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Hydrostatics, Hydrodinamics
Kristóf Karádi
Physics-Biophysics IDepartment of Biophysics
15. 10. 2019.
Basical properties of Fluids
Fluid:
•The volume is constant,
• Form changing, the form depends on the container
• It has shown resistance against the volume changing
force,
• Incompressible
𝜌 =𝑚
𝑉 𝑝 =𝐹
𝐴
𝑘𝑔
𝑚3
𝑁
𝑚2= 𝑃𝑎
density pressure
Units of pressure
1 𝑃𝑎 = 1𝑁
𝑚2
1 𝑏𝑎𝑟 = 105 𝑃𝑎
1 𝑎𝑡𝑚 = 1.013 ∙ 105 𝑃𝑎 (atmospheric pressure at sea level)
760 𝑇𝑜𝑟𝑟 = 1 𝑎𝑡𝑚
760 𝐻𝑔𝑚𝑚 = 1 𝑎𝑡𝑚 (this is the pressure of a mercury column with height of 760 mm)
Physics of Fluids
Fluidflow
HYDRODYNAMICS
Static fluids
HYDROSTATICS
Flow of Ideal Fluids Flow of Real Fluids
Laminar flow Turbulent flow
HYDROSTATICS
Hydrostatic pressure
hgA
Agh
A
Vg
A
mg
A
Fp
Hydrostatic pressure: the pressure coming from weight of fluids
The pressure changes linearly proportional with the height of the fluids.
Experiment: A glass tube which’s hole is covered by only plastic membrane is in a water filled
container.
At a given height, the same pressure pushes
the membrane from each direction.
F = G = mg h
direction: top-down, bottom-up and sidelong
The hydrostatic pressure is independent of the form of the
container, also proportional to the height (h) of the column
and the density (ρf) of the fluid.
Hydrostatic pressure
p=h ρf g
Problem:Is 760 mm column of mercury really have pressure equal to atmospheric pressure?
(density of mercury: 13595𝑘𝑔
𝑚3)
Is 760 mm column of mercury really have pressure equal to atmospheric pressure?
(density of mercury: 13595𝑘𝑔
𝑚3)
𝑝 = 𝜌 ∙ 𝑔 ∙ ℎ
𝑝 = 13595𝑘𝑔
𝑚3 ∙ 9.81𝑚
𝑠2∙ 0.76𝑚 = 101358,882 𝑃𝑎
ℎ = 760 𝑚𝑚 = 0.76 𝑚
𝜌 = 13595𝑘𝑔
𝑚3
𝑔 = 9.81𝑚
𝑠2(𝑑𝑜 𝑛𝑜𝑡 𝑟𝑜𝑢𝑛𝑑 𝑡ℎ𝑖𝑠 𝑖𝑛 𝑡ℎ𝑖𝑠 𝑝𝑟𝑜𝑏𝑙𝑒𝑚)
Communicating vessels
The pressure of fluid does not depend on the shape of vessel.
The height of the fluid columns are
independent of the form of the
holders, if the density of fluids are
the same.
Exp.: Fill an U – shaped tube, two holders with
two non mixing fluids with different densities.
In equilibrum:
𝑝1 = 𝑝2
ℎ1 ∙ 𝜌1∙ 𝑔 = ℎ2 ∙ 𝜌2∙ 𝑔
ℎ1ℎ2
=𝜌2 ∙ 𝑔
𝜌1 ∙ 𝑔
ℎ1ℎ2
=𝜌2𝜌1
Height of the non mixing fluids from
the common border is inversely
proportional to the density of fluids.
Propagation of pressure in fluids
F1
F2
Experiment: Two different piston connecting into a fluid filled container, push
the first with a given force.
