In this post, we shall be discussing two methods of material standardization, namely, Pearson Square method and Mass Balancing method. We'll start with the Pearson Square method and then later on delve into the mass balancing method. Our case studies will be based on dairy and juice processing.
The Pearson Square Method
Milk standardization is important in the dairy industry because it is used to ensure that every consumer gets milk with constant fat content and consistency. Primarily, milk may be separated into two products; cream with high fat content (about 40% fat) and skimmed milk (about 0.03% fat) using a centrifugal separator.
The process of standardization involves reduction of fat content in milk from the natural minimum of 3.25%. The cream extracted from the standardization process facilitates processing of other dairy products such as ice cream, butter, and ghee.
Milk standardization can take two forms i.e. either partial separation or mixing of skimmed and whole milk. Mass balancing plays a key role in successful standardization of milk.
Example 1
2000 kg of milk (with 87.6% water, 3.8% fat, 3.2% protein, 4.6% lactose, and 0.7% ash content) has to be reduced in fat content from 3.8% to 2.5% by removal of cream with 40% fat content from the milk. How much milk will have to be removed?
Solution.
TMB: 2000 = C + M
FMB: 2000×0.038 = 0.4×C + 0.025×M
2000×0.025 = 0.025×C + 0.025×M
Solving the equations will give the values of C = 69.3 kg and the remaining milk M = 1930.7 kg.
Example 2
If 3000 kg of the same milk used in the previous example is separated into cream with 45% fat and skimmed milk with 0.05% fat, how much cream and skimmed milk are expected assuming no losses?
Solution
TMB: 3000 = C + M
FMB: 3000×0.038 = 0.45×C + 0.0.0005×S
Solving the equations will give C = 250.3 kg and S = 2749.7 kg
Consider mixing W kg of whole milk with a fat content of fw and S kg of skimmed milk with a fat content of fs to obtain X kg of standardized milk with fx.
There is only one value, pw, for the fraction of W in X and one value, Ps = (1Pw) for the fraction of S in X.
The challenge to the processor is to determine these values.
Using mass balance:
TMB: W + S = X, and W + S = W + S
FMB: Wf_{w} + Sf_{s} = Xf_{x}
Xf_{x} = f_{x}(W + S)
Expanding the equations
Wf_{w} + Sf_{s} = Wf_{x} + Sf_{x}
W(f_{w} – f_{x}) = S(f_{x} – f_{s})
W/S = (f_{x} – f_{s})/(f_{w} – f_{s})
W/(W + S) = (f_{x} – f_{s})/[(f_{x} – f_{s}) + (f_{w} + f_{s})] = P_{w} = W/X
S/(W + S) = (f_{w} – f_{s})/[(f_{x} – f_{s}) + (f_{w} – f_{s})] = P_{s} = S/X
NB: (f_{x} – f_{s}) + (f_{w} – f_{s}) = (f_{w} – f_{s}).
Milk Standardization Using Pearson’s Square Method
Due to the complexity of the equations above, the Pearson’s Square method was developed. The square is drawn and then the immediate constituent content is placed at the center, which in this example is f_{x}.
Place the higher constituent content (f_{w} at the lefthand side top corner and the lower constituent content at the lefthand side bottom corner). Subtract diagonally the higher value from the smaller value as illustrated in the figure below.
Please note that the sum of the values on the righthand side of the square equals the value at the top left minus that at the bottom left side and this value represents X.
The value at the top right represents W and the value at the bottom right represents S. From the values, you can calculate the proportions of W and S, Pw and Ps respectively in X.
The Pearson’s Square is ideal for use when you have to use two streams/substances to standardize one constituent.
Example 3
How much whole milk with 3.9% fat and skimmed milk with 0.04% fat content will you need to produce 2000 kg of standardized milk with 2.5% fat?
Solution:
Using mass balance method:
TMB: W + S = 2000
FMB: 0.039×W + 0.0004×S = 0.025×2000
Solving for W = 1274.6 kg and S = 725.4 kg
Using Pearson’s Square method:
Proportion of the whole milk = 2.46/3.86
Amount of whole milk required = (2.46/3.86) ×2000 = 1274.6 kg
Proportion of skimmed milk = 1.4/3.86
Amount of skimmed milk required = (1.4/3.86) ×2000 = 725.4 kg (or 2000 – 1274.6)
Milk Standardization: Mass Balancing of Two Constituents in Food
Energy in Food
Nutritionally, energy from food is expressed in calories (Cal) or kilocalories (kCal). One calorie is the amount of heat energy required to raise the temperature of one gram of water by 1°C, say from 10°C to 11°C. This heat energy is equivalent to 4.1868 Joules. 1000 kCal is the amount of heat energy needed to raise the temperature of 1 kg of water by 1°C or 4186.8 J (4.1868kJ).
Body energy comes from fats (with 9kCal/g or 37683kJ/kg), carbohydrates (with 3.75kCal/g or 1570kJ/kg), and protein (with 4kCal/g or 1674kJ/kg).
Nutritionists express the nutrient and energy content of food as the amount contained in 100g of the food. Energy to protein ratio, expressed as kilocalories per gram of protein is an important nutritional quality aspect in food. Based on the composition and energy content, the nutritional content of food can be determined and expressed.
Protein Energy Ratio Calculation
The energy to protein ratio of food is used to define the nutritional quality of food. The recommended energy to protein ratio in human diet is approximately 40kCal/g protein (16742kJ/kg protein).
Example 4
Calculate the energy to protein ratio of milk given the composition of the sample is 87.6% water, 3.8% fat, 3.2% protein, 4.6% lactose, and 0.7% ash content.
Solution
Nutritional language: 
Constituent 
Content in 100g 
Energy contribution (kCal/100g) 

