ChE 131: Transport Processes
Outline
1.Class Policies
2.Introduction
3.Review
Course Assessment
3 Long Exams
Final Exam
3 Machine Problems
Classwork
60%
20%
15%
5%
Policies to Remember
Submit 12 sheets of colored pad paper at least the
day before an exam.
Get an official excuse slip from the College if you
miss an exam and you have a valid excuse.
No exemptions will be given for the final exam.
Policies to Remember
Quizzes may be given from time to time. All quizzes
shall be written in bluebooks. No makeup shall be
given to missed quizzes.
Outline
1.Class Policies
2.Introduction
3.Review
Transport Phenomena
What exactly are "transport phenomena"?
Transport phenomena are really just a fancy way that
Chemical Engineers group together three areas of study
that have certain ideas in common.
These three areas of study are:
• Fluid mechanics
• Heat transfer
• Mass transfer
Transport processes
Transport Processes
Momentum Transport – transfer of momentum which
occurs in moving media (fluid flow, sedimentation,
mixing, filtration, etc.)
Heat Transport – transfer of energy from one region to
another (drying, evaporation, distillation)
Mass Transport – transfer of mass of various chemical
species from one phase to another distinct phase
(distillation, absorption, adsorption, etc.)
Why Study Transport Phenomena?
Why Study Transport Phenomena?
Why Study Transport Phenomena?
Chemical
Engineering
Thermodynamics
Transport
Phenomena
PROCESS
EQUIPMENT
DESIGN
Chemical
Reaction
Kinetics
Materials
Science
Process
Economics
Levels of Analysis
MACROSCOPIC
MICROSCOPIC
MOLECULAR
Levels of Analysis
MACROSCOPIC
Use of macroscopic
balances
MICROSCOPIC
Overall assessment
of a system
MOLECULAR
Levels of Analysis
MACROSCOPIC
MICROSCOPIC
MOLECULAR
Small region/volume
element is selected
Use of equations of
change
Velocity, temperature,
pressure and
concentration profiles
are determined
Levels of Analysis
MACROSCOPIC
MICROSCOPIC
MOLECULAR
Molecular structure
and intermolecular
forces become
significant
Complex molecules,
extreme T and P,
chemically reacting
systems
Review
LET’S REVIEW!!!
Dimensional Analysis
Check the dimensional consistency of the following
empirical equation for heat transfer between a flowing fluid
and the surface of a sphere:
1
h  2.0kD  0.6D
G 
0.5
0.5
h – heat transfer coefficient (W/m2-K)
D – diameter of sphere (m)
k – thermal conductivity of fluid (W/m-K)
G – mass velocity of fluid (kg/m2-s)
μ – viscosity (kg/m-s)
cp – heat capacity (J/kg-K)
0.17 0.33 0.67
p
c
k
Dimensional Analysis
We use the following convention:
Energy unit – E
Mass unit – M
Length unit – L
Time unit – t
Temperature unit – T
Dimensional Analysis
E /t
E
 2
2
L T L  t T
E /t
E

L T L  t T
For the heat transfer coefficient:
For thermal conductivity:
For diameter: L
M
For viscosity:
Lt
Dimensional Analysis
M
For mass velocity: 2
L t
E
For heat capacity:
M T
Combining:
E
E  1 

  
  
2
L  t T
 L  t  T  L 
 1
 0.5
L
0.5
0.17 0.17
0.33
0.67





M
L
t
E
E


0.5 
0.17 
0.33 0.33  0.67
0.67
0.67 
  L  t  M
 M T
 L  t  T

Dimensional Analysis
E
E  1 

  
  
