Compiled Final Report Transport Phenomena

ChE 323 Transport Phenomena

Project 2

Group: ___11___

Student ID NameOverall

Contribution, %

Signature

14192289 Wang Mei Wen 33.3

14985599 Ahiamadu Jackson

Samuel

33.3

14423798 Seer Qiu Han 33.3

Type of submission: Project report Due date of submission: Friday, 27th May 2011 Due time of submission: 4 pm Is this a group submission: Yes

Supervisor: Dr Agus Saptoro

ChE323 Transport Phenomena | 2011

I declare that this project 1 report is solely our own work. All contributions made by others

have been duly acknowledged.

Signature : ____________________________________

Name : ____________________________________

Student ID : ____________________________________

Date : ____________________________________

Signature : ____________________________________

Name : ____________________________________

Student ID : ____________________________________

Date : ____________________________________

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Group 11 Page 1


Question 1:

Solution:

Assumptions:

Thin-walled copper tube, the thickness of tube is negligible.

The conductive heat transfer also is negligible.

Cross-sectional area of pipe Ai = Ao = A.

Assume mean vapor-condensate interface temperature (to be the first guess

that uses for iterative calculation).

Thermal resistance across the pipe wall is negligible.

Assume the difference between interface temperature and pipe surface temperature is

2ᵒC.

Group 11 Page 2


Given data:

Volumetric flow rate Q1 = 0.13 m3/s

Temperature of moist air T1 = 45ᵒC

Relative humidity ᶲ = 90% = 0.9

Pressure P = 100kPa = 1atm

Pipe diameter D = 1cm = 0.01m

Cold water flow rate Qc = 6L/min = 0.0001 m3/s

Length of pipe L = 1.7m

Cold water inlet temperature Tw,i = 8ᵒC

Question analysis:

There are three types of heat transfer resistances which are Rmass, Rc and Ri. Rmass is the

effective resistance associated with the mass transfer of vapor to the pipe wall. When there is

a temperature difference between free stream and pipe wall, condensable gas (water) will

condense outside the pipe wall to form a film. However, not condensable gas (air) is going to

accumulate at pipe surface. Therefore, the partial pressure will be affected by it at the vapor-

liquid interface. As a result, there can be a substantial temperature difference between the free

stream dew-point temperature (T1,dp) and the interface temperature (Ti ). Moreover, Rc is

condensation heat transfer of outside of n-cylinder tube. Furthermore, Ri is internal pipe flow

heat transfer resistance which is formed by forced convection. Finally, for calculate the

energy recovery by free stream system which will apply equation:

First iterative calculation

Group 11 Page 3


Rmass (Effective resistance):

For force and convective mass transfer external flow over cylinder we use Sherwood number

to build the relationship between Reynold’s number and Schmidt number which the equation

is shown below:

Moreover, for mass transfer from free stream to film interface of pipe mass transfer properties

can be determined under average temperature Tf.

Mass transfer properties:

Mass transfer diffusion coefficient DAB = 0.26x10-4 m2/s (Incropera et al. 2007, A-26).

Absolute viscosity (Incropera et al. 2007, A-15).

Kinematic viscosity (Incropera et al. 2007, A-15).

For find out mass convection coefficient, reynold’s number also has to be determined.

Therefore, density of vapour and air can be evaluated at both free stream condition and

interface film condition by applying ideal gas law.

For interface film condition, partial pressure of water at 15ᵒC, Pv,s= 1700Pa (Incropera et al.

2007, A-23).

For free stream condition, partial pressure of water at 45ᵒC, = 9590Pa (Incropera et al.

2007, A-23).

Group 11 Page 4


At this Reynolds number from 40-4000 for Hilpert’s correlation are C = 0.683 and m= 0.466

(Tait Sherman Pottebaum 2003).

At 15ᵒC, latent heat of vapour water hfg = 2465800J/kg (Incropera et al. 2007, A-23).

And when partial pressure in free stream = , from steam table dew

point temperature (Incropera et al. 2007, A-23).

Condensation heat transfer Rc:

To find out hc for cylinder pipe external flow:

(Incropera et al. 2007, 652)

Assume the difference between interface temperature and surface of pipe .

Therefore, water properties can be evaluated at 15-1=14ᵒC.

