bio science lecture sde

8/8/2019 Bio Science Lecture Sde

1/75

Basic Principles of Transport Phenomena

Dr. Sirshendu De

ProfessorDepartment of Chemical Engineering

IIT Kharagpur

Kharagpur - 721302


2/75

Applications in Bioprocesses

Modeling of bioprocesses

Design of bioreactors

Design of separation units

Design of downstream units


3/75

Non-dimensional Numbers

Re=Reynolds number= Inertial forces/viscous forces=

Pr=Prandtl number=momentum diffusivity/thermal diffusivity=

Sc=Schmidt number=momentum diffusivity/mass diffusivity=

Heat transfer coefficient:

Q=Heat flow rate=h*A*

Mass transfer coefficient=kA

Nusselt number=convective to conductive heat transfer=hL/k

Schmidt number=convective to diffusive mass transfer=kmL/D

udV

Q/pc kQ

/ DQ V

T(

c(


4/75

Fluid : A material that flows

Flow:

Bounded flow (Flow through a conduit)

Unbounded flow (free surface flow)

Flow Characterization:

To obtain velocity profile i.e. velocity components inx, y, z and t-coordinates/ t, r, , z co-ordinates.

Temperature profile as a function of time and space.

Concentration profile as a function of time and space.


5/75

Outcome:

1) Cross sectional average velocity

2) Shear force at the wall estimation of pressuredrop , friction factor and pump calculations

3) Prediction of flow field laminar or turbulent

4) Cross section average/ cup mixing,

temperature/concentration as well as point

variation

Physical properties associated in thecalculations:

Density, Viscosity, specific heat, thermal

conductivity, diffusivity, etc


6/75

Gross equations to be solved:

1. Overall mass balance/mass concentration equationknown as equation of continuity.

2. Momentum balance equations in three directions.

3. Overall energy equation.

4. Species conservation equation/mass balance equation

(n-1).

(1) + (2) Velocity profile

(1) + (2) + (3) Temperature profile

(1) + (2) + (4) Concentration profile

Coupled PDEs may be decoupled under simplistic cases.


7/75

Typical boundary conditions for fluid flow:

5 types of boundary conditions for may appear in

fluid flow.

They are:1. A solid surface (may be porous)

2. A free liquid surface

3. A vapor-liquid interface

4. A liquid-liquid interface

5. An inlet/outlet section


8/75

Condition at solid surface:

If it is a stationary/impervious wall then,

If it is a moving surface with velocity u0 inx-direction whichis known as NO-slipboundary condition,

Constant wall temperature (CWT): T=Tw as the surface

Constant wall flux (CHF):

Insulated Surface:

0 x y zv v v! ! !

0, 0 x y zv u v v! ! !

0 constantT

k qy

x

x

0 or 0 at the wallT Tky y

x x! !x x


9/75

Cooling/heating at the wall:

Permeable wall:

at the wall,

at the allcTk h T Ty

x x

tangential 0xv v! !

normal 0yv v! {


10/75

Mathematical terms of B.C.:

1. Dirichlet B.C.:

Constant valued B.C.

2. Neumann B.C.:

Derivative of dependent variable is specified.

3. Robin-mixed B.C.:

Dependent variable & its derivative are

specified through an algebraic equation.


11/75

2. Condition at liq-liq/liq-vapor interface:

At interface of two immiscible liquids:

At Liquid-vapor interface:

if 1 represents vapor

3. Inlet/outlet condition: May be specified

4. Physical B.C.:

1 2 1 2 1 2; ;v v T T X X! ! !

1 2Q Q

2 0dv

dy !

0 at 0 for flow through a tube

vrr

x

! !x

0 at or stokes irst problemv y! ! g

1 21 2

v v

y yQ Q

x x!

x x


12/75

Frame of references:

Before solving a fluid flow problem fix up the co-ordinate system.

Lagrangian Approach:Moving frame of reference.

Eulerian Approach:

Fixed frame of reference.

