food engineering lecture covering units and dimension

1. Sample Problem #1…..continuation
Step 2: State your assumptions: dry sugar is composed of 100% sugar.
Step 3: Write the total and component mass balances in the envelope around the process:
a) Total mass balance
100 + S2 = S3
B) Component mass balance: Soluble solids mass balance
(0.20 X 100) + (S2 X 1) = (0:50 X S3)
20 + S = 0.5S
Solving eqns (3.1) and (3.2) simultaneously, find S2=60 kg and S3=160 kg.
Therefore 60 kg of dry sugar per 100 kg of feed must be added to increase its
concentration from 20% to 50%.

2. FTECH 3310: Food
Engineering
ASSOC. PROF. JOANN DAVID DAR
DEPT. OF FOOD SCIENCE AND TECHNOLOGY, CHSI, CLSU
PGS FOOD SAFETY, QA SYSTEMS AND RISK ANALYSIS, UNIVERSITEIT GENT -BELGIUM
M.SC. FOOD SCIENCE AND TECHNOLOGY, ENSIA -SIARC - FRANCE
BS FOOD TECHNOLOGY, CLSU -PHILIPPINES

3. Unit 1: Review of Mathematical
Principles in Food Engineering

4. Unit 2: Units and
Dimensions

5. Sample Problem #1: Mixing
Example 3.1 How much dry sugar must be added in 100 kg of aqueous sugar solution in
order to increase its concentration from 20% to 50%?
S1
100 KG
20%
S2
50%
S3
100%
MIXING

6. Sample Problem #1...continuation.. Mixing
Step 2: State your assumptions: dry sugar is composed of 100% sugar.
Step 3: Write the total and component mass balances in the envelope around the process:
a) Total mass balance
100 + S2 = S3
B) Component mass balance: Soluble solids mass balance
(0.20 X 100) + (S2 X 1) = (0:50 X S3)
20 + S = 0.5S
Solving eqns (3.1) and (3.2) simultaneously, find S2=60 kg and S3=160 kg.
Therefore 60 kg of dry sugar per 100 kg of feed must be added to increase its
concentration from 20% to 50%.

7. Sample Problem #2: Evaporation and Mixing
Example 3.2 Fresh orange juice with 12% soluble solids content is concentrated to 60% in a
multiple effect evaporator. To improve the quality of the final product the concentrated juice is
mixed with an amount of fresh juice (cut back) so that the concentration of the mixture is 42%.
Calculate how much water per hour must be evaporated in the evaporator, how much fresh juice
per hour must be added back and how much final product will be produced if the inlet feed flow
rate is 10000 kg/h fresh juice. Assume steady state.
10,000
kg/h
12%
W
42% S3
100%
Evaporation
(I)
Mixing
X
60%
12% F
Y
I
II

8. Sample Problem #2…..continuation
Step 2: Write the total and component mass balances in envelopes I and II:
a) Total mass balance - in envelope I
10,000 = W+X
b) Component mass balance: Soluble solids mass balance in envelope I
0.12 X 10,000 = 0.60 X
c) Total mass balance in envelope II
X + F = Y
d) Component mass balance: Soluble solids mass balance in envelope II
(0:60 * X) + (0:12*F) =(0.42 * Y)

9. Sample Problem #2…..continuation
From equation (3.4) find X=2000 kg/h. Substituting X in eqn (3.3) and find W=8000 kg/h. Solve
eqns (iii) and (iv) simultaneously and Substitute X in eqn (3.3) and find=1200 kg/h and Y=3200
kg/h.
Therefore 8000 kg/h of water will be evaporated, 1200 kg/h of fresh juice will be
added back and 3200 kg/h of concentrated orange juice with 42% soluble solids
will be produced.

10. Sample Problem #3: Concentration
Exercise 3.3 A fruit juice with 1,000 kg/h with 10% solids is freeze-concentrated to 40% solids.
The dilute juice is fed to a freezer where the ice crystals are formed and then the slush is
separated in a centrifugal separator into ice crystals and concentrated juice. An amount of 500
kg/h of liquid is recycled from the separator to the freezer. Calculate the amount of ice that is
removed in the separator and the amount of concentrated juice produced. Assume steady state.
Solution: Step 1 - Draw the process diagram:
1,000
kg/h
10% 42%
Freezing
(I)
Separation
J
II
Ice
I

11. Sample Problem #3: Concentration…..continuation
Step 2: Write the total and component mass balances in the envelope around the process:
a) Total mass balance
1000 = I + J (Equation 3:7)
b) Component mass balance : Soluble solids mass balance
0:10 X 1000 = 0.40 X J (Equation 3.8)
From eqn (3.8), find J=250 kg/h and then from eqn (3.7) find I=750 kg/h.
Comment: Notice that the recycle stream does not affect the result. Only the streams that cut the
envelope take part in the mass balance.