Food Process Engineering Introduction to Food Process Engineering Dr. Muanmai Apintanapong Email: mmeang@gmail.com Tel: 081-844-0799

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

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

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

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 = 0.0078 m Q = A V = 20 x 0.0078 = 0.156 m / sec 3 m = Q = 1000 x 0.156 = 156 kg / sec

Temperature The Kelvin and Celsius scales are related by following function T ( K ) o T ( C ) + 273.15 = The Fahrenheit and Celsius scales are related by following function T ( C ) o [ T ( F) – 32 ] 1.8 =

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 =

Example2. How much 350 Kelvin degrees would be in Fahrenheit degrees Solution = 170.3 F Example 3. How much 60 Fahrenheit degrees would be in Kelvin degrees Solution = 288.7 K

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

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

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

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

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

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 =

Example 3. 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

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

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 + 2.5 kg dry solids

% 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%

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 (1743-1794)

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

Example 6. Unit Operation Wastes 20 kg/hr Feed 100 Kg /hr Product Assumption : the accumulation = 0 F = W + P 100 = 20 + P P = 100 - 20 P = 80 Kg / hr

Example 7. 10 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

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

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

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 + 2 = A

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

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 = 0.005W + 0.3 P P = 10 – 0.005 W 0.3 Step 2 Total Solid Balances Equation 2 F = W + P 100 = W + P P = 100 - W Equation 1 Step 1 Total mass Balances

Step 3 Determine Product rate Equation 1 = Equation 2 = 10 – 0.005 W 0.3 100 - W (0.3)(100) – 0.3 W = 10 – 0.005 W 30 - 10 = 0.3 W – 0.005W 20 = 0.295 W W = 20 / 0.295 = 67.8 kg /hr Step 4 Determine W P = 100 - W P = 100 – 67.8 P = 32.2 kg / hr

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

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

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 = 100 (0.3) + 0.005 W 0.1 ( 100+ W ) = 30 + 0.005 W 10 + 0.1 W = 30 + 0.005 W W = 210 .5 kg / min and F = 310.5 kg/min

W , 0.5 % TS Feed 10 % TS B 25 % TS R 2 % TS F +R = W +B 310.5 + R = 210.5 + B B = 100 +R 0.1 F + 0.02 R = 0.005 W + 0.25 B 0.1 (310.5) + 0.02 R = 0.005 (210.5) + 0.25 B 31.05 + 0.02 R = 1.0525 + 0.25 (100+R) R = 21.73 kg / min

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 =

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

Relationship between sensible Heat and latent Heat

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

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

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 = 2750 kJ/kg S = 4 kg s

Solution Select 100 kg of unpeeled potatoes as basis Mass balance F + S = W + P 100 + 4 = W + P W = 104 - P

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 = 11000 kJ Q F = F C (T – 0 ) P = 100 (3.7 kJ/kg K)( 17 -0) = 6290 kJ Q P = F C (T – 0 ) = P (3.5 kJ/kg K)( 35 -0) = 122.5 P kJ Q w = F C ( T – 0 ) P = W (4.2 kJ/kg K)( 60 -0) = 252 W kJ

Energy in from System = Energy out from system Energy balance Energy in from System = Energy out from system Q w p + s F = 6290 + 11000 = 122.5 P + 252 W 17290 = 122.5 P + 252 W

W = 104 - P Equation of mass balance 17290 = 122.5 P + 252 W Equation of energy balance = 122.5 P + 252 ( 104 –P ) = 122.5 P + 26208 – 252 P P = 68. 87 kg W = 104 – 68.87 = 35.14 kg

Reference: Prof. Athapol Noomhorm: Introduction toFood Process Engineering