We deal with mostly Eulerian approach.


13/75

Material Derivative:

Here,

IfQ is velocity

, , ,Q f x y z t !dQ Q Q d x Q dy Q dz

dt t x dt y dt z dt

x x x x!

x x x x

x y z

dQ Q Q Q Qv v v

dt t x y z

x x x x!

x x x x

. DQ Q v QDt tx! x rr

(Delta/gradient operator)i j k x y z

x x x !

x x x

r

v

r

Accumulation .Dv v

v vDt t

x! !

x

r r rr r

yx zDvDv Dv

i j kDt Dt Dt

!


14/75

Equation of motion for an ideal fluid (Zero Viscosity)

Consider a parallelopiped:

Let a fluid enters face ABCD with velocity vx and density.

Fluid leaves face EFGH, with velocity

So, rate of mass entering the CV through ABCD =

and densityxxv

v dx dxx x

VV

x x

x x

xv dydzV

z

x

d

x

H

G

F

E

D

C

B

Ad

y

d

z

y


15/75

Rate of mass leaving through EFGH:

Hence net rate of mass efflux (out - in)

Similarly, the net rate of mass efflux in,

y-direction is

x

x

vdx v dx dydz

x x

VV

xx !

x x

x xv v dx dydzx

V Vx

! x-

x x xv v dx dydz v dydzx

V V Vx ! x-

xv dxdydzx

Vx

!x

yv dxdydzx

Vx

!x


16/75

z-direction is

Net rate of accumulation in CV is

So, mass conservation equation is:

For a steady state flow,

For an incompressible flow,

For an incompressible, steady state flow:

zv dxdydzx

Vx

!x

dxdydzt

Vx!

x

( ) 0yx z

vv v

t x x x

VV

xx xx !

x x x x() 0

t

x!

x 0vV !r r

, , x y zV V{ 0vt

VV

x !

x

rr

. 0v !r r


17/75

EOC in cylindrical co-ordinates:

Conservation of momentum (EOM): for an ideal fluid

Xx, Yy, and Zz are components of body forces in three directions.

1

0r

r z

v

v v vt r r r zUV

V V V VU

x x x x

!x x x x

x x pF X dxdydzx

V x ! x

y y

pF Y dxdydz

yV

x!

x

z z pF Z dxdydzz

V x ! x

z

x

dydz

y

dx

p

p

p


18/75

Force = rate of change of momentum

For x-direction,

For an incompressible fluid

Similarly for y,z directions

Expanding material derivative:

x xD p

dxdydzv X dxdydz

Dt x

V Vx

!

x 1x

x

Dv p

Dt xV

x!

x

1yy

Dv pY

Dt yV

x!

x

1zz

Dv pZ

Dt zV

x!

x

1 x x x x x y z x

v v v v pv v v X

t x y z xV

x x x x x !

x x x x x

1 y y y y x y z y

v v v v pv v v Y

t x y z yV

x x x x x !

x x x x x

1 z z z z x y z z

v v v v pv v v Z

t x y z zV

x x x x x !

x x x x x


19/75

The above 3 are Eulers equation for inviscid flow (=0, ideal fluid)

Rotational Flow:

In a 3-D flow field:

Similarly,

Rotation Vector:

angular velocity (x-component)

1

2

x

yzvv

y z

[ !

x x!

x x

1

2

1

2

x zy

y xz

v v

z x

v v

x y

[

[

x x ! x x

x x!

x x

12 v[ ! v

rr r

url o

x y z

i j k

v v x y z

v v v

x x x v ! !

x x x

r r r


20/75

When components of rotation vector of

each point of flow field is equal to zero,flow is termed as Irrotational flow.

So, for irrotational flow.

2 v[; ! ! vr rr r

0; !r


21/75

Geometric description of the flow field:

Streamlines:

An imaginary line in the flow field such that tangent at every point

gives the direction or velocity vector.

Pathline:

Trajectory of a particular fluid particle in the flow field. Identity of aparticle, Tracer experiment.

