467 lines
14 KiB
Markdown
Executable File
467 lines
14 KiB
Markdown
Executable File
---
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author: Alvie Rahman
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date: \today
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title: MMME1048 // Fluid Mechanics
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tags: [ uni, nottingham, mechanical, engineering, fluid_mechanics, mmme1048 ]
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---
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# Lecture 1 // Properties of Fluids (2021-10-06)
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## What is a Fluid?
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- A fluid may be liquid, vapor, or gas
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- No permanent shape
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- Consists of atoms in random motion and continual collision
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- Easy to deform
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- Liquids have fixed volume, gasses fill up container
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- **A fluid is a substance for wich a shear stress tends to produce unlimited, continuous
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deformation**
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## Shear Forces
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- For a solid, application of shear stress causes a deformation which, if not too great (elastic),
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is not permanent and solid regains original positon
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- For a fluid, continuious deformation takes place as the molecules slide over each other until the
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force is removed
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- **A fluid is a substance for wich a shear stress tends to produce unlimited, continuous
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deformation**
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## Density
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- Density: $$ \rho = \frac m V $$
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- Specific Density: $$ v = \frac 1 \rho $$
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### Obtaining Density
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- Find mass of a given volume or volume of a given mass
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- This gives average density and assumes density is the same throughout
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- This is not always the case (like in chocolate chip ice cream)
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- Bulk density is often used to refer to average density
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### Engineering Density
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- Matter is not continuous on molecular scale
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- For fluids in constant motion, we take a time average
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- For most practical purposes, matter is considered to be homogenous and time averaged
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## Pressure
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- Pressure is a scalar quantity
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- Gases cannot sustain tensile stress, liquids a negligible amount
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- There is a certain amount of energy associated with the random continuous motion of the molecules
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- Higher pressure fluids tend to have more energy in their molecules
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### How Does Molecular Motion Create Force?
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- When molecules interact with each other, there is no net force
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- When they interact with walls, there is a resultant force perpendicular to the surface
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- Pressure caused my molecule: $$ p = \frac {\delta{}F}{\delta{}A} $$
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- If we want total force, we have to add them all up
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- $$ F = \int \mathrm{d}F = \int p\, \mathrm{d}A $$
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- If pressure is constant, then this integrates to $$ F = pA $$
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- These equations can be used if pressure is constant of average value is appropriate
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- For many cases in fluids pressure is not constant
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### Pressure Variation in a Static Fluid
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- A fluid at rest has constant pressure horizontally
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- That's why liquid surfaces are flat
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- But fluids at rest do have a vertical gradient, where lower parts have higher presure
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### How Does Pressure Vary with Depth?
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Let fluid pressure be p at height $z$, and $p + \delta p$ at $z + \delta z$.
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Force $F_z$ acts upwards to support the fluid, countering pressure $p$.
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Force $F_z + \delta F_z$acts downwards to counter pressure $p + \delta p$ and comes from the weight
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of the liquid above.
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Now:
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\begin{align*}
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F_z &= p\delta x\delta y \\
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F_z + \delta F_z &= (p + \delta p) \delta x \delta y \\
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\therefore \delta F_z &= \delta p(\delta x\delta y)
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\end{align*}
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Resolving forces in z direction:
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\begin{align*}
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F_z - (F_z + \delta F_z) - g\delta m &= 0 \\
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\text{but } \delta m &= \rho\delta x\delta y\delta z \\
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\therefore -\delta p(\delta x\delta y) &= g\rho(\delta x\delta y\delta z) \\
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\text{or } \frac{\delta p}{\delta z} &= -\rho g \\
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\text{as } \delta z \rightarrow 0,\, \frac{\delta p}{\delta z} &\rightarrow \frac{dp}{dz}\\
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\therefore \frac{dp}{dz} &= -\rho g\\
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\Delta p &= \rho g\Delta z
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\end{align*}
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The equation applies for any fluid.
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The -ve sign indicates that as $z$, height, increases, $p$, pressure, decreases.
