1.3 FLUID MECHANICS IN FOOD PROCESSING .1 V ISCOSITY OF F LUIDS.1 VISCOSITYOF FLUIDS
1.3.2 L AWS OF F LUID D YNAMICS
1.3 FLUID MECHANICS IN FOOD PROCESSING
The mass fl ow rate can be calculated from the volumetric fl ow rate or the fl uid velocity:
m r= V = rAu (1.59)
Therefore, Equation 1.58 can be re-written as
1A u1 1= 2A u2 2
r r (1.60)
In Equations 1.58 through 1.60, m· is the mass fl ow rate, (kg/s), V.
is the volumetric fl ow rate, (m3/s), r is the density, (kg/m3), A is the cross-sectional area of the conduit, and u- is the average velocity, (m/s).
Equation 1.60 is applicable to any fl uid fl ow. If the fl uid is incompressible (r1 = r2), Equation 1.60 can be simplifi ed as
1 1 2 2
A u =A u (1.61)
Fluid fl ow also follows the law of energy conservation. Energy within a system is neither created nor destroyed. However, it can be transformed from one form to another. There are three forms of energy that are considered: (1) pressure energy or fl ow work due to pressure imposed on the fl uid, (2) kinetic energy due to the velocity or momentum of the fl uid, and (3) potential energy due to the elevation of the fl uid. Bernoulli’s equation is used to describe the energy conservation of fl uid fl ow in a closed system without (1) work done on or by the fl uid, (2) heat transfer between the fl uid and its surroundings, and (3) frictional energy loss, which is expressed as
2 2
1 1 1 1 1 2 2 2 2 2
1 1
2 2
P+ ru +rgz = +P ru +rgz (1.62)
where
P is the pressure u is the velocity z is the elevation
r is the density of the fl uid g is the acceleration due to gravity
According to Bernoulli’s equation, if there is no heat added to the fl uid, no work done on the fl uid, and no frictional energy loss, the change in energy is zero.
That is
2 2
2 1
2 2 1 1
2 1
1 1
2 2 0
P P
E ⎛ u gz ⎞ ⎛ u gz⎞
∆ = ⎜⎝r + + ⎟ ⎜⎠ ⎝− r + + ⎟⎠= (1.63)
where DE is the energy change in the SI unit of J/kg.
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However, in case of a real fl uid fl ow, we cannot ignore the frictional energy loss. A pump or fan is used to increase the energy of the fl uid and overcome the frictional energy loss in a transport system. In this case, the energy change can be determined by
2 2
2 1
p f 2 2 1 1
2 1
1 1
E 2 2
P P
E E ⎛ u gz ⎞ ⎛ u gz⎞
∆ = − = ⎜⎝r + + ⎟⎠ − ⎜⎝r + + ⎟⎠ (1.64)
where
Ep is the energy imposed on the fl uid by a pump or fan (J/kg) Ef is the total frictional energy loss (J/kg)
Therefore, the energy requirement of a pump or fan per unit mass of fl uid is deter-mined by
2 2
2 1
p 2 2 1 1 f
2 1
1 1
2 2
P P
E u gz u gz E
ρ ρ
⎛ ⎞ ⎛ ⎞
= ⎜⎝ + + ⎟⎠ − ⎜⎝ + + ⎟⎠ + (1.65)
The head of the pump, hpump (m), which is usually used in the specifi cations of a pump, is defi ned as
p pump
h E
= g (1.66)
where g is the acceleration due to gravity (m/s2).
The required power output of a pump or fan is calculated by
P= mEp (1.67)
where m. is the mass fl ow rate (kg/s).
A pump is usually located in the pipeline used to deliver a fl uid. Careful attention is required to prevent vaporization of the liquid inside the pump. Therefore, the liquid pressure on the suction side of a pump should be higher than the saturation vapor pressure of the liquid at the operating temperature. That is
s 2 v
R s
NPSH 01
2
P P
g gu g
⎛ ⎞
=⎜⎝r + ⎟⎠ − r > (1.68)
Pump manufacturers specify the required net positive suction head (NPSHR). The designed net positive suction head should be higher than the specifi ed value.
