Week 5 (Intro and Static Pressure 1)

Question 1

What’s a fluid?

Answer 1

Anything that continuously deforms under a shear stress (applied on it’s surface)

Anything that flows
Any liquid or gas

Question 2

How is fluid mechanics relevant to biomedical engineering?

Answer 2

Cardiovascular system
Diagnostic tests
Reproduction
Some diseases (e.g sickle cell, gestational diabetes)

Fluid mechanics is relevant to all parts of engineering

Question 3

Fluids have many properties, the 2 we are concerned about are:

Answer 3

Density
Viscosity

Question 4

Viscosity

Answer 4

Measure how difficult it is to make it flow

Question 5

How does pressure act in fluids

Answer 5

In all directions
Tries to squash or crush submerged things
Always acts perpendicular (normal to surface)

Question 6

We generally got flow from high to low pressure, but pressure exists even when….

Answer 6

Fluid is not moving (static pressure)

Need to build structures to withstand

Question 7

What causes pressure?

Answer 7

• In a solid, weight solid above
• In a fluid, weight fluid above

Question 8

How do you derive the equation 𝑝= 𝜌𝑔ℎ

Answer 8

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

Question 9

the pressure equation for in a liquid isn’t dependent on what variable?

Answer 9

𝑝 =𝐹 / 𝐴 =𝜌𝐴ℎ𝑔 / 𝐴 = 𝜌𝑔ℎ

• Doesn’t depend on area!

Question 10

Pressure depends on:

Answer 10

Pressure = Fluid Density ×
Accel. due
to Gravity × Depth

𝑝 = 𝜌𝑔ℎ

• Doesn’t depend on area or shape!

Question 11

What if more than one layer of fluid? E.g. oil on water

Answer 11

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

Question 12

what measures atmospheric pressure?

Answer 12

• 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

Question 13

How do you calc atmospheric pressure using a barometer?

Answer 13

h, is the height from the bottom surface of the mercury from the barometer to the beginning of the vacuum

Question 14

In a barometer, why use Mercury and not water, to measure atmospheric pressure?

Answer 14

• 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.

Question 15

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:

Answer 15

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).

Question 16

What is the net pressure for the example of a leaky kettle?

Answer 16

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!

Question 17

Gauge Pressure

Answer 17

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

Question 18

‘Absolute pressure’ includes…

Answer 18

the atmospheric pressure

Question 19

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)?

Answer 19

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

Question 20

What causes static pressure?

Answer 20

caused by the weight of fluid above

Question 21

Why is static pressure important?

Answer 21

• Induces forces on submerged structures

Question 22

Tank of water – what is the force
experienced by the bottom surface?

Answer 22

• 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

Question 23

Tank of water

• What about horizontal force (𝐹x)

Answer 23

• Pressure now varies with depth
• Take average pressure

Fx = mean p . A

Same as taking the pressure at the mid-depth (depth = h/2)

Question 24

How to calculate using differential calculus (summary)

Answer 24

• Can also calculate forces by looking at force on infinitesimal
  element of the wall and integrating

Question 25

Slightly more complicated example – inclined plate

What are vertical and horizonal forces on
the inclined plate?

Answer 25

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)

Question 26

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

Answer 26

• 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!