Week 5 (Intro and Static Pressure 1) Flashcards
Introduction to Fluid Mechanics and Static Pressure
What’s a fluid?
Anything that continuously deforms under a shear stress (applied on it’s surface)
Anything that flows
Any liquid or gas
How is fluid mechanics relevant to biomedical engineering?
Cardiovascular system
Diagnostic tests
Reproduction
Some diseases (e.g sickle cell, gestational diabetes)
Fluid mechanics is relevant to all parts of engineering
Fluids have many properties, the 2 we are concerned about are:
Density
Viscosity
Viscosity
Measure how difficult it is to make it flow
How does pressure act in fluids
In all directions
Tries to squash or crush submerged things
Always acts perpendicular (normal to surface)
We generally got flow from high to low pressure, but pressure exists even when….
Fluid is not moving (static pressure)
Need to build structures to withstand
What causes pressure?
- In a solid, weight solid above
- In a fluid, weight fluid above
How do you derive the equation 𝑝= 𝜌𝑔ℎ
Consider the weight of fluid above him
*Take a cylinder with area, A
𝐹 = 𝑚𝑔
- What is mass of the cylinder? (Density x Volume) 𝑚 = 𝜌𝑉 = 𝜌𝐴ℎ
𝐹 = 𝜌𝐴ℎ𝑔 - Pressure is Force/Area 𝑝 =𝐹 /𝐴 =𝜌𝐴ℎ𝑔 /𝐴 = 𝜌𝑔ℎ
the pressure equation for in a liquid isn’t dependent on what variable?
𝑝 =𝐹 / 𝐴 =𝜌𝐴ℎ𝑔 / 𝐴 = 𝜌𝑔ℎ
- Doesn’t depend on area!
Pressure depends on:
Pressure = Fluid Density ×
Accel. due
to Gravity × Depth
𝑝 = 𝜌𝑔ℎ
- Doesn’t depend on area or shape!
What if more than one layer of fluid? E.g. oil on water
we can simply add up the pressures
* Total weight of fluid above
F = total mass x g
total mass = mass of water + mass of oil
p = F / A
using mass = density x volume (i.e A x h)
You get
p = rho1.g.h1 + rho2.g.h2
what measures atmospheric pressure?
- Barometer measures atmospheric pressure
- Mercury exposed to atmospheric pressure and a vacuum
- Impossible to make a perfect vacuum - Pressure in ‘vacuum’ is ‘vapour pressure’ (very close to zero)
- Difference in height is due to atmospheric pressure
How do you calc atmospheric pressure using a barometer?
h, is the height from the bottom surface of the mercury from the barometer to the beginning of the vacuum
In a barometer, why use Mercury and not water, to measure atmospheric pressure?
- How high would column of water be? ∆𝒑 = 𝝆𝒈𝒉
𝒉 = 𝒑 / 𝝆𝒈
𝒉 = 10.3m - Very high!
Alternatively the density of mercury is much higher.
Including atmospheric pressure
- Was our earlier calculation wrong? Do we need to think about atmospheric pressure too?
Just treat atmosphere as another
layer of fluid
Final Equation:
pressure = p (atm) + rho1.g.h1 + rho2.g.h2
Therefore whenever you solve a problem you need to always take into account atmospheric pressure (add it on).
What is the net pressure for the example of a leaky kettle?
What is the pressure driving the fluid out?
P out = Patm + rho.g.h
P in= P out
net pressure: P = P out - P in = rho.g.h
Atmospheric pressure cancels out anyway!
Gauge Pressure
We ignore atmospheric pressure – we measure the difference between pressure inside the kettle and outside it
Therefore if we use the gauge pressure don’t need to recalculate for e.g La Paz or anywhere!
- Similar to how we use degrees Celsius instead of Kelvin
- Except we can choose our ‘zero point’ depending on the context
- Gauge pressure doesn’t have to be measured with respect to atmosphere.
