transport phenomena

TRANSPORT PHENOMENA Giulio Lolli 10/31/08

Modern Methods in Heterogeneous Catalysis Research

Outline

10/31/08

2

Part I - Theory Why Transport? Fundamental equations Unified approach to transport phenomena Boundary Layer approach Dimensionless numbers

Part II - Practice Example: Methanol Synthesis

Transport Phenomena

Further Reading

10/31/08

3

Part I Bird, Steward, Lightfoot Transport Phenomena

Asano Welty Mass Transport Fundamentals of M,H&M Transfer

Part II Handout Web: Document DB # 15375

Perrys Chemical Engineering Handbook

Transport Phenomena

Theory

Part I

10/31/08

4

Transport Phenomena

Why Transport? Catalytic Process

Thermodynamic Reaction Equilibrium Phase equilibrium:

adsorption/desorption

Kinetic Reaction Kinetic

Transport Reactants/Product to/

from the active sites Heat to/from the

active site 10/31/08

5

Transport Phenomena

Transport Phenomena

Momentum Transport

Heat Transport

Mass Transport

Kinetic Energy

Newtons Law Thermal Energy Fouriers Law

Chemical Energy Ficks Law

Type of transport What is transferred

10/31/08

6

Transport Phenomena

Specific direction x

!xy = !uyx

qx = kTx

jA,x = DACAx

1st Law (Fluxes)

!!T = !!u !q = k!T !jA = DA!CA

! = !A2nd Law (variation in space-time)

D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+ArD

Dt= 2

Fundamental equations of the system

D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+Ar! (!u) = 0

1


!xy = !uyx

qx = kTx

jA,x = DACAx

1st Law (Fluxes)

!!T = !!u !q = k!T !jA = DA!CA


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+ArD

Dt= 2


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+Ar! (!u) = 0

1


!xy = !uyx

qx = kTx

jA,x = DACAx

1st Law (Fluxes)

!!T = !!u !q = k!T !jA = DA!CA


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+ArD

Dt= 2


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+Ar! (!u) = 0

1

First Law of transport

10/31/08 Transport Phenomena

7 Specific direction x

!xy = !uyx

qx = kTx

jA,x = DACAx

1st Law (Fluxes)

!!T = !!u !q = k!T !jA = DA!CA


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+ArD

Dt= 2


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+Ar! (!u) = 0

1


!xy = !uyx

qx = kTx

jA,x = DACAx

1st Law (Fluxes)

!!T = !!u !q = k!T !jA = DA!CA


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+ArD

Dt= 2


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+Ar! (!u) = 0

1


!xy = !uyx

qx = kTx

jA,x = DACAx

1st Law (Fluxes)

!!T = !!u !q = k!T !jA = DA!CA


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+ArD

Dt= 2


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+Ar! (!u) = 0

1

Math 101- Gradient


8 Math 101 Nabla, Gradient

! =

x

y

z

!S =

Sx

Sy

Sz

Math 102 Divergence, Laplacian

! =

x

y

z

!S =

Sx

Sy

Sz

! !V = Vxx + Vyy + Vzz2S = ! !S =

2S

x2+

2S

y2+

2S

z2

Math 103 Substantial Derivative

! =

x

y

z

!S =

Sx

Sy

Sz

! !V = Vxx + Vyy + Vzz2S = ! !S =

2S

x2+

2S

y2+

2S

z2

DS

Dt=

S

t+ !u !S = S

t+ ux

S

x+ uy

S

y+ uz

S

z

2

Second Law of Transport

Navier-Stokes

2nd Fouriers

2nd Fick

10/31/08

9

Transport Phenomena


!xy = !uyx

qx = kTx

jA,x = DACAx

1st Law (Fluxes)

!!T = !!u !q = k!T !jA = DA!CA


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+ArD

Dt= 2


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+Ar! (!u) = 0

1


!xy = !uyx

qx = kTx

jA,x = DACAx

1st Law (Fluxes)

