topic 7-aw prediction

Dr Ted Labuza University of Minnesota Department of Food Science and Nutrition St Paul MN 55108voice 612-624-9701 fax 651-483-3302 [email protected]

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tplabuza © 2008 1Slide #

Topic 7FScN 4342

Water Activity Prediction

Dr Ted P. Labuza

Department of Food Science and NutritionUniversity of Minnesota St Paul 55108 USA612-624-9701 fax 625-5272 [email protected]


Water Activity Predictionby

Raoult’s Law


1. Raoult’s law

Assumes water activity loweringdue to solutes (range 0.95 to 1)

Colligative effect based on # ofparticles

All solutes dissolved in all waterNo between solute interactionDilute solution ? How dilute


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aw= !

nH2O

nH2O+ n

so lute

= ! XH2O

XH2O= mole fraction water water

Xs= mole fraction solute

= 1" XH2O

! = effective activity coefficient


Raoults Law Example 1What is aw if you add 20 g propylene glycol to 100 gmeat at 50% moisture content (wb)?

Boundary conditions:100 g meat with 50 g wate+ 50 g solidsAssume initial aw = 1 before addition of soluteAssume meat solids no effectAssume all solute dissolvedAssume ideal ie γ =1

Ideal Calculation:Moles glycol =20/76= 0.263Moles water = 50/18 = 2.778Xwater = [ 2.778/(2.778 + 0.263) ] =0.9136 =aw= 0.914Xsolute = 1-0.9136=0.0864


Example 1 Solution #2position on real solute isotherm accounts for γ ≠ 1

1 .000 .950 .900 .851 0 0

2 0 0

3 0 0

4 0 0

5 0 0

6 0 0

7 0 0

8 0 0

Propylene glycol isotherm

water activity

y = - 2.4286e+5 + 8.4204e+5x - 9.7383e+5x^2 + 3.7593e+5x^3 R^2 =1.000


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Example 1 Solution #2position on real solute isotherm

1 .000 .950 .900 .851 0 0

2 0 0

3 0 0

4 0 0

5 0 0

6 0 0

7 0 0

8 0 0


water activity

y = - 2.4286e+5 + 8.4204e+5x - 9.7383e+5x^2 + 3.7593e+5x^3 R^2 =1.000

20 g glycol to 50 g water gives 50/20 = 2.5 g water/gglycol or 250 g water / 100 g glycolUse polynomial regressionaw = 0.906 thus less than ideal which gave 0.914From this γ = 0.906/0.914 = 0.991


questionsHow good is assumption ?

Can we estimate activity coefficientfrom base principals?

Van Laar equation

a =! Xwater

= ! [1" Xsolute

]

ln a = ln ! + ln [1" Xsolute

] at low Xs ln(1-X

s) =X

s

ln !H 2O

= K Xs

#$ %&2

Van Larr equation thus

ln a = K Xs

#$ %&2

+ ln [Xwater

]

lna

Xs

= K Xs

#$ %&2

To determine make solutions at different levels of Xs Plot ln [a/Xw ] vs Xs2


Single Solute correction fornon-ideality

Van Laar equation from heat of mixing (Hildebrandt and Scott)

ln! = KsX

2s

γ = activity coefficient

Ks = solute constant

Xs = solute mole fraction

ln a/Xw

2Xs

slope = K s

a

w= X

H2Oe

Ks

Xs

2

ln! = lna

w

Xw

= KsX

s

2


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Compound KPEG 600 -56.0 ± 2PEG 400 -26.6 ± 0.8Peg 200 -6.1 ± 0.3lactose -10.2lysine -9.3 ± 0.3lactulose -8.0 ± 0.3sucrose -6.47 ± 0.06ornithine -6.4 ± 0.4NaCl [2 - 26%] -6.26citric acid -6.17 ± 0.49DE 32.8 -5.97DE 42 -5.31DE 55 -5.18DE 64 -4.57DE 83 -3.78DE 91 -2.99


Compound K

tartaric acid -4.68 + 0.5maltose -4.54 + 0.02proline -3.9 + 0.1a-ABA -2.57 + 0.37alanine -2.52 + 0.37glucose/fructose-2.25 + 0.02malic acid -1.82 + 0.13sorbitol -1.65 + 0.14lactic acid -1.59 + 0.2xylose -1.54 + 0.02glycerol -1.16 + 0.01propylene glycol-1.0mannitol -0.91 + 0.27NaCl [0 to 2%] +17.91KCl [below 2%] +10.81alanine +2.59 + 0.37 Increases activityurea +2.02lactic acid +1.59 + 0.2glycine +0.87 + 0.11

}


substance MW in Daltons

water = 18.016propylene glycol = 76glycerol = 92.1ethanol = 46.1NaCl = 58.45KCl = 74.55sucrose = 342.3lactose = 342.3fructose = 180.2glucose = 180.2citric acid = 192.12


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Norrish 1 solute Example 1What is aw if you add 20 g propylene glycol to 100 gmeat of 50% moisture content?

