fluid & particulate systems fluid & particulatesystems...

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Fluid and Particulate systems 424521 /2018

EXTERNAL FLOWS /FLOW IN POROUS STRUCTURES

Ron ZevenhovenÅA Thermal and Flow Engineering

[email protected]

3F

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3.1 Flow around objects, drag forces, lift forces

Åbo Akademi University - Värme- och Strömningsteknik Biskopsgatan 8, FI-20500 Åbo / Turku Finland

RoNz 2februari 2018

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Flow Around BodiesF

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8

Basic concept

carlosscar wFP

212121 ,0, ppwwzz

Fw

wFP

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februari 2018 Åbo Akademi University - Värme- och Strömningsteknik Biskopsgatan 8, FI-20500 Åbo / Turku Finland

5/52

Fluid flow around objects In the cases of

– an object moving through a fluid– a fluid flow around an object

the velocity difference generates forces Forces acting parallel to the flow direction are drag

forces; forces acting perpendicular to the flow direction are lift forces

The flow field around an object can be divided in two parts: the boundary layer, where the viscous forces are active, and the free-stream velocity (or the stagnant surrounding fluid) P

ictu

re: h

ttp://

ww

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ages

/forc

es.jp

g

PTG / PRC

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6/52

Fluid flow over a surface

CRBH83

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Flow over a flat plate

F

Consider air flowing above a thin plate in x-direction with the undisturbed velocity w∞. Due to viscosity (fluid friction), the air layer closest to the plate has w = 0, and a velocity gradient in y-direction appear. The presence of the plate is hence “felt” up to a distance δv in y-direction; a boundary layer occur. Beyond, w∞ remains unchanged.

Since the air layers have a velocity gradient but are still attracted to each other, one layer will try to drag an adjacent layer along. Consequently, the plate is affected by a (drag) force F in the air flow direction. The friction drag force Fdivided by the area of the plate gives the shear stress.

februari 2018 RoNz 7Åbo Akademi University - Värme- och Strömningsteknik Biskopsgatan 8, FI-20500 Åbo / Turku Finland

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• For flow along a flat plate, the forces on the plate are frictionforces. The shear stress on each side of the surface is

with (laminar) boundary layer thickness δ and relative velocity vr

• The drag force on each side of a plate with length L and width b is then given by

• The pressure ½ρv2 is known as THRUST (sv: stöt)

Flow around a flat plate /1

5

fluid

fluidr 103η

xρv for ,

rfluid

½

fluid

fluidrrfluidwyyx v½ρ

η

ρxv.

δ

vηττ

5

fluid

fluidrL

2

103η

ρLv Re for

rfluid

½

fluid

fluidr

L

wDdrag

v½bLη

Lρv

dxbFF

331

0

.

Picture: BMH99

PTG / PRC

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Flow around a flat plate /2

This defines the (length-averaged) drag coefficient CD as

where A (m2) is the area (one side) of the plate For turbulent cases, experimental results give

For a flat surface with a laminar region followed by a turbulent region, a ”composite” drag composition can be calculated with

For a flate plate perpendicular to a fluid the drag coefficient equals CD~2 , largely independent of Re-number

9L

72.58

L10D

7L

51/5L

D 10Re10 for )log(Re

0.445 C ;10Re10 for

Re

0.074 C

5L

LD 103Re for

Re

1.33 C withvACF rDD

2½

Re

1740

Re

0.074 C

L1/5L

D Picture: KJ05


februari 2018 9/52

PTG / PRC

Transition regionRe ~ 5×105

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Calculation of total drag

F

2

2

Dprj

w

CAF

w∞

CD: drag coefficient

Aprj: projection area (area normal to the direction of the flow, datum area)

w∞: undisturbed fluid velocity


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11/52

For a general surface area A ┴

(m2) perpendicular to the flow, the drag force is

FD = CD·A┴·½ρvr2

(where ½ρvr2 is actually the pressure

difference between the front and the back of the object)

with drag coefficient CD

With increasing Re-numbers, boundary layer separationoccurs, and a wake region (sv: köl(vatten)) ariseswhere kinetic energy is only partlyconverted into pressure

Picture: KJ05

Flowaround a cylinder

Flow around cylinders, spheres /1PTG / PRC

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Flow around a cylinder

Fw∞

a

Consider air flowing across a cylinder. Air will attach to the cylinder and boundary layer and a friction drag force occur, as in the case of the plate.

221

0 wpp The flow will then divide into an upper and a lower flow, and the pressure starts to decrease as the air velocity increases. The boundary layer detaches from the cylinder at the separation point c, and a wake is formed in the rear. The pressure in the rear will hence be lower compared to the front, and an additional drag force, pressure drag, will further increase the total drag force acting on the cylinder.