!!T = !!u !q = k!T !jA = DA!CA


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+ArD

Dt= 2


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+Ar! (!u) = 0

1

Math 102 Divergence & Laplace


10

Math 101 Nabla, Gradient

! =

x

y

z

!S =

Sx

Sy

Sz


! =

x

y

z

!S =

Sx

Sy

Sz

! !V = Vxx + Vyy + Vzz2S = ! !S =

2S

x2+

2S

y2+

2S

z2


! =

x

y

z

!S =

Sx

Sy

Sz

! !V = Vxx + Vyy + Vzz2S = ! !S =

2S

x2+

2S

y2+

2S

z2

DS

Dt=

S

t+ !u !S = S

t+ ux

S

x+ uy

S

y+ uz

S

z

2



11

Math 101 Nabla, Gradient

! =

x

y

z

!S =

Sx

Sy

Sz


! =

x

y

z

!S =

Sx

Sy

Sz

! !V = Vxx + Vyy + Vzz2S = ! !S =

2S

x2+

2S

y2+

2S

z2


! =

x

y

z

!S =

Sx

Sy

Sz

! !V = Vxx + Vyy + Vzz2S = ! !S =

2S

x2+

2S

y2+

2S

z2

DS

Dt=

S

t+ !u !S = S

t+ ux

S

x+ uy

S

y+ uz

S

z

2

Equations of Motion


12


!xy = !uyx

qx = kTx

jA,x = DACAx

1st Law (Fluxes)

!!T = !!u !q = k!T !jA = DA!CA


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+ArD

Dt= 2


D!u

Dt= 2!u+!g !p

CpDT

Dt= k2T+u rHrxn

DCADt

= DA2CA+Ar! (!u) = 0

1

Example

Start-up in circular tube Cylindrical coordinates Only vz(r) 0


13

Example: startup in circular tube only uz(r) != 0

uzt

+!!!!uz

uzz

= p+ 1

r

r

(ruzr

)In S.S. conditions (/t = 0)

r

(ruzr

)= p

r

"ruzr

= p

r"2

2+"""A

rB.C.

uzr

= finite r = 0

uz = p

r2

4+B B.C. uz = 0 r = R

uz =p

4

(R2 r2)

uzt

+!!!!uz

uzz

= p+ 1

r

r

(ruzr

)In non S.S. conditions !?!?!

uz = C0e2 (C1J0 () + C2Y0 ())

J0 and Y0 are Bessel functions of the first and second kind of order 0.

uz = C0e2 (C1J0 () + C2Y0 ())

Jn(x) =k=0

(1)k (x2)n+2kk!(n+ k + 1)

Yn(x) = limpn

Jp(x) cos(ppi) Jp(x)sin(ppi)

3

z

r

Example


14

Example: startup in circular tube only uz(r) != 0

uzt

+!!!!uz

uzz

= p+ 1

r

r

(ruzr

)In S.S. conditions (/t = 0)

r

(ruzr

)= p

r

"ruzr

= p

r"2

2+"""A

rB.C.

uzr

= finite r = 0

uz = p

r2

4+B B.C. uz = 0 r = R

uz =p

4

(R2 r2)

uzt

+!!!!uz

uzz

= p+ 1

r

r

(ruzr

)In non S.S. conditions !?!?!

uz = C0e2 (C1J0 () + C2Y0 ())

J0 and Y0 are Bessel functions of the first and second kind of order 0.

uz = C0e2 (C1J0 () + C2Y0 ())

Jn(x) =k=0

(1)k (x2)n+2kk!(n+ k + 1)

Yn(x) = limpn

Jp(x) cos(ppi) Jp(x)sin(ppi)

3

z

r

10/31/08 Transport Phenomena 15

z

r

-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.0

0.5

1.0

vz/vmax(r=0)

r/R

0

0.0001

0.001

0.01

0.05

0.1

0.2

0.5

infinite

Numerical Solution

Finite elements approach Software packages Fluent CFD Physics

Set up the correct problem and boundary conditions Many times is not necessary to know the exact T,C,v

of every single point in your system


16

Boundary Layer

All the phenomena happen in the boundary layer Everything outside the boundary layer is in equilibrium

and constant

The boundary layer is very thin The size of the boundary layer () is smaller than the

characteristic size of the system Order of magnitude approach Taylor series


17

Reynolds Number


18

Boundary layer approach

Reinolds Number

D"u

Dt= 2"u

u!!u

x !

