# Fluid dynamics problem -- Filling linked cylinders with water

• ValeForce46
In summary, Stevin's law states that the pressure at a given altitude is equal to the pressure at a lower altitude plus the atmospheric pressure. In this case, the atmospheric pressure is 9.80 kilopascals and the pressure at the lower altitude is 8.20 kilopascals, so the pressure at the given altitude is 9.90 kilopascals.
ValeForce46
Homework Statement
A cylindrical container (section ##S= 4m^2##) is filled by water up to a height of ##h_1=3m##, the remaining part of satured vapour applies pressure on the water ##P=0.8 atm##. On the bottom there's a hole (section ##S_1=0.2m^2##), linked to a pipe of section ##S_1## which in the far end is closed by a tap ##R## (see picture (a)). In the pipe there's a vertical cylinder ##M##. The diametre of the pipe is negligible compared to ##h_1##. Determine the height ##h_2## which the water reaches in the cylinder.

Then, at the end of the pipe ##S_1## laid a new pipe (section ##S_2=0.1m^2##) (see picture (b)) and after short time the water is in steady state condition. Determine the new height ##h_3## of the water and the speed through ##S_2##.
Relevant Equations
Stevin's law: ##p=p_0+ ρgh##
Bernoulli's principle: ##p+ρgh+1/2ρv^2=constant##
Flow rate constant: ##v_1*S_1=v_2*S_2##

The first part of the problem I just used Stevin's law:
$$p_{atm}=P+ρg(h_1-h_2)=> h_2=(P-p_{atm}+ρgh_1)/(ρg) =>h_2=0.94m$$
Is this right? I considered ##ρ=10^3 {kg/m^3}##
About the second part... how can I be sure that ##h_1## remains unchanged? If it is unchanged, then can I use Bernoulli's principle in this way?
##P+ρgh_1=p_{atm}+1/2ρv_1^2## (##v_1## is the speed through section ##S_1##)
so I could find ##v_2## from ##S_1*v_1=S_2*v_2##

Last edited by a moderator:
It says "a short time later". I would take that as meaning the heights have not dropped significantly.
But you need to consider S2. Your Bernoulli equation is wrong because you omitted h2 (or whatever height it is there at steady state). Instead, apply Bernoulli from surface to S2.

Last edited:
You mean I should apply Bernoulli like this:
##P+ρgh_1=p_{atm}+1/2ρv_2## to find the speed? How do I find the new height ##h_3##?

ValeForce46 said:
You mean I should apply Bernoulli like this:
##P+ρgh_1=p_{atm}+1/2ρv_2## to find the speed? How do I find the new height ##h_3##?
Yes, sorry, I misread the subscripts on the S sections. I have edited my post.
Having found the flow speed v2 you can work back to find v1 and hence the new height.

Once I found ##v_2##, can I apply Bernoulli again for the cylinder##M## in this way?
##p_{atm}+ρgh_3=p_{atm}+1/2ρv_2^2## where ##h_3## is the new height at steady state and ##v_2## is the flow speed I just found?

ValeForce46 said:
Once I found ##v_2##, can I apply Bernoulli again for the cylinder##M## in this way?
##p_{atm}+ρgh_3=p_{atm}+1/2ρv_2^2## where ##h_3## is the new height at steady state and ##v_2## is the flow speed I just found?
No. Bernoulli is based on the assumption of a single set of like flows. The flow through S2 is dominated by the flow out of the main tank; the tiny flow down the cylinder can be ignored for the purpose of finding v2.
But having found v2 you can apply Bernoulli between surface and S1, or between S1 and S2 - either will do - to find the pressure at the bottom of h3. Simple hydrostatics then gives h3.

if I apply Bernoulli between surface and ##S_1## I get
##P+ρgh_1=P^*+1/2ρv_1^2##
and I have to find ##P^*##? Shouldn't it be ##p_{atm}## according to Pascal's Law?
However, if that is right, can I find ##v_1## from ##S_1*v_1=S_2*v_2## so I can find out ##P^*## from the previous equation?

ValeForce46 said:
if I apply Bernoulli between surface and ##S_1## I get
##P+ρgh_1=P^*+1/2ρv_1^2##
and I have to find ##P^*##? Shouldn't it be ##p_{atm}## according to Pascal's Law?
However, if that is right, can I find ##v_1## from ##S_1*v_1=S_2*v_2## so I can find out ##P^*## from the previous equation?
I do not understand how you are applying Pascal's law here. It is merely Bernoulli’s equation when no fluid is flowing.
Yes, you can find v1 and hence P* in the way you describe.

Yeah sorry, I had a stupid thought. Just for completeness, now I apply Stevin's Law:
##P^*=p_{atm}+ρgh_3## and I can find ##h_3## so the problem should be totally resolved.
Thank for you a lot :-)

## What is fluid dynamics?

Fluid dynamics is the study of how fluids, such as liquids and gases, flow and behave under different conditions. It involves understanding the properties of fluids, such as viscosity and density, and using mathematical equations to analyze and predict the behavior of fluids.

## What is a fluid dynamics problem?

A fluid dynamics problem is a specific situation or scenario that involves the flow or behavior of fluids. It can range from simple problems, such as the movement of water through a pipe, to complex problems, such as the behavior of air around an airplane wing.

Linked cylinders are two or more cylindrical containers that are connected to each other in some way, such as being stacked on top of each other or sharing a common wall. In fluid dynamics, linked cylinders may be used to model systems where fluid is being transferred between containers or to study the effects of fluid flow on different shapes and configurations.

## How is water typically used in fluid dynamics problems?

Water is a common fluid used in fluid dynamics problems because it has a relatively low viscosity and is readily available. It is also a good representation of many other fluids, such as air, in terms of its behavior and properties. Additionally, water is easy to manipulate and measure, making it a practical choice for experimentation and analysis.

## What are some real-world applications of fluid dynamics problems?

Fluid dynamics has a wide range of applications in various fields, including engineering, meteorology, and biology. Some examples of real-world applications of fluid dynamics problems include designing efficient pumps and turbines, predicting weather patterns, and understanding blood flow in the human body.

• Introductory Physics Homework Help
Replies
30
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
2K
• Introductory Physics Homework Help
Replies
47
Views
3K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
796
• Mechanical Engineering
Replies
5
Views
248
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K