Question: Using mass-balance solve following problems (include proof table in answer) Note: unless specified otherwise, assume that skim milk, condensed skim milk and skim milk powder
Using mass-balance solve following problems (include proof table in answer)
Note: unless specified otherwise, assume that skim milk, condensed skim milk and skim milk powder have 0% fat, and there is 9% milk snf in skim or in the skim fraction of liquid milk/cream and butter products.
a) Desired : 800 kg of ice cream mix @ 11% fat, 10.5% msnf, 11% sucrose, 5% corn syrupsolids, 0.25% stabilizer. Available : Butter @ 82% fat, skim milk, condensed skim milk @ 32% msnf, sucrose,corn syrup solids, stabilizer.
EXAMPLE PROBLEM 1 Desired : 100 kg mix @ 13% fat, 11% MSNF, 15% sucrose, 0.5% stabilizer, 0.15% emulsifier On hand: Cream @ 40% fat, 5.4% manf; skimmilk @ 9% manf; skimmilk powder @ 97% manf; sugar; stabilizer; emulsifier. Solution: (Note: only one source of fat, sugar, stabilizer, and emulsifier, but two sources of solids-not-fat) Cream 100 kg mix x 13 kg fat x 100 kg cream = 32.5 kg cream 100 kg mix 40 kg fat Sucrose 100 kg mix x 15 kg sucrose = 15 kg sucrose 100 kg mix Stabilizer 100 kg mix x 0.5 kg stabilizer = 0.5 kg stabilizer 100 kg mix Emulsifier 100 kg mix x 0.15 kg emulsifier = 0.15 kg 100 kg mixSkim milk and Skim powder, Note: two sources of 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 milkPROOF: Ingredient . . . Wt. of Total Solids (kg) Cream . . . 14.?5 Skim milk . . 4.20 Skim milk powder Note: 13% fat + 11% snf + 15% sucrose + 0.50% stab. + 0.15% emul. = 39.65% TS
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