Calculating the Flow Speed of Tea In a Tank

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The discussion focuses on calculating the flow speed of tea from a tank while simultaneously filling it with water from above. Participants suggest using Bernoulli's equation to analyze the situation, noting that the pressure and height differences will affect the flow speed. The key variables include the height of the tea (H), the height of the water being poured in (h), and the diameters of the tank and flow openings. The conversation emphasizes the need to consider how the added water influences the parameters in Bernoulli's equation. Overall, the application of fluid dynamics principles is central to solving the problem.
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Homework Statement

Imagine a large tank containg tea with a little opening at its bottom so that one can pour it into a glass. When open, the speed of the flow of tea from the tank is v0. How will this speed change if, while pouring a glass of tea, someone is filling the tank by pouring water into it from its top? Assume that the diameter of the tank is D, the diameter of the flow of tea into the tank is d, and that of the flow of tea out of the tank is much smaller than D. The tea level is height H above the lower opening, and the tank is being filled by pouring a water into it from height h above the tea level. You are free to neglect all friction.


The attempt at a solution

The current can be regarded as a cylinder - calculate its volume, according to the density of water calculate mass. With the gravitational acceleration - we can calculate the force of current. Force / area = pressure impact ... As a result of this pressure will increase speed. All I know. Do you have any idea to next solution?
I think it is related to the the Bernoulli equation.

Thanks very much and sorry for my bad English.
 
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Start with Bernoulli's equation and eliminate the terms that do not apply. Since you have a water being poured in, how does this affect the parameters of the equation?
 
Hmm, so that my way is wrong?
I don't know how starts with Bernulli's equation :-(
 
Bernuklli says p1 + (1/2)(rho) (V1)^2 + (rho) (g) (h1) = p2 + (1/2)(rho) (V2)^2 + (rho) (g) (h2)

Seems like p1 would be equal to p2, V1 = 0. h2 - h1 will be increasing because you are adding water to the tank.
Work the equation to see if you get something like gh = (1/2) V^2 then work on h as a function of time.
 
Hi Numeriprimi! :smile:

First, write out Bernoulli's equation.

(you should have done this already, as part of the homework template)

And remember that Bernoulli's equation must always be applied along a streamline

in this case the streamline will start at the height H+h
 
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