35.1.2.1. Pressure in Chamber

The pressure in this hydraulic system is determined by the following equation, based on the definition of the bulk modulus.

\[\begin{flalign} & \dot{p}=-\frac{E}{V}\dot{V} &\ \end{flalign}\]
where,
\(\dot{p}\) is pressure change rate
\(E\) is Bulk modulus
\(V\) is chamber volume
\(\dot{V}\) is the sum of all changes of the volume.

\(\dot{V}={{\dot{V}}_{chamber}}+{{Q}_{valve}}+{{Q}_{leakage}}\)

In this equation:
\({{\dot{V}}_{chamber}}\) is the pressure chamber volume change rate
\({{Q}_{valve}}\) is the oil flow through check valve
\({{Q}_{leakage}}\) is the oil flow through the leakage gap

35.1.2.1.1. Pressure Chamber Volume Change Rate

\[\begin{flalign} & {{\dot{V}}_{chamber}}={{S}_{p}}\times ({{\dot{y}}_{p}}-{{\dot{y}}_{c}}) & \end{flalign}\]
where,
\({{\dot{V}}_{chamber}}\): Pressure chamber volume change rate
\({{S}_{p}}\): Area of the plunger
\({{\dot{y}}_{p}}\): Velocity of the plunger
\({{\dot{y}}_{c}}\): Velocity of the cylinder

35.1.2.1.2. The Oil Flow through the Check Valve

According as the check ball moves between the plunger and cylinder by the plunger spring, the check valve opens or closes, and the oil flows.
When the check valve opens, the oil flow through the check valve is shown in Figure 35.9.
../_images/image01016.png

Figure 35.9 The oil flow rate through check valve

As the resistance of the oil through the check valve depends on the orifice area in this investigation, the dynamic resistance is considered for the turbulent flow of the oil flow through the check valve, which yields,

\(\begin{aligned} & {{Q}_{valve}}={{C}_{d}}\cdot A_{orifice}\cdot \sqrt{\frac{2}{\rho }({{p}_{o}}-p)}(if,{{p}_{o}}\ge p) \\ & {{Q}_{valve}}=0(if,{{p}_{o}}<p) \\ \end{aligned}\)

where,
\({{p}_{o}}\) is the supply pressure
\(p\) is the chamber pressure.
(If \({{p}_{o}}\) is less than \(p\), there is no oil flow through check valve.)
\(C_d\) is the discharge coefficient of check valve
\(A_{orifice}\) is the orifice area
\(\rho\) is the density of oil

\({{Q}_{valve}}\) represents the oil stream flowing rate through the opened check valve into the pressure chamber. The area of orifice can be obtained such as

\({{A}_{orifice}}=min(\pi d\sin \alpha \cos \alpha (2r_b+d\sin \alpha ), \pi {{r_{co}^2}})\)

where,
\(d\) is the ball displacement.
\(\alpha\) is the cylinder alpha.
\(r_b\) is the ball radius.
\({r}_{co}\) is the cylinder orifice radius.

35.1.2.1.3. The Oil Flow through the Leakage Gap

Since the pressure of chamber is different compared to the air pressure, the oil flow through the gap between the plunger and the cylinder is shown in Figure 35.10. The oil flowing between the plunger and the cylinder is a laminar flow.

../_images/image01115.png

Figure 35.10 Oil flow and through the leakage gap

As shown in Figure 35.10, the variation of oil speed is fully depended on the pressure difference between the chamber and atmosphere, and it is not affected by the plunger speed. The oil flow rate between the gap of plunger and cylinder, \({{Q}_{leakage}}\) can be written as

\[\begin{flalign} & {{Q}_{leakage}}=\frac{2{{r}_{p}}\pi {{h}^{3}}}{12\mu l}({{p}_{air}}-p) & \end{flalign}\]
where,

\(\mu\) is the coefficient of the viscosity of the oil.

35.1.2.1.4. Bulk Modules

The definition for the bulk modules is as follows:

\[\begin{flalign} & E=-\frac{dp}{dV}V & \end{flalign}\]

The resulting bulk modulus of the oil - air mix is:

\[\begin{flalign} & {{E}_{Res}}=\frac{1}{\frac{1-{{r}_{air}}}{{{E}_{oil}}}+\frac{{{r}_{air}}}{{{E}_{air}}}} & \end{flalign}\]
where,

\({{r}_{air}}\) is the air volume fraction of the mix. The air and the oil states are calculated separately.

In isothermal case, assume that the bulk module of the air formulated as:

\({{E}_{air}}=p\)

In isentropic case, assume that the bulk module of the air formulated as:

\({{E}_{air}}=kp\)

where,

\(k\) is the ratio of specific heat.

Assume that the bulk modulus of the oil formulated as:

\({{E}_{oil}}={{E}_{oil0}}+c(p-{{p}_{air}})\)

where,
\(c\) is the pressure coefficient
\({{E}_{oil}}\) is the bulk modulus at the reference pressure

The air volume can be calculated as follows:

\[\begin{flalign} & {{V}_{air}}={{\left( \frac{{{p}_{air}}}{p} \right)}^{1/k}}{{V}_{air0}} & \end{flalign}\]
where,

\(k\) is the ratio of specific heat.

The oil volume can be calculated as follows:

If \(c\ne 0\),

\[\begin{flalign} & {{V}_{oil}}=\frac{(1-{{r}_{air0}})V{}_{oil0}}{{{\left[ \frac{{{E}_{oil}}}{{{E}_{oil0}}} \right]}^{1/c}}} & \end{flalign}\]

If \(c=0\),

\[\begin{flalign} & {{V}_{oil}}=\frac{(1-{{r}_{air0}})V{}_{oil0}}{{{e}^{\left[ \frac{P-{{P}_{air}}}{{{E}_{oil0}}} \right]}}} & \end{flalign}\]

\({{r}_{air}}\) can be calculated as follows:

\[\begin{flalign} & {{r}_{air}}=\frac{{{V}_{air}}}{{{V}_{air}}+{{V}_{oil}}} & \end{flalign}\]