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+---
+author: Alvie Rahman
+date: \today
+title: MMME1048 // Fluid Dynamics
+tags: [ uni, nottingham, mechanical, engineering, fluid_mechanics, mmme1048, fluid_dynamics ]
+---
+
+\newcommand\Rey{\mbox{\textit{Re}}}
+\newcommand\textRey{$\Rey$}
+
+# Introductory Concepts
+
+These are ideas you need to know about to know what's going on, I guess?
+
+## Control Volumes
+
+A control volume is a volume with an imaginary boundry to make it easier to analyze the flow of a
+fluid.
+The boundry is drawn where the properties and conditions of the fluid is known, or where an
+approximation can be made.
+Properties which may be know include:
+
+- Velocity
+- Pressure
+- Temperature
+- Viscosity
+
+The region in the control volume is analyed in terms of enery and mass flows entering and leaving
+the control volumes.
+You don't have to understand what's going on inside the control volume.
+
+
+
+
+### Example 1
+
+The thrust of a jet engine on an aircraft at rest can be analysed in terms of the changes in
+momentum or the air passing through the engine.
+
+
+
+
+
+The control volume is drawn far enough in front of the engine that the air velocity entering can
+be assumed to be at atmospheric pressurce and its velocity negligible.
+
+At the exit of the engine the boundary is drawn close where the velocity is known and the air
+pressure atmospheric.
+
+The control volume cuts the material attaching the engine to the aircraft and there will be a force
+transmitted across the control volume there to oppose the forces on the engine created by thrust
+and gravity.
+
+The details of the flows inside the control volume do not need to be known as the thrust can be
+determined in terms of forces and flows crossing the boundaries drawn.
+However, to understand the flows inside the engine in more detail, a more detailed analysis would
+be required.
+
+
+
+## Ideal Fluid
+
+The actual flow pattern in a fluid is usually complex and difficult to model but it can be
+simplified by assuming the fluid is ideal.
+The ideal fluid has the following properties:
+
+- Zero viscosity
+- Incompressible
+- Zero surface tension
+- Does not change phases
+
+Gases and vapours are compressible so can only be analysed as ideal fluids when flow velocities are
+low but they can often be treated as ideal (or perfect) gases, in which case the ideal gas equations
+apply.
+
+## Steady Flow
+
+Steady flow is a flow which has *no changes in properties with respect to time*.
+Properties may vary from place to place but in the same place the properties must not change in
+the control volume to be steady flow.
+
+Unsteady flow does change with respect to time.
+
+## Uniform Flow
+
+Uniform flow is when all properties are the same at all points at any given instant but can change
+with respect to time, like the opposite of steady flow.
+
+## One Dimensional Flow
+
+In one dimensional (1D) flow it is assumed that all properties are uniform over any plane
+perpedenciular to the direction of flow (e.g. all points along the cross section of a pipe have
+identical properties).
+
+This means properties can only flow in one direction---usually the direction of flow.
+
+1D flow is never achieved exactly in practice as when a fluid flows along a pipe, the velocity at
+the wall is 0, and maximum in the centre of the pipe.
+Despite this, assuming flow is 1D simplifies the analysis and often is accurate enough.
+
+## Flow Patterns
+
+There are multiple ways to visualize flow patterns.
+
+### Streamlines
+
+A streamline is a line along which all the particle have, at a given instant, velocity vectors
+which are tangential to the line.
+
+Therefore there is no component of velocity of a streamline.
+
+A particle can never cross a streamline and *streamlines never cross*.
+
+They can be constructed mathematically and are often shown as output from CFD analysis.
+
+For a steady flow there are no changes with respect to time so the streamline pattern does not.
+The pattern does change when in unsteady flow.
+
+Streamlines in uniform flow must be straight and parallel.
+They must be parallel as if they are not, then different points will have different directions and
+therefore different velocities.
+Same reasoning with if they are not parallel.
+
+### Pathlines
+
+A pathline shows the route taken by a single particle during a given time interval.
+It is equivalent to a high exposure photograph which traces the moevement of the particle marked.
