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 $S_1*v_1=S_2*v_2$

 

[haruspex]

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.

 

[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 $h_3$?

 

[haruspex]

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?

 

[haruspex]

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?

 

[haruspex]

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 :-)