Transport phenomena
(이 동 현 상)
2009 학년
숭실대학교 환경화학공학과
Chap. 1 Introduction to Engineering
Principles and Units
1.1B. Fundamental Transport Processes
Imagine any plants in operation!!
(1) Momentum transfer
(2) Heat transfer
(3) Mass transfer
1.2. SI SYSTEM OF BASIC UNITS
- length : m
- mass : kg
- time : s (second)
* fps system
- length : ft (foot)
- mass : lbm (pound)
- time : s
* cgs system
- length : cm
- mass : g
- time : s
1.2A. SI system of units
- temperature: K (kelvin)
t (ºC) = T (K) – 273.15
t (ºC) = T (K)
- force: N = kgm/s2
- work, energy, heat: J = Nm=kgm2/s2
- power: W = J/s
- pressure: Pa = N/m2
105 Pa = 1 bar
 1 기압(atm)
* standard prefixes
- G : 109
- M : 106
- k : 103
- c : 10-2
- m : 10-3
-  : 10-6
- n : 10-9
1.3 TEMPERATURE AND COMPOSITION
1.3A. Temperature
ºF = 32 + 1.8  ºC
ºC = (ºF – 32)/1.8
ºR = ºF + 460
K = ºC + 273.15
Table1.3-1
1.3B. Mole, and Weight or Mass Units
xA 
nA
ntotal
xA: mole fraction of A
nA: moles of A
wA 
mA
mtotal
wA: weight (or mass) fraction of A
mA: mass of A
Ex 1.3-1)
1.4 GAS LAW
1.4A. Pressure
*1 atm = 760 mmHg (0 ºC)
= 760 torr
2
- Absolute pressure: 1.00 atm = 14.7 psia
*lbf/in (absolute)
= 14.7 psi
= 1.01325  105 Pa
= 1.01325 bar
- Gauge pressure: the pressure above the absolute pressure
Ex) 21.5 psig = 21.5 + 14.7 = 36.2 psia
(=1 atm)
1.4B. Ideal Gas Law
pV  nRT
Absolute Temperature (K)
R (gas (law) constant) = 8314.3 kgm2/kgmol  s2  K
= 8.3143 J/gmol K
1.4C. Ideal Gas Mixture
- Dalton’s law for ideal gas mixtures
P  pA  pB  pC  ...
P: total pressure
pi: partial pressure
- the number of moles 
nA
pA
xA 

ntotal
P
Ex)1.4-2
partial pressure
for ideal gas
1.5 CONSERVATION OF MASS AND MATERIAL BALANCES
* Conservation or Balance Equation
(accumulation)  (input)  (output)  ( generation consumption)
(reaction)
1.5A. Conservation of mass
(accumulation) = (input) – (output)
- steady-state  no accumulation
0
Ex 1.5-1)
= (input) – (output)
1.6 ENERGY AND HEAT UNITS
1.6A. Joule, Calorie, and Btu
*1 btu= 252.16 cal =1.05506 kJ
1.6B. Heat Capacity
- cp [J/kgmolK=J/kgmol ºC]
Q  Mcp T
Q : heat
Ex1.6-1)
*1.0 cal/g  ºC = 1.0 btu/lbm  ºF
1.6D. Heat of Reaction
(101.32 kPa, 298 K = 1 atm, 25 oC)
-standard heat of reaction: Hº
- Exothermic reaction: Hº < 0
- Endothermic reaction: Hº > 0
1.7 CONSERVATION OF ENERGY AND HEAT BALANCES
1.7B. Heat Balances
- at steady-state
o
0   H R  (H298
)  q   HP
(accumulation)  (input)  (output)  ( generation consumption)
Chap. 2 Principles of Momentum Transfer
and Overall Balances
2.2 FLUID STATICS
2.2A. Force, Units, and Dimensions
F  ma
-under the influence of gravity
F  mg
*1 lbf = 4.4482 N
1 lbm = 0.45359 kg
*1 N = 105 dyn (gcm/s2)
= 7.233 poundal (lbm ft/s2)
2.2B. Pressure in a Fluid
P0
m  V  hA
A
F  m a  ( hA)  g
F
P   hg (Gauge pressure)
A
p  hg  Po
Ex 2.2-2)
(Absolute pressure)
h
p
2.2C. Head of a fluid
- height of a fluid
- a common method of expressing pressures
h
P
g
2.2D. Devices to Measure Pressure and Pressure Differences
- Simple manometer
pa  pb  R(  A   B ) g
R : the reading of a manometer (the difference in height of fluid A (Fig. 2.2-4))
Ex 2.2-4)
Beginning of Transport Phenomena
2.3. GENERAL MOLECULAR TRANSPORT EQUATION
FOR MOMENTUM, HEAT, AND MASS TRANSFER
2.3A. General Molecular Transport Equation and General Property Balance
1. Introduction to transport processes
- a difference of concentration of a property (momentum, heat, or mass)
 transfer of the property by molecular movement
 net transport of the property
2. General molecular transfer equation
rate of a transfer process 
driving force
resistance
 d 
 z    
 dz 
 z : the flux of a property in z direction
= amount of transferred property per unit time per unit area
3. General property balance for unsteady state
(accumulation)  (input)  (output)  ( generation consumption)