!u

2(

x

)2

ux= 1/Re Re =

ux

Prandtl Number

uT = uT/

CpDT

Dt= k2T

CpuT!!!T

x k!

!!T

2T

CpuTx k

2T(T

)3 k

CpRe1/2

T Pr1/3 Pr = Cp

k

Schmidt Number

C Sc1/3 Sc = DA

4


Reinolds Number

D"u

Dt= 2"u

u!!u

x !

!u

2(

x

)2

ux= 1/Re Re =

ux

Prandtl Number

uT = uT/

CpDT

Dt= k2T

CpuT!!!T

x k!

!!T

2T

CpuTx k

2T(T

)3 k

CpRe1/2

T Pr1/3 Pr = Cp

k

Schmidt Number

C Sc1/3 Sc = DA

4

Prandtls Number


19

T


Reinolds Number

D"u

Dt= 2"u

u!!u

x !

!u

2(

x

)2

ux= 1/Re Re =

ux

Prandtl Number

uT = uT/

CpDT

Dt= k2T

CpuT!!!T

x k!

!!T

2T

CpuTx k

2T(T

)3 k

CpRe1/2

T Pr1/3 Pr = Cp

k

Schmidt Number

C Sc1/3 Sc = DA

4


Reinolds Number

D"u

Dt= 2"u

u!!u

x !

!u

2(

x

)2

ux= 1/Re Re =

ux

Prandtl Number

uT = uT/

CpDT

Dt= k2T

CpuT!!!T

x k!

!!T

2T

CpuTx k

2T(T

)3 k

Cp

T Pr1/3 Pr = Cp

k

Schmidt Number

C Sc1/3 Sc = DA

4

Schmidts Number

As usual everything is the same

10/31/08

20

Transport Phenomena

C


Reinolds Number

D"u

Dt= 2"u

u!!u

x !

!u

2(

x

)2

ux= 1/Re Re =

ux

Prandtl Number

uT = uT/

CpDT

Dt= k2T

CpuT!!!T

x k!

!!T

2T

CpuTx k

2T(T

)3 k

CpRe1/2

T Pr1/3 Pr = Cp

k

Schmidt Number

C Sc1/3 Sc = DA

4


Reinolds Number

D"u

Dt= 2"u

u!!u

x !

!u

2(

x

)2

ux= 1/Re Re =

ux

Prandtl Number

uT = uT/

CpDT

Dt= k2T

CpuT!!!T

x k!

!!T

2T

CpuTx k

2T(T

)3 k

CpRe1/2

T Pr1/3 Pr = Cp

k

Schmidt Number

C Sc1/3 Sc = DA

4

Nusselts Number

Considering the fluxes:

