Screen in Open Flow Channel

[Lonelywish]

Currently I'm designing an open flow channel to enable fix flow rates through the channel. Currently I have the width of the screen/channel, but the problem arise because I cant determine the water depth that the water need to pass through the screen. I've tried various ways to find them and the closest one which I've came across will be by using the specific energy profile of an open flow channel, where:
E = H + (V^2/H)
I only include in the headloss for flow through screen (by using kirschmer equation) as the only headloss in this calculation due to the reasons that the length of this channel is small and the friction loss from flow is very small and assuming negligible. But my minimal design height by calculation for an example of Q = 300 m3/hr at 0.3m width will be about 0.26m water depth which leads to a velocity above 1m/s which is over the limit according to the guidelines.
Anyone have any idea what I'm missing here?

 

[bimr]

You should design the channel for a minimum velocity of 0.6 m/sec to prevent solids deposition.

Use a flow control device such as a weir or parshall flume to control the depth of water in the flume.

Headloss through a screen =

(1/C)((VV-vv)/2g)

where C= headloss coefficient for clean screen = 0.7

V= velocity through clean screen in m/s

v= approach velocity through clean screen in m/s

C= headloss coefficient for clogged screen = 0.6

Assume that a clogged screen has 50% blockage or twice the velocity.

See Metcalf and Eddy Wastewater Engineering

 

[bimr]

The Manning equation for open channels is the one most commonly used for the design of sewers:

http://en.wikipedia.org/wiki/Open_channel_flow

 

[Lonelywish]

thanks for the tips. Will give it a try and hope it works.

 

[Lonelywish]

I've tested the way but my advisor says that it is not quite practical to put a weir or flume just to find or control the velocity. What he wants from me is to estimate what will be the water depth in the channel when a particular flow rate is run through it. The channel contains one screen. Therefore I'm trying to use the energy profile which use the sub, super, and critical flow as calculation where estimation is used as the amount of headloss through screen will be the only headloss. Therefore the energy level will increase due to the headloss and the height of the new energy will be estimated as the water depth, which after some editing I got a value as below:
Q = 300 m3/hr
w = 0.3m
depth = 0.29m
Velocity = 0.958m/s
which is still quite height because the flow through velocity will eventually exceeds 1m/s.
Do you have any idea what I'm missing out?
My channel will be short and therefore headloss due to friction should be able to be neglected.
By the way my channel will have no slope therefore I cant use manning's equation.
Hope anyone can help thanks

 

[Lonelywish]

I came across with this formula in the website:
Ecr/Ycr = 3/2 (b/B)^(-2/3)
where
Ecr = critical energy
Ycr = critical depth
b = reduced diameter due to the obstacle
(B - Diameter of obstacle)
B = channel width
but I dont seem to be able to find it in any reference book.
This formula is used for the flow pass a vertical circular cylinder which I think somehow it is quite familiar with my situation.
the website which I obtained this was

http://www.image.unipd.it/a.defina/paper_2006_02.pdf

What I really need to know is how do they justify this equation or is it just an experimental data they got for this particular case?
Is it applicable if I use it for the estimation of my water depth because I got my screen design which is the diameter of the openings.

 

[bimr]

"By the way my channel will have no slope therefore I can't use manning's equation"


Assume a slope. Water will not flow unless it is going to a lower elevation. The discharge point has to be lower than the influent point. The difference between the influent -effluent divided by the length is the slope. It does not have to be a continuous slope.


"I've tested the way but my advisor says that it is not quite practical to put a weir or flume just to find or control the velocity."

The slope sets the velocity. The weir or flume do not set the velocity. The backwater from the weir or flume sets the height of the water not the velocity.

Q = VA

(300 /60*60) / 0.6 = 0.1389 m2 Flume Area

where 0.6 = minimum velocity.

0.1389 m2 Area = heighth x width

0.1389 m2 Area/ .3m = .46m high flume

headloss across screen:

Assumes 50% block off due to solids, with velocity now 2 times, clean screen.

= (1/C)((VV-vv)/2g)

= (1/0.7) ((0.9*0.9)-(0.6*0.6))/2(9.8)

= 0.033 m headloss

 

[Lonelywish]

thanks bimr,
but i tried a different approach where I use excel to create a range of high with the formula of
E = H + V^2/2g
from there I used the Yc value from
Yc = ((Q/W)^2/g)^(1/3)
from this Yc I calculated the head loss through screen by using kirshmer equation (hope i dint spell wrong)
HL = K.(t/b)^(4/3)* sin (degree)* V^2/2g
then adding it up to the E at Ec to get the energy needed after headloss then find the energy needed before head loss.
from this equation I am able to get a desirable height for my coarse screen (screen in open flow channel) but in my rotary screen(fine screen) it seems a bit low on the height and therefore the flow rate able for the screen is low as well for the values.

Therefore I decided to reuse the bernouli's equation
H1 + V1^2/2g = H2 + V2^2/2g
and kirshmer equation assuming H1-H2 = HL
and I am able to get some values nearing desirable conditions.

but when i use bernouli in open channel flow n coarse screen it gives me a height which is quite low resulting a high velocity.

Anyone have any idea where did things went wrong?

 

[bimr]

"E = H + V^2/2g" is not an appropriate equation to use for a channel since it does not take into account the headloss from the friction of the pipe walls.