Fat 
3.8 
3.8×9 = 34.2 

Protein 
3.2 
3.2×4 = 13.8 

Lactose 
4.6 
4.6×3.75 = 17.25 
Total 
65.25 

Energy to protein ratio 
= 65.25/3.2 = 20.4kCal/g protein 



SI Engineering units: 
Constituent 
Content (fraction) 
Energy contribution (kJ/kg) 

Fat 
0.038 
0.038×1×37683 = 1432 

Protein 
0.032 
0.032×1×16748 = 535.9 

Lactose 
0.046 
0.046×1×15701 = 722.2 
Total 
2690.1 

Energy to protein ratio 
= 2690.1/(0.032×1) = 84065.6kJ/kg protein 
If you have to mix two ingredients in amounts of W and M (g) respectively so that the resulting mix has a stated energy to protein ratio, Rp, then one ingredient must have a greater ratio and the other smaller than the desired. Once you have selected, you can use the following equations to determine the proportions of the ingredients for the mix.
TMB: W + M = P
R_{p} = (We_{w}/Wp_{w}) + (Me_{m}/Mp_{m}) = (Pe_{p}/Pp_{p})
Where, e is the energy content and p is the protein content and the subscripts w and m represent ingredients W and M respectively.
Example 5
Assume you have the following foods and their composition
Ingredient 
Composition per 100g 


Protein content 
Energy content 
Whole milk powder 
27 
530 
Millet flour 
6.7 
330 
Soybean flour 
40 
433 
Using the information above, prepare a recipe for a 500kg mix containing 40kCal per gram of protein.
Solution.
Required energy/protein for the product, R_{p} = 40/1 = 40kCal/g
Energy/protein for whole milk powder, R_{w} = 530/27 = 19.6kCal/g <R_{p}
Energy/protein for millet flour, R_{m} = 330/6.7 = 49.25kCal/g >R_{p}
Energy/protein for soybean flour, R_{s }= 433/40 = 10.8kCal/g <R_{p}
Thus, a mixture of either millet and milk powder or millet and soybean flour can be used.
Using a mixture of W g of whole milk powder and M g of millet,
TMB: W + M = P
R_{p }= (530×W/27W) + (330M/6.7M) = 40/1
530W + 330M = 1080W + 268M
550W = 62M
W/M = 62/550
W/(W + M) = 62/612 = W/P
M/(W + M) = 550/612 = M/P
To prepare 500kg of the mix, amount of whole milk powder required, W = 500×(62/612) = 50.7 kg and the amount of millet flour, M = 500×(550/612) = 449.3 kg.
To verify the credibility of these figures,
R_{p }= (50.7×530 + 449.3×330) / (50.7×27 + 449.3×6.7) = 40
Example 6
Given whole milk powder, soybean flour and millet, prepare a recipe for 350 kg of a mix with 40 kCal per gram protein, which contains equal protein from soybean and whole milk powder.
Solution
From the constitution, it is not hard to deduce that 27g of soybean has the same amount of protein as 40g of whole milk powder and the amount is 27*0.4 = 40×0.27.
To begin with, we need to formulate a mixture of soybean flour and whole milk powder as given above and determine its composition.
Its energy content will be, e_{x} = [(27×433 + 40×530) / (27 + 40)] = 490.9 kCal/100g
Its protein content will be, p_{x} = [(27×40 + 40×27) / (27 + 40)] = 32.2 gprotein/100g
Energy to protein ratio of the mix, p_{x} = 490.9/32.2 = 15.24 < 40 hence you can mix it with millet flour to obtain the required energy to protein ratio.
TMB: X + M = 350
R_{p} = (491×X/32×X) + (330×M/6.7×M) = 40/1
491M + 330M =1280X + 268M
789X = 62M
X/M = 62/789
X/(X + M) = 62/851 = X/P
M/(X + M) = 789/851 = M/P
Amount of mix, X = (62/851)×350 = 25.5 kg
Amount of millet, M = 350 – 25.5 = 324.5 kg
In the mix;
The amount of whole milk powder, W = (25.5×40) / 67 = 15.2 kg
The amount of soybean flour, S = 25.5 – 15.2 = 10.3 kg
The final recipe is thus, 324.5 kg millet, 15.2 kg whole milk powder, and 10.3 kg soybean flour.
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