ρl = 1000kg/m3, µl = 1162.65x10-6N.s/m2, kl = 0.593w/m.k

Group 11 Page 5


Internal pipe flow Ri:

Water properties are evaluated at 8ᵒC, ρw = 1000kg/m3, µw = 1.377x10-3N.s/m2 (Engineering

toolbox).

(Incropera et al. 2007, 514)

And since this is a heating process, therefore n=0.4.

Overall calculation:

Where A is the surface area of pipe =

From formula:

Where

Therefore, the new interface temperature:

Group 11 Page 6


As the calculation above, we assume the temperature difference of interface film and pipe

surface is 2ᵒC, now we can use q and hconv to calculate:

For the iterative calculation was done by Microsoft excel (the excel file will be saved into CD

as reference), and the iterative results are shown below:

First iteration Second iteration Third iteration Fourth iteration

(kg/s)

0.0853 0.08574 0.0858

11365.17 11780.57 11907.85

5007.72 5007.72 5007.72

q (heat transfer

rate W)

378.6 377.17 376.767

8.902 8.899 8.898

10.49 10.46 10.452

0.62 0.60 0.592

Fifth iteration Constant solution

(kg/s)

0.0858 0.0858

11907.85 11907.85

5007.72 5007.72

q (heat transfer

rate W)

376.68 376.68

8.898 8.898

10.452 10.452

0.592 0.592

Therefore, using this heat transfer system, 376.68W of energy can be recovered.

Group 11 Page 7


Discussion: In real condition, there should be more energy recovered by this heat transfer

system. It is because in the question calculation we only focus on condensation, and other

sensible heat is ignored such as radiation heat transfer from surrounding environment.

Question 2:

Given information:

Ts = 50

Group 11 Page 8


Tsurr = 25 (for convection)

Tsurr = 20 (for radiation)

Widthbath = 1mLengthbath = 3.5m

Depthbath = 40cm =

Pressureaverage = 92

Assumption: Steady state. The surface emissivity and absorptivity are not equal. Constant properties. Assume air is dry since it is at room temperature.

a) Radiation

Heat loss by radiation occurs from the top of water surface to the surrounding surface.

Where

6

Area of bath =

b) Natural convection

Recall

The Thermophysical properties of water at 323K, is gotten from interpolation since the conditions of the question are not in standard state.

T

320 10.53 2390323 12.32 2382325 13.51 2378

It can be deduced from the question that the air at the surface is saturated, hence it can be concluded that the vapor pressure at the surface is equal to the saturation pressure of water at the surface temperature.

The vapor pressure ( ) of air at 298K =

Group 11 Page 9


Taking the water vapor and air as ideal gases, we derive the formula for calculating density at the surface of bath and density from far away the surface of bath.

Where: P = pressure V = volume n = number of moles R = gas constant T = temperature

Thus density here is said to be the number of molecules of the ideal gas in a certain volume, in this case a molar volume, which is expressed as

Where: = density

n = number of molecules V = volume

From the above two equation, the expression for the density becomes:

Where: = density, kg/m3

P = pressure, kPa R = gas constant T = temperature

Note: Total atmospheric pressure is the sum of water vapor pressure and dry air pressure.

Where

Gas constant for water vapor = =

Gas constant for dry air = =

Group 11 Page 10


Density at surface of bath

Using

Distance at far away from surface of bath

Using

From the above calculated densities, it can be seen that is greater than , hence we

deduce that the hot surface is facing up in this problem.Calculation of critical length

Area of top surface of bath (

Perimeter of top surface of the bath

In this problem, the mixture is not homogenous. Hence we use the below Grashoff number.

Where

Note that convection occurs naturally in this problem and that the surface is hot and facing up too. Hence, it is necessary to determine Nusselt conventional heat transfer coefficient.

Group 11 Page 11


Nusselt number calculation

Where

80.43

Convective heat transfer coefficient

c) Evaporation

Using to calculate the mass diffusivity of water vapor in air

with a calculated average velocity of 310.0K

From the heat and mass convection analogy, we use which analogous to

prandtl number which is the ratio of kinematic viscosity to the thermal diffusivity

Group 11 Page 12


Sherwood number: This analogous to prandtl number which is the ratio of kinematic viscosity to the thermal diffusivity

Mass transfer coefficient

Hence

Comments: From the above, it can be deduced that there will be a total heat loss from the top

surface to the water bath of about 4604.23W which is from the summation of the heat losses

by radiation, natural convection and evaporation. This implies that a resistance of this size is

also need to compensate for the heat loss from the top surface. Furthermore, it is important to

Group 11 Page 13


state clearly that there is need to supply water at a rate of about this is so

as to compensate for the water loss due to evaporation in the water bath

Question 3:

a) Evaluate the Sc number and the average Sh number for the solvent evaporation

process.

