Lecture 1 – Fundamentals
Difference between fluid and a solid:
Fluids deform continuously when subjected to shear stress (no matter how small it is).
τ =μ γ̇ for fluids
Solids resist shear stress by a static deformation (fluids cannot do this). τ =Gγ
Continuum hypothesis – “the actual molecular structure of fluids is replaced by a
hypothetical continuous medium called continuum”. This solves the problem of “voids”
between neighbouring molecules. (Not satisfied rarefied gases).
No-slip condition: when a fluid moves over a surface, the molecules
in contact with the wall stick to the wall. Consequently, flow velocity
immediately at a stationary wall is always 0.
Boundary layer: Flow velocity is 0 on a wall but non-zero further
away. Hence there must be a transition region, where flow increases
from 0 to the free-stream value. This is called the boundary layer.
Boundary layer thickness (δ): Usually taken as the distance from the wall where flow
speed is equal to 99% of undisturbed free stream velocity (normally U ∞ or U 0 ).
δ is typically a few mm / cm in mechanical/civil engineering problems
δ is absolutely crucial for determining nature of flow around bodies – major impact on
drag forces acting
Surface tension (Y or σ): at the interface between liquid-gas or two immiscible liquids,
forces develop in the liquid surface that cause the surface to behave as if it were a “skin” or
“membrane” stretched over the fluid mass.
Observable examples: needle floating on water, water insect walking on
water, droplets of mercury held together from surface tension
To measure surface tension, use a tensiometer:
Dip ring into liquid, then pull it out. Liquid curtain develops, which tries
to pull ring down. Surface tension is characterised as Force per length.
Length in the tensiometer is the circumference of the ring.
Typical values of surface tension: Air-Water is Y = 0.073 N/m and Air-Mercury is Y =
0.48 N/m
F
Pressure is defined as the normal force per unit area of boundary ( p= ). Fluids will exert a
A
force normal to a solid boundary. In each case, equate the tension force(s) Y with the
pressure increase.
Y
Liquid column: 2 RL∆ p=2YL which can be rearranged to ∆ p=
R
Spherical droplet: (Area × change in pressure = Circumference of droplet × Surface
2Y
tension), π R 2 ∆ ρ=2 πRY hence ∆ p=
R
4Y
Soap bubble: Approximately double that of the droplet, ∆ ρ ≈
R
Complex curved surface: Function of the principle radii of curvature ∆ p=Y ( R1 + R1 )
1 2
Lecture 2 – Pressure and manometers
,Pressure is the normal force divided by the area of boundary. This is in contrast to stress,
which is the force parallel to the area, divided by the area. Both
quantities have the same dimensions in Nm–2.
δF
Mean pressure is: p=
δA
To find the pressure at a point, take the limit of δA → 0,
δF dF
which gives: p= lim =
δA→ 0 δA dA
Pressure force at bottom = Pressure force at top + force due to
liquid in column
p1 A= p2 A + ρgA ( z 2−z 1 ), which can be rearranged to: p2− p1=−ρg ( z2 −z1 )
Since the RHS <0, this means that p2 < p1 (pressure is LESS at the top). This shows
that in any fluid under gravitational attraction, PRESSURE decreases with increase of
HEIGHT z.
One atmosphere = 101.325 x 103 Nm–2
The change in pressure for a diver going to a depth of 30m is 301.7 x 10 3 Nm–2. Hence,
the pressure increase corresponds to 3 times the atmospheric pressure.
1 atmosphere is also the pressure in 10m of water column.
Pressure DOES NOT CHANGE AT SAME LEVEL. The pressure of any 2 points at the same
level in body of fluid AT REST will be the same. The key point is the fluid body must be at
rest.
If the free surface of liquid in a container is datum (where pressure is atmosphere with
reference 0):
h=z 2−z 1, which can be used to write: p1=ρgh .
