2o8 MANAGEMENT OF DAIRY PLANTS 



cream required is equal to figure representing the pounds of 

 standardized cream multiplied by figure representing the differ- 

 ence between test of standardized and test of milk, this product 

 to be divided by figure representing the difference between test 

 of original cream and test of milk. 



Example 6: 



Prepare loo pounds of 20 per cent cream from 30 per cent 

 cream and 4 per cent milk. 



Ci = 100 X = 61 7/13 pounds of 30 per cent cream. 



30-4 



100 — 61 7/13 = 38 6/13 pounds of 4 per cent milk. 



To illustrate the similarity between the two methods suppose 

 Example 6 had been figured by the method of Pearson; the 

 formula would have been as follows: 



n \y 20 — 4 



Ci = 100 X 



(20—4)-!- (30—20) 

 Example 5 by that method would be solved as follows: 



r - Tr,^ V (20 -4)+ (30 -20) 



C2 = 100 X 



20 — 4 



For the purpose of facilitating the work of the ice cream de- 

 partment standardization tables may be used to advantage. 



The same rules and formulas as apply for standardization 

 of cream may also be used for standardization of milk. 



2. Homogenization of Cream. — By homogenization cream 

 is subjected to energetic mechanical treatment. The Gaulin 

 system is the oldest and possibly the best known in the United 

 States. 



The cream enters the machine at a high temperature. It 

 is forced through fine metal capillary tubes and against a conical 

 agate-valve disk at a pressure of from 2,000 pounds to 3,000 

 pounds per square inch. The fat globules in cream thus treated 

 are broken up to such an extent that they are removed with 

 difficulty with a cream separator if the homogenization is 

 properly done. 



For butter making this treatment is of no advantage but 



