Transport processes

7. Semester

Chemical Engineering

Civil Engineering

Course structure

Transport ProcessesChemical engineering

3 ECTS

Fluid and wave dynamics

Civil engineering5 ECTS

Course structure

Transport ProcessesChemical engineering

3 ECTS

Fluid and wave dynamics

Civil engineering5 ECTSFluid dynamics Fluid dynamics

Wave dynamicsHeat and mass transfer

6 lectures

Oral exam in Transport processes

Oral exam in Fluid and wave dynamics

My background

Literature

• Primary literature:– Geankoplis, Transport processes and separation Principles

• Supplementary literature:– (Brorsen, Strømningslære)

– Selected chapters from e-books

Course homepage

http://homes.et.aau.dk/mma/

On the course homepage you can find… – Slides

– Supplementary literature (e-books)

– Exercises/solutions

Motivation

• Why do I need to bother myself with this course? – You need to pass an exam!

– Knowledge about Fluid flow is essential for Chemical and Civil Engineers alike

– You can actually use the stuff you learn to make some practical calculations (well at least some of it…)

– This course forms the base for Numerical Fluid Dynamics (including heat and mass dynamics)

Exam

• 20 min Oral Exam

• 20 min preparation time

• Examination questions will be available 2 weeks before the exam

Course plan

1. Elementary Fluid Dynamics2. Fluid Kinematics3. Finite Control Volume Analysis4. Differential Analysis of Fluid Flow5. Viscous Flow and Turbulence6. Turbulent Boundary Layer Flow7. Principles of Heat Transfer8. Internal Forces Convection9. Unsteady Heat Transfer10. Boiling and Condensation11. Mass Transfer12. Porous Media Flow13. Non-Newtonian Flow

Today's lecture

• Elementary Fluid Dynamics– The Bernoulli Equation

– Static and Dynamic Pressure

What is a transport process?

• Momentum transfer (flow)

• Heat transfer

• Mass transfer

Diffusion

Convection

Radiation

Conduction

MomentumHeat

Mass

Truck!Goods!

Transport processes

• What is transport processes really?

Diffusion

Convection

Radiation

Conduction

Energy concepts

• Energy can be a lot of things, what is your understanding? – Heat

– Work

– Potential energy

– Kinetic energy

– Internal energy

– Enthalpy

PcρForce distance×

gzρ21

2 Vρ

u

h

14

Flow work• Flow work

– Flow through control volumes require work

– The required force is given by F=PA

flowW FL PAL PV= = =

Mechanical energy

• The mechanical energy is the flow work, the kinetic and the potential energy:

• Change in the mechanical energy:

15

[ ]2

2P Ve Pv ke pe gz J kgρ

= + + = + +

( )2 2

2 1 2 12 12

P P V Ve g z zρ− −

∆ = + + −

[ ]2

2mVE PV KE PE PV mgz J= + + = + +

Energy balance

• If the mechanical energy is conserved, then:

• This can be formulated as pressure:

• or as head:

( ) ( )2 2

1 1 2 21 22 2

P V P Vg z g zρ ρ+ + = + +

Did I hear somebody say that it was a good idea to always check the units?!?

2 21 2

1 1 2 22 2V VP gz P gzρ ρ ρ ρ+ + = + +

2 21 1 2 2

1 22 2P V P Vz zg g g gρ ρ+ + = + +

Example

• Fill out the blank spaces.

Energy and mass conservation• Mass balance for a system:

• Energy balance for a system:

18

in out sysm m m− = ∆

Mass which enters system boundaries Mass which exits

system boundaries

Change of the system mass (accumulation)

in out sysE E E− = ∆

Energy which enters system boundaries Energy which exits

system boundaries

Change of the systems energy

Conservation of Momentum

• Newtons 2 lov, momentumbalance

F ma=∑The sum of forces which act on a system Change of the

systems momentum

Bernoulli

• F=ma along a streamline

• Steady, incompressible flow:

2

constant along a streamline2

F ma

P V gzρ

=

⇓

+ + =

∑

2 21 1 2 2

1 22 2P V P Vgz gzρ ρ+ + = + +

Formal derivation

2

cst2

VP gzρ ρ+ + =

Name game…

Static pressureDynamic Pressure Hydrostatic Pressure

Name game…#2

• Stagnation Pressure :

• Total Pressure:

2

2stag static dynVP P P P ρ= + = +

2

2tot static dyn hydroVP P P P P gzρ ρ= + + = + +

Pitot-tube

• Assume that the hydrostatic pressure is negligible (the fluid is a gas)

• Static Pressure:

• Stagnations pressure:

• Combine:

2

3 2stagVP P P ρ= = +

1 4P P P= =

( )