Pressure: p = F / A Work: W = p ΔV
W1 = p1 ΔV = p1 A1 s1
The fluids are incompressible:
A1 s1 = A2 s2 = ΔV (one „side’s” ΔV equals the other’s)
W1 = W2
So p1 = p2
Then F1 / A1 = F2 / A2
Pascal’s law: Pressure applied to an
enclosed fluid at rest is transmitted
undiminished to every portion of the
fluid and to the walls of the containing
vessel.
F1 < F2
W2 = p2 ΔV = p2 A2 s2
(side-note: 𝑊 = 𝐹∆𝑠 = 𝐹∆𝑉
𝐴)
Arkhimedes’s law
An object, which has A surface h
height, dive into a given fluid.
in h1 deepness the F1 force is (pressing
down the object):
F1 =p1A= (h1ρg)A
In h2 deepness the F2 force is (pressing
upwards the object):
F2 = p2A=(h2ρg)A
Fnet = F2 -F1=ρ g(h2-h1)A (h2-h1)A=Vbody inside the fliuid
F = ρfluid g Vbody inside the fluid
All of the immersed objects are effected by
the buoyancy, which is equal with the
weight of the displaced fluid.
sinking
G > F
float,
levitation
G = F
rise up
G < F
the net force is:
Problems
A ball’s 10% is in water. What is the ball’s radius if it’s mass it 55g?
𝜌𝑤𝑎𝑡𝑒𝑟 = 1000𝑘𝑔
𝑚3𝑔 = 10
𝑚
𝑠2
m=55g=0.055kg
𝜌𝑤𝑎𝑡𝑒𝑟 = 1000𝑘𝑔
𝑚3
𝑉𝑠𝑝ℎ𝑒𝑟𝑒 =4
3𝑟3𝜋
Buoyancy problem
𝑉𝑖𝑛 =10
100𝑉𝑡𝑜𝑡𝑎𝑙
𝐹𝑏𝑢𝑜 = 𝜌𝑤𝑎𝑡𝑒𝑟 ∙ 𝑔 ∙ 𝑉𝑖𝑛
weight: 𝐺 = 𝑚𝑔
𝐹𝑏𝑢𝑜 = 𝐺
𝜌𝑤𝑎𝑡𝑒𝑟 ∙ 𝑔 ∙ 𝑉𝑖𝑛= 𝑚𝑔
𝜌𝑤𝑎𝑡𝑒𝑟 ∙ 𝑉𝑖𝑛= 𝑚
𝜌𝑤𝑎𝑡𝑒𝑟 ∙10
100𝑉𝑡𝑜𝑡𝑎𝑙 = 𝑚
𝜌𝑤𝑎𝑡𝑒𝑟 ∙10
100∙4
3𝑟3𝜋 = 𝑚
𝑟 =3 100
10
3
4
𝑚
𝜌𝑤𝑎𝑡𝑒𝑟𝜋= 0.0508 𝑚 = 5.08𝑐𝑚
A ball’s 10% is in water. What is the ball’s radius if it’s mass it 55g?
HYDRODYNAMICS
Flow: one directional motion of fluids
Fluids flowing with, and without friction
Depends on: pressure difference (Δp)
Volumetric Intensity of current or flow rate:
(It characterizes the flowing fluids)
[m3/s
v. liter/min]
In the aorta the flowrate is 6 l/min – cardiac output
The intensity of current is the quotient of the across flowing volume and the
required time.
𝐼𝑉 =∆𝑉
∆𝑡
𝑄 = 𝐼𝑉 =∆𝑉
∆𝑡
Note: the volumetric flow rate is often denoted as Q:
Continuity Equation
Fluids are incompressible, the intensity of current is constant in both
position and time. (conservation of matter!!!: what goes in should go out)
The cross section of the tube (A) is inversely proportional to the velocity of
flow (v).
