steady-state - 숭실대학교 화학공학과
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Transcript steady-state - 숭실대학교 화학공학과
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 = kgm/s2
- work, energy, heat: J = Nm=kgm2/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 kgm2/kgmol s2 K
= 8.3143 J/gmol K
1.4C. Ideal Gas Mixture
- Dalton’s law for ideal gas mixtures
P p A 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/kgmolK=J/kgmol ºC]
Q Mc p 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 (H 298
) 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 (gcm/s2)
= 7.233 poundal (lbm ft/s2)
2.2B. Pressure in a Fluid
P0
m V hA
A
F ma ( hA) g
F
P hg (Gauge pressure)
A
p hg 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
P
h
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
dz
z
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
z0
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 [(kgm/s2)/sm2]
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/sm2]
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/sm2]
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 : Pas=kg/m·s
- Viscous force:
1cp (centipoise)
= 0.01 g/cm·s
= 0.001 kg/m·s
- Newton’s law
F
v z
- F vz, A, 1/y
A
y
y 0
yz
dvz zx : flux of z-directed momentum in y direction
= shear stress [(kgm/s2)/sm2]
dy
dv z
: 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 mass efflux
rate of mass output rate of mass input
from C.V .
v cos dA (v n )dA
A
A
rate of mass accumulation
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 A2v2
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 energy efflux rate of energy acc. rate of energy gen. rate of energy cons.
from C.V .
in C.V .
in C.V .
in C.V .
rate of energy acc.
v2
EdV (U zg ) dV
t V
2
in C.V .
t V
net energy efflux
v2
( E pV ) v cos dA ( H zg ) v cos dA
from
C
.
V
.
2
A
A
H U pV
rate of energy gen.
q
in C.V .
rate of energy cons. &
Ws
in C.V .
&
v2
v2
(U zg ) dV q Ws
A ( H 2 zg ) v cos dA t
2
V
2.7C. Overall Energy Balance for Steady-State Flow System
H 2 H1
1 2
(v2 av v12av ) g ( z2 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
(more useful)
(v2 av v12av ) g ( z2 z1 ) 2
F Ws 0
2
(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 forces acting 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 forces acting
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
momentum in C.V . t V
A
A
F
v
(
v
n
)
dA
v
dV (Overall linear momentum balance, unsteady state)
t V
A
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
integrating with v x m / v x av
(v x22 ) av
(v x21 ) av
Fxg Fxp Fxs Rx m
m
v x 2 av
v x1 av
where (v x2 ) av
(v x2 ) av vx av
let
vx av
1
2
v
dA
x
A A
= 3/4
= 0.95 - 0.99
for laminar
for turbulent
Fxp p1 A1 p2 A2
Fxs 0
(frictional force is neglected)
Rx m
R m
vx 2 av
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 forces acting rate of momentum rate of momentum rate of acc. of
on
C
.
V
.
out
of
C
.
V
.
into
C
.
V
.
momentum
in
C
.
V
.
net momentum efflux
from
C
.V
.
sum of forces acting
pressure force [ pA] x [ pA] x x [ p(2rr )] x [ p(2rr )] x x
on
C
.
V
.
No body force (horizontal pipe)
Neglecting frictional force
net momentumefflux
( rx 2rx) r r ( rx 2rx) r
from C.V .
rate of acc. of
0
momentum in C.V .
Steady state
[ p(2rr )] x [ p(2rr )] xx ( rx 2rx) r r ( rx 2rx) r 0
Dividing by 2rx
(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
2L
Integrating using the boundary condition vx=0 at r=R
2
p0 pL 2 r
vx
R 1
4L
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
8L
32L
(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
4L
2
r 2
r 2
p0 pL 2 r
vx
R 1 vx max 1 2vx av 1
4L
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 forces acting rate of momentum rate of momentum rate of acc. of
on C.V .
out of C.V .
into C.V .
momentum in C.V .
W in the y direction
sum of forces acting
gravity force xWLg
on C.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
xWLg LW ( xz ) xx LW ( xz ) x 0
Dividing by LWx
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
dvz
dx
Incompressible, Newtonian
Laminar
g
dv z
x
dx
Integrating using the boundary conditions at x=, vz=0
2
g 2 x
vz
1
2
vx av
1
1
vz dA
A A
W
W
0
0
vz dxdy
* N Re
g 2
vz av
3
volumetric flow rate
for a pipe
g 2 3
vz av
2
2
g 3W
q Avz av Wvz av
3
mass flow rate per unit width of wall gvz av
Reynolds number
NRe < 2100 : laminar
NRe > 4000 : turbulent
vz=vz max at x=0
vz max
Dv
N Re
4
4vz 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 vL
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 Kc 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 sectional area of channel
wetted perimeter of channel
hydraulic radius
for a circular tube
2
D
4
D4 D
D
for an annular space
2
D
2
D
D
1 2
4
4 D D
D4
1
2
D1 D2
D2
for a rectangular duct
D4
ab
2ab
2a 2b a b
a
b
D1
2.10H. Entrance Section of a Pipe
Fully-developed flow