big rename

mechanical/mmme1048_fluid_dynamics.md

@@ -1,386 +0,0 @@
----
-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.
-
-<details>
-<summary>
-
-### 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.
-
-</summary>
-
-![](./images/vimscrot-2021-11-03T21:51:51,497459693+00:00.png)
-
-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.
-
-</details>
-
-## 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.
-
-![](./images/vimscrot-2021-11-17T14:14:05,079195275+00:00.png)
-
-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*}

mechanical/mmme1048_fluid_mechanics.md

@@ -1,506 +0,0 @@
----
-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?
-
-![From UoN MMME1048 Fluid Mechanics Notes](./images/vimscrot-2021-10-06T10:51:51,499044519+01:00.png)
-
-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
-
-![](./images/vimscrot-2021-10-13T09:09:32,037006075+01:00.png)
-
-$$p_{1,gauge} = \rho g(z_2-z_1)$$
-
-- 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)
-
-![](./images/vimscrot-2021-10-13T09:14:59,628661490+01:00.png)
-
-\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
-
-![](./images/vimscrot-2021-10-13T09:37:02,070474894+01:00.png)
-
-- 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
-
-![](./images/vimscrot-2021-10-13T09:56:15,656796805+01:00.png)
-
-- 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
-
-  $$p_1 = p_2 + \rho_wg(z_{C2} - z_{C1})$$
-
-<details>
-<summary>
-
-### Exercise Sheet 1
-
-</summary>
-
-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$^{-3}$.
-
-   ![](./images/vimscrot-2021-10-13T10:42:52,999793176+01:00.png)
-
-   > \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*}
-
-</details>
-
-## 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:
-
-![](./images/vimscrot-2021-10-20T10:01:30,080819382+01:00.png)
-
-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
-
-<details>
-<summary>
-
-##### Example 1
-
-Determine the location of the centroid of a rectangular lamina.
-
-</summary>
-
-###### Determining Location in $y$ direction
-
-![](./images/vimscrot-2021-10-20T10:14:17,688774145+01:00.png)
-
-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
-
-![](./images/vimscrot-2021-10-20T10:24:48,372189101+01:00.png)
-
-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 $$x_c = \frac{db^2}{2bd} = \frac b 2$$
-
-</details>
-
-### Horizontal Submereged Surfaces
-
-![](./images/vimscrot-2021-10-20T10:33:16,783724117+01:00.png)
-
-Assumptions for horizontal lamina:
-
-- Constant pressure acts over entire surface of lamina
-- Centre of pressure will coincide with centre of area
-- $total\,force = pressure\times area$
-
-![](./images/vimscrot-2021-10-20T10:36:12,520683729+01:00.png)
-
-### Vertical Submerged Surfaces
-
-![](./images/vimscrot-2021-10-20T11:05:33,235642932+01:00.png)
-
-- 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
-
-![](./images/vimscrot-2021-10-20T11:07:52,929126609+01:00.png)
-
-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
-
-![](./images/vimscrot-2021-10-20T11:15:51,200869760+01:00.png)
-
-\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.
-
-![](./images/vimscrot-2021-12-21T15:08:22,285753421+00:00.png)
-