+You could track pathlines with a drop of injected dye or inserting a buoyant solid particle which
+has the same density as the solid.
+
+Pathlines may cross.
+
+### Streaklines
+
+A streakline joins, at any given time, all particles that have passed through a given point.
+Examples of this are line dye or a smoke stream which is produced from a continuous supply.
+
+## Viscous (Real) Fluids
+
+### Viscosity
+
+A fluid offers resisistance to motion due to its viscosity or internal friction.
+The greater the resistance to flow, the greater the viscosity.
+
+Higher viscosity also reduces the rate of shear deformation between layers for a given shear stress.
+
+Viscosity comes from two effects:
+
+- In liquids, the inter-molecular forces act as drag between layers of fluid moving at different
+ velocities
+- In gases, the mixing of faster and slower moving fluid causes friction due to momentum transfer.
+ The slower layers tend to slow down the faster ones
+
+### Newton's Law of Viscosity
+
+Viscosity can be defined in terms of rate of shear or velocity gradient.
+
+
+
+Consider the flow in the pipe above.
+Fluid in contact with the surface has a velocity of 0 because the surface irregularities trap the
+fluid particles.
+A short distance away from the surface the velocity is low but in the middle of the pipe the
+velocity is $v_F$.
+
+Let the velocity at a distance $y$ be $v$ and at a distance $y + \delta y$ be $v + \delta v$.
+
+The ratio $\frac{\delta v}{\delta y}$ is the average velocity gradient over the distance
+$\delta y$.
+
+But as $\delta y$ tends to zero, $\frac{\delta v}{\delta y} \rightarrow$ the value of the
+differential $\frac{\mathrm{d}v}{\mathrm{d}y}$ at a point such as point A.
+
+For most fluids in engineering it is found that the shear stress, $\tau$, is directly proportional
+to the velocity gradient when straight and parallel flow is involved:
+
+$$\tau = \mu\frac{\mathrm{d}v}{\mathrm{d}y}$$
+
+Where $\mu$ is the constant of proportinality and known as the dynamic viscosity, or simply the
+viscosity of the fluid.
+
+This is Newton's Law of Viscosity and fluids that ovey it are known as Newtonian fluids.
+
+### Viscosity and Lubrication
+
+Where a fluid is a thin film (such as in lubricating flows), the velocity gradient can be
+approximated to be linear and an estimate of shear stress obtained:
+
+$$\tau = \mu \frac{\delta v}{\delta y} \approx \mu \frac{v}{y}$$
+
+From the shear stress we can calculate the force exerted by a film by the relationship:
+
+$$\tau = \frac F A$$
+
+# Fluid Flow
+
+## Types of flow
+
+There are essentially two types of flow:
+
+- Smooth (laminar) flow
+
+ At low flow rates, particles of fluid are moving in straight lines and can be considered to be
+ moving in layers or laminae.
+
+- Rough (turbulent) flow
+
+ At higher flow rates, the paths of the individual fluid particles are not straight but disorderly
+ resulting in mixing taking place
+
+Between fully laminar and fully turbulent flows is a transition region.
+
+## The Reynolds Number
+
+### Development of the Reynolds Number
+
+In laminar flow the most influentialfactor is the magnitude of the viscous forces:
+
+$$viscous\, forces \propto \mu\frac v l l^2 = \mu vl$$
+
+where $v$ is a characteristic velocit and $l$ is a characteristic length.
+
+In turbulent flow viscous effects are not significant but inertia effects (mixing, momentum
+exchange, acceleration of fluid mass) are.
+Interial forces can be represented by $F = ma$
+
+\begin{align*}
+m &\propto \rho l^3 \\
+a &= \frac{dv}{dt} \\
+&\therefore a \propto \frac v t \text{ and } t = \frac l v \\
+&\therefore a \propto \frac {v^2} l \\
+&\therefore \text{Interial forces} \propto \rho l^2\frac{v^2} l = \rho l^2v^2
+\end{align*}
+
+The ratio of internalforces to viscous forces is called the Reynolds number and is abbreviated to
+Re:
+
+$$\Rey = \frac{\text{interial forces}}{\text{viscous forces}} = \frac {\rho l^2v^2}{\mu vl} = \frac {\rho vl} \mu$$
+
+where $\rho$ and $\mu$ are fluid properties and $v$ and $l$ are characteristic velocity and length.
