# Problem 2

Desired : 100 kg mix @ 13% fat, 11% MSNF, 15% sucrose, 0.5% stabilizer, 0.15% emulsifier

On hand: Cream @ 40% fat, 5.4% msnf; skimmilk @ 9% msnf; skimmilk powder @ 97% msnf; sugar; stabilizer; emulsifier.

Solution: (Note: only one source of fat, sugar, stabilizer, and emulsifier, but two sources of serum solids)

Cream:

100 kg mix x 13 kg fat/100 kg mix x 100 kg cream/40 kg fat = 32.5 kg cream

Sucrose:

100 kg mix x 15 kg sucrose/100 kg mix = 15 kg sucrose

Stabilizer:

100 kg mix x 0.5 kg stabilizer/100 kg mix = 0.5 kg stabilizer

Emulsifier:

100 kg mix x 0.15 kg emulsifier/100 kg mix = 0.15 kg emulsifier

Algebraic method:

Skim milk and Skim powder, Note: two sources of the MSNF

Now, let x = skim powder, y = skim milk

MASS BALANCE (All the components add up to 100 kg)

x + y = 100 - (32.5 + 15 + 0.5 + 0.15) (1)

MSNF BALANCE (Equal to 11% of the mix and coming from the skim milk, the skim powder, and the cream)

0.97 x + 0.09 y = .11(100) - (.054 x 32.5) (2)

x + y = 51.85 so y = 51.85 - x from (1)

.97 x + .09 y = 9.245 from (2)

.97 x + .09 (51.85 - x) = 9.245 substituting

.97 x - .09 x + 4.67 = 9.245

.88 x = 4.58

x = 5.20 kg skim powder

y = 46.65 kg skim milk

The above shows the solution of a 2-unknown simultaneous equation. Likewise, if there were 3 unknowns, e.g., fat, msnf, and the total weight, then three equations could be written, one for each of fat, msnf, and weight. However, the above problem could also be solved with the Serum Point method, and the example and that solution along with the derivation of the equations follows. The Serum Point calculation assumes 9% msnf in skimmilk and the skim portion of all dairy ingredients. It then solves the calculation beginning with the most concentrated source of serum solids first.

Solution via the serum point method:

1. Amount of powdered skim milk needed is found by the following formula:

(SNF needed - (serum of mix X .09))/(% SNF in powder - 9) X 100 = kg skim powder

This is a generalized equation solved from a mas balance, that works in all situations where milk powder is the most concentrated source of serum solids, and all of the serum products are milk products (i.e., there is no water used in the recipe). An assumption is that the serum fraction of all dairy ingredients contains 9% solids-not-fat, e.g., 40% cream contains 60% skim (100-40), which contains 9% snf, so the snf content of the cream is .60 x .09 = 5.4%

The serum of the mix is found by adding the desired percentages of fat, sucrose, stabilizer and emulsifier together and subtracting from 100. In the present problem then,

100 - (13 + 15 + 0.5 + 0.15) = 71.35 kg serum.

Substituting in the formula we have:

(11 - (71.35 x .09))/(97 - 9) x 100 = 4.58/88 x 100 = 5.20 kg skim powder

2. The weight of cream will be 13 kg x 100 kg cream/40 kg fat = 32.5 kg cream

3. The sucrose will be 15 kg/ 100 kg mix.

4. The stabilizer will be 0.5 kg/ 100 kg mix.

5. The emulsifier will be 0.15 kg/ 100 kg mix.

6. The weight of mix supplied so far is,

Cream 32.50 kg

Skim powder 5.20 kg

Sucrose 15.00 kg

Stabilizer .50 kg

Emulsifier .15 kg

Total 53.35 kg

The skim milk needed therefore is 100 - 53.35 = 46.65 kg.

PROOF:

Ingredient | Total wt. (kg) | Wt. of Fat (kg) | Wt. of SNF (kg) | Wt. of Total Solids (kg) |
---|---|---|---|---|

Cream | 32.50 | 13.00 | 1.75 | 14.75 |

Skim milk | 46.65 | -- | 4.20 | 4.20 |

Skim milk powder | 5.20 | -- | 5.04 | 5.04 |

Sucrose | 15.00 | -- | -- | 15.00 |

Stabilizer | 0.50 | -- | -- | 0.50 |

Emulsifier | 0.15 | -- | -- | 0.15 |

Totals | 100.00 | 13.00 | 11.00 | 39.64 |

Note: 13% fat + 11% snf + 15% sucrose + 0.50% stab. + 0.15% emul. = 39.65% TS

DERIVATION OF THE SERUM POINT EQUATIONS: Let's resolve problem 2 again using simultaneous equations in a general way to show where the serum point equations come from.

On hand: cream @ 40% fat

(supplies fat, water, and serum solids, therefore can be thought of as a mixture of fat and skim milk)

skim milk @ 9% solids not fat

(supplies water and serum solids)

skim milk powder @ 97% solids not fat

(supplies water and serum solids)

sucrose

stabilizer

emulsifier

Solution

- Only one source of fat, sucrose, stabilizer, and emulsifier

kg fat = 100 kg mix x 13 kg fat/100 kg mix = 13 kg fat (The explanation for this assumption becomes clearer in a moment!)

kg sucrose = 100 kg mix x 15 kg sucrose/100 kg mix = 15 kg sucrose

kg stabilizer = 100 kg mix x 0.5 kg stab./100 kg mix = 0.5 kg stabilizer

kg emulsifier = 100 kg mix x 0.15 kg emul./100 kg mix = 0.15 kg emulsifier

- Two sources of serum solids

Let X = skim powder (kg)

Let Y = skim milk (kg) + skim milk in cream (kg)

MASS BALANCE X + Y = Total mix - components already added

X + Y = 100 - (13 + 15 + 0.5 + 0.15), (the "Serum of the Mix")

X + Y = 71.35

(so Y = 71.35 - X)

MSNF BALANCE 0.97X + 0.09Y = (0.11 x 100)

"Serum "Serum "Serum fraction

fraction fraction in mix"

in powder" in skim"

0.97 X + 0.09 (71.35 - X) = 11

0.97 X + (0.09 x 71.35) - 0.09 X = 11

0.97 X - 0.09 X = 11 - (0.09 x 71.35)

X = 11 - (.09 x 71.35)/ 0.97 - 0.09

Which is equal to:

kg skim powder = S.S. needed - (0.09 x serum of mix) x 100 % S.S. in powder - 9

X = 4.58/0.88 = 5.20 kg powder

kg cream = 13 kg fat x 100 kg cream/40 kg fat = 32.5 kg cream

kg skim = 100 - 32.5 - 15 - 0.5 - 0.15 - 5.2 = 46.65 kg