1 files changed, 13 insertions, 15 deletions
diff --git a/3819/CH3/EX3.19/Ex3_19.sce b/3819/CH3/EX3.19/Ex3_19.sce
index 790f78ed9..1fd7b5dd8 100644
--- a/3819/CH3/EX3.19/Ex3_19.sce
+++ b/3819/CH3/EX3.19/Ex3_19.sce
@@ -1,24 +1,22 @@
 // A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal
 // Chapter 3-Hydrostatic Forces on surfaces
 // Problem 3.19
-
 //Data given in the Problem
-theta=60
-AC=h/sin(theta*%pi/180)
-s=sin(theta/180*%pi)
-h=poly(0,"h")
-H=h/2
-b=1
-d=AC
-A=AC
-IG=b*d^3/12
+theta=60;
+s=sin(theta/180*%pi);
+hh=poly(0,"hh");
+AC=hh/s;
+H=hh/2;
+b=1;
+d=AC;
+IG=b*d^3/12;
 //COP=(IG/(A*H)+H)
 //COP=(h/sin(theta/180*%pi)^3/12/(h/sin(theta*%pi/180)/(h/2)+h/2
 //We know that COP is equal to (h-3),THAT IS ,the depth of centre of pressure
 //hence
-function f=F(h)
-f=((h/s)^3/12*s^2/(h/s*h/2))+(h/2)-(h-3);
+function f=F(hh)
+f=((hh/s)^3/12*s^2/(hh/s*hh/2))+(hh/2)-(hh-3);
 endfunction
-h=100;
-y=fsolve(h,F)
-mprintf("The height of wahter for tipping the gate is %f m",y)
+hh=100;
+y=fsolve(hh,F);
+mprintf("The height of water for tipping the gate is %f m",y)