Solution:

Known data:

The width of thin polymer is W=0.5m.

The length of thin polymer is L=2.5m.

The flowing air has a bulk velocity u = 1.5m/s.

Group 11 Page 14


The diffusion coefficient of the solvent in air Dab = 0.080cm2/s = 0.08x10-4 m2/s.

Kinematic viscosity ʋ = 1.5 x 10-5 m2/s.

Schmidt number Sc:

Find out Reynold’s number first to determine type of flow. And in this case, according to air

flow direction in above diagram, the Lc=W=0.5m.

For forced convection over a flat plate, Re = 50000 < 5x106 is laminar flow. Moreover Sc =

1.875 > 0.5. The formula shown below can be used for calculating average sherwoor number

(Agus, 2011).

b) What is the total evaporation rate of the solvent from the 0.5m by 2.5m polymer

film, recognising that both sides of the film are exposed to the following air.

Solution:

Assumption:

Thickness of this thin film is negligible, only up and down surface area will be used in

this section.

The vapour pressure of solvent in bulk condition is zero.

Molecular weight of solvent is mw=86g/gmol.

To find out mass convection coefficient:

Apply convection mass transfer equation:

For temperature at 20ᵒC and pressure at 1.0atm, ideal gas law can be applied to find out the

relationship between concentration and pressure.

Group 11 Page 15


Substitute this relationship into mass transfer convection formula:

Therefore, the total evaporation rate is 4.19g/s.

c) The solvent loading in the polymer film at the entrance of the drying process is

0.1g solvent per gram of dry polymer. The total mass flow rate of the polymer

film on a solvent free dry polymer basis is mo=50.0g dry polymer/s. What is the

solvent loading in the polymer film exiting the drying process (XAfinal)?

Solution:

Assumption:

Dry n-hexane polymer has density ρ=0.68kg/L = 680kg/m3 at 15ᵒC, in this question

we assume at 20ᵒC the density keeps constant. (Top Solvent Thaioil Group, 2011)

The thickness of polymer film is negligible, therefore, cross-section area just assume

unit area.

The mass ratio of solvent entering is XAo=0.1g solvent/g dry polymer.

Therefore, the solvent mass flow is ms = moXAo = 50x0.1 = 5g solvent/s.

According to answer of last section, 0.043g solvent will evaporate in this exposed surface area

per unit time.

Question 4

A spray cooling tower, as shown in Figure 4, is designed for melt granulation.

Group 11 Page 16


Figure 4 A spray cooling tower.

Determine the dimensions of the tower (height and diameter) and total processing time!

Note. Since the air and the melt particle flow in counter-current direction to each other, for

calculating the Reynold Number, velocity used is relative velocity where relative velocity (v r)

= air velocity (va) + terminal velocity of the particle (vt).

Use combined mass and heat balances to deal with this problem.

Given data:

Production rate = 3,000 kg/h

Desired particle size (Ds) = 2 mm = 0.002m

Particle density (ρs) = 1700 kg/m3

Average heat capacity of the particle (cp,s) = 1.46 kJ/kg.K

Inlet temperature of the melt = 110 °C

Solidification temperature = 70 °C

Cooling air velocity (va) = 2 m/s

Inlet air temperature = 10 °C

Outlet air temperature = 20 °C

The latent heat of fusion per unit mass (λ) = 186 kJ/kg

Properties:

By interpolation, the properties of air at 288 K are from Appendix I (Welty et al. 2008, 679).