+
+- During laminar flow, $\Rey$ is small as viscous forces dominate.
+- During turbulent flow, $\Rey$ is large as intertial forces dominate.
+
+\textRey is a non dimensional group.
+It has no units because the units cancel out.
+
+Non dimensional groups are very important in fluid mechancics and need to be considered when scaling
+experiments.
+
+If \textRey is the same in two different pipes, the flow will be the same regardless of actual
+diameters, densities, or other properties.
+
+#### \textRey for a Circular Section Pipe
+
+The characteristic length for pipe flow is the diameter $d$ and the characteristic velocity is
+mean flow in the pipe, $v$, so \textRey of a circular pipe section is given by:
+
+$$\Rey = \frac{\rho vd} \mu$$
+
+For flow in a smooth circular pipe under normal engineering conditions the following can be assumed:
+
+- $\Rey < 2000$ --- laminar flow
+- $2000 < \Rey < 4000$ --- transition
+- $\Rey > 4000$ --- fully turbulent flow
+
+These figures can be significantly affected by surface roughness so flow may be turbulent below
+$\Rey = 4000$.
+
+# Euler's Equation
+
+In a static fluid, pressure only depends on density and elevation.
+In a moving fluid the pressure is also related to acceleration, viscosity, and shaft work done on or
+by the fluid.
+
+$$\frac 1 \rho \frac{\delta p}{\delta s} + g\frac{\delta z}{\delta s} + v\frac{\delta v}{\delta s} = 0$$
+
+## Assumptions / Conditions
+
+The Euler euqation applies where the following can be assumed:
+
+- Steady flow
+- The fluid is inviscid
+- No shaft work
+- Flow along a streamline
+
+# Bernoulli's Equation
+
+Euler's equation comes in differential form, which is difficult to apply.
+We can integrate it to make it easier
+
+\begin{align*}
+\frac 1 \rho \frac{\delta p}{\delta s} + g\frac{\delta z}{\delta s} + v\frac{\delta v}{\delta s} &= 0
+ & \text{(Euler's equation)} \\
+\int\left\{\frac{\mathrm{d}p} \rho + g\mathrm{d}z + v\mathrm{d}v \right\} &= \int 0 \,\mathrm{d}s \\
+\therefore \int \frac 1 \rho \,\mathrm{d}p + g\int \mathrm{d}z + \int v \,\mathrm{d}v &= \int 0 \,\mathrm{d}s \\
+\therefore \int \frac 1 \rho \,\mathrm{d}p + gz + \frac{v^2}{2} &= \text{constant}_1
+\end{align*}
+
+The first term of the equation can only be integrated if $\rho$ is constant as then:
+
+$$\int \frac 1 \rho \,\mathrm{d}p = \frac 1 \rho \int \mathrm{d}p = \frac p \rho$$
+
+So, if density is constant:
+
+$$\frac p \rho + gz + \frac{v^2}{2} = \text{constant}_2$$
+
+## Assumptions / Conditions
+
+All the assumptions from Euler's equation apply:
+
+- Steady flow
+- The fluid is inviscid
+- No shaft work
+- Flow along a streamline
+
+But also one more:
+
+- Incompressible flow
+
+## Forms of Bernoulli's Equation
+
+### Energy Form
+
+This form of Bernoulli's Equation is known as the energy form as each component has the units
+energy/unit mass:
+
+$$\frac p \rho + gz + \frac{v^2}{2} = \text{constant}_2$$
+
+It is split into 3 parts:
+
+- Pressure energy ($\frac p \rho$) --- energy needed to move the flow against the pressure
+ (flow work)
+- Potential energy ($gz$) --- elevation
+- Kinetic energy ($\frac{v^2}{2}$) --- kinetic energy
+
+### Elevation / Head Form
+
+Divide the energy form by $g$:
+
+$$\frac\rho{\rho g} + z + \frac{v^2}{2g} = H_T$$
+
+where $H_T$ is constant and:
+
+- $\frac{p}{\rho g}$ --- static/pressure haed
+- $z$ --- elevation head
+- $\frac{v_2}{2g}$ --- dynamic/velocity head
+- $H_T$ --- total head
+
+- Each term now has units of elevations
+- In fluids the elevation is sometimes called head
+- This form of the equation is also useful in some applications
+
+### Pressure Form
+
+Multiply the energy form by $\rho$ to give the pressure form:
+
+$$p + \rho gz + \frac 1 2 \rho v^2 = \text{constant}$$
+
+where:
+
+- $p$ --- static pressure (often written as $p_s$)
+- $\rho gz$ --- elevation pressure
+- $\frac 1 2 \rho v^2$ --- dynamic pressure
+
+- Density is constant
+- Each term now has the units of pressure
+- This form is useful is we are interested in pressures
+
+### Comparing two forms of the Bernoulli Equation (Piezometric)
+
+$$\text{piezometric} = \text{static} + \text{elevation}$$
+
+Pressure form:
+
+\begin{align*}
+p_s + \rho gz + \frac 1 2 \rho v^2 &= \text{total pressure} \\
+p_s + \rho gz &= \text{piezometric pressure}
+\end{align*}
+
+Head form:
+
+\begin{align*}
+\frac{p_s}{\rho g} + z + \frac{v^2}{2g} &= \text{total head} \\
+\frac{p_s}{\rho g} + z &= \text{piezometric head}
+\end{align*}
diff --git a/uni/mmme/1048_thermodynamics_and_fluid_mechanics/fluid_mechanics.md b/uni/mmme/1048_thermodynamics_and_fluid_mechanics/fluid_mechanics.md
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+---
+author: Alvie Rahman
+date: \today
+title: MMME1048 // Fluid Mechanics Intro and Statics
+tags: [ uni, nottingham, mechanical, engineering, fluid_mechanics, mmme1048, fluid_statics ]
+---
+
+# Properties of Fluids
+
+## What is a Fluid?
+
+- A fluid may be liquid, vapor, or gas
+- No permanent shape
+- Consists of atoms in random motion and continual collision
+- Easy to deform
+- Liquids have fixed volume, gasses fill up container
+- **A fluid is a substance for wich a shear stress tends to produce unlimited, continuous
+ deformation**
+
+## Shear Forces
+
+- For a solid, application of shear stress causes a deformation which, if not too great (elastic),
+ is not permanent and solid regains original positon
+- For a fluid, continuious deformation takes place as the molecules slide over each other until the
+ force is removed
+- **A fluid is a substance for wich a shear stress tends to produce unlimited, continuous
+ deformation**
+
+## Density
+
+- Density: $$ \rho = \frac m V $$
+- Specific Density: $$ v = \frac 1 \rho $$
+
+### Obtaining Density
+
+- Find mass of a given volume or volume of a given mass
+- This gives average density and assumes density is the same throughout
+
+ - This is not always the case (like in chocolate chip ice cream)
+ - Bulk density is often used to refer to average density
+
+### Engineering Density
+
+- Matter is not continuous on molecular scale
+- For fluids in constant motion, we take a time average
+- For most practical purposes, matter is considered to be homogenous and time averaged
+
+## Pressure
+
+- Pressure is a scalar quantity
+- Gases cannot sustain tensile stress, liquids a negligible amount
+
+- There is a certain amount of energy associated with the random continuous motion of the molecules
+- Higher pressure fluids tend to have more energy in their molecules
+
+### How Does Molecular Motion Create Force?
