fluid flow in food processing · 8/3/2014 · fluid flow in food processing properties of non...
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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
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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
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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)
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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
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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)
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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)
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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
=
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4 Variable – Area Meters
5 Other Measurement Methods
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