Shell Momentum

8/6/2019 Shell Momentum

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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


2/16


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Rate of z momentum in across surface at x ( )( )x xz LW

Rate of z momentum out across surface at x+ x ( )( ) x x xz LW +

Rate of z momentum in across surface at z=0 ( )( ) 0= z z z vv xW

Rate of z momentum out across surface at z=L ( )( ) L z z z vv xW =

Gravity force acting on fluid ( )( ) cosg x LW

Thus, momentum balance:

( )( )x xz LW - ( )( ) x x xz LW + + ( )( ) 0= z z z vv xW - ( )( ) L z z z vv xW = +

( )( ) cosg x LW =0

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

and fourth terms just cancel one another.

Divided by ( ) x LW :

coslim0

g x

x xz x x xz

x

=

+

cosgdxd

xz = or 1cos C gx xz +=

At 0,0 == xz x cosgx xz =


4/16

Butdxdv z

xz =

So: xg

dx

dv z

= cos

Integration results in: 22

2cos

C xg

v z +

=

At 0, == zv x

=

22

12

cos

xgv z

Maximum velocity:

2

cos2max,

gv z =

Average velocity:

3cos

12

cos1 21

0

2

0

2

0 0

0 0 g xd xg

dxv

dydx

dydxv

v zW

W

z

z=

===

Volume rate of flow:

3cos2

0 0

gW vW dydxvQ z

W

z===


5/16

Film thickness:

32

3

cos

3

cos

3

cos

3

ggW

Q

g

v z ===

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

Force F of the fluid on the surface:

( )( )

coscos0 00 0

LW gg LW dzdydx

dvdzdyF L W

x z

x

L W

xz =

=== ==

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

Flow Through a Circular Tube


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Rate of z momentum in across cylindrical surface at r ( )( )r rzrL 2

Rate of z momentum out across cylindrical

surface at r+ r ( )( ) r r rzrL + 2

Rate of z momentum in across annular surface at z=0 ( )( ) 02 = z z z vvr r

Rate of z momentum out across annular surface at z=L ( )( ) L z z z vvr r = 2

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

Pressure force acting on annular surface at z=0 ( )( )02 pr r

Pressure force acting on annular surface at z=L ( )( ) L pr r 2

Momentum balance:

( )( )r rzrL 2 - ( )( ) r r rzrL + 2 + ( )( ) 02 = z z z vvr r - ( )( ) L z z z vvr r = 2 + ( )( )grLr 2

+ ( )( ) L p pr r 02 = 0

Divided by ( )r L 2 :

r g L

p p

r

r r Lr rzr r rz x

+=

+

0

0lim


7/16

( ) r L

r dr d L

rz

= 0 where gz p =

r C r

L L

rz10

2+

=

Because 0=rz at r =0 r L L

rz

=

20

Butdr

dv zrz =

So r Ldr

dv L z

=

20 or 2

20

4C r

Lv L z +

=

Because v z = 0 at r = R ( )

=

220 1

4 Rr

L

Rv L z

Maksimum velocity:

( ) L

Rv L z 4

20

max,

=

The average velocity:

( ) L

R

d rdr

d rdr v

v L R

R

z

z

8

20

2

0 0

2

0 0==


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Volume flow rate:

( ) L

RQ L

8

40 =

Force of the fluid on the wetted surface of pipe:

( ) ( ) ( ) g L R p p R Rdr

dv RLF L L Rr

z z

20

20

22 +==

= =


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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 k e ohm-1cm -1 with current density I amps cm -2. The rate of heat

production per unit volume ise

e k I

S2

=


10/16

Rate of thermal energy in across cylindrical surface at r ( )( )r r qrL 2

Rate of thermal energy out across cylindrical surface

at r+ r ( )( )r r r q Lr r ++ )(2

Rate of production of thermal energy by electrical dissipation ( ) eSrLr 2

Energy balance: ( )( )r r qrL 2 - ( )( )r r r q Lr r ++ )(2 + ( ) eSrLr 2 = 0

Divided by ( )r L 2 :

r Sr

rqrqe

r r r r r

x

=

+

lim0

or ( ) r Srqdr

d er

= , integration results in:r

C r Sq er

1

2

+=

at r = 0 q r is not infinite, so2

r Sq er =

Butdr dT

k q r = so 2r S

dr dT

k e=

Integration results in 22

4C

k

r ST e +=


11/16

At r = R T = T o

=

22

0 14 Rr

k

RST T e

Maximum Temperature:k

RST T e4

2

0max=

Average temperature:

( )

k

RS

d rdr

d rdr T r T

T T e R

R

8

)(2

2

0 0

2

0 00

0=

=

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

ee

Rr LS R RS

RLQ 22

2 ===


12/16

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


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( ) Bz Az A A AB Az N N x z x

cD N ++=

For 0= Bz N dzdx

xcD

N A

A

AB Az = 1

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

0= + z z Az z Az SN SN

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

dN Az

Or 01

=

dz

dx x

cDdzd A

A

AB

Simplified to: 01

1 =

dz

dx

xdzd A

A

Integration gives: 111

C dz

dx

x A

A

=

Second integration gives: ( ) 211ln C zC x A +=

Boundary conditions: at 11 A A x x z z ==

at 22 A A x x z z ==


14/16

12

1

1

2

1 11

11 z z

z z

A

A

A

A

x

x

x

x

=

12

1

1

2

1

z z z z

B

B

B

B

x x

x x

=

Average concentration;

( )

=

2

1

2

11

1

,

/

z

z

z

z B B

B

avg B

dz

x x

x

x

( )1212

, / ln B B B B

avg B x x x x

x=

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

( )

==

= ===1

2

121

11

11 ln1 B

B AB z z

B

B

AB z z

A

A

AB z z Az x

x

z z

cD

dz

dx

x

cD

dz

dx

x

cD N


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Mass balance:

0= + zSkcSN SN A z z Az z Az

k = first order rate constant for the reaction.

And then, 0=+ A Az kcdzdN

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

dz

dc D N A AB Az =

Then, 022

=+A

A AB kcdz

cd D

BC1: at z=0 c A = c A0

BC2: at z=L N Az = 0 or dc A /dz = 0

Integration with considering the Boundary conditions results in:

1

1

0 cosh1cosh

b

L z

b

cc

A

A

= where AB DkLb / 2

1=

Shell Momentum

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