0v dsv !r r


22/75

Streakline:

Astreakline at any given instant of time in thelocus of the temporary location of all particles

who have passed through a fixed point earlier inthe flow field.

Stream function ():To describe a flow field, often another scalar(known as stream function) is defined as,

=Constant along a streamline

for an irrotational flow (in 2-D):

;u vy x] ]x x! !

x x

2 0] !


23/75

Ex.1: Consider a flow field:

i. Check whether this represents an incompressible flow

ii. Derive the stream function

iii. Find out the equation of stream line

Solution:

(i): EOC for incompressible flow is

3 24 ; 3x yv x y v x y! !


24/75

So, EOC is satisfied. Velocity field represents an incompressible flow.

(ii)

or

2 23 3 0

yxvv

x xx x

xx ! !

x x

3 4xv x yy

]x! !

x

3 22 x y y f x] !

23df

x yx dx

]x!

x

23

dfx y

x dx

]x!

x

23y dfv x ydx!

2 23 3

dfx y x y

dx! 0 or constant

dff

dx! !


25/75

So, stream function is

(iii) Equation of stream line:

This has to be integrated by simpsons rule,y=f(x) is the equation of

streamline.

3 22 constant x y y] !

2

3 2 3

3or or

4 3 4x y

dx dy dx dy d y x y

v v x y x y d x x y! ! !


26/75

A double index notation is used to identify these stress components.

First index denotes the normal to the plane and the second denotes

the stress component itself.

yy normal stress on OA surface

yx shear stress on OA surface

yz shear stress on OA surface for a 3-D plane.

Shear stress is assumed to linearly proportional to strain and

proportionality constant is viscosity of fluid. This is calledN

ewtonsLaw of Viscosity.y x

xy yx

v v

x yX X Q

x x! ! x x-

VISCOUS FLOW


27/75

Equation of motion for viscous, incompressible flow:

For non-viscous (inviscid flow) flow:

Inertial term = pressure force term + body force term

For viscous flow:

Inertial term = pressure force term + body force term + viscous or

shear force terms

In terms of Velocity gradient:

X-Comp.:2 2 2

2 2 2

x x x x x x x

x y z

v v v v p v v vv v v

t x y z x x y zV Q

x x x x x x x x ! x x x x x x x x- -


28/75

Y-Component:

Z-Component:

Developing and fully developed flow: Consider flow through a pipe. At entrance the uniform velocity u0.

As the fluid enters the pipe, the velocity of fluid at the wall is zerobecause no-slip boundary.

The solid surface exerts retarding shear force on the flow. Thus, thespeed of fluid close to wall is reduced.

2 2 2

2 2 2

y y y y y y y

x y z

v v v v v v vp

v v vt x y z y x y zV Q

x x x x x x x x

! x x x x x x x x- -

2 2 2

2 2 2

z z z z z z z

x y z

v v v v p v v vv v v

t x y z z x y zV Q

x x x x x x x x ! x x x x x x x x- -


29/75

At successive sections, effects of solid wall is felt further into the

flow.

A boundary layer develops from both sides of the wall

After a certain length, boundary layers from both surfaces meet at

the center and the flow becomes fully viscous. This length isEntrance length.

For laminar flow:

0.06ReL

D !here, Re= vDV

Q


30/75

Beyond entrance length,

velocity profile does not change in shape and flow is termed as

Fully developed flow.

If flow is fully developed in x-direction, mathematically it is described

as,

For laminar flow, typically entrance length (L) is about a few cm.

0

xv

x

x!

x


31/75

Applications of EOM:

Ex1: S.S., fully developed, laminar flow through parallel plates,

flow driven be pressure gradient (Hagen Poiseulle flow)

EOM:

0 x y zv v vt x y z

V

V V V

x x x x !

x x x x


32/75

Assumptions:

i. Steady state flow,

ii. Incompressible flow, = constant.

iii. Z-dimension is too long; all variation in z-direction is very small.

iv. Fully developed flow in x-direction,

From EOC:

But aty=0,h vy = 0;

0t

x!

x

0xv

x

x!

x

0y

v

y

x!

x yv f y {

or constanty yv v x!

so 0, every here in loe ieldy

v !


