diff options
Diffstat (limited to '806/CH4')
| -rw-r--r-- | 806/CH4/EX4.5/45.sce | 11 | ||||
| -rw-r--r-- | 806/CH4/EX4.5/45.txt | 15 | ||||
| -rw-r--r-- | 806/CH4/EX4.6/46.sce | 16 | ||||
| -rw-r--r-- | 806/CH4/EX4.6/46.txt | 18 | ||||
| -rw-r--r-- | 806/CH4/EX4.7/47.sce | 19 | ||||
| -rw-r--r-- | 806/CH4/EX4.7/47.txt | 34 |
6 files changed, 113 insertions, 0 deletions
diff --git a/806/CH4/EX4.5/45.sce b/806/CH4/EX4.5/45.sce new file mode 100644 index 000000000..e76367382 --- /dev/null +++ b/806/CH4/EX4.5/45.sce @@ -0,0 +1,11 @@ +clc
+pathname=get_absolute_file_path('45.sce')
+filename=pathname+filesep()+'45.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\45.txt')
+disp("Water is flowing through an open channel at a depth of 2 m and a velocity of 3m/s.it then flows through a contracting chute into another channel where the depth is 1 m and the velocity is 10m/s.assuming frictionless flow,determine the difference in the elevation of the channel floors.")
+disp("Solution:")
+disp("The points 1 and 2 can be selected on the free surface and therefore p1=p2=0")
+disp("Therefore by bernoullis equation the difference in the elevation of the channel is y=")
+y=(v2^2)/(2*p)+h2-h1-(v1^2)/(2*p)
+disp("m",y)
\ No newline at end of file diff --git a/806/CH4/EX4.5/45.txt b/806/CH4/EX4.5/45.txt new file mode 100644 index 000000000..9717574b8 --- /dev/null +++ b/806/CH4/EX4.5/45.txt @@ -0,0 +1,15 @@ +
+ Water is flowing through an open channel at a depth of 2 m and a velocity of 3m/s.it then flows through a contracting chute into another channel where the depth i
+ s 1 m and the velocity is 10m/s.assuming frictionless flow,determine the difference in the elevation of the channel floors.
+
+ Solution:
+
+ The points 1 and 2 can be selected on the free surface and therefore p1=p2=0
+
+ Therefore by bernoullis equation the difference in the elevation of the channel is y=
+
+ 3.6381244
+
+ m
+
+-->exec('SCI/etc/scilab.quit','errcatch',-1);quit;
diff --git a/806/CH4/EX4.6/46.sce b/806/CH4/EX4.6/46.sce new file mode 100644 index 000000000..fc58bb642 --- /dev/null +++ b/806/CH4/EX4.6/46.sce @@ -0,0 +1,16 @@ +clc
+pathname=get_absolute_file_path('46.sce')
+filename=pathname+filesep()+'46.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\46.txt')
+disp("A venturimeter,consisting of a converging portion followed by a throat of constant diameter and then gradually diverging portion,is used to detremine the flow of the pipe.The diameter at section 1 is 6 in. and that of section 2 is 4 in.,Find the discharge through the pipe when p1-p2=3psi and oil of sp grvty 0.9 is flowing")
+disp("Solution:")
+disp("From the continuity equation Q=A1v1=A2v2")
+disp("Therefore,v1=Q*16/pi , v2=Q*36/pi")
+disp("And moreover Z1=Z2")
+disp("Therefore discharge Q =")
+a1=%pi/16//sq.ft
+a2=%pi/36//sq.ft
+q=sqrt(a*144*(%pi^2)*2*32.185/((s*62.4)*(36^2-16^2)))
+disp("cfs",q)
+diary(0)
\ No newline at end of file diff --git a/806/CH4/EX4.6/46.txt b/806/CH4/EX4.6/46.txt new file mode 100644 index 000000000..1933f4658 --- /dev/null +++ b/806/CH4/EX4.6/46.txt @@ -0,0 +1,18 @@ +
+ A venturimeter,consisting of a converging portion followed by a throat of constant diameter and then gradually diverging portion,is used to detremine the flow of
+ the pipe.The diameter at section 1 is 6 in. and that of section 2 is 4 in.,Find the discharge through the pipe when p1-p2=3psi and oil of sp grvty 0.9 is flo
+ wing
+
+ Solution:
+
+ From the continuity equation Q=A1v1=A2v2
+
+ Therefore,v1=Q*16/pi , v2=Q*36/pi
+
+ And moreover Z1=Z2
+
+ Therefore discharge Q =
+
+ 2.1677205
+
+ cfs
diff --git a/806/CH4/EX4.7/47.sce b/806/CH4/EX4.7/47.sce new file mode 100644 index 000000000..a665e9b42 --- /dev/null +++ b/806/CH4/EX4.7/47.sce @@ -0,0 +1,19 @@ +clc
+pathname=get_absolute_file_path('47.sce')
+filename=pathname+filesep()+'47.sci'
+exec(filename)
+diary('C:\users\Bhavesh\desktop\scilab\47.txt')
+disp("The water supply reservoir as shown in given figure has an average depth of 20m, a surface area of 20km^2, ans an outlet whose centerline is 15m below the water surface.If the otflow diameter is 1 m, what is the outflow and its associated velocity?What would be the drwa down during one week and one day periods?")
+disp("Solution:")
+V2=sqrt(2*g*z1)//Velocity of liquid from the pipe
+Q=V2*%pi*d2^2/4//Discharge per sec
+disp("m^3/s",Q,"Discharge")
+v1=Q*24*3600//Discharge per day
+v2=Q*7*24*3600//Dicharge per week
+v=A*H//Original volume of liquid
+disp("Hence drop down in level for one day and one week are:")
+h1=v1/v*H//draw down of one day
+h2=v2/v*H//draw down of one week
+disp("m",h1,"draw down of one day")
+disp("m",h2,"draw down of one week")
+diary(0)
diff --git a/806/CH4/EX4.7/47.txt b/806/CH4/EX4.7/47.txt new file mode 100644 index 000000000..4a8221360 --- /dev/null +++ b/806/CH4/EX4.7/47.txt @@ -0,0 +1,34 @@ +
+ The water supply reservoir as shown in
+ given figure has an average depth
+ of 20m, a surface area of 20km^2,
+ ans an outlet whose centerline is
+ 15m below the water surface.If the
+ otflow diameter is 1 m, what is t
+ he outflow and its associated velo
+ city?What would be the drwa down d
+ uring one week and one day periods
+ ?
+
+ Solution:
+
+ Discharge
+
+ 13.473642
+
+ m^3/s
+
+ Hence drop down in level for one day an
+ d one week are:
+
+ draw down of one day
+
+ 0.0582061
+
+ m
+
+ draw down of one week
+
+ 0.4074429
+
+ m
|