q = kTx

= hT

h!!!T k!!!T

T

Nu =hx

k x

Re1/2

TPr1/3

In general

Nu =hx

k= f (Re)Pr1/3 Sh =

KAx

DA= f (Re)Sc1/3

f (Re) =

Re1/2 Plate in laminar flow

2 +Re1/2 Sphere in laminar flow

0.005Re0.8 Packed bed in turbulent flow

Extension to macroscopic balances

Q =

A

q dA = AU T 1/U =

1/hi +

si/ki

1/U = 1/hint + s/kmetal + 1/hext

5

10/31/08

21

Transport Phenomena

T


q = kTx

= hT

h!!!T k!!!T

T

Nu =hx

k x

Re1/2

TPr1/3

Sh =KAx

DA x

Re1/2

CSc1/3

In general

Nu =hx

k= f (Re)Pr1/3 Sh =

KAx

DA= f (Re)Sc1/3

f (Re) =





Q =

A

q dA = AU T 1/U =

1/hi +

si/ki


5

Sherwoods Number


q = kTx

= hT

h!!!T k!!!T

T

Nu =hx

k x

Re1/2

TPr1/3

In general

Nu =hx

k= f (Re)Pr1/3 Sh =

KAx

DA= f (Re)Sc1/3

f (Re) =





Q =

A

q dA = AU T 1/U =

1/hi +

si/ki


5

10/31/08

22

Transport Phenomena

C


q = kTx

= hT

h!!!T k!!!T

T

Nu =hx

k x

Re1/2

TPr1/3

Sh =KAx

DA x

Re1/2

CSc1/3

In general

Nu =hx

k= f (Re)Pr1/3 Sh =

KAx

DA= f (Re)Sc1/3

f (Re) =





Q =

A

q dA = AU T 1/U =

1/hi +

si/ki


5

In General


23


q = kTx

= hT

h!!!T k!!!T

T

Nu =hx

k x

Re1/2

TPr1/3

In general

Nu =hx

k= f (Re)Pr1/3 Sh =

KAx

DA= f (Re)Sc1/3

f (Re) =





Q =

A

q dA = AU T 1/U =

1/hi +

si/ki


5

Extension to Macroscopic Balances


24

Overall Transfer Coefficient


25


q = kTx

= hT

h!!!T k!!!T

T

Nu =hx

k x

Re1/2

TPr1/3

In general

Nu =hx

k= f (Re)Pr1/3 Sh =

KAx

DA= f (Re)Sc1/3

f (Re) =





Q =

A

q dA = AU T 1/U =

1/hi +

si/ki


5

Log-Mean Temperature Difference


26


q = kTx

= hT

h!!!T k!!!T

T

Nu =hx

k x

Re1/2

TPr1/3

In general

Nu =hx

k= f (Re)Pr1/3 Sh =

KAx

DA= f (Re)Sc1/3

f (Re) =





Q =

A

q dA = AU T 1/U =

1/hi +

si/ki


5


q = kTx

= hT

h!!!T k!!!T

T

Nu =hx

k x

Re1/2

TPr1/3

Sh =KAx

DA x

Re1/2

CSc1/3

In general

Nu =hx

k= f (Re)Pr1/3 Sh =

KAx

DA= f (Re)Sc1/3

f (Re) =





Q =

A

q dA = AU T 1/U =

1/hi +

si/ki


Tln =T1 T2

ln (T1/T2)

5

Conclusions

Momentum, Heat and Mass Transfer Highly inter-correlated Similar physical principles Similar driving force Similar mathematical formulas

2nd order PDE numerical solutions Highly dependent from fluid-dynamic properties Simplifications Boundary layer Dimensionless numbers


27

Methanol Synthesis

PART II

10/31/08

28

Transport Phenomena

Methanol Production

Produced from SynGas Catalyst Cu/ZnO-Al2O3 Equilibrium limited P = 50 100 bar T = 200 300 C