Hydrostatic paradox: Pressure at the bottom of a liquid filled container depends ONLY on
the filling height of the container and the base area. The SHAPE of the container has NO
EFFECT.
Why is force not affected by shape? Force at bottom is : F= pA=ρghA
Even if weight of fluid is different in each shaped vessel, the force on the base of the
vessel is the same. It ONLY depends on the height h and base area A (independent of
total volume).
Archimedes principle: “A body totally or
partially immersed in a fluid is subject to an upward force
(buoyancy force) equal in magnitude to the weight of fluid it
displaces.”
, Net vertical pressure force due to the pressure difference top-bottom:
F p= p2 A− p1 A=ρ h2 gA−ρ h1 gA=ρ ( h2−h1 ) gA
Weight of liquid displaced is: F L =ρHAg=ρ ( h2−h 1) Ag
This demonstrates Archimedes principle mathematically
Gauge pressure is pressure measured relative to atmospheric pressure (taking P atm to be 0).
Absolute pressure takes into account the atmospheric pressure.
Absolute pressure = Gauge pressure + Atmospheric pressure, which is written as:
p= ρgh+ p atm
Pressure can be measured by manometer.
Simplest form is a pressure tube or piezometer. This is a single
vertical tube, open at the top, which is inserted into pipe/vessel
containing liquid under pressure – the liquid rises in the tube to a
height, depending on the pressure.
Pressure due to column: pcolumn =ρgh
The height rise is determined by the pressure rise inside pipe ( pB ). In
equilibrium, the column pressure must equal/balance the pipe pressure.
Hence, the pipe pressure can be measured by measuring height h.
U-tube manometer: used to measure pressure of gases or liquids. The pressure pB at B is
equal to the pressure pC at C. Any pressure increase at B is balanced by a pressure decrease
at C (given by ρgh )
Left limb: pB = p A + ρgh1
Right limb: pC = p atm + ρman g h2 (where patm is 0 due
to zero gauge)
Since pressures are balanced, equate:
p A + ρgh 1=ρman g h2
Hence: p A =ρman g h2 −ρg h1
Why does an inclined leg increase
sensitivity?
z
sin θ= , which can be rearranged to:
x
z
x=
sin θ
Since sin θ → 0 when θ → 0, this means that x → ∞
If the leg is vertical (θ=90), then meniscus inside the leg rises by z=x .
If the leg is inclined (θ< 90), then meniscus inside the leg rises by x > z .
, When pressure is small, the manometer can be made as sensitive as required by
adjusting the angle of inclination of the leg, and choosing manometer liquid with
suitable density value.
Inverted U-tube manometer: used for measuring pressure
differences between 2 vessels. First, must identify a point where the
pressure is the same.
Left limb: p xx = p A −ρga−ρman gh
Right limb: p xx = pB −ρg(b+ h)
Rearrange and simplify: pB − p A =ρg ( b−a ) + gh( ρ−ρman)
Lecture 3 – Newton’s law of viscosity
Viscosity is a measure of a fluid’s resistance to shear. Macroscopic viscosity results from
forces governing interactions of many molecules on microscopic level.
When plate pulled over a plane:
There is a narrow gap
(e.g. 1mm) between the
plate and plane which is
filled with fluid (e.g. oil)
The upper plate moves
with a velocity U
The top plate pulls the
fluid with it (it has to stick at plate surface and at plane
surface)
There is a linear velocity profile for the fluid (only for a
narrow gap)
x
Shear strain: γ=
y
x
x
Rate of shear strain: y x 1 u (since is
γ̇= = = t
t t y y
the velocity at which the plate is pulled)
du
Newton’s law of viscosity: τ =μ
dy
Interpreted as “Shear stress is directly proportional to velocity gradient”
μ is the dynamic viscosity (units Nsm–2 or kgs–1m–1)
μ
v is the kinematic viscosity, given by: v= , (units m2s–1)
ρ
If flow geometry involves a narrow gap, the fluid has a LINEAR velocity profile.
