dr. muanmai apintanapong email: mmeang@gmail.com mmeang@gmail.com tel: 081-844-0799

Dr. Muanmai ApintanapongDr. Muanmai Apintanapong

Email: Email: mmeang@gmail.com

Tel: 081-844-0799 Tel: 081-844-0799

Engineer Units

Parameter Symbol NameUnit Symb

ol

length l. metre m

mass m kilogram kg

time t second s

electric current I ampere A

thermodynamictemperature

T kelvin K

amount of substance n mole mol

luminous intensity Iv candela cd

SI Base Units

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

Density =Mass ( kg)

Volume ( m ) 3

Mass = Density x Volume

The density of food sample is defined as its mass per unit volume and is expressed as kg /m3

The density is influenced by temperature

Volumetric Flow rate Mass Flow rate Q = Volumetric flow rate

m = mass flow rateo

m = Density x Volume flow rateo

A1V1

A2V2

A1V1

A2V2Q = = m /sec3

m =o

Q Kg /sec

Example 1.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

D2 =

4 0.1

2= 0.0078 m

2

Q = A V = 20 x 0.0078 = 0.156 m / sec3

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 ) o + 273.15=

The Fahrenheit and Celsius scales are related by following function

T ( F ) o [ T ( C ) – 32 ] o 5

9=

PressurePressure 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.

FA

Pressure =

Example2. How much 350 Kelvin degrees would be in Fahrenheit degrees

Example 3. How much 60 Fahrenheit degrees would be in Kelvin degrees

Solution = 288.7 K

Solution = 170.3 F

System

System

Open system

Volumetric flow rate

Mass flow rate

System

Close system

Volume

Mass

surroundings

Moisture ContentMoisture Content expresses the amount of water present

in a moist sample.Two bases are widely used to express moisture content

Moisture content dry basis

MCdb

Moisture content wet basis

MCwb

Moisture content dry basis

MCdb

Moisture content wet basis

MCwb

MCdb

MCdb1 +

MCwb =

MCwb

MCwb1 -

MCdb =

Example 4.Example 4. Covert a moisture content of 85 % wet basis to moisture content dry basis

MCwb

MCwb1 -

MCdb =

0.851 -MC

db =0.85

MCdb = 5.67

= 567 % db

MCwb = 0.85

From equation

Example 5.Example 5. A food is initially at moisture content of 90 % dry basis . Calculate the moisture content in wet basis

MCdb

MCdb1 +

MCwb =

0.901 +MC

wb =0.90

MCwb = 0.4736

= 47.36 % wb

MCdb = 0.90

Food Sample =

Mass of product = Mass of water in food + Mass of dry solids

+ Food Liquid Food Solids

Mass of dry solid

Food Sample

Mass of water in food

Moisture Content , dry basis

kg water

kg dry solids

mass of water

mass of dry solids% Dry basis =

Moisture Content , wet basis

mass of water

mass of water +mass of dry solids

kg water

kg product

% Wet basis =

mass of water

mass of product=

Example 6.Example 6. The 10 kg of food sample at a moisture contents of 75 % wet basis

10 kg of product = 7.5 kg water + 2.5 kg dry solids

mass of water

mass of water +mass of dry solids% Wet basis =

0.75

1.00=

= 7.5 kg water + 2.5 kg dry solids10 kg of productat 75 % wet basis

25 % of total Solids 75 % of total water

% Dry basis = (75/25)*100 = 300%

Material BalanceThe principle of conservation of mass states

that

Mass can be neither created nor destroyed. However, its composition can altered from one from to another

Antoine Laurent Lavoisier (1743-1794)

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

=-

Unit Operation

Wastes

Mass in – Mass Out = Accumulation

F – (W+P) = Accumulation

Assumption: the accumulation = 0

F = W + P

Feed in raw product Product

Unit Operation

Wastes 20 kg/hr

Assumption : the accumulation = 0

Feed 100 Kg /hr Product

F = W + P

100 = 20 + P

P = 100 - 20

P = 80 Kg / hr

Example 10.Example 10.