Additionally, the undisturbed air has a pressure p∞ and a velocity w∞, at a point a before it acts on the cylinder. The air flow at the cylinder is stopped the stagnation point b, where the pressure increases according to


Detachment of Boundary layerStarts atRe ~ 10,Complete atRe ~100

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Friction and Pressure Drag


Pressure,forces

Shear, tangentialforces

Stress tensor (cartesian) τxxτyyτzz τxyτyxτxzτzxτyzτzy

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Friction and Pressure Drag


Picture: KJ05

τ forceshear P,force pressure with

cossin

sincos

AsurfaceAsurface

lift

AsurfaceAsurface

drag

dAdAPF

dAdAPF

Pressure and shear stresses acting on a cylinder (103 < ReD < 105).

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F w∞

Pressure drag and separation point location

F w∞

Low w∞ (Re<2ꞏ105), early separation, high pressure drag.

High w∞ (Re>2ꞏ105), late separation (transition to turbulent flow), low pressure drag.


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16/52

Flow around cylinders, spheres /2

For spherical particlesthe drag coefficientequals

For flow at Re <0.1 around a sphere, the relation CD=24/Re

follows also from Stokes’ law

Fdrag = 3πηvrd

for a sphere with diameter d and relative velocity vr = vsphere-vflow

Picture: KJ05

5

D

32

D

D

D

10Re800 for

0.44 C

800Re2 for

Re6

11

Re

24 C

2Re0.2 for

Re16

31

Re

24 C

0.2 or 1Re forRe

24 C

Picture: http://www.school-for-champions.com/science/friction_changing_fluid.htm

PTG / PRC

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Flow around cylinders, spheres /3

Roughness effects on the drag coefficient of a sphere.


RoNz 17

Picture: KJ05

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18/52

Boundary layer separation examples

Pic

ture

: http

://w

ww

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ospa

cew

eb.o

rg/q

uest

ion/

aero

dyna

mic

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215.

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: http

://w

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Med

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PTG / PRC

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19/52

A smooth (a) and roughened (b) ball entering water at 25 °C

CR83

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Drag coefficient CD

9.1, 9.2februari 2018 RoNz 20Åbo Akademi University - Värme- och Strömningsteknik

Biskopsgatan 8, FI-20500 Åbo / Turku Finland

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Lift - why?

The principle of conservation of angularmomentum demands that the angularmomenta of the two rotational patterns tobe equal and opposite.

9.3

w

p


Picture:KJ05

Pressure and shear acting on an airfoil.

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Example: drag on a flat plate

An advertising banner (1 m x 20 m) is towed behind an aeroplane at 90 km/h, in air at 32°C.

Calculate the power (in kW) needed to pull the banner.

Answer: The power P needed can be related to the drag force byP = FDꞏvr, where vr = 90 km/h = 25 m/s and FD = CDꞏAꞏ½ρvr

2, with CD depending on the Re-number

At 32°C (assuming p = 1 atm) ρair = 1.161 kg/m3 and η = 1.875×10-5 Pa.s. With a length scale L = 20 m → Re = 3.1×107

Calculating with gives CD = 0.0024

With surface (2 sides!) A = 40 m2 this gives FD = 35 N and P = 875 W = 0.875 kW

Re

1740

Re

0.074 C

L1/5L

D


februari 2018 RoNz 22

PTG / PRC

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3.2 Flow through or along tubebundles


RoNz 23februari 2018

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Flow in tube heat exchangers(outside the pipes)

Latitudinal direction

Longitudinal direction

Staggered orderIn-line order


parallel

perpendicular

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Flow in the longitudinal directionCross section, in-line order

daꞏd

bꞏd

The pressure loss for this case can becalculated by the theory for flow in non-circular ducts.

dbad

1

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9.4

Cross section, staggered order

bꞏd

aꞏd

cꞏdd


Hydraulic diameter

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Flow in the latitudinal direction

The pressure loss can be calculated according to

2

2max

Rloss

wNp

In-line order Staggered order

wmax: maximal velocity

NR: how many times the maximal velocity appear

ζ: dimensionless flow resistance coefficient


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Flow in the latitudinal direction

daꞏd

bꞏd

bꞏd

aꞏd

cꞏdd

In-line order

Staggered order


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In-line order

daꞏd

bꞏd

w∞

L

1max, a

aww a

For the control volume

The maximal velocity is reached between two pipes


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Staggered order

bꞏd

aꞏd

cꞏdd

w∞

The maximal velocity can occur at two places


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Staggered order

bꞏd

aꞏd

cꞏdd

w∞

Maximal velocity in the vertical passage

L

1max, a

aww a


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Staggered order

bꞏd

aꞏd

cꞏdd

w∞

L

Maximal velocity in the diagonal passage

12max, c

aww c


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Staggered orderac ww max,max,