Difference between fluid and a solid:
Fluids deform continuously when subjected to shear stress (no matter how small it is).
τ =μ γ̇ for fluids
Solids resist shear stress by a static deformation (fluids cannot do this). τ =Gγ
Continuum hypothesis – “the actual molecular structure of fluids is replaced by a
hypothetical continuous medium called continuum”. This solves the problem of “voids”
between neighbouring molecules. (Not satisfied rarefied gases).
No-slip condition: when a fluid moves over a surface, the molecules
in contact with the wall stick to the wall. Consequently, flow velocity
immediately at a stationary wall is always 0.
Boundary layer: Flow velocity is 0 on a wall but non-zero further
away. Hence there must be a transition region, where flow increases
from 0 to the free-stream value. This is called the boundary layer.
Boundary layer thickness (δ): Usually taken as the distance from the wall where flow
speed is equal to 99% of undisturbed free stream velocity (normally U ∞ or U 0 ).
δ is typically a few mm / cm in mechanical/civil engineering problems
δ is absolutely crucial for determining nature of flow around bodies – major impact on
drag forces acting
Surface tension (Y or σ): at the interface between liquid-gas or two immiscible liquids,
forces develop in the liquid surface that cause the surface to behave as if it were a “skin” or
“membrane” stretched over the fluid mass.
Observable examples: needle floating on water, water insect walking on
water, droplets of mercury held together from surface tension
To measure surface tension, use a tensiometer:
Dip ring into liquid, then pull it out. Liquid curtain develops, which tries
to pull ring down. Surface tension is characterised as Force per length.
Length in the tensiometer is the circumference of the ring.
Typical values of surface tension: Air-Water is Y = 0.073 N/m and Air-Mercury is Y =
0.48 N/m
F
Pressure is defined as the normal force per unit area of boundary ( p= ). Fluids will exert a
A
force normal to a solid boundary. In each case, equate the tension force(s) Y with the
pressure increase.
Y
Liquid column: 2 RL∆ p=2YL which can be rearranged to ∆ p=
R
Spherical droplet: (Area × change in pressure = Circumference of droplet × Surface
2Y
tension), π R 2 ∆ ρ=2 πRY hence ∆ p=
R
4Y
Soap bubble: Approximately double that of the droplet, ∆ ρ ≈
R
Complex curved surface: Function of the principle radii of curvature ∆ p=Y ( R1 + R1 )
1 2
Lecture 2 – Pressure and manometers
,Pressure is the normal force divided by the area of boundary. This is in contrast to stress,
which is the force parallel to the area, divided by the area. Both
quantities have the same dimensions in Nm–2.
δF
Mean pressure is: p=
δA
To find the pressure at a point, take the limit of δA → 0,
δF dF
which gives: p= lim =
δA→ 0 δA dA
Pressure force at bottom = Pressure force at top + force due to
liquid in column
p1 A= p2 A + ρgA ( z 2−z 1 ), which can be rearranged to: p2− p1=−ρg ( z2 −z1 )
Since the RHS <0, this means that p2 < p1 (pressure is LESS at the top). This shows
that in any fluid under gravitational attraction, PRESSURE decreases with increase of
HEIGHT z.
One atmosphere = 101.325 x 103 Nm–2
The change in pressure for a diver going to a depth of 30m is 301.7 x 10 3 Nm–2. Hence,
the pressure increase corresponds to 3 times the atmospheric pressure.
1 atmosphere is also the pressure in 10m of water column.
Pressure DOES NOT CHANGE AT SAME LEVEL. The pressure of any 2 points at the same
level in body of fluid AT REST will be the same. The key point is the fluid body must be at
rest.
If the free surface of liquid in a container is datum (where pressure is atmosphere with
reference 0):
h=z 2−z 1, which can be used to write: p1=ρgh .