- For example
- Inside an aeroplane, you might measure with respect to cabin pressure instead of atmosphere
- Inside the lungs, you might measure with respect to minimum pressure during a breath
‘Absolute pressure’ includes…
the atmospheric pressure
What if we want to measure gauge pressure in an engine?
Could hook engine up to a barometer . . . But this measures absolute pressure
(i.e. it includes atmospheric pressure)
What if we used two barometers – one for engine and one for atmospheric?
There is a simpler way:
Can we find 𝒑(𝒆𝒏𝒈𝒊𝒏𝒆)?
Use a manometer, through the use of a U-bend
Fluid heights are different, because they feel different pressures
𝒑(𝒆𝒏𝒈𝒊𝒏𝒆) = 𝜌𝑔Δℎ
note, this is for gauge pressure
where delta h, is the height difference between the 2 meniscus.
absolute pressure = atmospheric pressure + rho.g.delta h (gauge pressure).
Absolute pressure can be measured using a barometer
Gauge pressure for an engine by using a manometer
What causes static pressure?
caused by the weight of fluid above
Why is static pressure important?
- Induces forces on submerged structures
Tank of water – what is the force
experienced by the bottom surface?
- Absolute pressure on the bottom surface
includes 𝜌𝑔ℎ (due to water) and
atmospheric pressure - But atmospheric pressure also acts
upwards – so can cancel out - Only consider gauge pressure
Force = pressure × Area
Fy = p.A = rho.g.h.A
Tank of water
- What about horizontal force (𝐹x)
- Pressure now varies with depth
- Take average pressure
Fx = mean p . A
Same as taking the pressure at the mid-depth (depth = h/2)
How to calculate using differential calculus (summary)
- Can also calculate forces by looking at force on infinitesimal
element of the wall and integrating
Slightly more complicated example – inclined plate
What are vertical and horizonal forces on
the inclined plate?
The pressure always acts perpendicular
to the surface, so the force will have
horizontal and vertical components.
- Start by getting the magnitude of the total
force, F
Again we start by averaging the pressures
Average pressure = (Pa (surface) + Pb (bottom container)) / 2
Pa = 0
Pb = rho.g.h
F = average pressure . A
F = rho.g.h/2.A
What is the area?
𝐴 = 𝐿𝑤
(𝑤 is the direction coming out of the page) ℎ = 𝐿 cos 𝜃
F = average pressure . A = rho.g.h/2.(h/cos 𝜃.w)
F = rho.g.(w.h^2/2cos 𝜃)
Break down in horizontal and vertical components:
Fy = F.sin 𝜃
Fx = F.cos 𝜃
This method works well for simple cases, but sometimes we will need to be a more general approach (e.g. a curved wall, or getting moments instead of forces)
A more complex method:
This method works well for simple cases, but sometimes we will need to be a more general
approach (e.g. a curved wall, or getting moments instead of forces)
- Let’s look at the pressure and forces on a
single element of the wall - 𝑧 is measured along the direction of the wall
- Pressure on the fluid element is
𝑑𝐹 = 𝑝.𝑑A
Pressure at point z is
𝑝 = 𝜌𝑔ℎ
Where the depth is ℎ = 𝑧.cos 𝜃
Area of element is 𝑑𝐴 = 𝑤.𝑑z
So we get
𝑑𝐹 = 𝜌.𝑔.𝑧. cos 𝜃. 𝑤.𝑑z
We can find the total force, 𝐹, by integration:
F = Integral dF = Integral from L to 0 (𝜌𝑔𝑧 cos 𝜃 𝑤𝑑z)
F = 𝜌.𝑔. cos 𝜃. w. integral from L to 0 (z.dz)
F = 𝜌.𝑔. cos 𝜃. w. [z^2/z] from L to 0
F = 𝜌.𝑔. cos 𝜃. w. (L^2/2)
Remember that
ℎ = 𝐿 cos 𝜃 substitute
F = rho.g.(w.h^2/2cos 𝜃)
Same as we got with previous method!
