---
# System prepended metadata

title: 3.1 Hydrostatics and hydraulic applications
tags: [Mathematical modeling and optimization]

---

---
tags: Mathematical modeling and optimization
title: 3.1 Hydrostatics and hydraulic applications

---

## 3.1 Dynamics in hydrostatic and hydraulic relevant systems 


Following the transport phenomenon in the last chapter, the baseline governing equation for flow movement results to the Navier-Stokes Equation, which states the mass and momentum conservation for the [Newtonian fluid](https://en.wikipedia.org/wiki/Newtonian_fluid). 


$$
\begin{align}
\frac{\partial \rho}{\partial t}+\frac{\partial \rho u_i}{\partial x_i} &= 0 \\
\\
\frac{\partial \rho u_i}{\partial t} +\frac{\partial \rho  u_i u_j}{\partial x_j} &= -\frac{\partial p}{\partial x_i} + \frac{\partial}{\partial x_j}\left[ \mu\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} - \frac{2}{3}\frac{\partial u_k}{\partial x_k} \right)\right] + \rho g_i\\
\end{align}
$$

where $u_i$ is the flow velocity in all Cartesian directions. $\rho$ ,$p$, $\mu$ and $g_i$ are the density, pressure, kinetic viscosity and body force due to gravitation, repsectively. 

Setting $u_i = 0$ into the equation equation system refered to no flow movement, the pressure difference yields to a linear behavior to the displacement in the vertical direction ($z$).

$$
\displaystyle \Delta p = \rho g \Delta z
$$

The pressure difference can be utilized in the form of force, which can be found in different enginnering aspects. These applications are categorized under the discipline of hydrostatics and hydraulic devices.

### 3.1.1 Hydrostatics


Hydrostatics refers to the states when the flow movement is not present in the fluid system. Under this circumstance, pressure becomes the main character in the system. It can be applied to lift up heavy loads, such as [jack](https://en.wikipedia.org/wiki/Jack_(device)) while changing the tyre, or the [shock absorber](https://en.wikipedia.org/wiki/Shock_absorber) behind the door to resist the the bumping, etc. 
To realize the functions, certain amout of fluid are implemented in these equipments. Apart from the [pneumatic devices](https://en.wikipedia.org/wiki/Pneumatics), which apply gases to pass the pressure, the hydraulics stay in liquid regime. i.e. most of the medium is highy incompressible. 

Barometer(left) and manometer(right) are the intuitive examples. These devices measure the pressure of atmosphere and the target system repsectively. The pressure level is shown in the representation of fluid height. 

<img src="https://instrumentationtools.com/wp-content/uploads/2017/09/instrumentationtools.com_barometer-principle.png" width=55%> 
<img src="https://www.instrumentationtoday.com/wp-content/uploads/2011/09/U-Tube-Manometer.jpg" width = 35%>


### Buoyancy and floating object stability

When the pressure affects on a plane, force occurs. Thinking on pressure as a scalar (tensor 0th order), the force yields in the direction normal to the plane. Extending this idea of a immerse body in the fluid with infinitesimal normal planes on the body surface. The pressure varies the the depth, resulting in different force level. The integration of the force over the surface yields in a lifting tendence against the gravitation, which is [**buoyancy**](https://en.wikipedia.org/wiki/Buoyancy). 
<img src="https://live.staticflickr.com/65535/54655179395_fb4f077b2c_b.jpg" width=60%>

[Archimedes's buoyancy principle](https://en.wikipedia.org/wiki/Archimedes%27_principle) summarizes this force as the weight of displaced fluid in the opposite direction to the gravity. Thus,

$$\vec{F}_b = - \rho\  \nabla\  \vec{g} $$

where $\rho$ is the density of fluid, $\nabla$ is the displaced volume and the $\vec{g}$ is the gravitational acceleration.

Extedning Archimedes's principle with free body diagram of a floating object, where only partial of the entire volume is immersed in the liquid, it yields the object has smaller density than the liquid itself. Whereas, the immerse object shows identical density to the fluid, as well as the sinking object obsesses higher density than the fluid. I.e. by varying the density, the object is capable to float, sink or remain immerse. This is the fundamental working principle of a submarine. 

<img src="https://live.staticflickr.com/65535/54716141835_9e6f8a0dd8_b.jpg" width="85%">

While the buoyancy works on a floating body, **center of gravity** ($M$) and **center of buoyancy** ($B$) are not aligned on the same point. Under this circumstnace, if the two forces work on the same axis, equilibrium can be achieved, as the rectangular box left.



<img src="https://live.staticflickr.com/65535/54660714792_ffecc46ea0_b.jpg">

However, if the center of gravity ($G$) and center of buoyancy ($B$) are not on the same axis, the floating object tilts (box right). Depending on the position of $G$ and $B$, the floating object can turn back, capsizes or remain in the same pose.

**Metacenter** (M) is the concept to deal with this scenario. It is a fictional position on the intersection of the vertical line through B (when tilted) and the original vertical symmetry axis of the floating body.

