1. Food Process Engineering
Introduction to
Food Process Engineering
Dr. Muanmai Apintanapong
Tel:

2. Engineer Units Parameter Symbol Name Unit Symbol length l. metre m
SI Base Units
Parameter
Symbol
Name
Unit Symbol
length l.
metre m
mass
kilogram kg
time t
second s
electric current I
ampere A
thermodynamic
temperature T
kelvin K
amount of substance n
mole
mol
luminous intensity Iv
candela cd
The name Système International d'Unités (International System of Units) with the international abbreviation SI is a single international language of science and technology first introduced in 1960

3. Density
The density of food sample is defined as its mass per unit volume and is expressed as kg /m 3
Density =
Mass ( kg)
Volume ( m ) 3
Mass = Density x Volume
The density is influenced by temperature

4. Volumetric Flow rate Mass Flow rate
Q = Volumetric flow rate A 1 V 2 A 1 V 2
Q = =
m /sec 3
m = mass flow rate o
m = Density x Volume flow rate o
m = o Q
Kg /sec
Equation of Continuity

5. Example 1. Determine volumetric and mass flow rate of water ( density = 1000 kg /m3) , the diameter of pipe is 10 cm.
v = 20 m/s
A = 4 D 2 =
0.1
= m
Q = A V = 20 x = m / sec 3
m = Q = x = kg / sec

6. Temperature
The Kelvin and Celsius scales are related by following function
T ( K ) o
T ( C ) =
The Fahrenheit and Celsius scales are related by following function
T ( C ) o
[ T ( F) – 32 ]
1.8 =

7. Pressure
Pressure is the force on an object that is spread over a surface area. The
equation for pressure is the force divided by the area where the force is applied.
Although this measurement is straightforward when a solid is pushing on a
solid, the case of a solid pushing on a liquid or gas requires that the fluid be
confined in a container. The force can also be created by the weight of an
object. F A
Pressure =

8. Example2. How much 350 Kelvin degrees would be in Fahrenheit degrees
Solution = F
Example 3. How much 60 Fahrenheit degrees
would be in Kelvin degrees
Solution = K

9. System System Close system Volume Mass System Open system
Volumetric flow rate
Mass flow rate
surroundings

10. Moisture Content
Moisture Content expresses the amount of water present
in a moist sample.
Two bases are widely used to express moisture content
Moisture content dry basis MC db
Moisture content wet basis MC wb

11. Food Sample = Food Solids + Food Liquid
Mass of product = Mass of water in food + Mass of dry solids
Mass of dry solid
Food Sample
Mass of water in food

12. Moisture Content , dry basis
mass of water
mass of dry solids
% Dry basis =
kg water
kg dry solids

13. Moisture Content , wet basis
mass of water
mass of water +mass of dry solids
% Wet basis =
mass of water
mass of product =
kg water
kg product

14. MC = 1 - MC MC MC = 1 + wb db wb db Moisture content wet basis
1 - db =
Moisture content dry basis MC db
Moisture content wet basis MC wb MC db
1 + wb =

15. Example 3. Covert a moisture content of 85 % wet basis to moisture content dry basisMC wb =
0.85 MC wb
1 - db =
From equation
0.85
1 - MC db = MC db =
5.67 =
567 % db

16. Example 4. A food is initially at moisture content of 90 % dry basis
Example 4. A food is initially at moisture content of 90 % dry basis . Calculate the moisture content in wet basis MC db =
0.90 MC db
1 + wb =
0.90
1 + MC wb = MC wb =
0.4736 =
47.36 % wb

17. mass of water +mass of dry solids % Wet basis =
Example 5. The 10 kg of food sample at a moisture contents of 75 % wet basis
mass of water
mass of water +mass of dry solids
% Wet basis =
0.75
1.00 =
10 kg of product = 7.5 kg water kg dry solids

18. % Dry basis = (75/25)*100 = 300% = 7.5 kg water + 2.5 kg dry solids
10 kg of product
at 75 % wet basis
75 % of total water
25 % of total Solids
% Dry basis = (75/25)*100 = 300%

19. Material Balance The principle of conservation of mass states that = -
Mass can be neither created nor destroyed. However, its composition can altered from one from to another
Rate of mass entering through the boundary of system
Rate of mass exiting through the boundary of system
Rate of mass Accumulation through the boundary of system = -
Antoine Laurent Lavoisier ( )

20. Unit Operation Wastes Feed in raw product Product
Mass in – Mass Out = Accumulation
F – (W+P) = Accumulation
Assumption: the accumulation = 0
F = W + P

21. Example 6. Unit Operation Wastes 20 kg/hr Feed 100 Kg /hr Product
Assumption : the accumulation = 0
F = W + P
= P
P =
P = Kg / hr

22. Example kg of food at a moisture content of 80 % wet basis is dried to 30 % wet basis. weight is 5 kg. Calculate the amount of water removed and final product weight.
W= Water removed
F = 10 kg of raw product
(80 % w.b.)
Drying process
P =Product = ?? kg
(30 % w.b.)
30 % of total water
20 % of total Solids
80 % of total water
0.8 x 10 = 8 kg water
0.2 x 10 = 2 kg solid

23. Example 8. The 20 kg of food at a moisture content of 80 % wet basis is dried to 50 % wet basis. Calculate the amount of water removed
F = 20 kg of raw product
(80 % w.b.)
Product
(50 % w.b.)
Water removed
Drying process

24. Water = 20 kg product x 0.8 = 16 kg water
20 % of total Solids
80 % of total water
80 % w.b.
Water = 20 kg product x 0.8 = 16 kg water
Solid = 20 kg product x 0.2 = 4 kg dry solid