2

3 4

3 4

2

2

VP P

P PV

ρ

ρ

− =

−=

Example

• Determine an expression for h as a function of the flowrate Q

Bernolli:

Conservation of mass:

equilibrium:

Combine:

2 21 2

1 1 2 22 2V VP g z P g zρ ρρ ρ+ + = + +

1 1 2 2Q AV A V= =

( )

( ) ( )

2 1 1 2

1 2 2 1

SG

SG

g z z gl gh P gl gh P

P P g z z gh

ρ ρ ρ ρ ρ

ρ ρ ρ

− + + − = + −

− = − + −

( ) ( )( )

2 22 2 1

2 SG

Q A A Ah

gρ

ρ ρ−

=−

Example

• Flow from a large reservoir

P2

11 2

Vρ+ 1 2g z Pρ+ =

22

22V g zρ ρ+ +

2V gh=

Example

• Aerodynamic lift of a wing profile– Using bernoilli, explain how a lift is generated

– 2min, discus with the person next to you

Example

• Galloping of a stack (chimney)– http://www.youtube.com/watch?v=EgRBpeqPUEU&feature=player_e

mbedded

– http://media.efluids.com/galleries/all?medium=578

losses: (non-ideal flow)- short about pipe flow- very short about drag

Pipe flow

Laminar and turbulent flows

• Laminar flow:– Molecular diffusion

• Turbulent flow:– Macroscopic mixing

Reynolds number

• Characterization of flows

• ρ – density

• μ – viscosity

• D – diameter

• V – velocity

Re VDρµ

=Re 2300 laminar

2300 Re 10000 transitional10000 Re turbulent

≤≤ ≤≤

Example

• Is the flow in a water pipe laminar or turbulent? If we assume that it takes about 30 sekunder a one liter cup. The internal pipe diameter is assmed to be around 1 cm.

• The flow is in the transition interval!

• How would you calculate the pressure loss?

3

1000 / 3 0.42 / 0.01Re 37501.12 10 /

VD kg m m s mkg s m

ρµ −

⋅ ⋅= = =

⋅ ⋅

330.001 0.000033 /

30mQ m s

sek= =

2 0.42 /4

QV Q D m sA

π = = =

Loss...?

• We talk about pressure loss in pipes– Friction between fluid and wall, the fluid sticks to the wall

– The mechanical energy of the fluid is converted into heat

– A pressure difference which we cannot recover

• Pressure loss is calculate as:

• L – pipe length

• D – pipe diameter

• V – velocity

• f – friction factor

P1P2

ΔP=P1-P2

2

2L VP fDρ

∆ = ( )laminar: 64 / Returbulent: fε/D,Re

ff=

=

Energy equation with loss

2 21 2

1 1 2 22 2 lossV VP g z P g z Pρ ρρ ρ+ + = + + + ∆

Colebrook equation

• The relationship between roughness, Re and f:

1.111 6.91.8logRe 3.7

Df

ε = − +

Moody chart

Minor and major loss

• ”Minor” loss: Pressure loss in pipes

• ”Major” loss: pressure loss in pipe components

• Total pressure loss:

P1P2

ΔP=P1-P2

2

2L VP fDρ

∆ =

2

2LVP K ρ

∆ =

2 2

2 2tot LV L VP K f

Dρ ρ

∆ = +∑ ∑

Indløb og udløb

KL=1.0 KL=1.0

KL=1.0KL=1.0KL=0.5

KL=0.04KL=0.2

KL=0.8

Flow komponenter

Eksempel

• Bestem trykket P2:

KL findes for denne diffuser til 0.133

masse bevarelse:

komponent-tryktab:

Energibevarelse:

11 1 1 2 2 2 2 1

2

3.11 /AV A V A V V m sA

ρ ρ= ⇒ = =

21 0.00326bar

2LVP K ρ

∆ = =

2 21 2

1 1 2 2 , 1.06(empirisk konstant)2 2V VP g z P g z Pρ ρα ρ α ρ α+ + = + + + ∆ =

2 21 2

2 1 1682

V VP P P kPaαρ αρ−= + −∆ =

Drag

Drag

• Definition: The force acting on an object in a flow

• Mechanical energy of the fluid transferred to the object

• Friction on the surface of the object heat

Ideal fluid (no viscosity)

Drag Concepts

• The drag coefficient, CD:

• CD for low Re (Re<<1) flow:

• General:

212drag DF C V Aρ=

Pressure loss in pipes:

2

2LVP K ρ

∆ =

Pressure loss in pipes:

( )laminar: 64 / Returbulent: fε/D,Re

ff=

=

( )f shape,ε/L,Re,Ma,FrDC =

Standard drag curve

Effect of surface roughness

Excercises

• Time to wake up!