Boundary conditions100 g meat with 50 g water 50 g solidsAssume initial aw = 1 before additionAssume meat solids no effectAssume all solute dissolvedAssume non-ideal ie γ <1

Moles glycol =20/76= 0.263Moles water = 50/18 = 2.773Xwater = [ 2.778/(2.778 + 0.263) ] =0.9136Xsolute = 1-0.9136=0.0864K = -1 so γ = exp (-1 x 0.0.08642)= 0.9926 vs 0.991 obsaw = 0.9926 x 0.9136 = 0.907


comparison

Raoult’s law 0.914Graphical 0.906Van Lar-Norrish 0.907

Measured errror ± 0.003


What if add 20 g PEG 600?

Moles PEG =20/600= 0.03Moles water = 50/18 = 2.778Xwater = [ 2.778/(2.778 + 0.03) ] =0.989 =awXsolute = 1-0.989=0.011 K = -56 so γ = exp (-56 x 0.0.0112)= 0.993 aw = 0.993 x 0.989 = 0.982


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Example 2

How much PG is needed to reach an aw of 0.85 ?


Example 2 Solution 1How much PG is needed to reach an aw of 0.85 ?

nwater = 50/18 = 2.778 moles

aw= !

nwater

nwater

+ nsolute

= 0.85 =2.778

2.778 + nsolute

nsolute

= 0.490 moles = 37.26 grams


Example 2 solution 2How much PG is needed to reach an aw of 0.85 ?

From isotherm at aw =0.85 m= 147.5 g water/100 g PG

147.5

100=

50

x

x = 33.9 g PG

1 .000 .950 .900 .851 0 0

2 0 0

3 0 0

4 0 0

5 0 0

6 0 0

7 0 0

8 0 0


water activity

y = - 2.4286e+5 + 8.4204e+5x - 9.7383e+5x^2 + 3.7593e+5x^3 R^2 =1.000


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Example 3How much NaCl is needed to reach same aw of 0.85 ? Conc. (% w/w) aw

13.5 0.906 14.0 0.902 14.5 0.897 15.0 0.892 15.5 0.888 16.0 0.883 16.5 0.878 17.0 0.873 17.5 0.867 18.0 0.862 18.5 0.857 19.0 0.851 19.5 0.845

From Table ~ 19% salt w/w

Thus 19/81 = x/50

X= 11.7 grams


Example 4

How much PG is needed to reach an aw of 0.85 ?


Do by iteration

ln aw = ln Xw + Ks (1 - Xw)2

Assume value for Xw

Calculate aw

repeat and make plot aw vs Xw


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Example calculation

50 g water -> 2.778 moles water eg @ 10 g PG n2 = 10/76 = 0.132

Xw =2.778

2.778+0.132= 0.9546

Xs = 1- Xw = 0.045

ln aw = - 0.0464 - 1 [0.045]2

= - 0.0485

a w = 0.9527 @ 10g PG in 50 g water


Example calculation10 g PG aw = 0.95320 g PG aw = 0.90730 g PG aw = 0.86240 g PG aw = 0.820

1.00.9

0

10

20

30

40

50

Example solution propylene glycol

water activity

g g

lyco

l ad

ded

at aw = 0.85 ---->amount neededis 32.5 g PG vs33.9 grams fromisotherm and37.3 g fromRaoult’s Law


Water Activity Predictionby

Norrish Equation


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Multiple solute solution

lnaw= lnX

H2O+

Ki

Xi[ ]

i=1

n

!

Xi[ ]

2

i =1

n

!

2

1" XH2O[ ]

2

Xi = mole fraction of ingredient in all water + all solute = nsi/ (nH2O + Nsolute )

Ki = Van Laar constant for solute


Complex Food SystemPiglet Liquid Diet

Component Composition grams Canola Oil 40 Demineralized whey solids 10 (50% lactose, 50% protein)Gum Arabic 1.0 HFCS-solids 10 Potassium chloride 0.8 Sodium chloride 0.2 Glycerol 4.0 Propylene Glycol 8.0 Citric acid 2 Water 24

aw Food Calc 2.0


aw Food Calc 2.0

Sum solute moles = 0.2575 Moles water = 1.332

Example Xglycerol = 0.0434/(1.332 +0.2575) = 0.0273


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3. Grover Equation aw = 1.04 - 0.1 Es° + 0.0045 (Es°)2

Es

o=

Ei

mi

!