Example 7. 10 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.

F = 10 kg of raw product

(80 % w.b.)

Product = 2.86 kg

(30 % w.b.)

Water removed

Drying process

20 % of total Solids 80 % of total water

0.8 x 10 = 8 kg water 0.2 x 10 = 2 kg solid

30 % of total water

0.3 x 2.86 = 0.86 kg water

Mass of water of

raw product

= 8 kg water

Water removed

Drying process

Mass of water of

final product

= 0.86 kg water

8 = P + W

8 = 0.86 +W

W = 7.14 kg water

Example 8.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

Solid = 20 kg product x 0.2 = 4 kg dry solid

20 % of total Solids 80 % of total water

80 % w.b.80 % w.b.

F = 20 kg of

product (80 % w.b.)

Product

(50 % w.b.)

Water removed

Drying process

16 kg water

4 kg dry solid

A kg water

4 kg dry solid

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

0.5 A = 2

A = 20.5

= 4 kg water

F = 20 kg P = 8 kg

Water removed

Drying process

Total mass of product = 4 +4 = 8 kg

F = P + W

20 = 8 +W

W = 12 kg water

Example 9.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

MCdb1 +

MCwb =

3.201 +=

3.20= 0.7619

= 76.19 % w.b.

F = 10 kg of raw product

(76.2 % w.b.)Product

(50 % w.b.)

Water removed

Drying process

23.8 % of total Solids = 2.38 kg

76.2 % of total water = 7.62 kg Mass of total product

= A kg water + 2.38 kg

0.5 = A

A + 2.38

A = 2.38 kg water

F = P + W

7.62 = 2.38 + W

W = 7.62 -2.38 = 5.24 kg water

P = 4.76 kg

Unit Operation

Wastes0.5 % Total solid

Assumption: the accumulation = 0

Feed 100 kg /hr

10 % Total solid

Product

30 % Total solid

F = W + P

100 = W + P

P = 100 - W

Equation 1

Step 1 Total mass Balances

Example 11.Example 11.

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

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 3 Determine Product rate

Step 4 Determine W

P = 100 - W P = 100 – 67.8

P = 32.2 kg / hr

Example 12Example 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

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

Total product balance

Feed

10 % TS

B

25 % TS

R

2 % TS

W , 0.5 % 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

Total energy entering the system

Total energy leaving the system

Change in the total energy of system

=

The first law of thermodynamic states that energy can be neither created nor destroyed.

Sensible Heat

CP = Specific Heat at Constant pressure kJ/ kg K

Close System Open System

Q = m C P T m C P To

=

Latent HeatQ = 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 Cwater at

100 C

vapor 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 engineeringprocess can be viewed overall or as a series of units. Each unit is aunit operation.

Raw

materials

Unit Operation

Further Unit Operation

Previous Unit Operation

By-products By-products

Product Product

WastesWastes

EnergyEnergy

Wastes Energy

Capital

Energy

Labor

Control

Example 13Example 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 CP

F = 100 kg

T = 17 CF

o

W = ?

T = 60 CF

o

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

= 4 kg x 2750 kJ /kgQ s = S Hs= 11000 kJ

Q P = F C (T – 0 )P P

= P (3.5 kJ/kg K)( 35 -0)

= 122.5 P kJ

Q w = F C ( T – 0 )P w

= W (4.2 kJ/kg K)( 60 -0)

= 252 W kJ

Q F = F C (T – 0 )P F

= 100 (3.7 kJ/kg K)( 17 -0)

= 6290 kJ

Energy balance

Energy in from System = Energy out from system

Q wQ p +Q sQ 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

17290 = 122.5 P + 252 ( 104 –P )

17291 = 122.5 P + 26208 – 252 P

P = 68. 87 kg

W = 104 – 68.87

= 35.14 kg

Reference: