How to calculate the flow velocity at the bottom outlet of a sump?
Thread Content
This post was last edited by zhaohh3211 on 2017-5-18 23:17. How to calculate the flow velocity at the bottom outlet of the high-level tank? For example: there is 1 large tank with a liquid level of 10 meters; 2 openings are made at the bottom of the tank to connect pipes, one with a DN50 diameter and the other with a DN100 diameter. What are the flow rates at the outlets of these DN50 and DN100 pipes respectively? It would be best to explain how it’s calculated, as a form of popular science. (Please, moderators, move my post to the crafts section; thank you very much!) ）
Bernoulli’s equation, conversion of potential energy into kinetic energy
It’s too simple; I can show a few formulas to illustrate how things like static head, velocity head, potential head, and pipe losses are calculated. I’ll provide some basic explanations and give examples.
P1 + ρgh1 + (ρV1²/2) = P2 + ρgh2 + (ρV2²/2). On the left side of the equation, P1 = P2 (the pressures on both sides cancel each other out). ρgh1 = 1000*9.8*10; (ρV1²/2) = 1000*0²/2 = 0 (the initial velocity is zero). On the right side of the equation, P2 = P1 (the pressures on both sides cancel each other out). ρgh2 = 1000*9.8*0 = 0. Thus, (ρV2²/2) = ρgh1, and V2 = √(2gh1) = 14 m/s. The volumetric flow rate for DN50 is W50 = S1*V2 = 0.05²*3.14/4*14*3600 = 99 m³/h. The volumetric flow rate for DN100 is W100 = S2*V2 = 0.1²*3.14/4*14*3600 = 396 m³/h. These calculations are based on ideal fluid behavior; to determine pipe losses, it is necessary to have information such as the material of the pipes, the number of bends, and the types and quantities of valves involved.
After looking at the calculations above, there is only one conclusion: practice far exceeds theoretical calculations.