"

# #

$

%

& &

'

Ei = Grover value for component i

mi = g H2O/g solids of component i

limit Es° < 10Chorleywood-Camden Computer programBased on lowering relative to sucrose in baked goods


Grover constantsCompound Eiwater and fat 0.028 DE 0.742 DE 0.8gums 0.8starch 0.860 DE 1.0lactose 1.0sucrose 1.0glucose, fructose 1.3protein 1.3egg white 1.4glycerol < 20% 2.0glycerol >20% 4.0sorbitol 2.0acids 2.5propylene glycol 4.0ethanol 8.0Salts 9.0


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Example 1What is aw if you add 20 g propylene glycol to 100 gmeat of 50% moisture content?

Boundary conditions100 g meat with 50 g water 50 g solidsAssume initial aw = 1 before additionAssume meat solids no effectAssume all solute dissolved

Moles glycol =20/76= 0.263Moles water = 50/18 = 2.778


Example 1initial meat water activity

component grams mi Ei Ei/mi

moisture 50 protein 15 3.333 1.3 0.390 starch 1 50.0 0.8 0.016 fat 34 1.47 0 0.00

Σ(Ei/mi)= Es° = 0.406

mi = grams water total per gram componentget Ei values from Tableaw = 1.04 - 0.1 Es° + 0.0045 (Es°)2

initial aw = 1.04 - 0.1x0.406 + 0.0045 (0.406)2 = 1.0014


Example 120 g PG added to meat

component grams mi Ei Ei/mi

moisture 50 0.0protein 15 3.333 1.3 0.390starch 1 50.0 0.8 0.016fat 34 1.47 0 0.00PG 20 2.5 4 1.6

ΣEi/mi= Es° = 2.006

aw = 1.04 - 0.1 Es° + 0.0045 (Es°)2

aw = 1.04 - 0.1 x 2.006 + 0.0045 (2.006)2 =0.857


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comparison

Raoult's 0.914Norrish 0.907Isotherm 0.906Grover 0.857


Piglet Diet

aw Food Calc 2.0 Es°= Sum (Ei/mi)=3.302

aw = 1.04 - 0.1 Es° + 0.0045 (Es°)2

aw = 1.04 - 0.1 x 3.302 + 0.0045 (3.302)2 =0.759


4. Money and Born

aw=

1

1 + 0.27N

N = moles humectant/100 g H2O

Thus 20 g water in meat @ 50% water

So 40 g in 100 g water N=40/76=0.526

aw = 0.876ComparisonRaoult's 0.914Norrish 0.907Isotherm 0.906Money & Born 0.876Grover 0.857


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Piglet diet

aw Food Calc 2.0

aw = 1/[1 +0.27 N]

Sum N = 1.073

aw = 0.775


5. Ross Equation Assumes no interaction of solutes All water available to all ingredients Gibbs Duhem relationship applies

G = µ1N1

+ µ2N2

dG = µ1dN

1+ N

1dµ

1+ µ

2dN

2+ N

2dµ

2

for the addition or subtraction of N 1 or N2

dG =dG

dN1

!

" # #

$

% & & dN

1+

dG

dN2

!

" # #

$

% & & dN

2= µ

1dN

1+ µ

2dN

2

thus:

N1dµ

1+ N

2dµ

2= 0 dµ = RTd ln a

substitute in and divide both sides by RT /Ntotal

X1d ln a

1+ X

2d ln a

2= 0

if multiple solutes then and X 1 represents water:

lnaw

= ' X2

X1

(

) * *

+

, - - . d ln a

2' X

3

X1

(

) * *

+

, - - . d ln a

3

aw

= aw( )

2

0

aw( )

3

0

Note for starches, proteins, gums etc the individualisotherm is used to get the ingredient aw


Ross equation

a

w= ! a

H ifor example a

w= a

H1a

H2••a

H i

where aHi = activity of solution of component i assuming allof each ingredient is dissolved in all the water or interactswith all the water

For Example 1 20 g PG added to 100 g meat aw = 0.906

For example 2 assume initial meat at 0.99 from measuredaf = ai aH1 ai = 0.99 af = 0.85thus aH1 = 0.85 / 0.99 = 0.859

From isotherm mi = 155 g water / 100 g glycolor 32.3 g per 50 g water


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Comparison of methods

Comparison of all methods

Raoult's 37.3Norrish 32.5Isotherm 33.9Ross 32.3Money & Born 24.8Grover 21.2


Combined program aw 3.0


Piglet Diet Summary

Grover 0.759Money & Born 0.775Ross with Norrish 0.813Norrish 0.816

topic 7-aw prediction

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