1.3.3 ENERGY LOSSOF FLUID FLOW
Energy is lost because of the friction in the fl owing fl uid due to viscosity. The mag-nitude of the energy loss for a given fl uid is directly dependent on the characteristics
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of the fl uid fl ow. The fl ow characteristics could be laminar, turbulent, or in between (transitional). In a laminar fl ow, the fl uid appears to fl ow in layers in a smooth and regular manner while under the turbulent condition, the fl uid fl ows in random direc-tions within the fl ow stream. The fl ow characteristics of a fl uid are quantitatively described by the Reynolds number, which is defi ned as
Re
N =ruD
m (1.69)
where
D is the characteristic dimension (e.g., tube diameter) (m) r is the density (kg/m3)
u- is the average velocity (m/s) m is the dynamic viscosity (Pa s)
The fl ow characteristics can be determined by the Reynolds number:
NRe < 2100, laminar fl ow
2100 £ NRe £ 4000, transitional fl ow NRe > 4000, turbulent fl ow
During fl uid fl ow, a velocity gradient exists between the fl uid at the wall and the center of a conduit due to friction between the individual molecules of the fl uid and between the fl uid and the wall. The major energy loss due to the friction is usually expressed as
2
f,major 2 2 E f L u
= D (1.70)
where
Ef is the energy loss due to friction f is the loss coeffi cient or friction factor u- is the velocity of the fl uid
L is the length of the conduit D is the diameter of the conduit
The friction factor, f, varies with the fl ow characteristics described with the Reynolds number and surface roughness. The Moody chart shown in Figure 1.4 presents the friction factor as a function of the Reynolds number for various magnitudes of relative roughness of different pipes. The relative roughness of a pipe is the ratio of roughness, e (m), and diameter, D (m), of the pipe. The equiva-lent roughness, e, for new pipes made of different materials is (Singh and Heldman, 2001):
Cast iron: 259 × 10
• −6 m
Galvanized iron: 152 × 10
• −6 m
Steel or wrought iron: 45.7 × 10
• −6 m
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The various components used in pipelines, such as valves, tees, and elbows, and contraction and expansion of fl uid when it enters into and exits from a pipe cause minor frictional losses. Minor frictional energy loss can be calculated by
2 f,minor f,minor
2
E =C u (1.71)
The friction loss factor for sudden contraction is determined by (Singh and Heldman, 2001)
2 2
f,concentration
1 1
0.4 1.25 A if A 0.715
C A A
⎛ ⎞
= ⎜⎝ − ⎟⎠ < (1.72)
2 2
f,concentration
1 1
0.75 1 A if A 0.715
C A A
⎛ ⎞
= ⎜⎝ − ⎟⎠ > (1.73)
The friction loss factor for sudden expansion is determined by (Singh and Heldman, 2001)
2 2 f,expansion
1
1 A
C A
⎛ ⎞
=⎜⎝ − ⎟⎠ (1.74)
In Equations 1.72 through 1.74, A2 and A1 are small and large cross-sectional areas, respectively. Friction loss factors for standard fi ttings are given in Table 1.5 (Singh and Heldman, 2001).
10,000,000
Relative roughness (e/D)
0.00001 0.00005 0.0001 0.0004 0.0002 0.0006 0.001 0.002 0.005 0.01 0.020.030.040.05
1,000,000 100,000
Reynolds number 10,000
1,000 0.001
0.01
Friction factor (f)
0.1
Critical region f = 16/NRe
Smooth pipe
FIGURE 1.4 Moody diagram for the Fanning friction factor. (Reprinted from Singh, R.P.
and Heldman, D.R., Introduction to Food Engineering, Academic Press, San Diego 2001.
With permission.)