33/75

EOM in x-direction:

By sovling this,

But, at y=0, vx=0;

at y=h, vx=0;

SO finally,

2

20x

vp

x yQ

xx

! x x

2

1 2

1

2x

pv y c y c

xQ

x !

x

20c !

1

1

2

pc h

xQ

x !

x

21

2x

pv hy y

xQ

x ! - x


34/75

Volumetric flow rate (Q):

Substitute vx in the above equation and integrate it. We get,

We can write the above equation as,

Estimation of pressure drop

xAQ v dA

!

0

h

xQ v Wdy!

3

12

Wh p

Q xQ

x

! x 3

12

Wh pQ

LQ

!

3

12p Q

L Wh

Q(!


35/75

Average velocity:

Point of maximum velocity:

For maximum velocity,

We get maximum velocity at

31

12

Q Wh pv

Wh LQ

! !

2

12

h pv

LQ!

0xdv

dy!

at centerline2

hy !


36/75

Y-component EOM:

Z-component EOM:

0p gy

Vx

!x

hydrostatic pressurep

gy

Vx

! x

0 ,p

p p z xz

x! !

x


37/75

Definition:for flow over a flat surface, the boundary layer is defined as

locus of all points in the flow field such that velocity at each point is

99% of the free stream velocity.

EOC:

X-component EOM:

Laminar Boundary Layers

0 (1)yx

vv

x y

xx !

x x

2 2 2

2 2 2

1(2) x x x x x x x

x y z

v v v v v v vpv v v

t x y z x x y z

Q

V V

x x x x x x xx ! x x x x x x x x- -

u

u

u(x,y)(x)

L

y

x


38/75

Blasius flow over flat surface:

flow over a flat plate and u= constant.

Gov.Eq.:

B.C:

0 (1)yx

vv

x y

xx !

x x

2

2(2)

yx x

x y

vv vv v

x y yY

xx x !

x x x

0, 0, 0

,

x y

x

at y v v

at y v vg

! ! !

! g !


39/75

Local skin friction:

Total frictional force per unit width of the plate length L

Boundary layer thickness:

2

0.664

0.5 Re

w

fx

xu

X

V Vg

! !

2

00.664

L

w

LF dx u u

Y

X V gg

! !

21.328

Avg. skin friction0.5 Re

f

L

F

u LV

Vg

! ! !

0.99 at 5.0xv

uL

g

p !5

so,Re

x

xH !


40/75

Turbulent flow

Characterization of turbulent flow:

Irregular motion

Random fluctuation

Fluctuations due to disturbances, e.g., roughness of solid

surface

Fluctuations may be damped by viscous forces / may grow

by drawing energy from free stream


41/75

Critical Reynolds number:

Re < Recr:

Kinetic energy is not enough to sustain random fluctuations

against the viscous dampening. So, laminar flow continues

Re > Recr:

Kinetic energy of flow supports the growth of fluctuations and

transition to turbulence occurs.

Origin of Turbulence:

Frictional forces at the confining solid walls Wall turbulence

Different velocities of adjacent fluid layers Free turbulence


42/75

Turbulence results in better mixing of fluid and produces additional

diffusive effects Eddy diffusivity.

Velocity Profile:

The mean motion and fluctuations:

Axial velocity is written as,

Here in RHS the first term is time averaged component

second term is time dependent fluctuations.

', ,u y t u y u y t !

Fully developed

LaminarFully developed

TurbulentPlug flow


43/75

Incompressible flow:

EOC:

0yx zvv v x y z

xx x !x x x


44/75

X-component EOM:

The last three terms are the additional terms known as Reynolds

stress terms.

