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Food Process Engineering

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Presentation on theme: "Food Process Engineering"— Presentation transcript:

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

5 Example 1. Determine volumetric and mass flow rate of water ( density = 1000 kg /m^3) , 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 ( F ) o [ T ( C ) – 32 ] 5 9 =

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 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 =

12 Example 4. Covert a moisture content of 85 % wet basis to moisture content dry basis
MC wb = 0.85 MC wb 1 - db = From equation 0.85 1 - MC db = MC db = 5.67 = 567 % db

13 Example 5. A food is initially at moisture content of 90 % dry basis
Example 5. 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

14 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

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

16 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

17 mass of water +mass of dry solids % Wet basis =
Example 6. 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 10. 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. The final product weight is 5 kg. Calculate the amount of water removed. Water removed F = 10 kg of raw product (80 % w.b.) Drying process Product = 2.86 kg (30 % w.b.) 30 % of total water 0.3 x 2.86 = 0.86 kg water 20 % of total Solids 80 % of total water 0.8 x 10 = 8 kg water 0.2 x 10 = 2 kg solid

23 Drying process 8 = P + W 8 = 0.86 +W W = 7.14 kg water Water removed
Mass of water of raw product = 8 kg water Mass of water of final product = kg water 8 = P + W 8 = W W = kg water

24 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

25 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

26 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

27 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

28 Drying process MC = = 0.7619 = 76.19 % w.b. 3.20 1 +
Example 9. The 10 kg of food at a moisture content of 320 % dry basis is dried to 50 % wet basis. Calculate the amount of water removed F = 10 kg of raw product (320 % d.b.) Product (50 % w.b.) Water removed Drying process % d.b. change to % w.b. MC db 1 + wb = 3.20 = = % w.b.

29 Drying process Water removed F = 10 kg of raw product (76.2 % w.b.)
23.8 % of total Solids = 2.38 kg 76.2 % of total water = 7.62 kg Mass of total product = A kg water kg = A A A = 2.38 kg water F = P + W 7.62 = W W = = kg water P = 4.76 kg

30 Example 11. 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

31 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

32 Example 12. 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

33 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

34 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

35 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 =

36 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

37 Relationship between sensible Heat and latent Heat

38 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

39 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

40 Example 13 . 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

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

42 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

43 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

44 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

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


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