222

21 dcdbda

2221 bac

The maximal velocity in the diagonal passage is achieved if

Utilizing the theorem of Pythagorasbꞏd½ꞏaꞏd

cꞏdd

w∞gives

which also can be written as 112

a

aw

c

aw

The expression of c inserted in the inequality gives 1221 ab

The maximal velocity in the diagonal passage is hence achieved if

1221 ab

Otherwise, the maximal velocity occurs in the vertical passage

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How many times the maximal velocity appears

daꞏd

bꞏd

w∞

L

For in-line order

as many times NR as the number of pipe rows NB

NR = NB


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For staggered order when the maximal velocity occurs in the vertical

passage when

bꞏd

aꞏd

cꞏdd

w∞

L

awmax,

1221 ab

as many times NR as the number of pipe rows NB

NR = NB


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cwmax,

bꞏd

aꞏd

cꞏdd

w∞

L

For staggered order when the maximal velocity occurs in the diagonal

passage when 1221 ab

one time less NR than the number of pipe rows NB

NR = NB - 1


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2max

Rloss

wNp

Re

turblam

Re f

ff

dw

maxRe

),( bafflam

),(turb baff

Re),,(Re baff

from diagram

from diagram

from diagram

9.5

Dimensionless flow resistance coefficient

flam=f(a,b)

More correction factors are needed whenNB < 5, or if density, temperature or viscositychange significantly

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Dimensionless flow resistance coefficient

fRe=f(Re)

fturb= f(a,b)


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3.3 Excercises 9



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Exercises 9

9.1 A wire grating is placed in water (5ºC), which is flowing with a velocity of 4 m/s. The wire grating is made of steel wire that has a diameter of 2 mm, and the mesh has a length of 50 mm. Calculate the drag per m2 of the wire grating.

9.2 The total drag in the case when a fluid flows around a sphere can at low velocities (Rep<0.2) be described by the law of Stokes,

Derive an expression for the drag coefficient CD as a function of Re for fluid flow around spheres at low velocities.

9.3 The lift of an airplane wing was measured to 2000 N per m2 of the wing when the velocity of air below the wing was 120 m/s. Calculate the velocity of air above the wing.

februari 2018 Åbo Akademi University - Värme- och StrömningsteknikBiskopsgatan 8, FI-20500 Åbo / Turku Finland

RoNz 39

wdF 3

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Exercises 9

9.4 Derive the expression for dh for longitudinal flow in a tube heat exchanger that have the tubes in in-line order, and for a tube heat exchanger that have the tubes in staggered order.

9.5 Water of 80ºC is flowing in a tube heat exchanger in the latitudinal direction. The tubes have a staggered order (a=1.75, b=1.53, d=17 mm). At what velocity w∞ is the needed theoretical pump power equal to 1 kW per m3 of the heat exchanger?


RoNz 40

dbad

1

4h

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3.4 Flow in Porous Materials



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Porosity

Total porosity εtot: volume of all pores / total volume

Effective porosity ε : volume of open pores / total volume

8.1februari 2018 RoNz 42Åbo Akademi University - Värme- och Strömningsteknik


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Calculation concept

w

l

p~loss

Definition: Superficial flow (= fictive) velocity through a porous material

Acs is the whole cross sectional area, not only the area of the open pores.

Hypotheses:

2loss ~ wl

p

which is equivalent to laminar flow

which is equivalent to turbulent flow

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Law of Darcy and permeability

w

l

p~loss

Experiments have shown that agrees with flow in materials

with small pores when the velocity is low.

The relationship is called law of Darcy and can be written

wkl

p 1loss

The inverse value of the proportional constant k is called permeability, which means “how good a material let a fluid flow through it”.

8.2februari 2018 RoNz 44Åbo Akademi University - Värme- och Strömningsteknik


Often, vertical position changes must be considered:

)( 2121loss zzgppp

With l = z1-z2: )(

)(

21

2121

zz

zzgppkw

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Porosity and Permeability

Material Porosity ε Permeability k [m2]

loose sand 0.37 – 0.50 20 – 180 (ꞏ10-12)

soil 0.43 – 0.54 30 – 140 (ꞏ10-12)

sand (of stones) 0.08 – 0.38 0.0005 – 3 (ꞏ10-12)

lime stone 0.04 – 0.10 0.0002 – 0.0004 (ꞏ10-12)

brick 0.12 – 0.34 0.005 – 0.22 (ꞏ10-12)

leather 0.56 – 0.59 0.096 – 0.12 (ꞏ10-12)

fossil meal 0.37 – 0.49 0.013 – 0.05 (ꞏ10-12)

8.3

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Flow in packed beds

2loss ~ wl

p

When the pores are bigger and the velocity is higher, the flow will, at least partly, be turbulent. In such a case, following hypothesis is more agreeable than the law of Darcy.