(z  A)  ( z| z ) A  ( z| z  z ) A  R(z  A)
t
Fig.2.3-1
z0
  z

R
t
z

 2
 2  R
t
z
 d 

 dz 
 z   
Basic Relationships: Conservation law, rate, cond’n (IC&BC)
2.3B. Introduction to Molecular Transport
1. Momentum transport and Newton’s law
 zx  
d (v x  )
dz
 zx : flux of x-directed momentum in z direction
= shear stress [(kgm/s2)/sm2]

 d 
 z    
 dz 
: momentum diffusivity = kinematic viscosity [m2/s]



2. Heat transport and Fourier’s law
d (  c pT )
qz
 
A
dz
qz
: heat flux in z direction [J/sm2]
A

: thermal diffusivity [m2/s]
3. Mass transport and Fick’s law
*
J Az
  DAB
d cA
dz
* : flux of the molecule A in z direction [kgmol/sm2]
J Az
DAB : (molecular) diffusivity of the molecule A in B [m2/s]
*mathematical analogy
2.4 VISCOSITY OF FLUIDS
2.4A. Newton’s Law and Viscosity
- Fluid:
- Viscosity:
*the units of viscosity
SI : Pas=kg/m·s
- Viscous force:
1cp (centipoise)
= 0.01 g/cm·s
= 0.001 kg/m·s
- Newton’s law
F
vz
- F  vz, A, 1/y 
 
A
y
y 0

 yz   
dvz  zx : flux of z-directed momentum in y direction
= shear stress [(kgm/s2)/sm2]
dy
dvz
: velocity gradient
dy
Meaning:
= shear rate [1/s = s-1]
 : viscosity [kg/m·s]
A fluid contained between two infinite parallel plates
Linear velocity profile
Steady force (viscous drag)
Bottom plate
2.4B. Momentum Transfer in a Fluid
 zx
: the flux of z-directed momentum in the y direction
(per second per unit area) [(kg·m/s) /s·m2]
[(kg·m/s) /s·m2]
= the rate of flow of z-directed momentum per unit area
(per second) [(kg·m/s) /s]
[(kg·m/s) /s·m2]
= the amount of z-directed momentum transferred per second per unit area
[(kg·m/s) /s·m2]
- Momentum transfer: (molecular motion)
2.4C. Viscosities of Newtonian Fluids
- Newtonian Fluid
- viscosity is constant (at constant temp. and pressure)
- viscosity is independent of shear rate
 linear relation between shear stress and shear rate
- Non-Newtonian Fluid
- viscosity is not constant (even at constant temp. and pressure)
- viscosity is a function of shear rate
 non-linear relation between shear stress and shear rate
* gas: temperature or pressure   viscosity 
* liquid: temperature   viscosity (not affected by pressure)
2.5 TYPE OF FLUID FLOW AND REYNOLDS NUMBER
2.5B. Laminar and Turbulent Flow
- laminar flow
- the regime or type of flow where the layers of fluid slide by one
another without eddies or swirls
- no lateral mixing
- at low velocities
- turbulent flow
- the regime or type of flow where eddies are present giving the
fluid a fluctuating nature
- lateral mixing
- at higher velocities
2.5C. Reynolds Number
- flow regime in a tube (pipe)
= a function of fluid velocity, density, viscosity, and tube diameter
- Reynolds number
N Re 
Dv
D

