# Problem 3

Desired: 100 kg mix containing 18% fat, 9.5% SNF, 15% sucrose, 0.4% stabilizer, 1% frozen egg yolk.

On hand: Cream 30% fat, milk 3.5% fat, skim milk powder 97% solids, sucrose, stabilizer, and egg yolk.

The solution to this problem will be shown by simultaneous equations, since there are three sources of milk SNF, three sources of water, and two source of fat, which require three equations, and by the serum point method. Both produce the same results. Follow whichever method you prefer. Computer programs exist that solve simultaneous equations.

Solution via the algebraic method:

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

Stabilizer: 100 kg mix x 0.4 kg stabilizer/100 kg mix = 0.4 kg stabilizer

Egg yolk: 100 kg mix x 1 kg egg yolk/100 kg mix = 1 kg egg yolk

Now, let x = skim powder, y = milk, z = cream.

MASS BALANCE All the components add up to 100 kg

x + y + z = 100 - (15 + 0.4 + 1) (1)

MSNF BALANCE Equal to 9.5% of the mix and coming from the milk, the skim powder, and the cream; assume 9% in the skim portion of the milk and cream so that the msnf of the milk = .09 x (100 - 3.5) and of the cream = .09 x (100-30)

0.97 x + 0.08685 y + 0.063 z = .095 (100) (2)

FAT BALANCE Equal to 18% of the mix and coming from the milk and cream

.035 y + .3 z = .18 (100) (3)

Solution via the serum point method:

1. Find the amount of skim milk powder required by the following formula:

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

Substituting we have,

(9.5 - ( 65.6 x .09 ))/(97-9) x 100 = 3.596/88 x 100 = 4.08 kg powder

2. Amount of sucrose required is 15.0 kg.

3. Amount of stabilizer required is .4 kg.

4. Amount of egg required is 1.0 kg.

5. Find weight of milk and cream needed.

Materials supplied so far are 4.08 kg powder, 15 kg sucrose, 0.4 kg stabilizer, and 1 kg egg yolk, a total of 20.48 kg. 100 - 20.48 = 79.52 kg milk and cream needed.

6. Find the amount of cream by following formula:

((kg fat needed - (kg cream and milk needed x (% fat in milk/100)))/(% fat in cream - % fat in milk)) x 100

substituting we have,

(18 - ( 79.52 x 3.5/100 ))/(30-3.5) x 100 = 15.217/26.5 x 100 = 57.42 kg cream.

7. Amount of milk needed = 79.52 - 57.42 = 22.10 kg of milk.

PROOF:

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

Cream | 57.42 | 17.23 | 3.62 | 20.85 |

Milk | 22.10 | 0.77 | 1.92 | 2.69 |

Skim milk powder | 4.08 | -- | 3.96 | 3.96 |

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

Stabilizer | 0.40 | -- | -- | 0.40 |

Egg Yolk | 1.00 | -- | -- | 0.50 |

Totals | 100.00 | 18.00 | 9.50 | 43.40 |

Note: 18% fat + 9.5% snf + 15% sucrose + 0.40% stab. + 0.50% egg yolk solids (half the egg yolk) = 43.4% TS

If you wanted to make 3000 kg (for example) instead of 100 kg, you could multiply all of the numbers above by 30, or you could set up the equation to solve directly for 3000 kg, as shown below.

Solution via the serum point method for 3000 kg:

1. Find the amount of skim milk powder required by the following formula:

(MSNF needed - (serum of mix x .09))/(% snf in powder - 9) x 100 = skim powder

MSNF needed = 3000 x 9.5% = 285 kg; Serum of the mix = 3000 - 540 (fat) - 450 (sugar) - 12 (stab.) - 30 (egg yolk) = 1968 kg. Substituting we have,

(285 - ( 1968 x .09 ))/(97-9) x 100 = 107.88/88 x 100 = 122.59 kg powder

2. Amount of sucrose required is 3000 x 15% = 450.0 kg.

3. Amount of stabilizer required is 3000 x.4% = 12.0 kg.

4. Amount of egg required is 3000 x 1.0% = 30 kg.

5. Find weight of milk and cream needed.

Materials supplied so far are 122.59 kg powder, 450.0 kg sucrose, 12.0 kg stabilizer, and 30.0 kg egg yolk, a total of 614.59 kg. 3000 - 614.59 = 2385.41 kg milk and cream needed.

6. Find the amount of cream by following formula:

((kg fat needed - (kg cream and milk needed x (% fat in milk/100)))/(% fat in cream - % fat in milk)) x 100

substituting we have,

(540 - ( 2385.41 x 3.5/100 ))/(30-3.5) x 100 = 456.51/26.5 x 100 = 1722.68 kg cream.

7. Amount of milk needed = 2385.41 - 1722.68 = 662.73 kg of milk.

PROOF:

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

Cream | 1722.68 | 516.8 | 108.53 | 625.33 |

Milk | 662.73 | 23.2 | 57.56 | 80.76 |

Skim milk powder | 122.59 | -- | 118.91 | 118.91 |

Sucrose | 450.00 | -- | -- | 450.00 |

Stabilizer | 12.0 | -- | -- | 12.0 |

Egg Yolk | 30.0 | -- | -- | 15.0 |

Totals | 3000.0 | 540.0 | 285.0 | 1302.0 |

Note: 540/3000 = 18% fat; 285/3000 = 9.5% SNF; 1302/3000 = 43.4% Total solids