The distance from metacenter to center of gravity ($\overline{GM}$ ) is the moment arm for the buoyancy to the object. So the torque can be formulated in

$$
\tau_{\text{tilt}} = |\vec{F_B}| \cdot \overline{GM} \cdot \sin\theta  
$$


Depending on the metacenter position, the torque of buoyancy on the center of gravity can contribute to correct or intensify the tilted state. if 
- $M$ is above the $G$, i.e. $\overline{GM} > 0$, the bending moment works opposit to the tilted direction. The pose will be corrected (figure left). otherwise,
- $M$ is lower than $G$, i.e. $\overline{GM} < 0$, the bending moment works in the same direction to the tilt (figure right). Capsizing. 

<img src="https://live.staticflickr.com/65535/54661786954_d7b9f89656_b.jpg">

<br/>
<br/>

For small tilting angle $(\theta < 10°)$, the metacenter can be considered on the same position. This assumption simplifies certain practical analysis. 

A simplified rule of thumb finding out the metacenter refers to the equilibrium state reads:

<img src="https://live.staticflickr.com/65535/54674728998_836f0a82d7_c.jpg">

Distance between $G$ and $M$:  
 $$
 \overline{GM} = \overline{B_oM} - \overline{B_oG}
 $$ 

where:
- $B_o$ is the original center of buoyancy in equilibrium state.
- $\overline{B_oM} = \frac{I}{\nabla}$: *Radius of curvature* of B’s path.  
   - $I$: Moment of inertia of the waterline area, e.g., $(b\ h³) / 12$ for rectangular area.
   - $\nabla$: Displaced volume.  
- $\overline{BG}$: Distance between $B_o$ and $G$ at original equilibrium state.


Stability is a critical challenges in the buoyancy applications, which belongs to the core topics of marine engineering, referring to the design of ships or floating contructions. **Sink** and **tilt** are the main characteristics concerned, which raises the complexity to the dynamic level.

### Dynamics of floating object

Generalizing the issue of sinking and tilting, the floating object dynamics can be summarized in 6 degree of freedoms, including 3 directional displacement ($\Delta_x, \ \Delta_y, \ \Delta_z$ ) and 3 angular displacements ($\theta_x, \ \theta_y, \ \theta_z$ ).

Neglecting the horizontal displacement in $x, y$ axes and simplifying the rotation to one axis $(\theta)$, the dynamics along these two dimensions can be vividly illustrated by a free body diagram 

<img src="https://live.staticflickr.com/65535/54679036655_220abb553c_b.jpg" width="100%">

and the transport equations for momentum and angular momentum yield:

$$
\begin{align}
   m \frac{d u_z}{dt} = m \frac{d^2 \Delta_z}{dt^2} =& \ \sum F_{\text{excite}} - mg + F_{\text{buoyancy}} +  F_{\text{drag}}\
   \\
   \\
   I \frac{d \alpha}{dt} = I\ \frac{d^2 \theta}{dt^2}  = &\ \sum\tau_{\text{excite}} + \tau_{\text{buoyancy}} + \tau_{\text{drag}}
\end{align}
$$ 
   - $I$: Moment of inertia of the floating object
   - $\theta$, $\alpha$: tilted angle and angular velocity 
   - $\Delta_z$, $u_z$: Vertical displacement and velocity.  
   - $F_{\text{excite}}$, $\tau_{\text{excite}}$: External pertubating force and moment. 
   - $F_{\text{drag}}$, $\tau_{\text{drag}}$: hydrodynamic drag due to movement under water, usually modeled as $F_d = -c_z \ u_z$ and $\tau_d = -c_\theta\ \alpha$
   
This equation set can be solved numerically with given initial conditions, served as classical _**initial value problem**_.


__<ins>Scenario: Floating buoy with wave excitation__

A 1000 kg [cylindrical buoy](https://lh4.googleusercontent.com/proxy/geA7D7E5FooeEs6w0t5GLCepTQ2zeMe3CWoJqsojq19L4gPdarnEJUHxTy81rHugk3_fbDOuXOGB7z6i88a99phJgdCn6ci46bp_7yIcs1kQkTHB5w) (e.g., navigation buoy) of the dimension (1.25[m] radius, 2[m] height and 0.5[m] submerged height) subjected to ocean waves. The wave impact is given in the form of force as $F_{\text{wave}} = F_0 \sin(\omega t)$ with $F_0$ of 4000 [Nt], and $\omega$ as $\pi/8$ [rad/s], setting damping coefficient $c_z$ and $c_\theta$ as 500 [Nt s/m] and 300 [Nt m s/rad].

_Supposed the tilting angle is small $(\theta < 10°)$. Consider the **Tilting (Roll)** and **Vertical movement (Heave)** of the bouy, with the starting point of no vertical and angular displacement. What is the dynamics within the first 20 sec?_ 

#### **1. Vertical (Heave) Dynamics**  
$$
m \frac{d² \Delta_z}{dt²} + c_z \frac{d \Delta_z}{dt} + \underbrace{\rho g \pi R^2 \Delta_z}_{\text{buoyancy}} = F_{\text{wave}}(t)
$$ 
- $\Delta_z(t)$: Vertical displacement from equilibrium.  
- $c_z$: Damping coefficient (hydrodynamic drag).  

