Bio Eng 4316/7316 Chem Eng 4316/7316

Biomass Refinery Operations II Lecture 23

November 16th, 2015

Exam Review• Prob. 3Filtration

Premise #1: Biomass is our Carbon Resource

Premise #2: We Must Implement the Biomass Refinery Concept

# Agenda Reading Assignment

1 Monday, August 24, 2015 Introduction, Biomass Resource Billion Ton Report Exec. Summary et al.; Vasilev

Presentation / Article

Company Profile slide due 8/26

2 Wednesday, August 26, 2015 Biochhemisty & Bioproducts Review, Chapter 1 Analytical Method Slide,

Problems 1.4, 1.5, 1.6, 1.8,1.9, 1.10

due 9/2

3 Monday, August 31, 2015 Biochhemisty & Bioproducts Review, Bioprocess

Engineering Analysis, Analytical Methods

Chapter 2 Problems 2.1, 2.3,

4 Wednesday, September 2, 2015 Analytical Methods Size Reduction Slide, due 9/9

5 Wednesday, September 9, 2015 Analytical Methods, Size Reduction Biomass Comminution Energy Required IC&P

2011.pdf, Knife Mill Power Requirements

BioResTech 2009.pdf

6 Monday, September 14, 2015 Size Reduction Drying Slide, due 9/20

7 Wednesday, September 16, 2015 Size Reduction

8 Monday, September 21, 2015 Drying Chapter 11 Pr. 10.1, 10.4 (from 1st edition

drying chapter see on BB site), due

M, 9/28

9 Wednesday, September 23, 2015 Prof. David Brune, Aquaculture, Drying Feeding Slide, due 9/30

10 Monday, September 28, 2015 Drying Pr. 10.5, 10.7, 10.8 (from 1st

edition drying chapter see on BB

site), due 10/7

11 Wednesday, September 30, 2015 Drying

12 Monday, October 5, 2015 Drying

13 Wednesday, October 7, 2015 Drying

Monday, October 12, 2015 Class is cancelled

Wednesday, October 14, 2015 Class is cancelled

14 Monday, October 19, 2015 Drying Filtration Slide due 10/26

15 Wednesday, October 21, 2015 Lee Stevens, Monoclonal Antibody Production, Drying,

Feeding

16 Monday, October 26, 2015 Feeding Biomass Feeder Thesis, Dry Biomass Feed System

Report, Lock Hopper Desigh, Figleaf Feeder

Report, High Pressure Solids and Slurry Feeder Lit

Review

17 Wednesday, October 28, 2015 Cell Lysis & Flocculation Chapter 3 Problems 3.1, 3.2, 3.3, 3.4, 3.6, due

10/16

18 Monday, November 2, 2015 Cell Lysis & Flocculation

19 Wednesday, November 4, 2015 Exam Review, Problem Solving, Filtration Chapter 4 Problems 4.1, 4.2, 4.3, 4.5, 4.6,

4.7, 4.8, 4.11 due 11/16

20 Wednesday, November 4, 2015 Exam I 5-7 CL TBD

21 Monday, November 9, 2015 Go Over Exam 1, Filtration

22 Wednesday, November 11, 2015 Filtration

23 Monday, November 16, 2015 Filtration

24 Wednesday, November 18, 2015 Extraction Chapter 6 Problems 6.1, 6.4, 6.7, 6.9, due

12/2

25 Monday, November 23, 2015 Extraction Extraction Slide, due 12/2

26 Monday, November 30, 2015 Extraction

27 Wednesday, December 2, 2015 Sedimentation Chapter 23 Problems 5.4, 5.5, 5.6, 5.9,due

12/7

28 Monday, December 7, 2015 Exam II

29 Wednesday, December 9, 2015 Exam II Discussion & Corrections, Bioprocess Design

Biomass Refinery Operations I & II

I. Biomass and Carbonaceous ResiduesQuantity

ResiduesCrops

AgricultureAquaculture

QualityBiochemistryAnalytical Chemistry

Transportation and Storage of BiomassIndustrial Ecology & Life Cycle Analysis

II. Upstream ProcessingMechanical Processing

Size ReductionChoppingGrindingMillingFlaking

FeedingThermal Processing

DryingSteam Explosion

Chemical ProcessingChemical Pretreatments

Biological ProcessingCell LysesFlocculationEnzymatic Hydrolysis

(4315, 4316)

