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Hydraulic ball valve
The Ball Valve block models a variable orifice created by a spherical ball and a round sharp-edged orifice.
The flow rate through the valve is proportional to the valve opening and to the pressure differential across the valve. The flow rate is determined according to the following equations:
$$q={C}_{D}\cdot A(h)\sqrt{\frac{2}{\rho}}\cdot \frac{p}{{\left({p}^{2}+{p}_{cr}^{2}\right)}^{1/4}}$$
$$p={p}_{A}-{p}_{B}$$
$${p}_{cr}=\frac{\rho}{2}{\left(\frac{{\mathrm{Re}}_{cr}\cdot \nu}{{C}_{D}\cdot {D}_{H}}\right)}^{2}$$
$$h={x}_{0}+x\xb7or$$
$$A(h)=\{\begin{array}{ll}{A}_{leak}\hfill & \text{for}h=0\hfill \\ \pi \cdot {r}_{O}\left(1-\frac{{r}_{B}}{{D}^{2}}\right)\cdot D\hfill & \text{for}0h{h}_{\mathrm{max}}\hfill \\ {A}_{\mathrm{max}}+{A}_{leak}\hfill & \text{for}h={h}_{\mathrm{max}}\hfill \end{array}$$
$$D=\sqrt{{\left(\sqrt{{r}_{B}^{2}-{r}_{O}^{2}}+{h}^{2}\right)}^{2}+{r}_{O}^{2}}$$
$${D}_{H}=\sqrt{\frac{4A(h)}{\pi}}$$
$${A}_{\mathrm{max}}=\frac{\pi {d}_{O}^{2}}{4}$$
$${h}_{\mathrm{max}}={r}_{O}\cdot \left(\sqrt{\frac{{\left(1+\sqrt{1+4\frac{{d}_{B}^{2}}{{d}_{O}^{2}}}\right)}^{2}}{4}-1}-\sqrt{\frac{{d}_{B}^{2}}{{d}_{O}^{2}}-1}\right)$$
where
q | Flow rate |
p | Pressure differential |
p_{A}, p_{B} | Gauge pressures at the block terminals |
C_{D} | Flow discharge coefficient |
A(h) | Instantaneous orifice passage area |
x_{0} | Initial opening |
x | Ball displacement from initial position |
h | Valve opening |
d_{O} | Orifice diameter |
r_{O} | Orifice radius |
d_{B} | Ball diameter |
r_{B} | Ball radius |
ρ | Fluid density |
ν | Fluid kinematic viscosity |
p_{cr} | Minimum pressure for turbulent flow |
Re_{cr} | Critical Reynolds number |
D_{H} | Valve instantaneous hydraulic diameter |
A_{leak} | Closed valve leakage area |
A_{max} | Maximum valve open area |
h_{max} | Maximum valve opening |
The block positive direction is from port A to port B. This means that the flow rate is positive if it flows from A to B and the pressure differential is determined as $$p={p}_{A}-{p}_{B}$$. Positive signal at the physical signal port S opens the valve.
Fluid inertia is not taken into account.
The flow passage area is assumed to be equal to the side surface of the frustum of the cone located between the ball center and the orifice edge.
The diameter of the valve ball. It must be greater than the orifice diameter. The default value is 0.01 m.
The diameter of the orifice of the valve. The default value is 0.005 m.
The initial opening of the valve. Its value must be nonnegative. The default value is 0.
Semi-empirical parameter for valve capacity characterization. Its value depends on the geometrical properties of the orifice, and usually is provided in textbooks or manufacturer data sheets. The default value is 0.65.
The maximum Reynolds number for laminar flow. The transition from laminar to turbulent regime is assumed to take place when the Reynolds number reaches this value. The value of the parameter depends on the orifice geometrical profile. You can find recommendations on the parameter value in hydraulics textbooks. The default value is 10.
The total area of possible leaks in the completely closed valve. The main purpose of the parameter is to maintain numerical integrity of the circuit by preventing a portion of the system from getting isolated after the valve is completely closed. An isolated or "hanging" part of the system could affect computational efficiency and even cause simulation to fail. Therefore, MathWorks recommends that you do not set this parameter to 0. The default value is 1e-12 m^2.
Parameters determined by the type of working fluid:
Fluid density
Fluid kinematic viscosity
Use the Hydraulic Fluid block or the Custom Hydraulic Fluid block to specify the fluid properties.