# Fluid dynamics problem -- Filling linked cylinders with water

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##

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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.

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ValeForce46
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##?

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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.

ValeForce46
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?

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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.

ValeForce46
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?

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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.

ValeForce46
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 :-)