28/09/57

1

Fluid Flow in Food Fluid Flow in Food Fluid Flow in Food Fluid Flow in Food ProcessingProcessingProcessingProcessing

Properties of Non – Newtonian Liquids

� Time independent - respond immediately (flow) when stress is applied

� Relationship of stress and strain nonlinear

Chapter 2 Fluid Flow in Food Processing

0

n

dy

duK σ+

=σ (2.30)

Tomato ketchup require yield stress

After yield stress

flow as Newtonian -> Bingham liquids

Flow as shear thinning -> plastic

28/09/57

2

Fluid K n Typical examples

Herschel - Bulkley > 0 0 < n < > 0 minced fish paste, raisin paste

Newtonian > 0 1 0 water, fruit juice, honey, milk,

vegetable oil

Shear – thinning (pseudo plastic) > 0 0 < n < 1 0 applesauce, banana puree, orange

juice concentrate

Shear – thickening > 0 1 < n < 0 some types of honey, 40% raw

corn starch solution

Bingham Plastic > 0 1 > 0 toothpaste, tomato paste

TABLE 2.2 Values of Coefficients in Herschel – Bulkley Fluid Model

0σ

∞

∞

4. Handling Systems for Newtonian Liquids

• Flow Characteristics

Mass flow rate

Volumetric flow rate

Steady state flow

_.

uρAm =_.

uAV =

(kg/s)

(m3/s)

2

.

1

.

mm =

2

.

1

.

VV =

Ex. Water at 20 C is pumped through

4 cm φ pipe at 1.5 kg/s.

1. Determine u

2. New velocity when φ change to 8 cm

1. u = 1.195 m/s

2. u = 0.299 m/s

Constant m increase pipe φ 2 times

Decrease u 4 times

µρ

==uD

forces viscous

forces inertial N

Re(2.33)

Reynolds Number

dimensionless

As mass flow rate increase inertial forces (force of momentum) increase

But resisted by viscous forces

When viscous force is dominant NRe small

< 2100 -> laminar flow

> 10,000 -> turbulence flow

28/09/57

3

Larminar flow

( )22 r- RL4

P uµ∆

=

Used to calculate u at various r

• Laminar Flow

• Turbulent Flow

u = 0.8 umax

Ex. Calculate minimum mass flow rate required to establish fully developed turbulent flow of a cleaning solution

Given density = 1050 kg/m3

Diameter = 5 cm = 0.05 m

µ = 995 x 10-6 Pa.s

NRe = 10,000m = 0.391 kg/s

5. Mechanical Energy Balance

1. Potential Energy

2. Kinetic Energy

( )12

Z - Zg PE =∆

α=∆

2

u - u KE

1

2

2

2

(2.55)

(2.56)

(J/kg)

α= 0.5 laminar

α = 1.0 turbulent

ρ=

ρ∆ P - P

P

12

3. Pressure Energy

4. Friction Energy

ρ∆

=P

Ef

(2.58)

(2.57)

28/09/57

4

Friction energy

For straight pipe

D

Lu2

ρ

∆PE

2_

ff f==

For sudden contraction

f = friction factor

2

uK

ρ

∆P_2

ff =

)D

D0.4(1.25K

21

22

f −=

< 0.715

)D

D0.75(1K

21

22

f −=

> 0.715

For sudden increase

2

2

121f )]

A

A([1

2

u

ρ

∆P−=

For fittings (elbows, tees, valves)

Equivalent length of straight pipe, Le

Added to actual L

f - PE - KE - dV P W ∫ ∆∆= (2.61)

fpE

P KE PE E +

ρ∆

+∆+∆= (2.62)

Bernoulli equation

( )p

.

Em =Power (2.64)

f

2

22

2p

1

21

1E

p

2

u gZ E

p

2

u Zg +

ρ+

α+=+

ρ+

α+

(2.63)

Example

Apple juice is pumped from open tank through 1 in. pipe to a 2nd tank. Mass flow rate is 1 kg/s through 30 m pipe with 2 x 90 elbows, 1 angle valve. Compute power requirements of the pump.

Given µ = 2.1 x 10-3 Pa.s

ρ = 997.1 kg/m3

D = 0.02291 m (Table 2.4 p. 75)

M = 1 kg/s Z1 = 3 m, Z2 = 12 m

L = 30 m

f = 0.006

28/09/57

5

90 standard elbows Le/D = 32

Angle valve Le/D = 170

Find u from u = m/PA, NRe

u = 2.433 m/s

Sudden contraction

Kf = 0.4(1.25 – 0) = 0.5

J/kg 1.482

(2.433)0.5

ρ

∆P 2f ==

2 Std elbow Le/D = 32

Le = 32 (0.02291) x 2 = 1.466 m

Angle valve Le/D = 170

Le = 170 (0.02291) = 3.895 m

3.895m)1.466m(30m0.02291

(2.433)2(0.006)

D

Lu2f

ρ

∆P 22_

++==

= 109.63 J/kg

From Bernoulli equation

1.48)(109.632

(2.433)3)9.81(12Ep

2

+++−=

= 202.36 J/kg

Power = Ep . M = 202.36 J/kg x 1 kg/s

= 202.36 J/s (W)

With 60 % efficiency, power = 202.36/0.6 = 337 W

6. Pump Selection and Performance Evaluation

1. Characteristic Diagram of Pumps

2. Net Positive Suction Head

fsvpaA h- h- h- h PSHN = (2.65)

28/09/57

6

3. Affinity Laws

( )1212

N/N V V = (2.64)

( )3

1212N/N h h =

( )3

1212N/NP P ′=′

(2.65)

(2.66)

7. Flow Measurement Manometer

( )12

21 ZZg P - P

−=ρ

(2.69)

m

m

caba Zg P - P

P - P

∆=ρ

=ρ (2.67)

m

m

caba Zg P - P

P - P

∆=ρ

=ρ

(2.70)

( )mmcb

Zg - P - P ∆ρρ= (2.71)

m

mcbZg 1 -

P - P∆

ρρ

=ρ

(2.72)

ρρ

∆= 1 - Z Z - Z m

m12 (2.73)

28/09/57

7

1. The Pitot Tube

ρ+=

ρ2

2

21P

2

u

P (2.74)

( ) 21

21

2

P - P2C u

ρ

= (2.75)

( )2

1

mm2Z -

g2C u

∆ρρ

ρ= (2.76)

Connected to manometer

2. The Orifice Meter

ρ+=

ρ+ 22

2

11

2P

2

u

P

2

u(2.77)

22

1

2

22

1

21 u

D

D u

A

A u == (2.78)

ρ+

=

ρ+ 1

2

2

4

1

22

2

2 P

2

u

D

D

P

2

u(2.79)

( )( )( )

21

4

1

4

2

212

DD - 1

P - P2C u

ρ= (2.80)

( )[ ]

21

4

12

m

m

2

DD - 1

Z 1 - 2g

C u

∆

ρρ

= (2.81)

Connected to manometer

3. The Venturi Meter

( )

21

4

12

m

m

2

DD - 1

Z 1 - g2

C u

∆

ρρ

=

28/09/57

8

4 Variable – Area Meters

5 Other Measurement Methods