1 files changed, 30 insertions, 0 deletions
diff --git a/914/CH7/EX7.10/ex7_10.sce b/914/CH7/EX7.10/ex7_10.sce
new file mode 100755
index 000000000..eb0a0a1a5
--- /dev/null
+++ b/914/CH7/EX7.10/ex7_10.sce
@@ -0,0 +1,30 @@
+clc;
+warning("off");
+printf("\n\n example7.10 - pg298");
+// given
+d=0.03;  //[m] - diameter
+g=9.784;  //[m/sec] - acceleration due to gravity
+deltaz=-1;
+// using the equation (1/2)*(U3^2/alpha3-U1^2/alpha1)+g*deltaz=0
+// assuming
+alpha1=1;
+alpha3=1;
+// also since the diameter of the tank far exceeds the diameter of the hose , the velocity at point 1 must be negligible when compared to the velocity at point 3
+U1=0;
+U3=(-2*g*deltaz+(U1^2)/alpha1)^(1/2);
+p=1000;  //[kg/m^3] - density of water
+S3=(%pi/4)*(d)^2
+w=p*U3*S3;
+printf("\n\n the mass flow rate is \n w=%fkg/sec",w);
+// the minimum pressure in the siphon tube is at the point 2. Before the result of 3.13 kg/sec is accepted as the final value, the pressure at point 2 must be calcilated in order to see if the water might boil at this point
+// using deltap=p*((U3^2)/2+g*deltaz)
+deltap=p*((U3^2)/2+g*deltaz);
+p1=1.01325*10^5;  //[N/m^2] - is equal to atmospheric pressure
+p2=p1+deltap;
+vp=0.02336*10^5;
+if p2>vp then
+    printf("\n\n the siphon can operate since the pressure at point 2 is greater than the value at which the liquid boils");
+else
+    printf("\n\n the siphon cant operate since the pressuer at point 2 is less than the value at which the liquid boils");
+    
+end
+\ No newline at end of file