Water and Wastewater Engineering - Sciences Club

16-22 WATER AND WASTEWATER ENGINEERING
d. Calculate the area of each orifice.
A � � � �
p( 005 . m)
4
3
1. 963 10
2
m
e. Calculate the headloss for one orifice using a coefficient of discharge of 0.60.
1 ⎛ 225 . 10 m /s
h �
2
2981 ( . m/s ) ⎝
⎜
( 060 . )( 196 . 3�10 � 0. 022 m or 2. 2 cm or 2 cm
2
� �4
3
�3 2
m
Because the orifices are all at the same elevation, this is the headloss for water flowing
out of the sedimentation tank into the launder. A sketch of the cross section of the launder
and the water levels in the tank and launder are shown below.
Orifice
Tank
Launder
2 cm
Comment. This is a theoretical estimate. Corrosion or encrustation of the orifice will change
the diameter and, thus, the headloss. A safety factor of 2 in the estimated headloss is not unreasonable.
This would yield an estimate of 4 cm headloss.
The headloss in the fall of the water from the orifice to the surface level of the water in the
launder requires an estimate of the depth of water in the launder. This is, in essence, a design of
the launder width and depth. This is an open channel flow problem. The flow in an open channel
may be estimated using Manning’s equation:
/ /
Qchannel
� A R S
n
100 .
2 3 12
( )( )
(16-2)
where Q channel � flow rate in channel, m 3 / s
n � Manning’s coefficient, dimensionless
A � cross-sectional area of flow, m 2
R � hydraulic radius, m
� A/P
P � wetted perimeter, m
S � slope of bottom of the channel, m/m
The coefficient 1.00 has implicit units of m 1/3 / s. The wetted perimeter is the perimeter where the
water is in contact with walls and floor of the channel. Manning’s n i s taken as 0.012 for finished
concrete and 0.018 for steel. The estimate of the surface water level in a launder is illustrated in
Example 16-2.
⎞
) ⎠
⎟
12 /