I = A v = constant, Stacionary flow
𝑐𝑜𝑛𝑠𝑡 = 𝐼𝑉 =∆𝑉
∆𝑡=𝐴 ∙ 𝑑
∆𝑡=𝐴 ∙ 𝑣 ∙ ∆𝑡
∆𝑡= 𝐴 ∙ 𝑣
*criterion: tubes with rigid walls, stationary flow, ideal (without-friction forces) liquid
Bernoulli’s law
.22
2
2
221
2
11 consthg
vphg
vp
static pressure
dynamic pressure
hydrostatic pressure
𝑝 +𝜌 ∙ 𝑣2
2+ 𝜌 ∙ 𝑔 ∙ ℎ
Simplified background:The equation comes from the conservation of energy!!! (by multiplying the equation with V we get energies: )
𝑉𝑝 +𝜌𝑉 ∙ 𝑣2
2+ 𝜌𝑉 ∙ 𝑔 ∙ ℎ
𝑉𝑝 +𝑚 ∙ 𝑣2
2+ 𝑚 ∙ 𝑔 ∙ ℎ = constant
Problems 𝜌𝑤𝑎𝑡𝑒𝑟 = 1000𝑘𝑔
𝑚3𝑔 = 10
𝑚
𝑠2
In a tube 3 cm3 water travels through per secundum. What is the velocity of the water where the tube’s diameter is 0.5 cm and where 0.8 cm?
Water flows in a tube. It’s velocity is 3 m/s at one point. At another point being 1 m
higher it is 4 m/s. What is the pressure here, if at the lower point it is 20kPa?
Continuity equation
c𝑜𝑛𝑠𝑡 = 𝐼𝑉 =∆𝑉
∆𝑡= 𝐴 ∙ 𝑣
𝑣 =𝐼𝑣𝐴=
𝐼𝑣𝑟2𝜋
𝑣𝑟1 = 15.29𝑐𝑚
𝑠
𝑣𝑟2 = 5.97𝑐𝑚
𝑠
As expected: 𝑣𝑟1 > 𝑣𝑟2
because: 𝑟1 < 𝑟2
In a tube 3 cm3 water travels through per secundum. What is the velocity of the water where the tube’s diameter is 0.5 cm and where 0.8 cm?
𝑑12= 𝑟1 = 0.25𝑐𝑚
𝑑22= 𝑟2 = 0.4𝑐𝑚
𝐼𝑉 = 3𝑐𝑚3
𝑠
Bernoulli
v (m/s) h (m) p (Pa)
A 3 0 20000
B 4 1 ?
𝑝𝐴 +𝜌𝑣𝐴
2
2+ 𝜌𝑔ℎ𝐴 = 𝑝𝐵 +
𝜌𝑣𝐵2
2+ 𝜌𝑔ℎ𝐵
𝑝𝐵 = 𝑝𝐴 +𝜌𝑣𝐴
2
2+ 𝜌𝑔ℎ𝐴 −
𝜌𝑣𝐵2
2− 𝜌𝑔ℎ𝐵 = 6500𝑃𝑎
Water flows in a tube. It’s velocity is 3 m/s at one point. At another point being 1 m
higher it is 4 m/s. What is the pressure here, if at the lower point it is 20kPa?
𝜌𝑤𝑎𝑡𝑒𝑟 = 1000𝑘𝑔
𝑚3𝑔 = 10
𝑚
𝑠2
Venturi effect
Flowing gas or liquid has a kind of suction effect (this is called the Venturi effect): based on Bernoulli’s law the higher the flow velocity (for example because of the tube becoming narrower) the less will be the static pressure. For example in the case of an oxygen mask where oxygen flows with high velocity the pressure inside the mask is less so it can suck air from the outside through holes.
Laminar flow of Real (not ideal) Fluids
Newton’s law of friction
h
vAF
**
Viscosity: η
[Pa*s]
Viscosity depends on:
• Quality of material
• Concentration
• Temperature (↑temp , η ↓)
• Pressure
η ≠ 𝜌 !!!
e.g.:
𝜌𝐻𝑔 > 𝜌𝐻2𝑂 η𝐻𝑔 < η𝐻2𝑂
Liquids for which Newton’s law of friction is valid (constant viscosity independent of stress) are called Newtonian fluids. Fluids found in joints (knee) are not Newtonian fluids. They are called synovial fluids and their viscosity decreases with pressure, working more efficient as greases.