Density (ρa) = 1.228 kg/m3

Group 11 Page 17


Heat capacity of (cp,a) = 1.006 kJ/kg.K

Viscosity (μa) = 1.789×10-5 Pa.s

Thermal conductivity (ka) = 2.530×10-2 W/m.K

Prandtl number (Pra) = 0.711

Assumptions:

1. The properties of particles are same at solid or melt state.

2. The solid particles reach the bottom at solidification temperature, 70 °C.

3. Solid particles drop at a constant terminal velocity.

4. The temperature of the melt particle is spatially uniform at any instant.

Solution:

Take the tower as a system. Assume no heat generation, no heat accumulation and well

insulated tower. Thus, no heat losses from the tower to the surrounding.

Energy balance in the tower.

In this case, energy in refer to the energy gained by the cooling air and energy out refer to

energy lost by the melt particles. Thus,

Tower diameter is determined.

By using the given formula, Archimedes number is determined to calculate the Reynolds

number, which velocity used is relative velocity where relative velocity (v r) = air velocity (va)

Group 11 Page 18


+ terminal velocity of the particle (vt) as the air flow and the melt particles flow are in

countercurrent direction. Thus, terminal velocity of the particle (vt) can be obtained.

There is forced convection over external surfaces of melt feed drops. Assume that the falling

melt feed drops modeled as spheres. The recent correlation proposed by Whitaker is

0.71<Pr<380, 3.5<Re<7.6×104, and . Thus, Whitaker correlation can be

used to calculate average Nusselt number in this case by makingassumption .

Average convective heat transfer coefficient is determined.

The total processing time include two parts, which are the cooling period and solidification

period. Cooling period refers to the temperature of melt particle decreases from the

temperature at the inlet to solidification temperature but no phase change occurs. On the other

hand, solidification period refers to the phase change of melt particle to solid phase from at

solidification temperature.

For the cooling period, consider the particle as a system. Assume no heat generation.

Work done is negligible in this case. . The equation reduced as below.

Group 11 Page 19


For the solidification period, consider the particle as a system. Assume no heat generation.

Work done is negligible in this case. . The equation reduced as below.

Since and is constant.

Total processing time is determined.

The height of the tower is determined.

Question 5

Group 11 Page 20


(a) As an employee of the Miri Air Quality Commission, you have been asked to develop a

model for computing the distribution of NO2 in the atmosphere. The molar flux of NO2 at

ground level is presumed known. This flux is attributed to automobile and smoke stack

emissions. It is also known that the concentration of NO2 at a distance well above ground

level is zero and that NO2 reacts chemically in the atmosphere. In particular, NO2 reacts with

unburned hydrocarbons (in a process that is activated by sunlight) to produce PAN

(peroxyacetylnitrate), the final product of photochemical smog. The reaction is first order, and

the local rate at which it occurs may expressed as NA= -k1CA where k1 = 0.03 s-1.

Assuming steady-state conditions and a stagnant atmosphere, obtain an expression for the

vertical distribution CA(x) of the molar concentration of NO2 in the atmosphere!

Given data:

1. k1 = 0.03 s-1.

Assumptions:

1. One dimensional and steady state transient diffusion in a stagnant atmosphere.

2. Ideal gas.

3. Uniform total molar concentration.

4. Constant properties of temperature (T = 300K) and pressure (1 atm) of air.

Schematic diagram:

Solution:

The molar flux within the film is described as below.

For one dimensional steady state mass transfer, the general differential equation is shown as

below.

The diffusion with homogenous chemical reactions is considered in this case. For Cartesian

coordinates, the molar form of the species diffusion equation is shown as below (Saptoro

2011).

Group 11 Page 21

x = 0x

NO2

Air


The differential equation can be rewritten as below.

Where m= (k1/DAB) ½

The general solution of the above ordinary differential equation is shown as below.

The integration constants C1 and C2 are determined using the boundary conditions.

When , , , , .

When , , .

Since,

Hence,

Where m= (k1/DAB) ½.

(b) If an NO2 partial pressure of pA = 2 x 10-6 bar is sufficient to cause pulmonary damage,

what is the value of the ground level molar flux for which you would issue a smog alert?

Solution:

Group 11 Page 22


At ground level,

For ideal gas,

The best estimate for the diffusion coefficient of NO2 to air was found to be 0.154 cm2s-1 =

0.154 ×10-4m2s-1 (Para n.d.).

Where m= (k1/DAB) ½ = (0.03/0.154×10-4) ½ = 44.14 m-1.

