c)
+//Using Eq.(6.45) with f_Lmax_rh = 400
+
+err = 1;
+eps = 10^-3;
+Ma_ac = rand(1,1);
+i =1;
+while((err > eps))  
+  A = 2*(1+((gama-1)/2)*Ma_ac^2)/((gama+1)*Ma_ac^2);
+  B = gama*400+1+(gama+1)*log(A)/2;
+  Ma_anew  = sqrt(1/B);
+  err = Ma_ac-Ma_anew;
+  Ma_ac = Ma_anew;
+end
+Ma_ac;
+uac  = Ma_ac*ua/Ma_a //[ft/s]
+Gc = uac*0.795 //[lb/ft^2-s]
\ No newline at end of file
-- 
cgit