III. ConversionsMechanical ProcessingThermal Processing

CombustionGasificationPyrolysisLiquefaction

Chemical ProcessingTransesterification

Biological ProcessingFermentationsEnzyme Catalysis

IV. FractionationsMechanical Processing

Expulsion Thermal Processing

DistillationChemical Processing

ExtractionBiological Processing

V. Downstream ProcessingMechanical Processing

FiltrationSedimentation

Thermal ProcessingDrying

Chemical ProcessingCrystallization

Biological Processing

V. Plant-wide ProcessesSafetyOdor Measurement and MitigationPermitting

In the News

Questions?Stéphane Sarrade of Crainte Espoir

muengrbe2180hwexams@missouri.edu

Start at 3:27 Done at 3:42 Elapsed time = 15

1 2 3 4 5

10 0.060 0.439 0.733 0.865 0.929

20 0.526 0.642 0.783 0.880 0.928

30 0.679 0.738 0.824 0.894 0.937 obs pred

40 0.754 0.793 0.848 0.906 0.939 0.060 0.060 0.068

50 0.807 0.827 0.867 0.909 0.942 0.526 0.526 0.518

60 0.833 0.851 0.882 0.921 0.949 0.679 0.679 0.676

0.754 0.754 0.756

0.807 0.807 0.805

0.833 0.833 0.837

0.439 0.439 0.433

1 2 3 4 5 9.7430418 0.642 0.642 0.638

10 0.068 0.433 0.729 0.867 0.928 0.990 0.738 0.738 0.735

20 0.518 0.638 0.787 0.882 0.933 k = 10.594 0.793 0.793 0.791

30 0.676 0.735 0.824 0.895 0.937 a = 3.055 0.827 0.827 0.828

40 0.756 0.791 0.851 0.905 0.941 b = 1.016 0.851 0.851 0.854

50 0.805 0.828 0.871 0.913 0.944 0.733 0.733 0.729

60 0.837 0.854 0.886 0.920 0.947 0.783 0.783 0.787

829.62919 0.824 0.824 0.824

R = 0.956 0.848 0.848 0.851

omega = 5 0.867 0.867 0.871

1 2 3 4 5 tau = 94.990337 0.882 0.882 0.886

10 -0.008 0.006 0.004 -0.001 0.001 0.865 0.865 0.867

20 0.008 0.003 -0.004 -0.002 -0.005 0.880 0.880 0.882

30 0.003 0.003 0.000 -0.001 0.000 0.894 0.894 0.895

40 -0.002 0.002 -0.003 0.001 -0.002 0.906 0.906 0.905

50 0.002 -0.001 -0.004 -0.004 -0.002 0.909 0.909 0.913

60 -0.004 -0.003 -0.003 0.001 0.001 0.921 0.921 0.920

0.929 0.929 0.928

resid = 0.000 0.928 0.928 0.933

0.937 0.937 0.937

F = 10 liter/min = 0.1666667 liters/s 0.939 0.939 0.941

tau = 10 s = 0.942 0.942 0.944

V = 1.6666667 liters 0.949 0.949 0.947

Ccdw = 100 g/liter

mdotidcw = 71 g/min of intact cells

mdotlyscdw = 929 g/min of lysed cells

tau (s)

w (rotation rate, rpm)

tau (s)

w (rotation rate, rpm)

tau (s)

w (rotation rate, rpm)

ba

kR

w 1

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000

a) (2/1 pts): If you operated a residence time of 10 s, estimate the rotation rate required to

achieve 0.99 cell disruption

w = _______ rpm

b) (2/1 pts): If you operated a rotation rate of 5 rpm, estimate the residence time required to

achieve 0.99 cell disruption

= _____ rpm

d) (2/1 pts): The volumetric feed flow rate the bead mill is 10 liters/minute. The cell mass

concentration is 100 g CDW/liter. Operating at a rotation rate of 5 rpm and a residence time of

10 s, estimate the volume of the bead mill required.