2
L  t T
 L  t  T  L 
 1
 0.5
L
0.5
0.17 0.17
0.33
0.67





M
L
t
E
E


0.5 
0.17 
0.33 0.33  0.67 0.67 0.67 
  L  t  M
 M T
 L t T

Simplifying:
E
E
  2
2
L t T
L t T
M


0.5 0.170.33
t
L
0.17 0.5 1 0.67
0.5 0.17 0.67
T
E

0.33  0.67
0.33 0.67

Dimensional Analysis
Simplifying:
E
E

  2
2
L  t T
L  t T
0.5 0.170.33
0.17 0.5 1 0.67
0.33 0.67
M
L
E




E
E
E
 2
  2
2
L  t T
L  t T L  t T
0.5 0.17 0.67
0.33  0.67
t
T



Material Balance
An evaporator is fed continuously with 25 metric
tons/h of a solution consisting of 10% NaOH, 10% NaCl,
and 80% H2O. During evaporation, water is boiled off,
and salt precipitates as crystals, which are settled and
removed from the remaining liquor. The concentrated
liquor leaving the evaporator contains 50% NaOH, 2%
NaCl, and 48% H2O.
Calculate the MT of water evaporated per hour, the
MT of salt precipitated per hour, and MT of liquor
produced per hour.
Material Balance
25 MT/h
0.1 NaOH
0.1 NaCl
0.8 H2O
EVAPORATOR
H (water)
1.0 H2O
M (mother liquor)
0.5 NaOH
0.02 NaCl
0.48 H2O
C (crystals)
1.0 NaCl
NaOH bal: 0.10(25) = 0.5M + 0C + 0H
NaCl bal: 0.10(25) = 0.02M + 1.0C + 0H
H2O bal: 0.80(25) = 0.48M + 0C + 1.0H
Material Balance
25 MT/h
0.1 NaOH
0.1 NaCl
0.8 H2O
EVAPORATOR
H (water)
1.0 H2O
M (mother liquor)
0.5 NaOH
0.02 NaCl
0.48 H2O
C (crystals)
1.0 NaCl
H = water evaporated per hour = 17.6 MT/h
C = salt precipitated per hour = 2.4 MT/h
M = liquor produced per hour = 5 MT/h
Material Balance
Dry gas containing 75% air and 25% NH3 vapor enters the
bottom of a cylindrical packed absorption tower that is 2 ft
in diameter. Nozzles in the top of the tower distribute
water over the packing. A solution of NH3 in H2O is drawn
at the bottom of the column, and scrubbed gas leaves the
top. The gas enters at 80°F and 760 mm Hg. It leaves at
60°F and 730 mm Hg. The leaving gas contains, on the
solute-free basis, 1.0% NH3.
If the entering gas flows through the empty bottom of the
column at velocity (upward) of 1.5 ft/s, how many ft3 of
entering gas are treated per hour? How many pounds of
NH3 are absorbed per hour?
Material Balance
G (scrubbed gas)
0.01 NH3
(solute-free)
W (water)
1.0 H2O
SCRUBBER
D (dry gas)
0.75 air
0.25 NH3
S (water + ammonia)
x H2O
y NH3
Volume of gas entering = velocity  diameter of tower
ft   
ft 3
2  3600 s 

=  1.5    2 ft  
s  4 


1h
  16964.6 h

Material Balance
Convert solute-free basis percentage to mass fraction:
0.01 NH3 , solute-free =
xNH3
1  xNH3
xNH3  0.0099
xair  0.9901
We now rewrite our diagram:
Material Balance
G (scrubbed gas)
0.0099 NH3
0.9901 air
W (water)
1.0 H2O
SCRUBBER
D (dry gas)
0.75 air
0.25 NH3
S (water + ammonia)
x H2O
y NH3
Determine the number of moles of dry gas entering
the scrubber. Assuming ideal gas behavior,
Material Balance
Determine the number of moles of dry gas entering
the scrubber. Assuming ideal gas behavior and a basis
of 1 hour:
 P  V   TSTP 
n2  
nSTP



 PSTP  VSTP   T 
 760 mm Hg   16964.6 ft 3   460  32R 
n2  


 1 lbmol

3
 760 mm Hg   359 ft
  460  32R 
n2  42.35 lbmol
Material Balance
G (scrubbed gas)
0.0099 NH3
0.9901 air
W (water)
1.0 H2O
SCRUBBER
D (dry gas) = 42.35 lbmol
0.75 air
0.25 NH3
S (water + ammonia)
x H2O
y NH3
Air balance: 0.75(42.35) = 0.9901G
G = amount of dry gas = 32.08 lbmol dry gas
Material Balance
G (scrubbed gas)
0.0099 NH3
0.9901 air
W (water)
1.0 H2O
SCRUBBER
D (dry gas) = 42.35 lbmol
0.75 air
0.25 NH3
S (water + ammonia)
x H2O
y NH3
NH3 balance: 0.25(42.35) = 0.0099(32.08) + xS
xS = amount of NH3 absorbed = 10.27 lbmol NH3
Material Balance
G (scrubbed gas)
0.0099 NH3
0.9901 air
W (water)
1.0 H2O
SCRUBBER
D (dry gas) = 42.35 lbmol
0.75 air
0.25 NH3
Pounds of NH3 absorbed:
S (water + ammonia)
x H2O
y NH3
 17 lb NH3 
lb NH3
10.27 lb mol NH3 
  174.58
hr
 lb mol NH3 
Energy Balance
Air is flowing steadily through a horizontal heated
tube. The air enters at 40°F and at a velocity of 50
ft/s. It leaves the tube at 140°F and 75 ft/s. The
average specific heat of air is 0.24 Btu/lb-°F.
How many Btu’s per pound of air are transferred
through the wall of the tube?
Energy Balance
Energy Balance:
v
H 
 g z  Q  Ws
2
2
Simplifying:
v 2
cp T 
Q
2
Energy Balance
Energy Balance:
v 2
cp T 
Q
2

Btu 
 0.24
 140  40F  
lbm  F 

 752  502

2

Btu
Q  24.06
lbm




1 lbf
ft 2  
Btu

Q
2 
ft
s   32.174 lb
778.2 lbf  ft 



m 2 
s 

Differential Equation
Solve the following differential equation:
dy
 y tan( x)  cos( x)
dx
Differential Equation
dy
 y tan( x)  cos( x)
dx
This equation follows the form:
dy
 yP ( x)  Q( x)
dx
whose solution is:
y e 
 P ( x ) dx

P ( x ) dx

Q ( x )e
dx  C
Differential Equation
dy
 y tan( x)  cos( x)
dx
sin( x )
 P ( x)dx   tan( x)dx   cos( x) dx
Let u = cos(x), du =  sin(x)dx
sin( x )
1
 1 
1 
 cos( x) dx    u du   ln(u)  ln  u   ln  cos( x)   lnsec( x)
Differential Equation
dy
 y tan( x)  cos( x)
dx
y e 
 P ( x ) dx
P ( x ) dx

Q ( x )e
dx  C

y e
dx  C
 cos( x)e
y e
 cos( x) sec( x)dx  C
y  cos( x )   dx   C
 ln(sec( x ))
ln(cos( x ))
y  x cos( x )  C
ln(sec( x ))