The velocity profile has a parabola shape:(the friction is biggest between the fluid and the wall of the tube: so the velocity is the lowest there)
Consequence: the red blood cells are concentrated along the axis of the vessels, because the pressure increases towards the walls of the vessels (Bernoulli’s law).
Flow with friction
Laminar flow Turbulent flow
• if the velocity of the flow (v) is
small
• No mixing
• Smooth surface
• if the velocity of the flow (v) is higher
• Rough surface
• eddies
dvR
Reynolds number 1160
1160
R
R laminar
turbulent
laminar
turbulent
v: velocityρ: liquid densityd: diameter of the tubeη: viscosity of fluid
Hagen-Poiseuille equation
12
4
8pp
l
rI
pI
lI
I
rI
~
1~
1~
~ 4
(for laminar flow in a tube)
*criterion: tube with rigid wall, stationary, laminar flow, also it considers friction of fluids
Important:e.g.:3 times bigger radius:34=81 times bigger flow rate!!!
so the body most easily regulates the blood flow by regulating the diameter of the blood vessels
but be careful! the global cross section area what matters: the capillary vessels are small but there are many, so the flow is the slowest there
A1
p1
v1 p2v2
A2 A1
p1
v1
Aneurysm: devil’s circle
12
12
12
pp
vv
AA
Continuity’ s law
Bernoulli’s law
This equation describes the friction force that affects a spherical body with radius r that travels with velocity v in a fluid of viscosity η (considering a small Reynolds number)
Stokes’ law:
𝐹𝑆 = 6 ∙ 𝜋 ∙ η ∙ 𝑟 ∙ 𝑣
We want to infuse the patient with a solution of 20 cm3 that has 10-3 Pa s viscosity. We have to do this in 4 minutes against 1600 Pa vein pressure. How much pressure should we use? (the syringe that we use has a 10 cm long needle that has a 1 mm diameter)
Problem
d=1mm --> r=0,5mm = 0,0005m
L=10cm = 0,1m
η=0,001Pa s
V=20cm3 = 0,000 02m3
t=4 min = 240s
p1=1600Pa
p2=?
Hagen-Poiseuille
I=∆𝑉
∆𝑡=
𝜋∙∆𝑝∙𝑅4
8∙𝜂∙𝐿
∆𝒑 =𝟖 ∙ 𝜼 ∙ 𝑳 ∙ 𝑽
𝝅 ∙ 𝑹𝟒 ∙ 𝒕=
=8 ∙ 0,001 ∙ 0,1 ∙ 0, 𝟎𝟎𝟎02
3,14 ∙ 0,0005 4 ∙ 240𝑃𝑎 = 339,7𝑃𝑎
𝑝2 − 𝑝1 = ∆𝑝 → 𝑝2 = ∆𝑝 + 𝑝1 = 1939,7𝑃𝑎
We want to infuse the patient with a solution of 20 cm3 that has 10-3 Pa s viscosity. We have to do this in 4 minutes against 1600 Pa vein pressure. How much pressure should we use? (the syringe that we use has a 10 cm long needle that has a 1 mm diameter)
Sources:
-Dr. Andrea Leipoldné Vig’s and Veronika Takács-Kollár’s physical basis of biophysics slides (PTE ÁOK Biophyics Institute)
-Dr. András Lukács’s pharmacy materials (PTE ÁOK Biophyics Institute)
-Elek Telek’s pharmacy lecture slide (PTE ÁOK Biophyics Institute)
-https://forums.studentdoctor.net/threads/aamc-fl2-cp-25.1275134/