Semi empirical expressions for Reynolds stresses:

1. Boussinesqs eddy viscosity:

Analogies to Newton's law of viscosity

Here, (t) is turbulent coefficient of viscosity / eddy viscosity which

is a strong function of position.

2 ' '2 '2 2 ' '

2

2 2 2

x y x x x x x x z

x y z x

v vv v v v v v vpv v v vt x y z x x y z

V Q V x x x x x x xx ! x x x x x x x x- -

( ) ( )t t x

yx

dv

dy

X Q!


45/75

2. Prandtls mixing length:

Assuming eddies move around like gas molecules, analogous

to mean free path of gas in kinetic theory:

where,

3. Von Karmans Hypothesis:

where, k= 0.3 - 0.4, an empirical constant.

( ) 2t x x

yx

dv dvl

dy dyX V!

; y is distance from solid and 0.4l ky k! !

( ) 2eddy viscosityt xdv

ldy

Q V

! !

3

( ) 2

2 2

/

/

xt x

yxx

dv dy dvk

dyd v dyX V!


46/75

4. Diesselers relationnear the wall:

where, n = empirical constant = 0.124

Nikuradses Experimental results:

These are also known as Universal Velocity Profile

( ) 2 21 exp /t x yx x x dvn v y n v y rdy

X V ! -

or 0 5

5ln 3 or 5 30

2.5ln 3 or 30

u y y

y y

y y

! e e

! e e

! "


47/75

Friction factor for turbulent flow inside smooth pipe:

Friction factor:

Estimation of pressure drop along the two sections of pipe.

This will estimate in turn the pump rating.

definition:

Here, um =cross section averaged velocity

L=Length

D = pipe diameter

compared with experimental data,

2

.2

muLp fD

V!

1 2.035log Re 0.91ff !

1

2log Re 0.8ff

!


48/75

So,

So, for specific flow (um, specified) fcan be calculated and pas

well.

2

2

muL

p fD

V!


49/75

Convective Heat transfer for internal flow

Flow through a pipe:

Thermal Boundary Layer:

If entry temperature,the convection of heat occurs.

Wall condition:

CWT = constant wall temperature TS = constant

CHT = constant Heat flux qS

= constant


50/75

In both cases, fluid temperature increases compared ro inlet

temperature.

If Pr > 1:

xt> xh hydrodynamic BL grows earlier than thermal BL

If Pr < 1:

xt< xh thermal BL grows faster.

0.05 Re Thermal entry length

0.05 Re Hydrodynamic entry length

t

D t

Lam

h

D h

Lam

x Pr x

D

xx

D

! !

!;

t

h

x Prx

!


51/75

For oils, Pr > 100:

Hydrodynamic BL is very small compared to entrance length

forxth and hence, for such cases, velocity is assumed to be fully

developed in the developing thermal BL.

Mean Temperature (Tm

):

Absence of any free stream temperature requires to define amean temperature.

Defined based on rate of transport of energy.

For incompressible fluid, = constant, cp= constant

p

p

uc TdT

mc

V!

&

20

2ulk or ean or up-mixing temperature

R

m

m

T uTrdru R

!


52/75

Newtons law of cooling:

estimation of heat transfer at any local point.

If a heat transfer occurs, Tm increases withx, (ifTS > Tm)

Fully developed condition:

Since there is convective heat transfer between surface andbulk, fluid temperature changes withx.

Condition for fully developed velocity profile,

For fully developed temperature profile,

" local heat trans er coe .S S mq h T T h! !

0du

dx!

0 so, 0mdT T

dx x

x{ {

x


53/75

Condition for fully developed temperature profile:

Condition for fully developed temp. profile:

Relative shape of temperature profile does not change.IfqS = constant, then TS=TS(x)

IfTS = constant, then qS=qS(x)

Under FDF,

Heat transfer coefficient h is constant

Ratio: is independent ofS

S m

T T xT T

. . 0S

S m

T Tdi e

dx T T

!