25. A kg water 16 kg water Drying process 4 kg dry solid 4 kg dry solid A
Water removed
A kg water
4 kg dry solid
16 kg water
4 kg dry solid
F = 20 kg of
product (80 % w.b.)
Product
(50 % w.b.)
Drying process
50 % w.b. = A
A + 4 kg dry solids
0.5 = A
A + 4 kg dry solids
0.5 A +(4 x 0.5) = A
0.5 A = A

26. Total mass of product = 4 +4 = 8 kg
0.5
= 4 kg water
Total mass of product = = 8 kg
F = 20 kg
P = 8 kg
Water removed
Drying process
F = P + W
20 = 8 +W
W = 12 kg water

27. Example 9. Unit Operation Wastes 0.5 % Total solid Feed 100 kg /hr
Product
30 % Total solid
Unit Operation
Assumption: the accumulation = 0
F (0.1) = W(0.005) + P (0.3)
100 kg /hr (0.1) = W(0.005) + P (0.3)
10 kg/hr = W P
P = – W
0.3
Step 2 Total Solid Balances
Equation 2
F = W + P
= W + P
P = W
Equation 1
Step 1 Total mass Balances

28. Step 3 Determine Product rate
Equation 1 = Equation 2
= – W
0.3
100 - W
(0.3)(100) – 0.3 W = 10 – W
= 0.3 W – 0.005W
20 = W
W = 20 / = kg /hr
Step 4 Determine W
P = W
P = – 67.8
P = kg / hr

29. Example 10. 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 10,000 kg/h fresh juice. Assume steady state. Ans Product from evaporator = 2000 kg/h Water removed from evaporator = 8000 kg/h Cut back = 1200 kg/h, final product = 3200 kg/h

30. Example 11. A membrane separation system is used to concentrate the liquid food from 10 % to 30 % total solid (TS). The product is accomplished in two stages, in the first stage, a low total solid liquid stream is obtained. In the second stage, there are two streams, the first one is final product stream with 30% TS and the second is recycled to the first stage. Determine the magnitude of the recycle stream when the recycle contains 2 % TS , the waste stream from first stage contains 0.5 % TS and the stream between stages 1 and 2 contains 25 % TS . The final product is 100 kg/min with 30 % TS.
Feed
10 % TS B
25 % TS R
2 % TS
100 kg/ min of
product
30 % TS
W , 0.5 % TS
first stage
Second stage

31. Total product balance first stage Second stage F = P + W
Feed
10 % TS
100 kg/ min of
product
30 % TS
W , 0.5 % TS
first stage
Second stage
F = P + W
0.1 F = (0.3) W
0.1 ( 100+ W ) = W
W = W
W = kg / min and F = kg/min

32. W , 0.5 % TS
Feed
10 % TS B
25 % TS R
2 % TS
F +R = W +B
R = B
B = R
0.1 F R = W B
0.1 (310.5) R = (210.5) B
R = (100+R)
R = kg / min

33. Energy Balance
The first law of thermodynamic states that energy can be neither created nor destroyed.
Total energy entering the system
Total energy leaving the system
Change in the total energy of system =

34. Sensible Heat Latent Heat m Q = C T Q = m L = L = latent heat o P C
Close System
Open System
Q = m C P T o = C P
= Specific Heat at Constant pressure kJ/ kg K
Latent Heat
Q = m L
L = latent heat

35. Relationship between sensible Heat and latent Heat

36. Relationship between Sensible Heat and Latent Heat
ICE at -50 C
ICE at 0 C
water at 0 C
water at
100 C
vapor at
150 C
Q1 = sensible heat
Q3 = sensible heat
Q5 = sensible heat
Q2 = Latent heat
Of Fusion
Q 4 = Latent heat
Of vaporization

37. Overall View of an Engineering Process
Using a material balance and an energy balance, a food engineering
process can be viewed overall or as a series of units. Each unit is a
unit operation.
Raw
materials
Unit Operation
Further Unit Operation
Previous Unit Operation
By-products
Product
Wastes
Energy
Wastes Energy
Capital
Labor
Control

38. Example 12 . Steam is used for peeling of potatoes in a semi-continuous operation . Steam is supplied at the rate of 4 kg per 100 kg of unpeeled potatoes. The unpeeled potatoes enter system with a temperature of 17 C and the peeled potatoes leave at 35 C . A waste stream from the system leaves at 60 C . The specific heats of unpeeled potato, waste stream and peeled potatoes are 3.7 , 4.2 and 3.5 kJ/ (kg K ) , respectively. If the heat content of steam is 2750 kJ /kg , determine the quantities of the waste stream and the peeled potatoes from the process
P = ?
T = 35 C P
F = 100 kg
T = 17 C F o
W = ?
T = 60 C
H = kJ/kg
S = 4 kg s

39. Solution Select 100 kg of unpeeled potatoes as basis Mass balance
F + S = W + P
= W + P
W = P

40. Energy balance Q = S H Q = F C (T – 0 ) Q = F C (T – 0 ) Q
= 4 kg x 2750 kJ /kg Q s
= S H
= kJ Q F
= F C (T – 0 ) P
= 100 (3.7 kJ/kg K)( 17 -0)
= kJ Q P
= F C (T – 0 )
= P (3.5 kJ/kg K)( 35 -0)
= P kJ Q w
= F C ( T – 0 ) P
= W (4.2 kJ/kg K)( 60 -0)
= 252 W kJ

41. Energy in from System = Energy out from system
Energy balance
Energy in from System = Energy out from system Q w p + s F =
= P W
= P W

42. W = P
Equation of mass balance
= P W
Equation of energy balance
= P ( 104 –P )
= P – 252 P
P = kg
W = 104 – 68.87
= kg

43. Reference:
Prof. Athapol Noomhorm: Introduction toFood Process Engineering