updated the code

1 files changed, 60 insertions, 93 deletions

diff --git a/3136/CH6/EX6.9/Ex6_9.sce b/3136/CH6/EX6.9/Ex6_9.sce
index 7e8d593bd..1451dc707 100755
--- a/3136/CH6/EX6.9/Ex6_9.sce
+++ b/3136/CH6/EX6.9/Ex6_9.sce
@@ -1,93 +1,60 @@
-clear all; clc;

-

-disp("The inlet specific volume is calculated from the ideal gas equation")

-R=53.3

-T1=580

-p1=65

-nu1=(R*T1)/(144*p1)

-printf(" nu1=%0.3f ft^3/lbm=",nu1)

-

-disp("Q1=m*v1=4594 cfm")

-

-R=53.3

-T1=580

-//let y=(n/(n-1))

-y=2.625

-p2=250

-p1=65

-H_oa=R*T1*(y)*[(p2/p1)^(1/y)-1]

-printf(" The overall adiabetic head is calculated as H_o/a= %0.0f ft-lbf/lbm",H_oa)

-

-y=(0.75*1.4)/(1.4-1)

-printf("\n Where n/(n-1)=(Eta_p*k)/(k-1)=%0.3f",y)

-

-disp("From figure 6.7 , a centrifugal compressor with speed N=10000rpm is appropriate for the present application")

-

-disp("To use figure 6.7,the specific speed can be calculated from Ns=N*(V1^0.5)/(H_ad^0.75)")

-V1=4954/60

-printf(" Where V1= %0.1f cfs",V1)

-disp("H_ad=(H_o/a)/Eta_s is the head for each stage .")

-disp("Selecting the number of stages to be 2,4, and 6,the head for each stage and specific speed  can be calculated,then the total-to=total adiabetic efficiencies can be read ")

-disp("The required shaft horse power can be calculated with the volumetric and mechanical efficiencies assumed to be 0.98 and 0.95 respectively")

-disp("That is Ps=(m*(H_o/a)/(33000*Eta_ad*Eta_v*Eta_m)")

-

-H_oa=54417

-V1=82.6

-Eta_v=0.98

-Eta_m=0.95

-N=10000

-

-StageNo=[2 4 6];

-Eta_ad=[0.72 0.83 0.87];

-

-

-H_ad= zeros(1,length(StageNo));

-Ns = zeros(1,length(StageNo));

-Ps = zeros(1,length(StageNo));

-

-for i = 1: length(StageNo)

-   

-

-    H_ad(i) =H_oa/(StageNo(i));

-    Ns(i) =N*(V1^0.5)/(H_ad(i)^0.75);

-    Ps(i) = (m*H_oa)/(33000*Eta_ad(i)*Eta_v*Eta_m);

-end

-

-disp("StageNo   H_ad          Ns       Eta_ad      Ps")

-disp("       (ft-lbf/lbm)                         (hp)")

-table = [StageNo' H_ad' Ns' Eta_ad' Ps'];

-disp(table)

-

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+clear; clc;
+
+m = 100;
+disp("The inlet specific volume is calculated from the ideal gas equation")
+R=53.3
+T1=580
+p1=65
+nu1=(R*T1)/(144*p1)
+printf(" nu1=%0.3f ft^3/lbm=",nu1)
+
+disp("Q1=m*v1=4594 cfm")
+
+R=53.3
+T1=580
+//let y=(n/(n-1))
+y=2.625
+p2=250
+p1=65
+H_oa=R*T1*(y)*[(p2/p1)^(1/y)-1]
+printf(" The overall adiabetic head is calculated as H_o/a= %0.0f ft-lbf/lbm",H_oa)
+
+y=(0.75*1.4)/(1.4-1)
+printf("\n Where n/(n-1)=(Eta_p*k)/(k-1)=%0.3f",y)
+
+disp("From figure 6.7 , a centrifugal compressor with speed N=10000rpm is appropriate for the present application")
+
+disp("To use figure 6.7,the specific speed can be calculated from Ns=N*(V1^0.5)/(H_ad^0.75)")
+V1=4954/60
+printf(" Where V1= %0.1f cfs",V1)
+disp("H_ad=(H_o/a)/Eta_s is the head for each stage .")
+disp("Selecting the number of stages to be 2,4, and 6,the head for each stage and specific speed  can be calculated,then the total-to=total adiabetic efficiencies can be read ")
+disp("The required shaft horse power can be calculated with the volumetric and mechanical efficiencies assumed to be 0.98 and 0.95 respectively")
+disp("That is Ps=(m*(H_o/a)/(33000*Eta_ad*Eta_v*Eta_m)")
+
+H_oa=54417
+V1=82.6
+Eta_v=0.98
+Eta_m=0.95
+N=10000
+
+StageNo=[2 4 6];
+Eta_ad=[0.72 0.83 0.87];
+
+
+H_ad= zeros(1,length(StageNo));
+Ns = zeros(1,length(StageNo));
+Ps = zeros(1,length(StageNo));
+
+for i = 1: length(StageNo)
+   
+
+    H_ad(i) =H_oa/(StageNo(i));
+    Ns(i) =N*(V1^0.5)/(H_ad(i)^0.75);
+    Ps(i) = (m*H_oa)/(33000*Eta_ad(i)*Eta_v*Eta_m);
+end
+
+disp("StageNo   H_ad          Ns       Eta_ad      Ps")
+disp("       (ft-lbf/lbm)                         (hp)")
+table = [StageNo' H_ad' Ns' Eta_ad' Ps'];
+disp(table)
\ No newline at end of file