#### **2. Tilting (Roll) Dynamics**  
$$
I_{\text{roll}} \frac{d² \theta}{dt²} + c_{\theta} \frac{d \theta}{dt} + \underbrace{\rho g \pi R^2 \cdot GM (d-\Delta_z)}_{\text{buoyancy contribution}} \sin\theta = F_{\text{wave}}(t) \cdot R
$$ 
- $\theta(t)$: Roll angle.  
- $I_{\text{roll}} = \frac{m}{12}(3R^2 + H^2)$: Moment of inertia.  
- $GM = \frac{R^2}{4d} - \left(\frac{H}{2} - \frac{d}{2}\right)$: Metacentric height.  


[<ins>__Sample Solution__](https://cocalc.com/share/public_paths/559ec31fa6670c99961e8774caeaf062f5ed5610)


<img src="https://live.staticflickr.com/65535/54680898938_c3e9bc7f24_b.jpg">


For floating object of more complex geometry, CFD used to be the proper tool which resolves the flow dynamics and is capable to evaluate the submerged volume as well as buoyancy driven torque in detail. The sepcific discipline in the CFD framework involves passive FSI (flow structure interaction) with dynamic meshing technique. 

---
### 3.1.2 Hydraulic systems

Instead of persuing the equilibrium state of hydrostatics, hydraulic system applies the fluid to transmit pressure and create force. 

<img style="float: right;"  src="https://www.aplusphysics.com/courses/honors/fluids/images/PascalPrinciple.png" width = 50%>

<br/>

A simplfied sketch shows the working principle of hydraulic devices. Since

$$
\frac{F_1}{A_1}= p_1 = p_2 = \frac{F_2}{A_2}\quad,
$$

the force acting on the platform is amplified by the ratio of effecting area $A_2$ to $A_1$ and realize the effect of a __hydraulic jack__.


<img  src="http://t3.gstatic.com/licensed-image?q=tbn:ANd9GcQEGPSwy_skL5La5ZXlOj8tczkhrsMFwsPx_RjaQ7qOEIy8YxmXEDJgMh3EIE0YkKnw2Ta9_wvROuWFODJPh0o" width = 46%>
<img src="https://www.researchgate.net/publication/309235120/figure/fig1/AS:418888862580739@1476882295607/Principle-of-hydraulic-jack.png" width = 50.2%>

#### Hydraulic press machine

The same principle is also applied in the hydraulic pressure machine. Within the system, a [hydraulic pump](https://en.wikipedia.org/wiki/Hydraulic_pump) is applied to pressurize hydraulic fluid and delivers it to the hydraulic cylinder through control valves and piping.

<img src="https://redmount.ie/wp-content/uploads/2023/04/Redmount-Press-50-Tonne.jpg" width =32%>
<img style="float: right;" src="https://i.ytimg.com/vi/JxJUPD-Ajnc/maxresdefault.jpg" width= 55%>

Onece the pressurized hydraulic fluid enters the hydraulic cylinder, exerting force on the piston. The force exerted by the hydraulic cylinder is transmitted to the workpiece, doing compressing, shaping, or forming. 

After the desired compression or forming operation is complete, the control valves are adjusted to release hydraulic pressure from the hydraulic cylinder.
The hydraulic fluid returns to the hydraulic reservoir, and the hydraulic cylinder retracts to its initial position.

The hydraulic press machine can perform multiple compression or forming operations in a cyclical manner by pressurizing and releasing hydraulic fluid as needed.

__<ins>Scenario:  Hydrostatic Lift with Load Sensing__

An excavator’s boom cylinder is controlled by a [load-sensing pump](https://www.youtube.com/watch?v=PbXB2VnhpXw). The pressure feedback signal has a time lag $(\tau)$ due to sensor processing and hydraulic line delays, causing oscillations during rapid movements.  
<img src="https://hwpartstore.com/cdn/shop/articles/Excavator-Cylinder-Identification-Diagram-Resized-Compressed_1024x.jpg?v=1628266835" width="80%">

The load force $F_{\text{load}}(t)$ is equal to the weight of boom $(mg)$ and  the digging force $(F_{\text{dig}}(t))$. 
$$F_{\text{load}}(t) =mg +  F_{\text{dig}}(t)$$
However, due to the lag time, the responsing force refers to the prior time, as well as the pressure 
$$P_{\text{response}}(t) = \frac{F_{\text{response}}(t)}{A_p} = \frac{F_{\text{load}}(t-\tau)}{A_p}$$ 

Taking the fluid inside the cylinder as imcompressible. The governing equation yields:
   $$
   m \frac{d² \Delta_x}{dt²} = A_p P_{\text{response}}(t) - F_{\text{load}}(t) - c \frac{d\Delta_x}{dt}    $$ 
   where $c$ is the damping coefficient. we'll dive into this mechanism in the later sections.
   
[<ins>__Sample solution__](https://cocalc.com/share/public_paths/c4b6a4f33f29544705719bfb4a9080446f15fc86)
<img src="https://live.staticflickr.com/65535/54691724466_5af7179243_c.jpg">


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