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-rw-r--r--629/CH6/EX6.9/ex6_9.txt5
-rw-r--r--629/CH6/EX6.9/example6_9.sce25
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diff --git a/629/CH6/EX6.9/ex6_9.txt b/629/CH6/EX6.9/ex6_9.txt
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+
+ The flow rate under the sluice gate = 2008 ft^3/s.
+
+ The force on the sluice gate = 66.6 tons.
+ \ No newline at end of file
diff --git a/629/CH6/EX6.9/example6_9.sce b/629/CH6/EX6.9/example6_9.sce
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+clear
+clc
+//Example 6.9 FORCE ON A SLUICE GATE
+g=32.2; //[ft/s^2]
+d1=20; //[ft]
+d2=3; //[ft]
+w=20; //gate width[ft]
+v2=sqrt((2*g*(d1-d2))/(1-(d2/d1)^2)) //[ft/s]
+v1=d2*v2/d1 //[ft/s]
+rho=1.94; //[slugs/ft^3]
+Q=v2*d2*w //discharge[ft^3/s]
+printf("\n The flow rate under the sluice gate = %.f ft^3/s.\n",Q)
+m=rho*Q //mass flow rate[slugs/s]
+Gamma=62.4; //[lbf/ft^3]
+F1=Gamma*w*(d1^2)/2
+F2=Gamma*w*(d2^2)/2
+//momentum inflow
+mi=m*v1 //[lbf]
+//momentum outflow
+mo=m*v2 //[lbf]
+Fx=mo-mi //[lbf]
+//Sum of forces in x-direction, Fx=F1-F2-Fg
+//1ton=2000 lbf
+Fg=(F1-F2-Fx)/2000 //tons
+printf("\n The force on the sluice gate = %.1f tons.\n",Fg) \ No newline at end of file