Estimating discharge at the outflow structure of a lake

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SUMMARY

The discussion focuses on estimating discharge at the outflow structure of a lake, specifically using a box cement outlet structure governed by two adjacent 5ft by 0.5ft dam boards. The participant seeks to manipulate the lake level by six inches and is exploring the relationship between the impounded water pressure and the average velocity of water exiting the structure. Key insights include the application of Bernoulli's equation, which states that velocity is proportional to the square root of the depth of water, and the formula for calculating discharge as cfs = ft/sec * length * width of the opening.

PREREQUISITES
  • Understanding of Bernoulli's equation
  • Knowledge of fluid dynamics principles
  • Familiarity with discharge calculations
  • Basic concepts of pressure in fluid systems
NEXT STEPS
  • Study Bernoulli's equation in detail
  • Research methods for calculating discharge in open channel flow
  • Learn about pressure dynamics in fluid systems
  • Explore techniques for measuring water velocity without flow meters
USEFUL FOR

Fisheries biologists, hydrologists, civil engineers, and anyone involved in water management and lake level manipulation will benefit from this discussion.

salconflu
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I am a fisheries biologist interested in manipulating a lake level by about six inches via a box cement outlet structure on a small lake. The outlet structure is currently governed by two adjacent 5ft by 0.5ft dam boards,which have historically been removed one at a time in the spring and replaced in the fall. However pulling whole boards releases too much water at once so I am trying to figure out whether incrementally lower boards would make a meaningful difference in Q.

As a starting point I am trying to figure out how much pressure is behind the 70 acre lake and what the average velocity (ft/sec) would be across the 2.5 square ft opening without the benefit of a flow meter. Is there a mathematical relationship between the impounded water and the velocity at the outlet? If there is then I could figure out the discharge for different square openings in the outlet structure, ie: cfs = ft/sec* length*width of the opening...
 
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Welcome to PF!

Hi salconflu! Welcome to PF! :smile:

Pressure depends only on depth, and not on horizontal area …

the pressure from the whole lake is exactly the same as the pressure at the same height in your bath. :wink:

From Bernoulli's equation the velocity is proportional to the square-root of the depth: v2 = 2ρgh.

And obviously the total flow is also proportional to the vertical area of the hole.
 

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