Main Commodity 20 Million ton/year

Typical plant size is 1 Million ton/year 12 m3/s


29

Catalyst specs


30

KATALCO is a trade mark of the Johnson Matthey Group of Companies

Johnson Matthey CatalystsPO Box 1BillinghamCleveland

TS23 1LBEnglandTel: +44 1642 522888Fax: +44 1642 522542

Johnson Matthey CatalystsTwo TransAm Plaza DriveOakbrook TerraceIllinois 60181

USATel: +1 630 268 6300Fax: +1 630 268 9797

Johnson Matthey Catalysts1100 HerculesHoustonTexas 77058

USATel: +1 281 488 0638Fax: +1 281 488 7055

Johnson Matthey CatalystsGlobal Business Park6

th Floor, Tower B

Mehrauli-Gurgaon Road

Gurgaon 122,002 (Haryana)IndiaTel: +91 124 2359836/7/8Fax: +91 124 2359844

Tracerco AsiaTracerco/Dialog Alliance109, Block G,Phileo Damansara 1

46350 Petaling JayaSelangorMalaysiaTel: + 603 79551199Fax: + 603 79546953

Johnson Matthey JapanIncorporatedImperial Tower 14F1-1-1, Uchisaiwai-cho,

Chiyoda- ku,Tokyo 100-0011 JAPANPhone +81-3-5511-8556Fax +81-3-5511-8561

2003 Johnson Matthey www.jmcatalysts.com

KATALCO 51-8Methanol Synthesis Catalyst

Product Benefits Long life as a result of the optimized formulation incorporating a patenteduse of a fourth component MgO

Low operating temperatures minimize the catalyst sintering rate

High catalyst selectivity gives very low impurities in the crude product

Close approach to equilibrium is achieved and maintained

Easy to reduce and start-up

Product Uses Synthesis of methanol from H2, CO and CO2 mixtures arising from steamreforming of hydrocarbons, coal gasification or POx

GeneralDescription

KATALCO 51-8 is a copper catalyst on a ZnO-Al2O3 support with a MgO

promoter

Physical Properties (Typical) Chemical Composition (Typical)

Form

Diameter

Length

Bulk Density

Crush Strength

(axial)

KATALCO 51-8

Cylindrical pellet

5.4 mm

5.2 mm

1250 kg/m3

80 kgf

CuO:

Al2O3:

ZnO:

MgO:

64 wt%

10 wt%

24 wt%

2 wt%

Shipping &Handling

Avoid contact with skin and clothing. Avoid breathing dust. Do not takeinternally. Please refer to the relevant Material Safety Data Sheet forfurther information.

KATALCO 51-8 catalyst is available in non-returnable polythene lined

mild steel drums or bulk bags for easy loading.

Reactor types

10/31/08

31

Transport Phenomena

Multi-Bed Reactor


32

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Reactor dimensioning

1 m/s wanted gas velocity 12 m2 section approx 4 m diameter Approx 7 m tall


33

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544

644

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Pressure Drop


36

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External properties

Nusselt and Sherwood Numbers Colburn Factors

10/31/08

37

Transport Phenomena

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External Limitations


38

No external limitations in terms of gas composition and temperature profile Due to the turbulent flow regime

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Heat Exchanger

1st bed 580 490 K 35 MW power to take out Boiling water to produce low pressure steam 16 kg/s steam produced

Shell tube heat exchanger: 2 (5cm) tubes hout=220 W/m2K U 200 W/m2K


39

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Evaporation/condensation

hl /kl!l1/2 = 0.065(NPr)l1/2Fvc1/2 (5-99a)where Fvc = fG2vm /2!v (5-99b)

Gvm = ! "1/2

(5-99c)

and f is the Fanning friction factor evaluated at(NRe)vm = DiGvm /v (5-99d)

and the subscripts vi and vo refer to the vapor inlet and outlet, respec-tively. An alternative formulation, directly in terms of the friction fac-tor, is

h = 0.065 (c!kf/2!v)1/2Gvm (5-99e)expressed in consistent units.

Another correlation for vapor-shear-controlled condensation is theBoyko-Kruzhilin correlation [Int. J. Heat Mass Transfer, 10, 361(1967)], which gives the mean condensing coefficient for a streambetween inlet quality xi and outlet quality xo:

= 0.024 ! "0.8

(NPr)l0.43 (5-100a)

where GT = total mass velocity in consistent units

! "i= 1 + xi (5-100b)

and ! "o= 1 + xo (5-100c)

For horizontal in-tube condensation at low flow rates Kernsmodification (Process Heat Transfer, McGraw-Hill, New York, 1950)of the Nusselt equation is valid:

hm = 0.761 # $1/3

= 0.815 # $1/4

(5-101)

where WF is the total vapor condensed in one tube and " t is tsv # ts .A more rigorous correlation has been proposed by Chaddock [Refrig.Eng., 65(4), 36 (1957)]. Use consistent units.

