MOMENTUM TRANSFER (VELOCITY DISTRIBUTION IN LAMINAR

FLOW)

Shell Momentum Balances: Boundary Conditions

(Rate of momentum in) – (Rate of momentum out) + (Some of forces acting

on system) = 0

Flow of A Falling Film

Rate of z momentum in across surface at x ( )( ) xxzLW τ

Rate of z momentum out across surface at x+∆x ( )( ) xxxzLW ∆+τ

Rate of z momentum in across surface at z=0 ( )( ) 0=∆ zzz vvxW ρ

Rate of z momentum out across surface at z=L ( )( ) Lzzz vvxW =∆ ρ

Gravity force acting on fluid ( )( )βρ cosgxLW∆

Thus, momentum balance:

( )( ) xxzLW τ - ( )( ) xxxzLW ∆+τ + ( )( ) 0=∆ zzz vvxW ρ - ( )( ) Lzzz vvxW =∆ ρ +

( )( )βρ cosgxLW∆ =0

Because zv is the same at 0=z as it is at Lz = for each value of x, the third

and fourth terms just cancel one another.

Divided by ( )xLW∆ :

βρττ

coslim0

gx

xxzxxxz

x=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∆

−∆+

→∆

βρτ cosgdxd

xz = or 1cos Cgxxz += βρτ

At 0,0 == xzx τ βρτ cosgxxz =

But dxdvz

xz µτ −=

So: xgdxdvz

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

µβρ cos

Integration results in: 22

2cos Cxgvz +⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

µβρ

At 0, == zvx δ

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛−=

22

12

cosδµ

βδρ xgvz

Maximum velocity: µ

βδρ2

cos2

max,gvz =

Average velocity:

µ

βδρδδµ

βδρδ

δ

δ

δ

3cos1

2cos1 21

0

2

0

2

0 0

0 0 gxdxgdxvdydx

dydxvv zW

W

z

z =⎟⎠⎞

⎜⎝⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=== ∫∫

∫ ∫

∫ ∫

Volume rate of flow:

µ

βδρδδ

3cos2

0 0

gWvWdydxvQ z

W

z === ∫ ∫

Film thickness:

32

3cos

3cos

3cos

3βρ

µβρ

µβρ

µδ

ggWQ

gvz Γ

===

Where, the mass flow rate per unit width of wall zvρδ=Γ

Force F of the fluid on the surface:

( )( ) βδρµ

βδρµµτ δδ coscos

0 00 0

LWggLWdzdydxdv

dzdyFL W

xz

x

L W

xz =⎟⎟⎠

⎞⎜⎜⎝

⎛−−=−== ∫ ∫∫ ∫ ==

The above correlations aplly for laminar flow: Re < 1000, µρδ /4Re zv=

Flow Through a Circular Tube

Rate of z momentum in across cylindrical surface at r ( )( ) rrzrL τπ2

Rate of z momentum out across cylindrical

surface at r+∆r ( )( ) rrrzrL ∆+τπ2

Rate of z momentum in across annular surface at z=0 ( )( ) 02 =∆ zzz vvrr ρπ

Rate of z momentum out across annular surface at z=L ( )( ) Lzzz vvrr =∆ ρπ2

Gravity force acting on cylindricall shell ( )( )grLr ρπ ∆2

Pressure force acting on annular surface at z=0 ( )( )02 prr∆π

Pressure force acting on annular surface at z=L ( )( )Lprr∆π2

Momentum balance:

( )( ) rrzrL τπ2 - ( )( ) rrrzrL ∆+τπ2 + ( )( ) 02 =∆ zzz vvrr ρπ - ( )( ) Lzzz vvrr =∆ ρπ2 + ( )( )grLr ρπ ∆2

+ ( )( )Lpprr −∆ 02π = 0

Divided by ( )rL∆π2 :

rgL

ppr

rr Lrrzrrrz

x⎟⎠⎞

⎜⎝⎛ +

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∆

−∆+

→∆

ρττ 0

0lim

( ) rL

rdrd L

rz ⎟⎠⎞

⎜⎝⎛ Ρ−Ρ

= 0τ where gzp ρ−=Ρ

r

CrL

Lrz

10

2+⎟

⎠⎞

⎜⎝⎛ Ρ−Ρ

=τ

Because 0=rzτ at r=0 rL

Lrz ⎟

⎠⎞

⎜⎝⎛ Ρ−Ρ

=2

0τ

But drdvz

rz µτ −=

So rLdr

dv Lz⎟⎟⎠

⎞⎜⎜⎝

⎛ Ρ−Ρ−=

µ20 or 2

20

4Cr

Lv L

z +⎟⎟⎠

⎞⎜⎜⎝

⎛ Ρ−Ρ−=

µ

Because vz = 0 at r = R ( )⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

Ρ−Ρ=

220 1

4 Rr

LR

v Lz µ

Maksimum velocity:

( )L

Rv L

z µ4

20

max,Ρ−Ρ

=

The average velocity:

( )L

R

drdr

drdrvv L

R

R

z

z µθ

θ

π

π

8

20

2

0 0

2

0 0 Ρ−Ρ==

∫ ∫

∫ ∫

Volume flow rate:

( )L

RQ L

µπ

8

40 Ρ−Ρ

=

Force of the fluid on the wetted surface of pipe:

( ) ( ) ( ) gLRppRRdrdvRLF LLRr

zz ρπππµπ 2

02

022 +−=Ρ−Ρ=⎟

⎠⎞

⎜⎝⎛−= =

HEAT TRANSFER (TEMPERATURE DISTRIBUTIONS)

Shell Energy Balances: Boundary Conditions

(Rate of thermal energy in) – (Rate of thermal energy out) + (Rate of

thermal energy production) = 0

Heat Conduction With An Electrical-Heat Source

An electrical wire of circular cross section with radius R and electrical

conductivity ke ohm-1cm-1 with current density I amps cm-2. The rate of heat

production per unit volume is e

e kIS

2

=

Rate of thermal energy in across cylindrical surface at r ( )( )rrqrLπ2

Rate of thermal energy out across cylindrical surface

at r+∆r ( )( )rrrqLrr ∆+∆+ )(2π

Rate of production of thermal energy by electrical dissipation ( ) eSrLr∆π2

Energy balance: ( )( )rrqrLπ2 - ( )( )rrrqLrr ∆+∆+ )(2π + ( ) eSrLr∆π2 = 0

Divided by ( )rL∆π2 :

rSr

rqrqe

rrrrr

x=⎟⎟

⎠

⎞⎜⎜⎝

⎛

∆

−∆+

→∆lim

0

or ( ) rSrqdrd

er = , integration results in: r

CrSq e

r1

2+=

at r = 0 qr is not infinite, so 2rS

q er =

But drdTkqr −= so

2rS

drdTk e=−

Integration results in 2

2

4C

krS

T e +−=

At r = R T = To ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=−

22

0 14 R

rkRS

TT e

Maximum Temperature: kRS

TT e

4

2

0max =−

Average temperature:

( )

kRS

drdr

drdrTrTTT e

R

R

8

)(2

2

0 0

2

0 00

0 =−

=−

∫ ∫

∫ ∫π

π

θ

θ

Heat flow at the surface (for length L of wire):

ee

Rr LSRRS

RLQ 2

22 ππ ===

MASS TRANSFER (CONCENTRATION DISTRIBUTION)

Shell Energy Balances: Boundary Conditions

(Rate of mass of A in) – (Rate of mass of A out) + (Rate of production of

mass A by homogeneous chemical reaction) = 0

Diffusion Through a Stagnant Gas Film

( )BzAzAA

ABAz NNxz

xcDN ++∂∂

−=

For 0=BzN dz

dxx

cDN A

A

ABAz −

−=1

Mass balance (S= cross section area of the column) :

0=− ∆+ zzAzzAz SNSN

Divided by S∆z and ∆z approaches zero, gives: 0=−dz

dN Az

Or 01

=⎟⎟⎠

⎞⎜⎜⎝

⎛− dz

dxx

cDdzd A

A

AB

Simplified to: 01

1=⎟⎟

⎠

⎞⎜⎜⎝

⎛− dz

dxxdz

d A

A

Integration gives: 111 C

dzdx

xA

A

=⎟⎟⎠

⎞⎜⎜⎝

⎛−

Second integration gives: ( ) 211ln CzCxA +=−−

Boundary conditions: at 11 AA xxzz ==

at 22 AA xxzz ==

12

1

1

2

1 11

11 zz

zz

A

A

A

A

xx

xx −

−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

=⎟⎟⎠

⎞⎜⎜⎝

⎛−−

12

1

1

2

1

zzzz

B

B

B

B

xx

xx −

−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

Average concentration;

( )

∫

∫= 2

1

2

11

1

,/

z

z

z

zBB

B

avgB

dz

xx

xx

( )12

12, /ln BB

BBavgB xx

xxx −=

The rate of mass transfer at the liquid-gas interface:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

==−

−= ===1

2

121

11

11 ln

1 B

BABzz

B

B

ABzz

A

A

ABzzAz x

xzz

cDdz

dxx

cDdz

dxx

cDN

Or ( )( ) ( )21ln12

1 AAB

ABzzAz xx

xzzcD

N −−

==

Or ( )( )

( )( )( ) ( )21

ln121

2

121

/ln/AA

B

AB

B

BABzzAz pp

pzzRTpD

pp

zzRTpDN −

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

==

Diffusion With Homogeneous Chemical Reaction

Here, gas A dissolves in liquid B and diffuses into the liquid phase. As it

diffuses, A also performs an irreversible chemical reaction: A + B AB.

Mass balance:

0=∆−− ∆+ zSkcSNSN AzzAzzAz

k = first order rate constant for the reaction.

And then, 0=+ AAz kc

dzdN

If A and AB are present in small concentrations, then we may use:

dz

dcDN AABAz −=

Then, 02

2

=+− AA

AB kcdz

cdD

BC1: at z=0 cA = cA0

BC2: at z=L NAz = 0 or dcA/dz = 0

Integration with considering the Boundary conditions results in:

1

1

0 cosh

1cosh

bLzb

cc

A

A⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−

= where ABDkLb /21 =