ChE 131: Transport Processes

ChE 131: Transport Processes

Outline

1.Class Policies

2.Introduction

3.Review

Course Assessment

3 Long Exams 60%Final Exam 20%3 Machine Problems 15%Classwork 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?

Transport Phenomena

Chemical Engineering

Thermodynamics

PROCESSEQUIPMENT

DESIGN

MaterialsScience

ProcessEconomics

ChemicalReactionKinetics

Why Study Transport Phenomena?

Levels of Analysis

MACROSCOPIC

MICROSCOPIC

MOLECULAR

Levels of Analysis

MACROSCOPIC

MICROSCOPIC

MOLECULAR

Use of macroscopic balances

Overall assessment of a system

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:

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)

1 0.5 0.5 0.17 0.33 0.672.0 0.6 ph kD D G 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

For the heat transfer coefficient:

For thermal conductivity:

For diameter:

For viscosity:

2 2/E t E

L T L t T

/E t E

L T L t T

L

ML t

Dimensional Analysis

For mass velocity:

For heat capacity:

Combining:

2ML tE

M T

2

0.5 0.17 0.17 0.33 0.67

0.5 0.5 0.17 0.33 0.33 0.67 0.67 0.67

1

1

E EL t T L t T L

M L t E EL L t M M T L t T

Dimensional Analysis

Simplifying:

2

0.5 0.17 0.17 0.33 0.67

0.5 0.5 0.17 0.33 0.33 0.67 0.67 0.67

1

1

E EL t T L t T L

M L t E EL L t M M T L t T

0.5 0.17 0.33 0.17 0.5 1 0.67 0.33 0.67

2 2 0.5 0.17 0.67 0.33 0.67M L EE E

L t T L t T t T

Dimensional Analysis

Simplifying:

0.5 0.17 0.33 0.17 0.5 1 0.67 0.33 0.67

2 2 0.5 0.17 0.67 0.33 0.67M L EE E

L t T L t T t T

2 2 2E E E

L t T L t T L 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

NaOH bal: 0.10(25) = 0.5M + 0C + 0HNaCl bal: 0.10(25) = 0.02M + 1.0C + 0HH2O bal: 0.80(25) = 0.48M + 0C + 1.0H

EVAPORATOR25 MT/h0.1 NaOH0.1 NaCl0.8 H2O

M (mother liquor)0.5 NaOH0.02 NaCl0.48 H2O

H (water)1.0 H2O

C (crystals)1.0 NaCl

Material Balance

H = water evaporated per hour = 17.6 MT/hC = salt precipitated per hour = 2.4 MT/hM = liquor produced per hour = 5 MT/h

EVAPORATOR25 MT/h0.1 NaOH0.1 NaCl0.8 H2O

M (mother liquor)0.5 NaOH0.02 NaCl0.48 H2O

H (water)1.0 H2O

C (crystals)1.0 NaCl

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

Volume of gas entering = velocity diameter of tower

SCRUBBER

D (dry gas)0.75 air0.25 NH3

S (water + ammonia)x H2Oy NH3

G (scrubbed gas)0.01 NH3 (solute-free)

W (water)1.0 H2O

32 3600 s= 1.5 2 ft 16964.6 4 1 h

ft fts h

Material Balance

Convert solute-free basis percentage to mass fraction:

We now rewrite our diagram:

3

3

3

30.01 NH , solute-free = 10.0099

0.9901

NH

NH

NH

air

xx

xx

Material Balance

Determine the number of moles of dry gas entering the scrubber. Assuming ideal gas behavior,

SCRUBBER

D (dry gas)0.75 air0.25 NH3

S (water + ammonia)x H2Oy NH3

G (scrubbed gas)0.0099 NH3

0.9901 air

W (water)1.0 H2O

Material Balance

Determine the number of moles of dry gas entering the scrubber. Assuming ideal gas behavior and a basis of 1 hour:

2

3

2 3

2

760 mm Hg 16964.6 ft 460 32 R 1 lbmol760 mm Hg 359 ft 460 32 R42.35 lbmol

STPSTP

STP STP

P V Tn nP V T

n

n

Material Balance

Air balance: 0.75(42.35) = 0.9901GG = amount of dry gas = 32.08 lbmol dry gas

SCRUBBER

D (dry gas) = 42.35 lbmol0.75 air0.25 NH3

S (water + ammonia)x H2Oy NH3

G (scrubbed gas)0.0099 NH3

0.9901 air

W (water)1.0 H2O

Material Balance

NH3 balance: 0.25(42.35) = 0.0099(32.08) + xSxS = amount of NH3 absorbed = 10.27 lbmol NH3

SCRUBBER

D (dry gas) = 42.35 lbmol0.75 air0.25 NH3

S (water + ammonia)x H2Oy NH3

G (scrubbed gas)0.0099 NH3

0.9901 air

W (water)1.0 H2O

Material Balance

Pounds of NH3 absorbed:

SCRUBBER

D (dry gas) = 42.35 lbmol0.75 air0.25 NH3

S (water + ammonia)x H2Oy NH3

G (scrubbed gas)0.0099 NH3

0.9901 air

W (water)1.0 H2O

3 33

3

17 lb NH lb NH10.27 lb mol NH 174.58 lb mol NH hr

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:

2

2

2

Simplifying:

2

s

p

vH g z Q W

vc T Q

Energy Balance

Energy Balance:

2

2 2 2

2

2

20.24 140 40

1 75 50 2 778.2 32.174

24.06

p

m

f

fm

m

vc T Q

Btu Flb F

lbft Btu Qfts lb ftlbs

BtuQlb

Differential Equation

Solve the following differential equation:

tan( ) cos( )dy y x xdx

Differential Equation

This equation follows the form:

whose solution is:

tan( ) cos( )dy y x xdx

( ) ( )dy yP x Q xdx

( ) ( )( )P x dx P x dxy e Q xe dx C

Differential Equation

tan( ) cos( )dy y x xdx

sin( )( ) tan( ) cos( )

Let u = cos(x), du = sin(x)dxsin( ) 1 1 1ln( ) ln ln lnsec( )cos( ) cos( )

xP xdx xdx dxx

x dx du u xx u u x

Differential Equation

tan( ) cos( )dy y x xdx

( ) ( )

ln(sec( )) ln(sec( ))

ln(cos( ))

( )cos( )

cos( )sec( )cos( )

cos( )

P x dx P x dx

x x

x

y e Q xe dx C

y e xe dx C

y e x xdx C

y x dx C

y x x C