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### Absolute and Gauge Pressure
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- Absolute Pressure is measured relative to zero (a vacuum)
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- Guage pressure = absolute pressure - atmospheric pressure
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- Often used in industry
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- If abs. pressure = 3 bar and atmospheric pressure is 1 bar, then gauge pressure = 2 bar
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- Atmospheric pressure changes with altitude
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## Compressibility
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- All fluids are compressible, especially gasses
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- Most liquids can be considered **incompressible** most of the time (and will be in MMME1048, but
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may not be in future modules)
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## Surface Tension
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- In a liquid, molecules are held together by molecular attraction
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- At a boundry between two fluids this creates "surface tension"
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- Surface tension usually has the symbol $$\gamma$$
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## Ideal Gas
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- No real gas is perfect, although many are similar
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- We define a specific gas constant to allow us to analyse the behaviour of a specific gas:
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$$ R = \frac {\tilde R}{\tilde m} $$
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(Universal Gas Constant / molar mass of gas)
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- Perfect gas law
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$$pV=mRT$$
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or
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$$ p = \rho RT$$
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- Pressure always in Pa
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- Temperature always in K
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## Units and Dimentional Analysis
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- It is usually better to use SI units
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- If in doubt, DA can be useful to check that your answer makes sense
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# Lecture 2 // Manometers (2021-10-13)
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$$p_{1,gauge} = \rho g(z_2-z_1)$$
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- Manometers work on the principle that pressure along any horizontal plane through a continuous
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fluid is constant
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- Manometers can be used to measure the pressure of a gas, vapour, or liquid
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- Manometers can measure higher pressures than a piezometer
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- Manometer fluid and working should be immiscible (don't mix)
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\begin{align*}
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p_A &= p_{A'} \\
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p_{bottom} &= p_{top} + \rho gh \\
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\rho_1 &= density\,of\,fluid\,1 \\
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\rho_2 &= density\,of\,fluid\,2
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\end{align*}
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Left hand side:
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$$p_A = p_1 + \rho_1g\Delta z_1$$
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Right hand side:
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$$p_{A'} = p_{at} + \rho_2g\Delta z_2$$
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Equate and rearrange:
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\begin{align*}
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p_1 + \rho_1g\Delta z_1 &= p_{at} + \rho_2g\Delta z_2 \\
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p_1-p_{at} &= g(\rho_2\Delta z_2 - \rho_1\Delta z_1) \\
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p_{1,gauge} &= g(\rho_2\Delta z_2 - \rho_1\Delta z_1)
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\end{align*}
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If $\rho_a << \rho_2$:
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$$\rho_{1,gauge} \approx \rho_2g\Delta z_2$$
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## Differential U-Tube Manometer
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- Used to find the difference between two unknown pressures
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- Can be used for any fluid that doesn't react with manometer fluid
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- Same principle used in analysis
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\begin{align*}
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p_A &= p_{A'} \\
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p_{bottom} &= p_{top} + \rho gh \\
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\rho_1 &= density\,of\,fluid\,1 \\
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\rho_2 &= density\,of\,fluid\,2
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\end{align*}
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Left hand side:
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$$p_A = p_1 + \rho_wg(z_C-z_A)$$
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Right hand side:
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$$p_B = p_2 + \rho_wg(z_C-z_B)$$
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Right hand manometer fluid:
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$$p_{A'} = p_B + \rho_mg(z_B - z_a)$$
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\begin{align*}
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p_{A'} &= p_2 + \rho_mg(z_C - z_B) + \rho_mg(z_B - zA)\\
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&= p_2 + \rho_mg(z_C - z_B) + \rho_mg\Delta z \\
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\\
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p_A &= p_{A'} \\
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p_1 + \rho_wg(z_C-z_A) &= p_2 + \rho_mg(z_C - z_B) + \rho_mg\Delta z \\
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p_1 - p_2 &= \rho_wg(z_C-z_B-z_C+z_A) + \rho_mg\Delta z \\
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&= \rho_wg(z_A-z_B) + \rho_mg\Delta z \\
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&= -\rho_wg\Delta z + \rho_mg\Delta z
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\end{align*}
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## Angled Differential Manometer
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- If the pipe is sloped then
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$$p_1-p_2 = (\rho_m-\rho_w)g\Delta z + \rho_wg(z_{C2} - z_{C1})$$
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- $p_1 > p_2$ as $p_1$ is lower
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- If there is no flow along the tube, then
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$$p_1 = p_2 + \rho_wg(z_{C2} - z_{C1})$$
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<details>
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<summary>
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## Exercise Sheet 1
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</summary>
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1. If 4 m$^3$ of oil weighs 35 kN calculate its density and relative density.