+
+- When molecules interact with each other, there is no net force
+- When they interact with walls, there is a resultant force perpendicular to the surface
+- Pressure caused my molecule: $$ p = \frac {\delta{}F}{\delta{}A} $$
+- If we want total force, we have to add them all up
+- $$ F = \int \mathrm{d}F = \int p\, \mathrm{d}A $$
+
+ - If pressure is constant, then this integrates to $$ F = pA $$
+ - These equations can be used if pressure is constant of average value is appropriate
+ - For many cases in fluids pressure is not constant
+
+### Pressure Variation in a Static Fluid
+
+- A fluid at rest has constant pressure horizontally
+- That's why liquid surfaces are flat
+- But fluids at rest do have a vertical gradient, where lower parts have higher presure
+
+### How Does Pressure Vary with Depth?
+
+
+
+Let fluid pressure be p at height $z$, and $p + \delta p$ at $z + \delta z$.
+
+Force $F_z$ acts upwards to support the fluid, countering pressure $p$.
+
+Force $F_z + \delta F_z$acts downwards to counter pressure $p + \delta p$ and comes from the weight
+of the liquid above.
+
+Now:
+
+\begin{align*}
+F_z &= p\delta x\delta y \\
+F_z + \delta F_z &= (p + \delta p) \delta x \delta y \\
+\therefore \delta F_z &= \delta p(\delta x\delta y)
+\end{align*}
+
+Resolving forces in z direction:
+
+\begin{align*}
+F_z - (F_z + \delta F_z) - g\delta m &= 0 \\
+\text{but } \delta m &= \rho\delta x\delta y\delta z \\
+\therefore -\delta p(\delta x\delta y) &= g\rho(\delta x\delta y\delta z) \\
+\text{or } \frac{\delta p}{\delta z} &= -\rho g \\
+\text{as } \delta z \rightarrow 0,\, \frac{\delta p}{\delta z} &\rightarrow \frac{dp}{dz}\\
+\therefore \frac{dp}{dz} &= -\rho g\\
+\Delta p &= \rho g\Delta z
+\end{align*}
+
+The equation applies for any fluid.
+The -ve sign indicates that as $z$, height, increases, $p$, pressure, decreases.
+
+### Absolute and Gauge Pressure
+
+- Absolute Pressure is measured relative to zero (a vacuum)
+- Guage pressure = absolute pressure - atmospheric pressure
+
+ - Often used in industry
+
+- If abs. pressure = 3 bar and atmospheric pressure is 1 bar, then gauge pressure = 2 bar
+- Atmospheric pressure changes with altitude
+
+## Compressibility
+
+- All fluids are compressible, especially gasses
+- Most liquids can be considered **incompressible** most of the time (and will be in MMME1048, but
+ may not be in future modules)
+
+## Surface Tension
+
+- In a liquid, molecules are held together by molecular attraction
+- At a boundry between two fluids this creates "surface tension"
+- Surface tension usually has the symbol $$\gamma$$
+
+## Ideal Gas
+
+- No real gas is perfect, although many are similar
+- We define a specific gas constant to allow us to analyse the behaviour of a specific gas:
+
+ $$ R = \frac {\tilde R}{\tilde m} $$
+
+ (Universal Gas Constant / molar mass of gas)
+
+- Perfect gas law
+
+ $$pV=mRT$$
+
+ or
+
+ $$ p = \rho RT$$
+
+ - Pressure always in Pa
+ - Temperature always in K
+
+## Units and Dimentional Analysis
+
+- It is usually better to use SI units
+- If in doubt, DA can be useful to check that your answer makes sense
+
+# Fluid Statics
+
+## Manometers
+
+
+
+$$p_{1,gauge} = \rho g(z_2-z_1)$$
+
+- Manometers work on the principle that pressure along any horizontal plane through a continuous