v


 ratio of kinetic or inertia force to viscous force
N Re
inertia force
v 2


viscous force v / D
* NRe < 2100 : laminar
(for a straight circular pipe)
NRe > 4000 : turbulent
2.6. OVERALL MASS BALANCE AND CONTINUITY EQUATION
2.6A. Introduction and Simple Mass Balances
(accumulation)  (input)  (output)  ( generation consumption)
- no generation
- steady state no accumulation
0
= (input) – (output)  (input) = (output)
  V  1 A1v1  2 A2v2 (simple mass balance, steady state, 1-D)
 m
m
V
G  v
*2:output
1: input
2.6B. Control Volume for Balances
- a system:
- a control volume:
- control surface
2.6C. Overall Mass-Balance Equation
(accumulation)  (input)  (output)  ( generation consumption)
(rate of mass output from C.V.)-(rate of mass input from C.V.)+(rate of acc. in C.V.)=0
 net m assefflux

  rate of m assoutput  rate of m assinput
 from C.V .

 
  v cos dA    (v  n )dA
A
A
 rate of massaccumulation 
dM

   dV 
dt
 in C.V .
 t V
 
  (v  n)dA 
A
m 2  m 1 

dV  0

t V
dM
0
dt
(overall mass balance, unsteady state)
- 1-D
 v cos dA   v cos
A
2
dA   v cos1 dA
A2
A1
 1 A1v1   2 A2 v2
 2 A2 v2  1 A1v1 
dM
 0 (overall mass balance, unsteady state, 1-D)
dt
- 1-D, steady state
2 m
1  0
m
1 A1v1  2 A2v2
(overall mass balance, steady state, 1-D)
- component i
m i 2  m i1 
Ex 2.6-2)
dM i
 Ri
dt
(mass balance for component i, unsteady state, 1-D)
2.6D. Average Velocity to Use in Overall Mass Balance
-If the velocity is not constant but varies across the surface area
vav 
Ex 2.6-3)
1
v dA
A 
A
2.7 OVERALL ENERGY BALANCE
2.7A. Introduction
E  Q  W
2.7B. Derivation of Overall Energy-Balance Equation
(accumulation)  (input)  (output)  ( generation consumption)
 net energyefflux  rate of energyacc.  rate of energygen.  rate of energycons.

  
  
  

 from C.V .
  in C.V .
  in C.V .
  in C.V .

 rate of energyacc. 

v2

   EdV   (U   zg ) dV
t V
2
 in C.V .
 t V
 net energyefflux
v2

   ( E  pV ) v cos dA   ( H   zg ) v cos dA
from
C
.
V
.
2

 A
A
H  U  pV
 rate of energygen.

  q
 in C.V .

 rate of energycons. &

  Ws
 in C.V .

&
v2

v2
(
H


zg
)

v
cos

dA

(
U


zg
)

dV

q

W
s
A
2
t 
2
V
2.7C. Overall Energy Balance for Steady-State Flow System
H 2  H1 
1 2
(v2 av  v12av )  g ( z 2  z1 )  Q  Ws (overall energy balance, steady-state, 1-D)
2
3
av
3
(not very often used)
v
:kinetic-energy velocity correction factor
(v ) av
1
vav   v dA
A A
1
(v 3 ) av   (v 3 ) dA
A A

- laminar flow :  = 0.5
- turbulent flow :   1
2.7E. Applications of Overall Energy-Balance Equation
Ex2.7-1)
2.7F. Overall Mechanical-Energy Balance
p2 dp
1 2
2
(v2 av  v1av )  g ( z2  z1 )  
  F  Ws  0
p1 
2
(overall mechanical-energy balance, steady-state, 1-D)
1 2
p  p1
(v2 av  v12av )  g ( z2  z1 )  2
  F  Ws  0
2

