The hydraulic mill is consists of two main part; servo valve and hydraulic cylinder.
We will consider flow-control servo valve. The dynamics of the servo valve is \begin{equation} \label{1}\tag{1} Q_s=C_Q\pi dx\sqrt{\frac{2}{\rho}(P_s-P_1)} \end{equation} Here,
- $Q_s$=supply flow,
- $C_Q$=flow coefficient,
- $\rho$=oil density,
- $x$=servo-valve displacement,
- $P_s$=supply pressure,
- $P_1$=output pressure of the valve.
The equation for the flow of oil to the hydraulic cylinder, \begin{equation} \label{2}\tag{2}Q_s=a\dot{y}+\frac{V_1}{\beta}\dot{P_1} \end{equation} Here,
- $a$=area of the cylinder,
- $y$=hydraulic piston displacement,
- $V_1$=volume of the primary side of the cylinder,
- $\beta$=bulk modulus of the oil,
- $P_1$=cylinder pressure on primary side.
Schematic: Hydraulic servo system
Equation \eqref{1} needs to be linearized for both input-flow & output-flow to the cylinder according to valve displacement $x$.
Help me find the ODE for rate of change in cylinder Pressure $\dot{P_1}$ and $\dot{P_2}$.
Assuming an operating point at $x_0,P_{s_0},P_{1_0}$ we have
regarding $K_0x\sqrt{P_s-P_1}$
$$ K_0 x \sqrt{P_s-P_1} = K_0 x_0\sqrt{P_{s_0}-P_{1_0}}+K_0\sqrt{P_{s_0}-P_{1_0}} (x-x_0)-\frac{K_0 x_0 (P_1-P_{1_0})}{2 \sqrt{P_{s_0}-P_{1_0}}}+\frac{K_0 x_0 (P_s-P_{s_0})}{2 \sqrt{P_{s_0}-P_{1_0}}}+O_1(|x-x_0|^2)+O_2(|P_s-P_{s_0}|^2)+O_3(|P_1-P_{1_0}|^2) $$
so near the operating point we have
$$ a\dot{y}+\frac{V_1}{\beta}\dot{P_1} = K_0 x_0\sqrt{P_{s_0}-P_{1_0}}+K_0\sqrt{P_{s_0}-P_{1_0}} (x-x_0)-\frac{K_0 x_0 (P_1-P_{1_0})}{2 \sqrt{P_{s_0}-P_{1_0}}}+\frac{K_0 x_0 (P_s-P_{s_0})}{2 \sqrt{P_{s_0}-P_{1_0}}} $$