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Example 1.7
A pumping system as shown in Figure 1.5 is designed to pump milk at 20°C from an open tank through a steel pipe 20 m in length and 2.5 cm inside diameter to a second tank at a higher level. The mass fl ow rate of milk is 2 kg/s. The supply tank maintains a liquid level of 2 m from the fl oor. The milk leaves the system at an elevation of 15 m above the fl oor (reference elevation). Determine the major and minor energy losses due to friction. Determine the required pump head and power requirement of the
TABLE 1.5
Friction Loss Factors for Standard Fittings
Fittings Friction Loss Factor
Elbows
Long radius 45° and 90°, fl anged 0.2
Long radius 90°, threaded 0.7
Regular 45°, threaded 0.4
Regular 90°, threaded 1.5
Regular 90°, fl anged 0.3
180° return bend, threaded 1.5
180° return bend, fl anged 0.2
Tees
Branch fl ow, fl anged 1.0
Branch fl ow, threaded 2.0
Line fl ow, fl anged 0.2
Line fl ow, threaded 0.9
Union, threaded 0.8
Valve
Angle, fully open 2
Ball valve, 1/3 closed 5.5
Ball valve, 2/3 closed 210
Ball valve, fully open 0.05
Diaphragm valve, ¼ closed 2.6
Diaphragm valve, ½ closed 4.3
Diaphragm valve, open 2.3
Gate, ¾ closed 17
Gate, ¼ closed 0.26
Gate, ½ closed 2.1
Gate, fully open 0.15
Globe, fully open 10
Swing check, backward fl ow ∞
Swing check, forward fl ow 2
Source: From Singh, R.P. and Heldman, D.R., in Introduction to Food Engi-neering, Academic Press, San Diego, 2001. With permission.
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pump. (Suppose that the density and viscosity of milk are 1000 kg/m3 and 1.99 m Pas, respectively.)
Solution 1.7
1. From Equation 1.59, the mean velocity of milk in the pipe is
( )
= = =
× π
0.0252 3
2[kg/s]
4.08 m/s 1000 [kg/m ]
4 u m
A r
2. From Equation 1.69, the Reynolds number is
× ×
= = 3 =
Re
1,000 [kg/m ] 4.08 [m/s] 0.025[m]
51,256 0.00199 [Pa s]
N ruD m
The relative roughness of the pipe is
− −
= 45.7 10 [m]× 6 = × 3 1.828 10 0.025 [m]
D e
Using the Reynolds number and relative roughness, the frictional factor can be found from Figure 1.4: f = 0.0065. Using Equation 1.71, the major frictional energy loss is
= 2 = × × × 2 =
f,major
20 4.08
2 2 0.0065 86.6 J/ kg
2 0.025 2
E f L u D
FIGURE 1.5 Typical pumping system.
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3. There are two 90° standard elbows and one angle valve in the pipeline. The entrance and exit of the pipeline are the concentration and expansion types, respectively. From Table 1.5, the energy loss factor for elbow and angle valve is 1.5 and 2, respectively.
The entrance loss factor due to sudden contraction (A2/A1 ≈ 0) is
( )
⎛ ⎞
= ⎜⎝ − ⎟⎠= × − =
2 f,entrance
1
0.4 1.25 A 0.4 1.25 0 0.5
C A
The exit loss factor due to sudden expansion (A1/A2 ≈ 0) is
⎛ ⎞
= −⎜⎝ ⎟⎠ = − =
2 1 2 f,entrance
2
1 A (1 0) 1
C A
The total minor energy loss factor is thus
= + + + = + × + + =
f,minor f,entrance 2 f,elbow f,valve f,exit 0.5 2 1.5 2 1 6.5
C C C C C
The total minor energy loss is
= 2 = × 2 =
f,minor f,minor
6.5 4.08 54.1 J/kg
2 2
E C u
4. From Equation 1.65, the energy requirement of the pump is
( )
⎛ ⎞ ⎛ ⎞
=⎜⎝ + + ⎟ ⎝⎠−⎜ + + ⎟⎠+
⎛ ⎞ ⎛ ⎞
=⎜⎝ + × + × ⎟ ⎜⎠ ⎝− + × + × ⎟⎠+ +
=
2 2
2 1
p 2 2 1 1 f
2 1
2 2
1 1
2 2
101325 1 101325 1
0 9.81 15 0 9.81 2 86.6 54.1
1000 2 1000 2
268.2 J/kg
P P
E u gz u gz E
r r
5. From Equation 1.66, the pump head is
= p = =
pump
268.2
27.4 m 9.8
h E g
6. From Equation 1.67, the power requirement of the pump is
= p = ×2 268.2 536.5 W= P mE
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