At high condensing loads, with vapor shear dominating, tube orien-tation has no effect, and Eq. (5-100a) may also be used for horizontaltubes.

Condensation of pure vapors under laminar conditions in the pres-ence of noncondensable gases, interfacial resistance, superheating,variable properties, and diffusion has been analyzed by Minkowyczand Sparrow [Int. J. Heat Mass Transfer, 9, 1125 (1966)].

BOILING (VAPORIZATION) OF LIQUIDS

Boiling Mechanisms Vaporization of liquids may result fromvarious mechanisms of heat transfer, singly or combinations thereof.

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For example, vaporization may occur as a result of heat absorbed, byradiation and convection, at the surface of a pool of liquid; or as aresult of heat absorbed by natural convection from a hot wall beneaththe disengaging surface, in which case the vaporization takes placewhen the superheated liquid reaches the pool surface. Vaporizationalso occurs from falling films (the reverse of condensation) or from theflashing of liquids superheated by forced convection under pressure.

Pool boiling refers to the type of boiling experienced when theheating surface is surrounded by a relatively large body of fluid whichis not flowing at any appreciable velocity and is agitated only by themotion of the bubbles and by natural-convection currents. Two typesof pool boiling are possible: subcooled pool boiling, in which the bulkfluid temperature is below the saturation temperature, resulting incollapse of the bubbles before they reach the surface, and saturatedpool boiling, with bulk temperature equal to saturation temperature,resulting in net vapor generation.

The general shape of the curve relating the heat-transfer coefficientto " tb, the temperature driving force (difference between the walltemperature and the bulk fluid temperature) is one of the few para-metric relations that are reasonably well understood. The familiarboiling curve was originally demonstrated experimentally by Nukiyama[J. Soc. Mech. Eng. ( Japan), 37, 367 (1934)]. This curve points outone of the great dilemmas for boiling-equipment designers. They arefaced with at least six heat-transfer regimes in pool boiling: naturalconvection (+), incipient nucleate boiling (+), nucleate boiling (+),transition to film boiling (#), stable film boiling (+), and film boilingwith increasing radiation (+). The signs indicate the sign of the deriv-ative d(q/A)/d " tb. In the transition to film boiling, heat-transfer ratedecreases with driving force. The regimes of greatest commercialinterest are the nucleate-boiling and stable-film-boiling regimes.

Heat transfer by nucleate boiling is an important mechanism inthe vaporization of liquids. It occurs in the vaporization of liquids inkettle-type and natural-circulation reboilers commonly used in theprocess industries. High rates of heat transfer per unit of area (heatflux) are obtained as a result of bubble formation at the liquid-solidinterface rather than from mechanical devices external to the heatexchanger. There are available several expressions from which reason-able values of the film coefficients may be obtained.

The boiling curve, particularly in the nucleate-boiling region, is sig-nificantly affected by the temperature driving force, the total systempressure, the nature of the boiling surface, the geometry of the system,and the properties of the boiling material. In the nucleate-boilingregime, heat flux is approximately proportional to the cube of the tem-perature driving force. Designers in addition must know the minimum"t (the point at which nucleate boiling begins), the critical "t (the "tabove which transition boiling begins), and the maximum heat flux (theheat flux corresponding to the critical "t). For designers who do nothave experimental data available, the following equations may be used.