SS m

T T f xT T

{


54/75

CHF (constant heat flux):

"

,

S

m m in

p

q

T x T xmc

V

! &

"

, , S

m o m in

p

qT T L

mc

V !

&


55/75

Constant wall temperature (CWT):

TS = constant

By integrating,

we get,

For overall length,

m S mp

dT hT T

dx mc

V!

&

,

ln S mx

S m in p

T T xh

T T mc

V!

&

,

,

ln S m oL

S m in p

T T Lh

T T mc

V!

&

expoL

i p

T Lh

T mc

V (!

( &

Avg. heat trans er coe icientL

h !


56/75

Laminar flow in circular tubes:

Fully developed region:

Both velocity and temperature profiles are fully developed.

From EOC: v = 0;

From EOM:

From Energy balance:

2

22 1m ru u R

!

2

22 1m

T r T

u r x R r r r

Ex x x !

x x x


57/75

Case 1:Constant heat flux

Temperature profile:

Tm= mixing cup temperature

For constant heat flux,

By simplifying this we get,Nu = 4.36 for constant heat flux

4 222 3 1 1

16 16 4

m m

s

u R dT r rT r T dx R RE

! -

2

0

2 R

m

uT r dru R

!

2211

48

m m

m s

u R dTT T

dxE

!

"

m S

p

dT q

dx mc

V!

&


58/75

Case 2: Constant wall temperature

TS = constant,

we get, Nu = 3.66

For entry region:

For constant wall temperature,

For combined entry region (both velocity and temp. profile

developing), Seider tale correlation:

? A

2 / 3

0.0688 / Re r3.66

1 0.04 / Re r

D

D

D

D LhDNu

k D L! !

0.141/ 3Re Pr

1.86/

D

D

w

NuL D

Q

Q

!

sHere, T constant, 0.5 Pr 16,700! e e


59/75

External flow:

For liquid metals,Pr


60/75

For turbulent flow over flat plate:

In the mixed boundary layer case:

Where, Rex,c= 5*105

1 1

530.03Re Pr 0.6 Pr 60x xNu !

Laminar region: 0

Turbulent region:

c

c

x x

x x L

e e

e e

1/ 2 4 / 5 4 / 5 1/ 3, ,0.664 Re 0.037 Re Re PrL x c L x cNu ! -


61/75

Flow past a cylinder:

Zhukauskas correlation is

All properties at T

1

40.36 r

Re rr

m

D D

s

Nu c

!

6for 0.7 Pr 500 and 1 Re 10

D

ReD 0.4-4 4-40 40-4000 4000-40,000 >40,000

c 0.989 0.911 0.683 0.193 0.027

m 0.33 0.385 0.466 0.618 0.805


62/75

For non- circular cylinders:

based on geometry & ReD range, c and m are specified

Flow past a sphere:

Turbulent flow inside a conduit:

The above relation is called Dittus-Boelter relation.

1/ 421

0.4322 0.4 Re 0.06 Re Pr D D D

s

NuQ

Q

!

0.8 0.33

0.023 Re r Nu !


63/75

Free Convection

Forced convection:flow is induced by external source, pump/ compressor.

Free Convection:

No forced fluid velocity.Ex: Heat transfer from pipes/ steam radiators/ coil of

refrigerator to surrounding air

Consider, two plates at different temperatures, T1

& T2

and T2

> T1

2 < 1 means Density decreases in the direction of gravity

(Buyoant force)


64/75

If Buyoant force overcomes the viscous forces, instability occurs

and fluid particles start moving from bottom to top.

Gravitational force on upper layer exceeds that at the lower one and

fluid starts circulating.

Heavier fluid comes down from top, warms up and becomes lighterand moves up.

In the case, T1 > T2;

Density no longer decreases in the direction of gravity and there is

no bulk motion of fluid.

0 & 0dT d

dx dx

V

"


65/75

In case a Free convection

In case b Only conduction

Boundary layer development on a heated vertical plate:

Fluid close to the plate is heated and becomes less dense.