V = ______ liters

e) (2/1 pts): For the conditions of “d” above, estimate the delivery rate of intact (un-lysed) cells

(dry basis).

ṁicdw = ______ g/minute

Filtration

Filtrate: a liquid that has passed through a filter.

Permeate: a liquid that has passed through a filter.

Retentate: solid or high MW material that is retained by a filter.

Cake: solid material that is retained by and builds up on a filter.

See Example 4_2 worksheet.xlxs

HW Hint:

4.3: The flux through the porous media will be NQ. Generate expressions for J using the Hagen-Poiseulle equation & the Darcy Equation (4.2.1)

4.5: No Excel necessary, start with integrated form of Darcy’s law.

4.11: No Excel necessary, use eqn 4.2.23 for the transmembrane flux and set J = 0, sigma = 1.

Chapter 4: Filtration

Overall Mass, Hi

MW or particle

species,

Continus Phase kg mol J

shown

Else

Differential

m = 2 chemical species present

s = 3 distinct flowing streams

Balance Equations = 3

Independent Balance Equations = 2

Unkowns = 6

Unk - Eqn = 4

Steady-state Dynamic

Open Closed

Reactive Non-reactive

y ≡

System ≡

Surroundings ≡

Algebraic Integral

0 p (valuable product)

+ -

= 0 overall mass

= 0 cont phase

Example 4.5

J is transmembrane flux c is concentration of solute

Correlations for mass transfer coefficient for laminar flow:

Correlations for mass transfer coefficient for turbulent flow:

Shear at wall to 1/3 power

See Example 4_2.xlsx

k = D/d

MW1 = 12,000 Da MW2 = 120,000 Da

eqn 4.2.17 (see also 4.2.23)

eqn 1.8.9

UF of Protein Solutions, Monday filter a low MW protein (1) and Tuesday filter a high MW protein (2).Same molar conc, same membrane & transmembrane flux and same boundary layer thickness.Which has higer polarization modulus? What is the relationship of the polarization moduli between the two cases?

ln(cw1/cb)*D1 = Jd = constant

ln(cw2/cb)*D2 = Jd = constant

ln(cw1/cb) /ln(cw2/cb) = D2/ D1

ln(cw1/cb ) = D2/D1 ln(cw2/cb)

cw1/cb = (cw2/cb)^(D2/D1)D1 = kT/(6pm1a1)

D2 =kT/(6pm2a2)D1/D2 = m2a2/(m1a1)

cw1/cb = (cw2/cb)^(m1a1/m2a2)

The viscosity & particle size are greater for the high MW material, therefore cw2 > cw1.

In order to achieve equal J and delta, how would you have to operate differently?

Problem 4.8

k = Boltzman constanta = molec. diameter

T is abs temp

Crossflow Filtration Media: Membranes

See Example 4..3 and 4.4 worksheet, to explore washing concept in context of rotary vacuum filter.

Washing to remove product, constant mu and delPWrite eqn 4.2.4 after filtering Vf in time tf.

See Example 4..3 and 4.4 worksheet, it will help with Problem 4.5

X (1/(1-X))Mass CDW/area cake x Mass total cake/mass CDW

Conventional Filtration Media•Woven fabrics (10 um), metal fabrics (5 um), rigid porous media•Sterile filtrations: membrane or depth, asymmetric membrane (0.22 or 0.45 um)•Air: HEPA

Filter Aids: Cake Engineering