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## Lesson 22. STANDARDIZATION OF MILK

Module 6. Common dairy operations

Lesson 22

STANDARDIZATION OF MILK

STANDARDIZATION OF MILK

22.1 Introduction

Many dairy processes require standardization of the chemical composition of milk meant for market purpose or milk products manufacture. Standardizing milk might require control of only one component (usually fat) while allowing the others to vary or control two or more components simultaneously.

22.2 Standardization

It may be defined as the adjustment of one or more of the milk constituents to a nominated level. In market milk industry, this normally involves reducing the butterfat content by addition of skim milk or through the removal of cream.

22.2.1 Objectives

- To comply with the legal requirements for particular milk/dairy products.

- To provide the consumer with a uniform product.

- To ensure economics in production.

22.3 Methods of Calculation

For standardization of milk or cream for product manufacture, usually the proportions of the various ingredients of known composition to be mixed, is required to be estimated. This can be done by:

- Pearson's Square method

- Algebraic equations

Draw a square and place in the centre of it the desired fat percentage. Place at the left hand corners of the square, the fat percentage of the materials to be mixed. Next, subtract the number in the centre from the larger number at the left hand side of the square and place the remainder at the diagonally opposite right hand corners. The number on the right hand side now represents the number of parts of each of the original materials that must be blended to have the desired fat content in resultant mix. The number at the upper right corner refers to the parts of material whose fat test was placed at the upper left corner and the number at the lower right corner refers to the parts of material whose fat test was placed at the lower left corner. If the numbers on the right are added, the sum obtained will represent the parts of the finished product.

Examples

Problem 1

600 kg of cow milk testing 4% fat is to be standardized to toned milk by removing 33% fat cream. Calculate the amount of toned milk.

Solution

A = Fat percentage of cow milk = 4%

B = Fat percentage of cream = 33%

E = Fat percentage of toned milk = 3%

B = Fat percentage of cream = 33%

E = Fat percentage of toned milk = 3%

Calculation

30 kg of cow milk requires removal of 1 kg of cream

So, 1 kg of milk requires removal of 1/30 kg of cream

Therefore, 600 kg milk will require removal of (1/30 x 600) = 20 kg cream

So, the amount of toned milk will be (600 – 20) = 580 kg.

So, 1 kg of milk requires removal of 1/30 kg of cream

Therefore, 600 kg milk will require removal of (1/30 x 600) = 20 kg cream

So, the amount of toned milk will be (600 – 20) = 580 kg.

Proof Sheet

In order to check the accuracy of calculations made, put the calculated values of various raw materials in the below given format and check:

In the product, fat required = 580 * (3/100) = 17.4 kg

Hence proved, because the required value is equal to the calculated value (in the table).

Problem 2

1000 kg of double toned milk (DTM) is to be prepared by mixing whole milk, testing 5.5% fat and skim milk testing 0.2% fat. Calculate the amount of whole milk and skim milk required.

Solution

A = Whole milk fat content = 5.5 %

B = Skim milk fat content = 0.2 %

E = Required percentage of fat in double toned milk = 1.5%

So, C = A – E = 5.5 – 1.5 = 4.0 kg skim milk

D = E - B = 1.5 – 0.2 = 1.3 kg whole milk

B = Skim milk fat content = 0.2 %

E = Required percentage of fat in double toned milk = 1.5%

So, C = A – E = 5.5 – 1.5 = 4.0 kg skim milk

D = E - B = 1.5 – 0.2 = 1.3 kg whole milk

Calculation

DTM quantity = 4 kg + 1.3 kg = 5.3 kg

To prepare 5.3 kg of DTM, whole milk required = 1.3 kg.

Therefore, to prepare 1000 kg of DTM, whole milk required will be = (1.3/5.3) x 1000 = 245.28 kg whole milk

To prepare 5.3 kg of DTM, skim milk required is 4 kg.

So, to prepare 1000 kg of DTM, we require - (4/5.3) x 1000 = 754.72 kg skim milk

The required quantity of whole milk is 245.28 kg and the skim milk is 754.72 kg for preparation of 1000 kg of DTM.

To prepare 5.3 kg of DTM, whole milk required = 1.3 kg.

Therefore, to prepare 1000 kg of DTM, whole milk required will be = (1.3/5.3) x 1000 = 245.28 kg whole milk

To prepare 5.3 kg of DTM, skim milk required is 4 kg.

So, to prepare 1000 kg of DTM, we require - (4/5.3) x 1000 = 754.72 kg skim milk

The required quantity of whole milk is 245.28 kg and the skim milk is 754.72 kg for preparation of 1000 kg of DTM.

Problem 3

200 kg milk testing 7.2% fat is to be standardized to 4.5% fat by addition of skim milk testing 0.1% fat. Calculate the amount of skim milk required?