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Relative density is a term used to define the density of a fluid relative
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> $$ \frac{35000}{9.81\times4} = 890 \text{ kgm}^{-3} $$
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>
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> $$1000 - 891.9... = 108 \text{ kgm}^{-3}$$
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2. Find the pressure relative to atmospheric experienced by a diver
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working on the sea bed at a depth of 35 m.
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Take the density of sea water to be 1030 kgm$^{-3}$.
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> $$
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> \rho gh = 1030\times 9.81 \times 35 = 3.5\times10^5
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> $$
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3. An open glass is sitting on a table, it has a diameter of 10 cm.
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If water up to a height of 20 cm is now added calculate the force exerted onto the table by
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the addition of the water.
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> $$V_{cylinder} = \pi r^2h$$
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> $$m_{cylinder} = \rho\pi r^2h$$
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> $$W_{cylinder} = \rho g\pi r^2h = 1000\times9.8\pi\times0.05^2\times0.2 = 15.4 \text{ N} $$
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4. A closed rectangular tank with internal dimensions 6m x 3m x 1.5m
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high has a vertical riser pipe of cross-sectional area 0.001 m2 in
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the upper surface (figure 1.4). The tank and riser are filled with
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water such that the water level in the riser pipe is 3.5 m above the
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Calulate:
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i. The gauge pressure at the base of the tank.
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> $$\rho gh = 1000\times9.81\times(1.5+3.5) = 49 \text{ kPa}$$
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ii. The gauge pressure at the top of the tank.
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> $$\rho gh = 1000\times9.81\times3.5 = 34 \text{ kPa}$$
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iii. The force exercted on the base of the tank due to gauge water pressure.
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> $$F = p\times A = 49\times10^3\times6\times3 = 8.8\times10^5 \text{ N}$$
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iv. The weight of the water in the tank and riser.
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> $$V = 6\times3\times1.5 + 0.001\times3.5 = 27.0035$$
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> $$W = \rho gV = 1000\times9.8\times27.0035 = 2.6\times10^5\text{ N}$$
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v. Explain the difference between (iii) and (iv).
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*(It may be helpful to think about the forces on the top of the tank)*
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> The pressure at the top of the tank is higher than atmospheric pressure because of the
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> riser.
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> This means there is an upwards force on the top of tank.
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> The difference between the force acting up and down due to pressure is equal to the
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> weight of the water.
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6. A double U-tube manometer is connected to a pipe as shown below.
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Taking the dimensions and fluids as indicated; calculate
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the absolute pressure at point A (centre of the pipe).
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Take atmospheric pressure as 1.01 bar and the density of mercury as 13600 kgm$^{-3}$.