Buoyant force induces a free convection BL in whichheated fluid rises at vertically entraining the fluids fromsurroundings

Velocity is zero at the wall andy =.


66/75

GrashofNumber:

Expected:

Both free and forced convection are important if

if

if

2

0

2

0

3

2

rasho number

uoyancy orce

iscous orce

s

s

g T T L u L

u

g T T L

F

Y

F

Y

g

g

!

! !

Re , ,Pr L L L LNu Nu Gr!

21

ReL

L

Gr;

2

1 free convection is small, Re,PrRe

L

L L

L

GrNu Nu p !

2

1 forced convection is small, ,PrRe

L

L L L

L

GrNu Nu Gr" p !


67/75

Correction for free convection from a vertical plate; uniform wall

temperature:

For entire range ofRaL

n

L LhLNu cRak

! !

3Rayligh number

r

L

s

L

Ra

g T T LGr

F

YE

g

!

! !

9

9

10.59, or r 10

4

10.1, or r 10

3

L

L

c n Gr

c n Gr

! !

! ! "2

1

6

89 27

16

0.3870.0825

0.4921

r

L

L

RaNu

!

-


68/75

For long horizontal cylinder:

For Sphere:

2

1

6

89 27

16

0.3870.6

0.5591

r

L

D

RaNu

!

-

1

4

49 9

16

0.5892

0.469

1 r

L

D

RaNu !

-


69/75

Uniform wall heat flux:

Vertical plate:

Modified Grashof NumberGrx* is defined as

Avg. Nusselt number:

Physical properties are at film temperature.

1* 5 * 115

0.22* 13 * 16

0.6 Pr for 10 Pr 10

0.568 Pr for 2 10 Pr 10

x x x

x x

Nu Gr Gr

Gr Gr

!

! v

*

4

2

x x x

w

Gr GrNu

g q xKFY

!

!

? A

? A

5 * 11

13 * 16

1.25 , for 10 Pr 10

1.136 , for 2 10 Pr 10

m x xx L

x xx L

Nu Nu Gr

Nu Gr

!

!

!

! v


70/75

Free convection on horizontal plate:

Uniform wall temperature: Pr

n

mNu c Gr!

Orientation of plate Range of GrL.Pr C n

Hot surface facing up or

cold surface facing down

105 to 2*1017 0.54

2*107 to 3*1010 0.14 1/3

Hot surface facing down or

cold surface facing up

3*105 to 3*1010 0.27 1/4


71/75

Uniform wall heat flux:

For horizontal plate with heated surface facing up:

For horizontal plate with the heated surface facing down:

Physical properties at mean temperature

1 / 3 8

1 / 3 8 11

0.13 Pr for Pr 2 10

0.16 Pr for 5 10 Pr 10

m L L

L L

Nu Gr Gr

Gr Gr

! v

! v

1/ 5 6 11

0.58 Pr or 10 Pr 10m L L

Nu Gr Gr!


72/75

Ex1: Given Data: Vertical plate of L=1.5 m, W=1.5 m.

it is insulated on one side. Wall temp. is Tw= 400 k

and also T= 300 k.

calculate the total rate of heat loss from plate?

Sol: 1

400 300 350 k2

fT ! !

6 2 0

r20.75 10 m / s; 0.697; 0.03 w/m ckY V

! v ! !

3 -11 2.86 10 k f

TF ! ! v

number at 5m;Gr

311

5 28.137 10w

L

g T T LGr

F

Y

g

!

! ! v


73/75

2

1

6

89 27

16

0.387 Pr0.0825

0.4921

Pr

L

m

GrhLNu

k

! !

-

2 0

5.51 /m . ch !

Heat trans er rate

5.51 5 1.5 400 300

4.133 k

whA T T

g!

! v v

!


74/75

Mass Transfer operations

Nu is replaced by Sh

Pr is replaced by Sc


75/75

Thank you

bio science lecture sde

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