A = Fat in whole milk = 7.2%

B = Fat in skim milk = 0.1%

C = A - E = 7.2 - 4.5 = 2.7 kg skim milk

D = E - B = 4.5 – 0.1 = 4.4 kg whole milk

E = Fat percentage of standardized milk, 4.5%

B = Fat in skim milk = 0.1%

C = A - E = 7.2 - 4.5 = 2.7 kg skim milk

D = E - B = 4.5 – 0.1 = 4.4 kg whole milk

E = Fat percentage of standardized milk, 4.5%

Solution

Here,

fat % of given milk A = 7.2

Fat % of skim milk B = 0.1%

Fat % of standardized product E = 4.5%

So, C = A - E = 7.2 - 4.5 = 2.7 kg of skim milk

D = E - B = 4.5 – 0.1 = 4.4 kg of whole milk

So, here standardized milk is prepared by mixing whole milk and skim milk

E = C + D = 2.7 + 4.4 = 7.1 kg of standardized milk.

To prepare 7.1 kg of standardized milk, given whole milk required is 4.4 kg.

So, to prepare 200 kg of standardized milk, it requires (4.4/7.1) x 200 = 123.94 kg whole milk.

7.1 kg of standardized milk requires 2.7 kg skim milk.

So, to prepare 200 kg of standardized milk, it requires (2.7/7.1) x 200 = 76.00 kg skim milk.

The required quantity of skim milk is 76.00 kg for preparation of 200 kg of standardized milk testing 4.5% fat.

Fat % of skim milk B = 0.1%

Fat % of standardized product E = 4.5%

So, C = A - E = 7.2 - 4.5 = 2.7 kg of skim milk

D = E - B = 4.5 – 0.1 = 4.4 kg of whole milk

So, here standardized milk is prepared by mixing whole milk and skim milk

E = C + D = 2.7 + 4.4 = 7.1 kg of standardized milk.

To prepare 7.1 kg of standardized milk, given whole milk required is 4.4 kg.

So, to prepare 200 kg of standardized milk, it requires (4.4/7.1) x 200 = 123.94 kg whole milk.

7.1 kg of standardized milk requires 2.7 kg skim milk.

So, to prepare 200 kg of standardized milk, it requires (2.7/7.1) x 200 = 76.00 kg skim milk.

The required quantity of skim milk is 76.00 kg for preparation of 200 kg of standardized milk testing 4.5% fat.

22.3.2 Algebraic equations

In this method, we should know the composition of the products to be mixed, the final product and the quantity of any one product. Mass balance equations are formed and solved.

The formula for determining the quantities of skim milk and raw milk on this basis is as follows:

- Skim milk (kg) = kg standard milk required x (% fat in raw milk - % fat in standard milk)

- Whole milk (kg) = kg standard milk required x (% fat in standard milk - % fat in skim milk)

Prepare 500 kg milk testing 3.0% fat and 8.5% SNF. You are provided with whole milk having 5.0% fat and 9.0% SNF and skim milk powder having 0.5% fat and 96.0% SNF.

Solution

Let the quantity of the whole milk = X kg

Quantity of SMP = Y kg

Quantity of water = Z kg

Fat equation:

X + Y + Z = 500

Now, solving equations (1) and (2) by 5, we get,

SNF required = 500 * (8.5/100) = 42.5 kg

Hence proved, because the required values are equal to the calculated values (in the table).

22.4 Methods of Standardization

There are three methods for standardization. These are batch, continuous and automatic standardization. They all involve the separation of whole milk into skim milk and cream and then proceeding for blending the required quantities only.

22.4.1 Batch standardization

It is a process most commonly used in the dairies. Raw milk is held in a silo and its fat content is evaluated. Some quantity of milk is removed and separated into skim milk and cream. The amount of skim milk or cream required is determined by the calculation (or from charts) and then added to the bulk milk under continuous agitation. The bulk milk is retested to check whether the fat content is as per the desired figure or not. If it is not, further adjustments are made until the batch is standardized correctly. The demerits of batch standardization are the time taken for agitation, testing and final mixing.

22.4.2 Continuous standardization

Continuous standardization employs an inline sampler in association with a testing device, which samples, measures and displays the fat content every 20 seconds. The operator observes the fat content displayed and adjusts the values to blend skim milk or cream into the milk line, before the sampling point, to alter the fat content to the required level.

22.4.3 Automatic standardization

It is an extension of the continuous process. The separator is replaced by a microprocessor/controller unit linked to the sampler/tester system. The microprocessor / controller unit has information about the desired fat content and flow rates of the whole and skim milk. It receives signals from the sampler/tester system and responds by opening or closing a valve, which regulates the amount of skim milk added to the whole milk. The merits of this automatic process are time and labour savings and ensure more accurate standardization than other methods. Standardization depends on correct sampling, accurate testing of fat content, efficient separation and the correct amount of skim milk or cream needed.

22.5 Tri-process Machine

Tri-process machine is designed to clarify, separate, standardize milk in a single unit. The general construction is similar to that of standard cream separator. The tri-process separator has external valves in the discharge lines of cream and skim milk. A precise needle valve is fixed in the outlet for cream, which controls the cream flow rate. There is a bypass line connected from the cream discharge line to the skim milk discharge line. This bypass line has a needle valve which would control the flow of the cream coming in the bypass line. A cream meter is installed in the cream outlet line.

For standardization of milk to the desired fat content, the needle valve in the bypass line is adjusted to such a position that the bypass cream when mixed with the skim milk would result in the desired fat % in the standardized milk.

Last modified: Tuesday, 6 November 2012, 6:56 AM