Determining Water Depth

Have you been drinking? If so, it's okay. If not, I am far dumber than I thought and need to leave here. AND STOP CALLING ME BOY!

No, alcohol is banned on the Empire's ships and camps anyway. Okay, what would you prefer me to call you then?

Bitch. Thank you.

Give us more information, draw a sketch, what is the source of the water coming in, flow rate, depth, head, velocity. How long is the channel, cross section dimensions/geometry, what is the type of surface, concrete, smooth, rough, steel, corrugated, plastic, earthen, etc.. and lastly, what does it dump into, open drop, culvert?

Please refer to the sketch.

The water is pumped from a river. The flow is 1.6m³/sec @ 20 m-head. The channels and tank are made of smooth-walled reinforced concrete. The water will discharge into a lake 2000m³ with a depth of 4 meters from a drop of 2 meters from outlet channel.

Thanks for your help.

What you provided helps, but still does not give us the needed info to proceed. If you are pumping, how is it being fed to the channel (pipe, chute, chamber, free fall)? What is important is that you have a flow rate coming in, you take it from there. Therefore the hydraulic gradient is not zero, because you have a velocity head starting at the discharge point #1, then it enters the channel, which is the crux of your inquiry. the formula is Q=1.486/nXAXR^(2/3)/S^1/2 =AV. Refer to Open Channel hydraulics, Chow or Handbook of Hydraulics, Brater & King. You can also get some free spreadsheets or low cost programs from Haestads. Your diagrams does not say if its plan view, section or what I'm looking at. It resembles both. It looks like a chamber that you are dumping into from a source (what, where). Is the water building up, it has to build up to build a head to push it through the baffle and up the lift to a delivery chute. Again, I'm speculating from the limited information an crude drawing. Do you have pictures of the sources and delivery, you can email me too.

Okay, thanks for the advice. The top drawing is the plan view while the lower one is the side view.

I thought so, how does the water get delivered to the inlet on the left? Is it 3.5 m from the baseof the tank? I need to know how much the water has to raise to the outlet on the right side, it appears that it may be 3.2m (5.0-1.8)? For the sake of brevity, we can assume that the water equalizes and flows out on the right with a head as in a reservoir in a chute 5 m long and 2m wide and determine the depth that way. YOu apply bernoulis equation for conservation of energy from the reservoir to the outlet 5m to the right for the depth with a zero channel slope. Z1+D1+fV1^2/(2g) = Z2+fV2^2/(2g)+H Z1=Z2, CANCEL f=LOSS FACTOR V=VELOCITY D=DEPTH OF FLOW g=GRAVITY CONSTANT, 32.2 FPS, 9.2 MPS H=ENERGY SLOPE, DIFFERENCE IN DROP BETWEEN 1 & 2 I calc out 56.5' or 17.2m, too much flow for a small chute

Okay, thanks for your help. I'll let my friend know that this is an unfeasible design.