Boiling Coefficients For the nucleate-boiling coefficient theMostinski equation [Teplenergetika, 4, 66 (1963)] may be used:

h = bPc0.69 ! "0.7

#1.8 ! "0.17

+ 4 ! "1.2

+ 10 ! "10

$ (5-102)where b = (3.75)(10#5)(SI) or (2.13)(10#4) (U.S. customary), Pc is thecritical pressure and P the system pressure, q/A is the heat flux, and his the nucleate-boiling coefficient. The McNelly equation [J. Imp.Coll. Chem. Eng. Soc., 7(18), (1953)] may also be used:

h = 0.225 ! "0.69

! "0.31

! # 1"0.33

(5-103)

where cl is the liquid heat capacity, $ is the latent heat, P is the systempressure, kl is the thermal conductivity of the liquid, and & is the sur-face tension.

An equation of the Nusselt type has been suggested by Rohsenow[Trans. Am. Soc. Mech. Eng., 74, 969 (1952)].

hD/k = Cr(DG/)2/3(c/k)#0.7 (5-104a)

in which the variables assume the following form:

# $1/2

= Cr # ! "1/2

$2/3

! "#0.7

(5-104b)c!k

W!A

gc&!!g(!L # !v)

'(!

gc&!g(!L # !v)

h'(!

k

!l!!v

Pkl!&

qcl!A$

P!Pc

P!Pc

P!Pc

q!A

5-22 HEAT AND MASS TRANSFER

FIG. 5-10 Dukler plot showing average condensing-film coefficient as a func-tion of physical properties of the condensate film and the terminal Reynoldsnumber. (Dotted line indicates Nusselt theory for Reynolds number < 2100.)[Reproduced by permission from Chem. Eng. Prog., 55, 64 (1959).]


40

Pr water 7

Heat Exchanger

1st bed 580 490 K 35 MW power to take out Boiling water to produce low pressure steam 16 kg/s steam produced

Shell tube heat exchanger: 2 (5cm) tubes hout=220 W/m2K U 200 W/m2K

Low pressure steam 1-2 atm 120C Tln= 140K Exchange Area 1300 m3 (about 2000 tubes) 40 banks of 50 tubes (4-5 m in size)


41

Further Reading

10/31/08

42

Part I Bird, Steward, Lightfoot Transport Phenomena

Asano Welty Mass Transport Fundamentals of M,H&M Transfer

Part II Handout Web: Document DB # 15375

Perrys Chemical Engineering Handbook

Transport Phenomena

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>/>>,0< >/>>0, 0/=,.>= => >/>>4/>>?3,< >/0 >/4344, >/3=333, >/3?00,0 ,,/4?,>/>>,0, >/>>0, 0/=,.>= => >/>>4/>>?3,= >//4344/3=333 >/3?00, ,,/4?,>3

>/>>,0 >/>>0, 0/=,.>= => >/>>4/>>?3,= >//43440 >/3=333 >/3?00 ,,/4?,=

>/>>,=3 >/>>0,> 0/=,.>= => >/>>4/>>?3,= >//43443 >/3=33?4 >/3?00>0 ,,/4?,=

>/>>,=0 >/>>0,> 0/=,.>= => >/>>43>3 >/34?> >/30,?? ,=/>?4/>>,= >/>= => >/>>/>>,/4,?=, >/==?0? >/34343 >/30=3,, ,=/>3/>>< >/>.>= => >/>>,,43 .>/>>043 >/0/34, ,=/=?

>/>>3? >/>= => >/>>,>3? .>/>>/0>>= >//3?=,4? >/34=4, ,=/00

>/>>3 >/>= => >/>>?,< .>/>>/=3,0 >//3?0=3 >/344,>4 ,=/4

>/>>?, >/>=>, /4?.>= => >/>>=?0 .>/>>>= >/=004>< >/= >/3?44, >/343< ,=/==3

>/>>44 >/>=>0 /44.>= => >/>>,== .>/>>?0? >/=? >//3???>, >/3?>3 ,=/30

>/>>4 >/>=>?0 /44.>= => >/>>? .>/>>4, >/=>/,3403 >/3?30, >/3?,3, ,=/,=3

>/>>0? >/>= /40.>= => >/>>>>0 .>/>>=3 >/? >/,?/33>/3?,44 ,=/,==3

=(7 )0$

II;

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