Modified the code

4 files changed, 0 insertions, 81 deletions

diff --git a/806/CH8/EX8.2/82.sce b/806/CH8/EX8.2/82.sce
deleted file mode 100644
index 9b4886aa5..000000000
--- a/806/CH8/EX8.2/82.sce
+++ /dev/null
@@ -1,15 +0,0 @@
-clc

-pathname=get_absolute_file_path('82.sce')

-filename=pathname+filesep()+'82.sci'

-exec(filename)

-diary('C:\users\Bhavesh\desktop\scilab\82.txt')

-disp("A submarine moves through water at 30 ft/s.At a point A on the submarine 5 ft above the nose,the velocity of the submarine relative to the water is 50 ft/s.Determine the dynamic-pressure difference between this point and the nose,and determine the difference in total pressure between the two points.")

-disp("If the submarine is sationary and the water is movinf past it,The velocity at the nose is zero and the velocity at A is 50ft/s.By selecting the dynamic pressure at infinity as zero")

-E=q^2/2+g*hn

-p1=p*E//Pressure at nose

-p2=p*(E-q1^2/2)//Pressure at point A

-P1=p2-p1//Pressure difference at point A and nose

-P2=p*(g*hn-g*ha+(qn^2-q1^2)/2)

-disp("lb/ft^2",P1,"Hence dynamic pressure difference between point A and nose is=")

-disp("lb/ft^2",P2,"Hence total pressure difference between point A and nose is=")

-diary(0)

diff --git a/806/CH8/EX8.2/82.txt b/806/CH8/EX8.2/82.txt
deleted file mode 100644
index e4947d3bd..000000000
--- a/806/CH8/EX8.2/82.txt
+++ /dev/null
@@ -1,31 +0,0 @@
- 

- A submarine moves through water at 30 f 

-      t/s.At a point A on the submarine  

-      5 ft above the nose,the velocity o 

-      f the submarine relative to the wa 

-      ter is 50 ft/s.Determine the dynam 

-      ic-pressure difference between thi 

-      s point and the nose,and determine 

-       the difference in total pressure  

-      between the two points.            

- 

- If the submarine is sationary and the w 

-      ater is movinf past it,The velocit 

-      y at the nose is zero and the velo 

-      city at A is 50ft/s.By selecting t 

-      he dynamic pressure at infinity as 

-       zero                              

- 

- Hence dynamic pressure difference betwe 

-      en point A and nose is=            

- 

-  - 2418.75  

- 

- lb/ft^2   

- 

- Hence total pressure difference between 

-       point A and nose is=              

- 

-  - 2730.0334  

- 

- lb/ft^2   

diff --git a/806/CH8/EX8.3/83.sce b/806/CH8/EX8.3/83.sce
deleted file mode 100644
index 90666a4f9..000000000
--- a/806/CH8/EX8.3/83.sce
+++ /dev/null
@@ -1,13 +0,0 @@
-clc

-pathname=get_absolute_file_path('83.sce')

-filename=pathname+filesep()+'83.sci'

-exec(filename)

-diary('C:\users\Bhavesh\desktop\scilab\83.txt')

-disp("A source with strength 0.2m^3/s and a vortex with strength 1 m^2/s are located at the origin.Determine the equations for velocity potential and stream function.What are the velocity components at x=1 m,y=0.5 m?")

-disp("Solution:")

-r=sqrt(x^2+y^2)

-Vr=1/(10*%pi*r)

-Vt=1/(2*%pi*r)

-disp("m/s",Vr,"Vr=")

-disp("m/s",Vt,"Vt=")

-diary(0)

diff --git a/806/CH8/EX8.3/83.txt b/806/CH8/EX8.3/83.txt
deleted file mode 100644
index 44a79db57..000000000
--- a/806/CH8/EX8.3/83.txt
+++ /dev/null
@@ -1,22 +0,0 @@
- 

- A source with strength 0.2m^3/s and a v 

-      ortex with strength 1 m^2/s are lo 

-      cated at the origin.Determine the  

-      equations for velocity potential a 

-      nd stream function.What are the ve 

-      locity components at x=1 m,y=0.5 m 

-      ?                                  

- 

- Solution:   

- 

- Vr=   

- 

-    0.0284705  

- 

- m/s   

- 

- Vt=   

- 

-    0.1423525  

- 

- m/s