+ fluid is constant
+- Manometers can be used to measure the pressure of a gas, vapour, or liquid
+- Manometers can measure higher pressures than a piezometer
+- Manometer fluid and working should be immiscible (don't mix)
+
+
+
+\begin{align*}
+p_A &= p_{A'} \\
+p_{bottom} &= p_{top} + \rho gh \\
+\rho_1 &= density\,of\,fluid\,1 \\
+\rho_2 &= density\,of\,fluid\,2
+\end{align*}
+
+Left hand side:
+
+$$p_A = p_1 + \rho_1g\Delta z_1$$
+
+Right hand side:
+
+$$p_{A'} = p_{at} + \rho_2g\Delta z_2$$
+
+Equate and rearrange:
+
+\begin{align*}
+p_1 + \rho_1g\Delta z_1 &= p_{at} + \rho_2g\Delta z_2 \\
+p_1-p_{at} &= g(\rho_2\Delta z_2 - \rho_1\Delta z_1) \\
+p_{1,gauge} &= g(\rho_2\Delta z_2 - \rho_1\Delta z_1)
+\end{align*}
+
+If $\rho_a << \rho_2$:
+
+$$\rho_{1,gauge} \approx \rho_2g\Delta z_2$$
+
+### Differential U-Tube Manometer
+
+
+
+- Used to find the difference between two unknown pressures
+- Can be used for any fluid that doesn't react with manometer fluid
+- Same principle used in analysis
+
+\begin{align*}
+p_A &= p_{A'} \\
+p_{bottom} &= p_{top} + \rho gh \\
+\rho_1 &= density\,of\,fluid\,1 \\
+\rho_2 &= density\,of\,fluid\,2
+\end{align*}
+
+Left hand side:
+
+$$p_A = p_1 + \rho_wg(z_C-z_A)$$
+
+Right hand side:
+
+$$p_B = p_2 + \rho_wg(z_C-z_B)$$
+
+Right hand manometer fluid:
+
+$$p_{A'} = p_B + \rho_mg(z_B - z_a)$$
+
+\begin{align*}
+p_{A'} &= p_2 + \rho_mg(z_C - z_B) + \rho_mg(z_B - zA)\\
+ &= p_2 + \rho_mg(z_C - z_B) + \rho_mg\Delta z \\
+\\
+p_A &= p_{A'} \\
+p_1 + \rho_wg(z_C-z_A) &= p_2 + \rho_mg(z_C - z_B) + \rho_mg\Delta z \\
+p_1 - p_2 &= \rho_wg(z_C-z_B-z_C+z_A) + \rho_mg\Delta z \\
+&= \rho_wg(z_A-z_B) + \rho_mg\Delta z \\
+&= -\rho_wg\Delta z + \rho_mg\Delta z
+\end{align*}
+
+### Angled Differential Manometer
+
+
+
+- If the pipe is sloped then
+
+ $$p_1-p_2 = (\rho_m-\rho_w)g\Delta z + \rho_wg(z_{C2} - z_{C1})$$
+
+- $p_1 > p_2$ as $p_1$ is lower
+- If there is no flow along the tube, then
+
+ $$p_1 = p_2 + \rho_wg(z_{C2} - z_{C1})$$
+
+
+
+
+### Exercise Sheet 1
+
+
+
+1. If 4 m$^3$ of oil weighs 35 kN calculate its density and relative density.
+ Relative density is a term used to define the density of a fluid relative
+
+ > $$ \frac{35000}{9.81\times4} = 890 \text{ kgm}^{-3} $$
+ >
+ > $$1000 - 891.9... = 108 \text{ kgm}^{-3}$$
+
+2. Find the pressure relative to atmospheric experienced by a diver
+ working on the sea bed at a depth of 35 m.
+ Take the density of sea water to be 1030 kgm$^{-3}$.
+
+ > $$
+ > \rho gh = 1030\times 9.81 \times 35 = 3.5\times10^5
+ > $$
+
+3. An open glass is sitting on a table, it has a diameter of 10 cm.
+ If water up to a height of 20 cm is now added calculate the force exerted onto the table by
+ the addition of the water.
+
+ > $$V_{cylinder} = \pi r^2h$$
+ > $$m_{cylinder} = \rho\pi r^2h$$
+ > $$W_{cylinder} = \rho g\pi r^2h = 1000\times9.8\pi\times0.05^2\times0.2 = 15.4 \text{ N} $$
+
+4. A closed rectangular tank with internal dimensions 6m x 3m x 1.5m
+ high has a vertical riser pipe of cross-sectional area 0.001 m2 in
+ the upper surface (figure 1.4). The tank and riser are filled with
+ water such that the water level in the riser pipe is 3.5 m above the
+
+ Calulate:
+
+ i. The gauge pressure at the base of the tank.
