diff options
Diffstat (limited to '839/CH6/EX6.2/Example_6_2.sce')
-rwxr-xr-x | 839/CH6/EX6.2/Example_6_2.sce | 114 |
1 files changed, 57 insertions, 57 deletions
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]
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