Help me calculate venturi flow rate

[declanka]

Hi I have a challenge (Its over 30 years since I did physics). I have the following information and I want to calculate the flow rate change .

Opening diameter 2m , narrows smoothly to diameter 1m. Seawater, tidal flow of 2.5 m/s into the opening, what is the effective flow rate in the throat of the venturi

 

[tiny-tim]

welcome to pf!

hi declanka! welcome to pf!

water is incompressible, so the flow rate (volume per second) must be constant, so the speed (metres per second) must be inversely proportional to cross-section area

(and in this case, the area is 1/4, so the speed must be x 4)

 

[declanka]

Thanks for this response, it logically confirms the calculations I came up with. Is the formula always true , even in the scenario where its just an open venturi (narrowing) pipe facing into the tidal flow .

 

[tiny-tim]

Is the formula always true , even in the scenario where its just an open venturi (narrowing) pipe facing into the tidal flow .

yes, it's conservation of mass …

the water can't bunch up anywhere, or thin out, so the flow into any part of the pipe must be the same as the flow out of it … ie the flow rate (volume per second) past any two cross-sections must be the same

 

[Nickie]

.

Sure, I would be happy to help you calculate the venturi flow rate. The venturi effect is a principle in fluid dynamics that describes the relationship between the velocity of a fluid and the cross-sectional area of a pipe. In this case, we can use the following equation to calculate the flow rate in the throat of the venturi:

Q = A1V1 = A2V2

Where:
Q = flow rate
A1 = cross-sectional area of the opening diameter (2m)
V1 = velocity of the fluid entering the venturi (2.5 m/s)
A2 = cross-sectional area of the narrowed diameter (1m)
V2 = velocity of the fluid in the throat of the venturi (to be calculated)

Using this equation, we can rearrange it to solve for V2:

V2 = (A1V1)/A2

Plugging in the values given, we get:

V2 = (2m * 2.5 m/s)/1m = 5 m/s

Therefore, the effective flow rate in the throat of the venturi would be 5 m/s. I hope this helps you with your calculation. Let me know if you have any further questions.