+
+ > $$\rho gh = 1000\times9.81\times(1.5+3.5) = 49 \text{ kPa}$$
+
+ ii. The gauge pressure at the top of the tank.
+
+ > $$\rho gh = 1000\times9.81\times3.5 = 34 \text{ kPa}$$
+
+ iii. The force exercted on the base of the tank due to gauge water pressure.
+
+ > $$F = p\times A = 49\times10^3\times6\times3 = 8.8\times10^5 \text{ N}$$
+
+ iv. The weight of the water in the tank and riser.
+
+ > $$V = 6\times3\times1.5 + 0.001\times3.5 = 27.0035$$
+ > $$W = \rho gV = 1000\times9.8\times27.0035 = 2.6\times10^5\text{ N}$$
+
+ v. Explain the difference between (iii) and (iv).
+
+ *(It may be helpful to think about the forces on the top of the tank)*
+
+ > The pressure at the top of the tank is higher than atmospheric pressure because of the
+ > riser.
+ > This means there is an upwards force on the top of tank.
+ > The difference between the force acting up and down due to pressure is equal to the
+ > weight of the water.
+
+6. A double U-tube manometer is connected to a pipe as shown below.
+
+ Taking the dimensions and fluids as indicated; calculate
+ the absolute pressure at point A (centre of the pipe).
+
+ Take atmospheric pressure as 1.01 bar and the density of mercury as 13600 kgm$^{-3}$.
+
+ 
+
+ > \begin{align*}
+ P_B &= P_A + 0.4\rho_wg &\text{(1)}\\
+ P_C = P_{C'} &= P_B - 0.2\rho_mg &\text{(2)}\\
+ P_D = P_{at} &= P_A + 0.4\rho_wg &\text{(3)}\\
+ \\
+ \text{(2) into (3)} &\rightarrow P_B -0.2\rho_mg-0.1\rho_wg = P_{at} &\text{(4)}\\
+ \text{(1) into (4)} &\rightarrow P_A + 0.4\rho_wg - 0.2\rho_mg - 0.1\rho_wg = P_{at} \\
+ \\
+ P_A &= P_{at} + 0.1\rho_wg + 0.2\rho_mg - 0.4\rho_wg \\
+ &= P_{at} + g(0.2\rho_m - 0.3\rho_w) \\
+ &= 1.01\times10^5 + 9.81(0.2\times13600 - 0.3\times1000)\\
+ &= 124.7\text{ kPa}
+ > \end{align*}
+
+
+
+## Submerged Surfaces
+
+### Prepatory Maths
+
+#### Integration as Summation
+
+#### Centroids
+
+- For a 3D body, the centre of gravity is the point at which all the mass can be considered to act
+- For a 2D lamina (thin, flat plate) the centroid is the centre of area, the point about which the
+ lamina would balance
+
+To find the location of the centroid, take moments (of area) about a suitable reference axis:
+
+$$moment\,of\,area = moment\,of\,mass$$
+
+(making the assumption that the surface has a unit mass per unit area)
+
+$$moment\,of\,mass = mass\times distance\,from\,point\,acting\,around$$
+
+Take the following lamina:
+
+
+
+1. Split the lamina into elements parallel to the chosen axis
+2. Each element has area $\delta A = w\delta y$
+3. The moment of area ($\delta M$) of the element is $\delta Ay$
+4. The sum of moments of all the elements is equal to the moment $M$ obtained by assuing all the
+ area is located at the centroid or:
+
+ $$Ay_c = \int_{area} \! y\,\mathrm{d}A$$
+
+ or:
+
+ $$y_c = \frac 1 A \int_{area} \! y\,\mathrm{d}A$$
+
+ - $\int y\,\mathrm{d}A$ is known as the first moment of area
+
+
+
+
+##### Example 1
+
+Determine the location of the centroid of a rectangular lamina.