(c) In problems (a) and (b), NO2 transport by diffusion in a stagnant atmosphere was

considered for steady-state conditions. However, the problem is actually time dependent and a

more realistic approach would account for transient effects. Consider the ground level

emission of NO2 to begin in the early morning (at t = 0), when the NO2 concentration in the

atmosphere is everywhere zero. Emission occurs throughout the day at a constant flux and

NO2 again experiences a first order photochemical reaction in the atmosphere NA= -k1CA

where k1 = 0.03 s-1. For a differential element in the atmosphere, derive a differential equation

that could be used to determined the molar concentration CA(x,t). State appropriate initial and

boundary conditions!

Solution:

For unsteady state diffusion, the species conservation equation for rectangular is shown as

below.

Since there is one dimensional diffusion and no bulk motion, the species conservation

equation reduced as below.

Group 11 Page 23


For a semi-infinite medium,

The initial condition, t=0.

The boundary condition, .

As it’s very far, no mass transfer to that point.

(d) Obtain an expression for CA(x,t) under the special condition for which photochemical

reactions may be neglected. For this condition what are the molar concentration of NO2 at

ground level and at 100-m elevation 3 h after the start of the emission, if a constant emission

flux is equal to 3 x 10-11 kmol/s.m2.

Solution:

The heat transfer for semi-infinite solid under constant surface heat flux can be determined as

below (Incropera et al. 2007, 286).

The concept of heat transfer can be applied for mass transfer, the mass transfer in a semi-

infinite medium under constant diffusive flux can be determined by changing k and to

and .

At the ground level, x = 0 and t = 3 hr,

At the 100m, x = 100 m and t = 3 hr =10800 s,

Group 11 Page 24


Since much greater than 1, the will be nearly to 1.

Reference list:

Group 11 Page 25


Agus Saptoro. (2011). Mass transport lecture 3 : Mass Transport: Part 3 Convection Mass

Transfer. PowerPoint slides. Retrieved from Curtin University Sarawak Campus Moodle Web

site: http://moodle.curtin.edu.my/course/view.php?id=453 (accessed on 10th May 2011)

Air properties http://www.engineeringtoolbox.com/dry-air-properties-d_973.html (accessed on 5th May 2011)

Dewitt., P. D. Bergman., L. T. Lavine, S. A. Fundamentals of Heat and Mass Transfer 6th edition ,Incropera., John Wiley and Sons (Asia) Pte Ltd

Emissivity of water http://www.infrared-thermography.com/material-1.htm (accessed on 5th May 2011)

F.P.Incropera, D.P.Dewitt, T.L.Bergman, A.S.Lavine. 2007. Fundamentals of Heat and Mass

Transfer. 6th edition. John Wiley & Sons(Asia) Pte Ltd.

Gas constant for water vapor http://amsglossary.allenpress.com/glossary/search?id=gas-constant1 (accessed 6th May 2011)

Gases-Densities. n.d. http://www.engineeringtoolbox.com/gas-density-d_158.html (accessed May 22, 2011).

Grashof number http://web2.clarkson.edu/projects/subramanian/ch302/notes/Natural%20Convection.pdf (accessed 6th May 2011)

Incropera, F.P., D. P. DeWitt, T. L. Bergman, and A. S. Lavine. 2007. Fundamentals of Heat and Mass Transfer. 6th Edition. New York: John Wiley & Sons, Inc.

McCabe, L. W; Smith, C. J; and Harriott, P. 2005. Unit Operation of Chemical Engineering.

7th ed. McGraw Hill.

Para, J. n.d. Performance and Application of an Inexpensive method for Measurement of Nitrogen Dioxide. http://www.mcnair program.pdx.edu/Online%20Journal%20200405/Jeremy%20Para--Final.pdf (accessed May 22, 2011).

Richard, S. An Introduction to Air Density and Density Altitude Calculations. updated: Mar 21, 2011 copyright 1998 - 2010 http://wahiduddin.net/calc/density_altitude.htm(accessed 6th May 2011)

Saptoro, A. 2011. Mass Transport: Part2 Mass Diffusion with Reaction & Transient Diffusion. http://moodle.curtin.edu.my/mod/resource/view.php?id=34511 (accessed May 22, 2011).