Hydrostatic paradox: Pressure at the bottom of a liquid filled container depends ONLY on
the filling height of the container and the base area. The SHAPE of the container has NO
EFFECT.
Why is force not affected by shape? Force at bottom is : F= pA=ρghA
Even if weight of fluid is different in each shaped vessel, the force on the base of the
vessel is the same. It ONLY depends on the height h and base area A (independent of
total volume).
Archimedes principle: “A body totally or
partially immersed in a fluid is subject to an upward force
(buoyancy force) equal in magnitude to the weight of fluid it
displaces.”
, Net vertical pressure force due to the pressure difference top-bottom:
F p= p2 A− p1 A=ρ h2 gA−ρ h1 gA=ρ ( h2−h1 ) gA
Weight of liquid displaced is: F L =ρHAg=ρ ( h2−h 1) Ag
This demonstrates Archimedes principle mathematically
Gauge pressure is pressure measured relative to atmospheric pressure (taking P atm to be 0).
Absolute pressure takes into account the atmospheric pressure.
Absolute pressure = Gauge pressure + Atmospheric pressure, which is written as:
p= ρgh+ p atm
Pressure can be measured by manometer.
Simplest form is a pressure tube or piezometer. This is a single
vertical tube, open at the top, which is inserted into pipe/vessel
containing liquid under pressure – the liquid rises in the tube to a
height, depending on the pressure.
Pressure due to column: pcolumn =ρgh
The height rise is determined by the pressure rise inside pipe ( pB ). In
equilibrium, the column pressure must equal/balance the pipe pressure.
Hence, the pipe pressure can be measured by measuring height h.
U-tube manometer: used to measure pressure of gases or liquids. The pressure pB at B is
equal to the pressure pC at C. Any pressure increase at B is balanced by a pressure decrease
at C (given by ρgh )
Left limb: pB = p A + ρgh1
Right limb: pC = p atm + ρman g h2 (where patm is 0 due
to zero gauge)
Since pressures are balanced, equate:
p A + ρgh 1=ρman g h2
Hence: p A =ρman g h2 −ρg h1
Why does an inclined leg increase
sensitivity?
z
sin θ= , which can be rearranged to:
x
z
x=
sin θ
Since sin θ → 0 when θ → 0, this means that x → ∞
If the leg is vertical (θ=90), then meniscus inside the leg rises by z=x .
If the leg is inclined (θ< 90), then meniscus inside the leg rises by x > z .
, When pressure is small, the manometer can be made as sensitive as required by
adjusting the angle of inclination of the leg, and choosing manometer liquid with
suitable density value.
Inverted U-tube manometer: used for measuring pressure
differences between 2 vessels. First, must identify a point where the
pressure is the same.
Left limb: p xx = p A −ρga−ρman gh
Right limb: p xx = pB −ρg(b+ h)
Rearrange and simplify: pB − p A =ρg ( b−a ) + gh( ρ−ρman)
Lecture 3 – Newton’s law of viscosity
Viscosity is a measure of a fluid’s resistance to shear. Macroscopic viscosity results from
forces governing interactions of many molecules on microscopic level.
When plate pulled over a plane:
There is a narrow gap
(e.g. 1mm) between the
plate and plane which is
filled with fluid (e.g. oil)
The upper plate moves
with a velocity U
The top plate pulls the
fluid with it (it has to stick at plate surface and at plane
surface)
There is a linear velocity profile for the fluid (only for a
narrow gap)
x
Shear strain: γ=
y
x
x
Rate of shear strain: y x 1 u (since is
γ̇= = = t
t t y y
the velocity at which the plate is pulled)
du
Newton’s law of viscosity: τ =μ
dy
Interpreted as “Shear stress is directly proportional to velocity gradient”
μ is the dynamic viscosity (units Nsm–2 or kgs–1m–1)
μ
v is the kinematic viscosity, given by: v= , (units m2s–1)
ρ
If flow geometry involves a narrow gap, the fluid has a LINEAR velocity profile.
