Pressure Difference after Throttle/Orifice

Question

I am helping a student with her exams, but since I have a little to no idea regarding fluid dynamics, I would like to ask here about a force. Consider the system of the following figure

The goal is to derive the transfer function of θ to P, i.e. $\frac{\Theta(s)}{P(s)}$.

By dividing the system into three parts

• pressured fluid P->P2
• piston head - damper - spring
• vertical rod

it is easy to analyze the applied forces. The latter two are trivial, but I don't know how to derive the force applied from the pressured fluid to the piston head.

It is given that

$Q = C_d A \sqrt{\frac{2(P-P_2)}{\rho}} $

The force will be $F = P_2 \cdot A$, but how we calculate $P_2$?

P.S. I know that this stupid thing will not work. It's just an exercise.

Answer 1

Ok, so I think I found the right answer

Given (but I think we can derive it from Bernoulli's Eq)

$Q=C_dA\sqrt{\frac{2(P-P_2)}{ρ}} \Rightarrow Q = C_dA\sqrt{\frac{2}{ρ}}\sqrt{ΔP} = K \sqrt{ΔP}$

Linearization

$Q \approx K\sqrt{ΔP_0} + \frac{\partial{Q}}{\partial{ΔP}}|_{@ΔP_0}\cdot(ΔP-ΔP_0)$

Since linearization takes place near 0, $P=P_0=ΔP_0 = 0$, so

$Q \approx \frac{\partial{Q}}{\partial{ΔP}}|_{@ΔP_0}\cdotΔP=\frac{K}{2\sqrt{ΔP_0}}ΔP=K_NΔP=K_NP-K_NP_2$

But, $Q=Av=A\dot{x}=K_NP-K_NP_2$, so $P_2=P-\frac{A}{K_N}\dot{x}$

Free body diagram on piston head

$F_{P_2}-2F_D-F_S=M\ddot{x}\Rightarrow \\AP-\frac{A^2}{K_N}\dot{x}-2D\dot{x}-kx=M\ddot{x} \Rightarrow \\ AP(s)=(Ms^2+(\frac{A^2}{K_N}+2D)s+k)X(s)$

Using $X(s)=RΘ(s) \Rightarrow $

$AP(s)=(MRs^2+R(\frac{A^2}{K_N}+2D)s+Rk)Θ(s)\Rightarrow \\ \frac{Θ(s)}{P(s)}=\frac{A}{MRs^2+R(\frac{A^2}{K_N}+2D)s+Rk}$

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It was quite arbitrary to linearize Q using as variable ΔP, but I think it's the most quick way to derive the solution.