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/