18451_lec 10-18451_lec 10-18451_lec 10.ppt

Design

1. Tail Escape
Waterway
Road
Water Body

2. •Tail escape is a structure constructed at
the end of a waterway to evacuate the
water to a water body.
•Tail escape consists of a well where its
crest level at the high water level. It is
also equipped with an orifice at its bottom
to evacuate all the water in 24 hrs if
necessary.
•A pipe evacuates the water to the water
body under the road.

3. D
d
o
d
Road
Tail Escape

4. Hydraulic Design
•Hydraulic design of the tail escape consists
of three parts:
1. Automatic evacuation (Weir)
2. Controlled evacuation (Orifice)
3. Discharging to Sea (Pipe)

5. 1.Automatic Evacuation:
The excess water is evacuated over the
weir's crest.
The maximum acceptable rise in water
level is 25 cm.
the discharge of water over the weir can
be calculated as follows:
Q
w= T
s. h . V
s

6. Where:
Q
wis the weir’s discharge in m
3
/sec,
T
sis the top width of the water
surface in m,
h is the rise in water level =0.25 m,
V
sis the surface water velocity in
m/sec.
V
s= 1.17 v
w

7. Where v
wis the mean water velocity of the
water way.
25 cm
T
s
H.W.L.

8. To design the size of the well, the weir
equation is used,
Where
Cd is the discharge coefficient = 0.55,
B is the crest length, and5.1
2
3
2
hgBCQ
dw DB
4
3


9. 2. Controlled Evacuation:
Orifice is designed to evacuate the
waterway in 24 hrs. The following
equation is used to determine the size of
the orifice,
Where:
T is the time to empty the waterway
= 24 . 60 .60 secgyaC
yLb
T
d
e
2
)(2


10. L is the waterway length in m,
b
eis the waterway average width in m,
y is the depth of the waterway,
Cd is the discharge coefficient = 0.6
T
s
H.W.L.
b
e
b
y

11. a is the area of the orifice in m
2
and,
Notice: The orifice is equipped with a gate
so the water in the waterway can be used
for any purpose. The gate is open to
evacuate the water if needed.2
4
oda



12. 3. Pipe:
The pipe is designed to the maximum
discharge for emergency as:
Where:
Q
pis the pipe discharge,
Q
ois the orifice dischargeowp
QQQ 

13. Where:
And,
Where V
pis the pipe velocity (2.0 m/sec)
and D is the pipe diameter.0
2ghaCQ
do
 25.0yh
o 2
4
DVQ
pp



14. •Example:
A tail escape is required to be constructed
at the end of a waterway according to the
following data;
Q = 4.6 m
3
/sec, Waterway length =
3000 m, V
c=0.45 m/sec
7.6 m
5.8 m
4 m
1.8 m

15. •Sol.
a. Automatic evacuation
Q
w= 7.6 * 0.15 * (1.17*0.45)
= 0.6 m
3
/sec
0.6 =
(2/3)*0.55*B*(√2*9.81)*(0.15)
1.5
B = 6.36 m
D = 2.7 m
b. Controlled Evacuation
T = 24*60*60

16. a = 0.2 m
2
d
o= 0.5 m
c. Pipe
Q
p=Q
w+Q
o
Q
o=0.6*0.2*(√2*9.81*2.05)
= 0.76 m
3
/sec
Q
p= 1.36 m
3
/sec
Assume V
p= 2 m/sec

17. 1.36 = 2 * A
p
D = 0.93 m take D = 1.0 m
V
p= 1.73 m/sec