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> \begin{align*}
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P_B &= P_A + 0.4\rho_wg &\text{(1)}\\
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P_C = P_{C'} &= P_B - 0.2\rho_mg &\text{(2)}\\
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P_D = P_{at} &= P_A + 0.4\rho_wg &\text{(3)}\\
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\\
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\text{(2) into (3)} &\rightarrow P_B -0.2\rho_mg-0.1\rho_wg = P_{at} &\text{(4)}\\
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\text{(1) into (4)} &\rightarrow P_A + 0.4\rho_wg - 0.2\rho_mg - 0.1\rho_wg = P_{at} \\
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\\
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P_A &= P_{at} + 0.1\rho_wg + 0.2\rho_mg - 0.4\rho_wg \\
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&= P_{at} + g(0.2\rho_m - 0.3\rho_w) \\
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&= 1.01\times10^5 + 9.81(0.2\times13600 - 0.3\times1000)\\
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&= 124.7\text{ kPa}
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> \end{align*}
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</details>
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# Lecture 3 // Submerged Surfaces
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## Prepatory Maths
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### Integration as Summation
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### Centroids
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- For a 3D body, the centre of gravity is the point at which all the mass can be considered to act
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- For a 2D lamina (thin, flat plate) the centroid is the centre of area, the point about which the
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lamina would balance
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To find the location of the centroid, take moments (of area) about a suitable reference axis:
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$$moment\,of\,area = moment\,of\,mass$$
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(making the assumption that the surface has a unit mass per unit area)
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$$moment\,of\,mass = mass\times distance\,from\,point\,acting\,around$$
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Take the following lamina:
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1. Split the lamina into elements parallel to the chosen axis
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2. Each element has area $\delta A = w\delta y$
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3. The moment of area ($\delta M$) of the element is $\delta Ay$
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4. The sum of moments of all the elements is equal to the moment $M$ obtained by assuing all the
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area is located at the centroid or:
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$$Ay_c = \int_{area} \! y\,\mathrm{d}A$$
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or:
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$$y_c = \frac 1 A \int_{area} \! y\,\mathrm{d}A$$
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- $\int y\,\mathrm{d}A$ is known as the first moment of area
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<details>
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<summary>
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#### Example 1
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Determine the location of the centroid of a rectangular lamina.
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</summary>
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##### Determining Location in $y$ direction
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1. Take moments for area about $OO$
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$$\delta M = y\delta A = y(b\delta y)$$
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2. Integrate to find all strips
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$$M = b\int_0^d\! y \,\mathrm{d}y = b\left[\frac{y^2}2\right]_0^d = \frac{bd^2} 2$$
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($b$ can be taken out the integral as it is constant in this example)
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but also $$M = (area)(y_c) = bdy_c$$
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so $$y_c = \frac 1 {bd} \frac {bd} 2 = \frac d 2$$
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##### Determining Location in $x$ direction
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1. Take moments for area about $O'O'$:
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$$\delta M = x\delta A = x(d\delta x)$$
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2. Integrate
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$$M_{O'O'} = d\int_0^b\! x \,\mathrm{d}x = d\left[\frac{x^2} 2\right]_0^b = \frac{db^2} 2$$
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but also $$M_{O'O'} = (area)(x_c) = bdx_c$$
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so $$x_c = \frac{db^2}{2bd} = \frac b 2$$
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</details>
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## Horizontal Submereged Surfaces
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Assumptions for horizontal lamina:
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- Constant pressure acts over entire surface of lamina
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- Centre of pressure will coincide with centre of area
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- $total\,force = pressure\times area$
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## Vertical Submerged Surfaces
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- A vertical submerged plate does experience uniform pressure
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- Centroid of pressure and area are not coincident
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- Centroid of pressure is always below centroid of area for a vertical plate
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- No shear forces, so all hydrostatic forces are perpendicular to lamina
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Force acting on small element:
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\begin{align*}
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\delta F &= p\delta A \\
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&= \rho gh\delta A \\
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&= \rho gh w\delta h
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\end{align*}
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Therefore total force is
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$$F_p = \int_{area}\! \rho gh \,\mathrm{d}A = \int_{h_1}^{h_2}\! \rho ghw\,\mathrm{d}h$$
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### Finding Line of Action of the Force
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\begin{align*}
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\delta M_{OO} &= \delta Fh = (\rho gh\delta A)h \\
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&= \rho gh^2\delta A = \rho gh^2w\delta h \\
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\\
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M_{OO} &= F_py_p = \int_{area}\! \rho gh^2 \,\mathrm{d}A \\
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&= \int_{h1}^{h2}\! \rho gh^2w \,\mathrm{d}h \\
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\\
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y_p = \frac{M_{OO}}{F_p}
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\end{align*}
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