+
+
+
+###### Determining Location in $y$ direction
+
+
+
+1. Take moments for area about $OO$
+
+ $$\delta M = y\delta A = y(b\delta y)$$
+
+2. Integrate to find all strips
+
+ $$M = b\int_0^d\! y \,\mathrm{d}y = b\left[\frac{y^2}2\right]_0^d = \frac{bd^2} 2$$
+
+ ($b$ can be taken out the integral as it is constant in this example)
+
+ but also $$M = (area)(y_c) = bdy_c$$
+
+ so $$y_c = \frac 1 {bd} \frac {bd} 2 = \frac d 2$$
+
+###### Determining Location in $x$ direction
+
+
+
+1. Take moments for area about $O'O'$:
+
+ $$\delta M = x\delta A = x(d\delta x)$$
+
+2. Integrate
+
+ $$M_{O'O'} = d\int_0^b\! x \,\mathrm{d}x = d\left[\frac{x^2} 2\right]_0^b = \frac{db^2} 2$$
+
+ but also $$M_{O'O'} = (area)(x_c) = bdx_c$$
+
+ so $$x_c = \frac{db^2}{2bd} = \frac b 2$$
+
+
+
+### Horizontal Submereged Surfaces
+
+
+
+Assumptions for horizontal lamina:
+
+- Constant pressure acts over entire surface of lamina
+- Centre of pressure will coincide with centre of area
+- $total\,force = pressure\times area$
+
+
+
+### Vertical Submerged Surfaces
+
+
+
+- A vertical submerged plate does experience uniform pressure
+- Centroid of pressure and area are not coincident
+- Centroid of pressure is always below centroid of area for a vertical plate
+- No shear forces, so all hydrostatic forces are perpendicular to lamina
+
+
+
+Force acting on small element:
+
+\begin{align*}
+\delta F &= p\delta A \\
+&= \rho gh\delta A \\
+&= \rho gh w\delta h
+\end{align*}
+
+Therefore total force is
+
+$$F_p = \int_{area}\! \rho gh \,\mathrm{d}A = \int_{h_1}^{h_2}\! \rho ghw\,\mathrm{d}h$$
+
+#### Finding Line of Action of the Force
+
+
+
+\begin{align*}
+\delta M_{OO} &= \delta Fh = (\rho gh\delta A)h \\
+&= \rho gh^2\delta A = \rho gh^2w\delta h \\
+\\
+M_{OO} &= F_py_p = \int_{area}\! \rho gh^2 \,\mathrm{d}A \\
+&= \int_{h1}^{h2}\! \rho gh^2w \,\mathrm{d}h \\
+\\
+y_p = \frac{M_{OO}}{F_p}
+\end{align*}
+
+## Buoyancy
+
+### Archimedes Principle
+
+> The resultant upwards force (buoyancy force) on a body wholly or partially immersed in a fluid is
+> equal to the weight of the displaced fluid.
+
+When an object is in equilibrium the forces acting on it balance.
+For a floating object, the upwards force equals the weight:
+
+$$mg = \rho Vg$$
+
+Where $\rho$ is the density of the fluid, and $V$ is the volume of displaced fluid.
+
+### Immersed Bodies
+
+As pressure increases with depth, the fluid exerts a resultant upward force on a body.
+There is no horizontal component of the buoyancy force because the vertiscal projection of the body
+is the same in both directions.
+
+### Rise, Sink, or Float?
+
+- $F_B = W$ \rightarrow equilirbrium (floating)
+- $F_B > W$ \rightarrow body rises
+- $F_B < W$ \rightarrow body sinks
+
+### Centre of Buoyancy
+
+Buoyancy force acts through the centre of gravity of the volume of fluid displaced.
+This is known as the centre of buoyancy.
+The centre of buoyancy does not in general correspond to the centre of gravity of the body.
+
+If the fluid density is constant the centre of gravity of the displaced fluid is at the centroid of
+the immersed volume.
+
+
+
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