(more useful)
(overall mechanical-energy balance, steady-state, 1-D, incompressible liquid)
 F: all frictional losses per unit mass
Ex 2.7-4)
2.7G. Bernoulli Equation for Mechanical-Energy Balance
- no mechanical energy: Ws=0
- no friction:  F
- turbulent:  
v12 p1
v22 p2
z1 g    z2 g  
2 
2 
Ex 2.7-6)
2.8 OVERALL MOMENTUM BALANCE
2.8A. Derivation of General Equation


P  Mv

P : total linear momentum vector
M:

v:

 dP
 F  dt
(accumulation)  (input)  (output)  ( generation consumption)
 sum of forcesacting  rate of momentum  rate of momentum  rate of acc. of


  
  
  

on
C
.
V
.
out
of
C
.
V
.
into
C
.
V
.
momentum
in
C
.
V
.

 
 
 


 sum of forcesacting

   F
on
C
.
V
.


 rate of momentum  rate of momentum  net momentumefflux

  
  

out
of
C
.
V
.
into
C
.
V
.
from
C
.
V
.

 
 


  
  v v cos dA   v  (v  n) dA
 rate of acc. of
 


   v dV
 momentumin C.V . t V
A
A

  


F

v

(
v

n
)
dA

v
dV (Overall linear momentum balance, unsteady state)
 


t
A
V
 Fx   vx v cos dA 
A

vx dV (x component)

t V
 Fx  Fxg  Fxp  Fxs  Rx   vx v cos dA 
A
Fxg
Fxp
Fxs
Rx

vx dV
t 
V
2.8-B. Overall Momentum Balance in Flow System in One Direction
 Fx  Fxg  Fxp  Fxs  Rx   vx v cos dA 
A
steady state 

vx dV  0
t 
V
only x direction  v  vx cos   1

vx dV
t 
V
 / vx av
integrating with vx  m
(vx22 ) av
(vx21 ) av


Fxg  Fxp  Fxs  Rx  m
m
vx 2 av
vx1 av
1
where (vx2 ) av   vx2 dA
A A
(vx2 ) av vx av

let
vx av

 = 3/4
 = 0.95 - 0.99
for laminar
for turbulent
Fxp  p1 A1  p2 A2
Fxs  0
(frictional force is neglected)

Rx  m
vx 2 av
R  m
v2 av



m
 m
vx1 av

v1 av

(Overall linear momentum balance, steady state
1-D(horizontal), neglecting frictional force)
2.9 SHELL MOMENTUM BALANCE AND VELOCITY PROFILE IN
LAMINAR FLOW
2.9A. Introduction
For laminar flow at steady state
Shell Momentum Balance  shear stress
 velocity profile (Newton’s law of viscosity)
 average velocity
 maximum velocity
2.9B. Shell Momentum Balance Inside a Pipe
For the flow of fluids inside a horizontal circular pipe
incompressible Newtonian fluid
one-dimensional, steady-state, fully-developed, laminar flow
 sum of forcesacting  rate of momentum  rate of momentum  rate of acc. of


  
  
  

 onC.V .
  outof C.V .
  into C.V .
  momentumin C.V .
 net m om entumefflux


 from C.V .

 sum of forcesacting

  pressure force  [ pA] x  [ pA] xx  [ p(2rr )] x  [ p(2rr )] x  x
on
C
.
V
.


No body force (horizontal pipe)
Neglecting frictional force
 net momentumefflux

  ( rx 2rx) r  r  ( rx 2rx) r
 fromC.V .

