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and /dev/null differ diff --git a/mechanical/images/vimscrot-2021-12-21T21:43:44,776710459+00:00.png b/mechanical/images/vimscrot-2021-12-21T21:43:44,776710459+00:00.png deleted file mode 100644 index 189dfd5..0000000 Binary files a/mechanical/images/vimscrot-2021-12-21T21:43:44,776710459+00:00.png and /dev/null differ diff --git a/mechanical/mmme1048_fluid_dynamics.md b/mechanical/mmme1048_fluid_dynamics.md deleted file mode 100755 index 45e9e58..0000000 --- a/mechanical/mmme1048_fluid_dynamics.md +++ /dev/null @@ -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. - -
- - -### 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. - - - -![](./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. - -
- -## 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*} diff --git a/mechanical/mmme1048_fluid_mechanics.md b/mechanical/mmme1048_fluid_mechanics.md deleted file mode 100755 index 5eb4f1b..0000000 --- a/mechanical/mmme1048_fluid_mechanics.md +++ /dev/null @@ -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 - - $$p_1 = p_2 + \rho_wg(z_{C2} - z_{C1})$$ - -
- - -### 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$^{-3}$. - - ![](./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*} - -
- -## 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 - -
- - -##### Example 1 - -Determine the location of the centroid of a rectangular lamina. - - - -###### 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 $$x_c = \frac{db^2}{2bd} = \frac b 2$$ - -
- -### 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) - diff --git a/mechanical/mmme1026_maths_for_engineering/calculus.md b/uni/mmme/1026_maths_for_engineering/calculus.md similarity index 100% rename from mechanical/mmme1026_maths_for_engineering/calculus.md rename to uni/mmme/1026_maths_for_engineering/calculus.md diff --git 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