Group 11 Page 26

http://wahiduddin.net/calc/density_altitude.htm(accessed

http://web2.clarkson.edu/projects/subramanian/ch302/notes/Natural%20Convection.pdf

http://web2.clarkson.edu/projects/subramanian/ch302/notes/Natural%20Convection.pdf

http://amsglossary.allenpress.com/glossary/search?id=gas-constant1

http://amsglossary.allenpress.com/glossary/search?id=gas-constant1

http://www.infrared-thermography.com/material-1.htm

http://www.engineeringtoolbox.com/dry-air-properties-d_973.html

http://moodle.curtin.edu.my/course/view.php?id=453


Tait Sherman Pottebaum. 2003. The relationship between near-wake structure and heat

transfer for an oscillating circular cylinder in cross-flow. Accessed 10 th May 2011,

http://thesis.library.caltech.edu/1886/1/Pottebaum_thesis.pdf

Top Solvent Thaioil Group. 2011.

http://www.topsolvent.com/index.php/en/products/hidrocarbon-solvent/hexane-polymer.html

(accessed on 11th May 2011).

Vapor Pressure of water vapor at 293K http://firstyear.chem.usyd.edu.au/bridging_course/Questions/PartialPressure.htm (accessed 18th May 2011)

Welty, J.R., C.E. Wicks, R.E. Wilson, and G.L. Rorrer. 2008. Fundamentals of Momentum, Heat, and Mass Transfer. 5th ed. United States of America: John Wiley & Sons, Inc.

ChE Transport Phenomena

Minutes of Meeting Form for Group Project

Group 11 Page 27

http://firstyear.chem.usyd.edu.au/bridging_course/Questions/PartialPressure.htm

http://www.topsolvent.com/index.php/en/products/hidrocarbon-solvent/hexane-polymer.html

http://thesis.library.caltech.edu/1886/1/Pottebaum_thesis.pdf


Meeting Minutes 1

Meeting date: 3 May 2011

Participants: Wong Mei Wen, Seer Qiu Han, Ahiamadu Jackson Samuel.

Apologies (those who were absent): -

What work did each member complete, based on targets that were set in the previous meeting?

Member 1:Wong Mei Wen

Member 2:Seer Qiu Han

Member 3:Ahiamadu Jackson Samuel

- - -

Summary of discussions in the meeting:

During first meeting, we discussed problems of assignment question paper. In the end, five questions were distributed equally to every member. Ahiamadu Jackson Samuel is in charge of question 2. Wong Mei Wen is in charge of question 1 and 3. Seer Qiu Han is in charge of question 4 and 5.

List the target work for each member that should be completed by the next meeting:

Member 1 Member 2 Member 3

To present the solving steps and result of question 2 by handwritten.

Any problem from assignment question will be discussed.

To show the solving steps and result of question 1 by handwritten.

Look through question 3. Any difficulty from

assignment question will be discussed.

To show the solving steps and result of question 4 by handwritten.

Look through question 5. Any problem will be

discussed.

Planned date of next meeting: 11 May 2011

Any further comments:

After the handwritten solutions from each group member have been presented, the solution paper will be exchanged within the group members for double checking.

Meeting Minutes 2

Group 11 Page 28









Question 2 had been done by handwritten.




During second meeting, all group members completed their job based on the targets that were set in the previous meeting. Team members discussed and confirmed the answer for question 1, 2 and 4. The ways on doing the question 4 and 5 were discussed.

List the target work for each member that should be completed by the next meeting:

Member 1 Member 2 Member 3

Type in the handwritten solution of question 2.

Any problem from assignment question will be discussed.


To show the solving steps and result of question 3 by handwritten and type in the solution.

Any difficulty from assignment question will be discussed.


To show the solving steps and result of question 5 by handwritten and type in the solution.

Any problem will be discussed.

Planned date of next meeting: 23 May 2011

Any further comments:

The type in solutions will be sent within the group members.

Meeting Minutes 3

Group 11 Page 29









Question 2 had been double checked.

Question 1 and 3 had been typed in.

Question 4 and 5 had been typed in.


All the member has accomplish their jobs base on the target which was set on last meeting. We brought in the problem occurs in our last phase. All problems was discussed and solved in the meeting. We also discussed and done the compilation together during the meeting, therefore, the group project has completed in the meeting. The softcopy had been saved among the group members for checking the finalised assignment.

Group 11 Page 30

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Compiled Final Report Transport Phenomena

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