A dimensionless flow resistance number K is used in the same way as ζ = 4 f in calculation of the pressure loss in pipe flow.

p

2loss

2 d

wK

l

p

The length quantity dp is defined as the diameter of an equivalent sphere, which have the same volume as the non-spherical particle that the bed is made up of.

dpV

V


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Flow in packed bedsDarcy’s Law:

with permeability K

Kozeny - Carman equation:

Sv = specific surface = surface/volume

Sv = 6 /dp for a sphere with diameter dp

L

LS

pu

fluidv

2

3

15 )(

L

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Ergun equation

5.31

Re

3001

p3

K

The dimensionless flow resistance number K can be described by the Ergun equation,

Ψ: shape factor, which is the surface area of the equivalent sphere divided byThe surface of the actual particle that the bed is made up of (≤1)

Rep: Reynolds number for the particles

p

pRedw

ε: bed porosity

8.4


Note: also fluidised beds for gas flowsbelow fluidisation velocity are packed bedsAT the point of minimum fluidisation_Δppacked bed = Δpfluidised bed

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3.5 Excercises 8



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Exercises 8

8.1 A 150 cm3 sample of a porous plastic material have a dry weight of 104 g. The weight of it, when soaked in water (10ºC), is 147 g (the air has first been removed with vacuum). The compact density of the material is 1150 kg/m3.a) Calculate the total porosity.b) Calculate the effective porosity.

8.2 Calculate the volumetric flow of water (20ºC) through the sand filter below. The cross sectional area of the filter is 3 m2 and the permeability of the sand is 20·10-12 m2.


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Exercises 8 8.3 The permeability is sometimes given in the unit of darcy = cP cm2 / (s atm).

What is the factor to convert darcy into m2.

8.4 a) Calculate the pressure loss when 3.0 m3/s of a gas (ρ=0.52 kg/m3, η=34·10-6 kg/ms) is flowing through a packed bed with a height of 0.60 m. The packed bed is made up of cylindrical catalyst particles with a length of 8 mm and a diameter of 5 mm. Experiments have showed that the porosity ε is 0.34. The cross sectional area of the bed is 3.8 m2.b) Calculate the superficial gas velocity w at which the pressure loss is equal to the value of (the bed gravity / cross sectional area). The compact density of the cylindrical catalyst particles is 2100 kg/m3.c) The particle bed starts to expand (→fluidized bed) when the gas velocity reach values over the velocity calculated in b). The pressure loss is beyond this superficial velocity constant. Calculate the bed porosity ε when the gas velocity is 50% higher than the value calculate in b).


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Further reading BMH99: Beek, W.J., Muttzall, K.M.K., van Heuven, J.W. ”Transport phenomena”

Wiley, 2nd edition (1999) BSL60: R.B. Bird, W.E. Stewart, E.N. Lightfoot ”Transport phenomena” Wiley (1960) CR83: Coulson, J.M., Richardson, J.F., Backhurst, J.R., Harker, J.H. “Chemical

Engineering, Vol. 2 : Unit Operations” Pergamon Press, Oxford (1983) Chapter 3 CEWR10: C.T. Crowe, D.F. Elger, B.C. Williams, J.A. Roberson ”Engineering Fluid

Mechanics”, 9th ed., Wiley (2010) FZ98: L-S Fan, C Zhu “Principles of gas-solid flows” Cambridge Univ. Press (1998) KJ05: D. Kaminski, M. Jensen ”Introduction to Thermal and Fluids Engineering”, Wiley

(2005) Chapter 4, chapter 10 T06: S.R. Turns ”Thermal – Fluid Sciences”, Cambridge Univ. Press (2006) vD82 van Dyke, M. “An album of fluid motion”, The Parabolic Press, Stanford (CA)

(1982) Z13: R. Zevenhoven ”Introduction to process engineering / Processteknikens

grunder”, Chapter 6: Fluid mechanics: fluid statics, fluid dynamics. Course material ÅA (version 2013) available on line: http://users.abo.fi/rzevenho/PTG%20Aug2013.pdf

ÖS96: G. Öhman, H. Saxén ”Värmeteknikens grunder”, Åbo Akademi University (1996)

Ö96: G. Öhman ”Teknisk strömningslära” Course compendium ÅA (1996) sections8 and 9 (p. 27-34)

februari 2018 Åbo Akademi University - Värme- och StrömningsteknikBiskopsgatan 8, FI-20500 Åbo / Turku Finland

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