 rate of acc. of


  0
 momentumin C.V .
Steady state
[ p(2rr)] x  [ p(2rr)] xx  ( rx 2rx) r r  ( rx 2rx) r  0
Dividing by 2rx
(r rx ) r  r  (r rx ) r
r

r ( p x  p x  x )
x
( p x  p x  x )
x

p
L
Fully developed
r  0
d (r rx )  p 
  r
dr
 L 
Integrating
 p  r C1
 
 L 2 r
 rx  
When r = 0, rx should not be infinite.  C1=0
 p  r p0  pL
r
 
2L
 L 2
 rx  
Newton’s law of viscosity
 rx   
dv x
dr
Incompressible, Newtonian
Laminar
dvx
p  pL
 0
r
dr
2L
Integrating using the boundary condition vx=0 at r=R
2
p0  pL 2   r  
vx 
R 1    
4L
  R  
vx av 
1
1
v
dA

x
A 
R 2
A
vx av
2
 
0
R
0
vx rdrd
( p0  pL ) R 2 ( p0  pL ) D 2


8L
32L
(Hagen-Poiseulle equation)
a horizontal circular pipe
incompressible Newtonian fluid
one-dimensional, steady-state, fully-developed, laminar flow
vx=vx max at r=0
vx max 
p0  pL 2
R  2vx av
4L
2
  r 2 
  r 2 
p0  pL 2   r  
vx 
R 1      vx max 1      2vx av 1    
4L
  R  
  R  
  R  
2.9C. Shell Momentum Balance for Falling Film
For a fluid as a film down a vertical surface
incompressible Newtonian fluid
one-dimensional, steady-state, fully-developed, laminar flow
 sum of forcesacting  rate of momentum  rate of momentum  rate of acc. of


  
  
  

 onC.V .
  outof C.V .
  into C.V .
  momentumin C.V .
W in the y direction
 sum of forcesacting

  gravity force  xWLg
 onC.V .

mass
down a vertical surface
 net momentumefflux

  LW ( xz ) xx  LW ( xz ) x
from
C
.
V
.


 rate of acc. of


  0
momentum
in
C
.
V
.


Steady state
xWLg  LW ( xz ) xx  LW ( xz ) x  0
Dividing by LWx
g 
 xz x  x   xz x
x
x  0
d xz
 g
dx
Integrating using the boundary conditions at x=0, xz=0 and at x=x, xz= xz
 xz  gx
Newton’s law of viscosity
 xz   
dv z
dx
Incompressible, Newtonian
Laminar
 g 
dvz
   x
dx
  
Integrating using the boundary conditions at x=, vz=0
g 2
vz 
2
  x 2 
1    
    
vx av
1
1
  vz dA 
A A
W
vz av
W

0
0
 
vz dxdy
* N Re 
g 2

3
Dv

for a pipe
NRe < 2100 : laminar
NRe > 4000 : turbulent
vz=vz max at x=0
g 2 3
vz max 
 vz av
2
2
volumetric flow rate
q  Avz av  Wv z av
g 3W

3
mass flow rate per unit width of wall   gvz av
Reynolds number
N Re 
4


4vz av

a fluid as a film down a vertical surface
(NRe < 1200 : laminar)
2.10 DESIGN EQUATIONNS FOR LAMINAR AND TURBULENT FLOW IN
PIPES
2.10A. Velocity Profiles in Pipes
2.10B. Pressure Drop and Friction Loss in Laminar Flow
Friction loss
Ff 
p f

p f  ( p1  p2 ) f 
32 vL
D2
L v 2
p f  4 f
D 2
(for laminar, Hagen-Poiseulle equation)
(for both laminar and turbulent flow
f: Fanning friction factor)
*Fanning friction factor :
f 
16
16

Dv /  N Re
(for laminar)
2.10E. Effect of Heat Transfer on Friction Factor
heating  f 
cooling  f 
2.10F. Friction Losses in Expansion, Contraction and Pipe Fittings
friction losses in mechanical-energy-balance equation
L v 2
v12
v22
v12
 F  4 f D 2  Kex 2  K c 2  K f 2
straight pipe
(Fanning friction)
expansion
contraction
*2:outlet
1: inlet
fittings or valves
2.10G. Friction Loss in Noncircular Conduits
equivalent diameter D  4rH  4
cross  sectionalarea of channel
wetted perim eterof channel
hydraulic radius
for a circular tube
2
D
 
4
D4   D
D
for an annular space
2
D
2
D 
D 
  1    2 
4
 4   D D
D4  
1
2
D1  D2
D2
for a rectangular duct
D4
ab
2ab

2a  2b a  b
a
b
D1
2.10H. Entrance Section of a Pipe
Fully-developed flow