6 Similarities Between Heat, Momentum and Mass Transfer

78



5



The Fundamentals of Rate Processes



Table 5.1 The similarity between heat, momentum and mass transfer

Flux



Diffusivity



Gradient



Heat (W m−2 )



Thermal (m2 s−1 )



Momentum (kg m−1 s−2 )

(= shear stress, Pa)

Molar (kmol m−2 s−1 )



Kinematic (m2 s−1 )



Heat per unit volume per unit length

(J m−4 )

Momentum per unit volume per unit

length (N s m−4 )

Molar mass per unit volume per unit

length (kmol m−4 )

(= Molar concentration per unit length)



(Mass) diffusivity (m2 s−1 )



developed to stress the profound similarity between the three rate equations which form the basis of

the study of fluid behaviour in food process engineering.



Further Reading

Bird, R. B., Stewart, W. E., and Lightfoot, E. N. 2007. Transport phenomena. New York, NY: Wiley.

Geankoplis, C. J. 1993. Transport processes and unit operations. Engelwood Cliffs, NJ: Prentice Hall.

Kay, J. M. and Nedderman, R. M. 1985. Fluid mechanics and transfer processes. Cambridge: Cambridge University

Press.

Welty, J. R., Wicks, C. E., and Wilson, R. E. 1984. Fundamentals of momentum, heat, and mass transfer. New York,

NY: Wiley.



Chapter 6



The Flow of Food Fluids



Nomenclature

A

Ao

cf

CD

C1

C2

C3

d

de

di

do

De

e

g

G

h

hF

hm

h

k1 , k2 , k3

K

l

L

m

n

N

NPSH

p

P

r

ra

R

R0

RY

Re

t



Cross-sectional area

Cross-sectional area of orifice or venturi throat

Friction factor

Discharge coefficient

Parameter in Ellis model

Parameter in Ellis model

Parameter in Casson model

Diameter

Hydraulic mean diameter

Inner diameter of annulus

Outer diameter of annulus

Deborah number

Magnitude of surface roughness

Acceleration due to gravity

Shear modulus

Head loss

Head loss due to friction

Manometer height difference

Height

Coefficients in power series

Consistency coefficient

Separation of layers in simple shear

Length; characteristic length

Mass flow rate

Flow behaviour index

Parameter in Ellis model

Net positive suction head

Pure component vapour pressure

Pressure

Radius

Arithmetic mean radius

Shear stress

Initial shear stress

Yield stress

Reynolds number

Time



P.G. Smith, Introduction to Food Process Engineering, Food Science Texts Series,

DOI 10.1007/978-1-4419-7662-8_6, C Springer Science+Business Media, LLC 2011



79



80



u

ur

ut

V

Z



6 The Flow of Food Fluids



Mean velocity

Velocity at a radius r

Terminal falling velocity

Volumetric flow rate

Height above datum



Greek Symbols

γ

γ

˙

γwall

˙

ΔP

Δh

θ

μ

μa

μp

ρ

ρav

υ

τ



Displacement angle in simple shear

Shear rate

Wall shear rate

Pressure drop

Total head

Cone angle

Viscosity

Apparent viscosity

Plastic viscosity

Density

Average density

Specific volume

Torque; time constant

Angular velocity



Subscripts

A

B

i

o

s



Fluid A

Fluid B

Inner

Outer

Solid



6.1 Introduction

This chapter develops the concept of momentum transfer into a treatment of the flow of fluids over

surfaces and in pipes and ducts. A later chapter will deal with the behaviour of food particles in a fluid

stream. Many foods are of course liquid and the study of fluid flow, or fluid mechanics, is necessary to

understand how fluids are transported, how they can be pumped, mixed and so on. The high viscosity

of many liquid foods means that laminar flow is particularly important. However, very many foodstuffs

are non-Newtonian and later sections of this chapter cover a wide variety of rheological models; these

are treated as mathematical descriptions of physical behaviour with the objective of enabling the

reader to apply models to experimental data in order to determine whether or not they can be used

predictively.



6.2 Fundamental Principles

6.2.1 Velocity and Flow Rate

As we saw in Section 5.3, when a viscous fluid flows under the influence of a shear stress a velocity

gradient develops across the flow channel. However, it is convenient to use a mean velocity u which

is equal to the volumetric flow rate V divided by the cross-sectional area of the conduit A. Thus



6.2



Fundamental Principles



81



V=uA



(6.1)



The mean velocity is often referred to as the superficial velocity.

Example 6.1

A cylindrical vessel, 2 m in diameter, is to be filled to a depth of 1.5 m. The liquid discharges through

a 2.5 cm diameter pipe with a velocity of 3.0 m s−1 . How long will it take to fill the vessel? If the

outlet valve is accidentally left partially open, and the velocity in the outlet pipe of 2.5 cm diameter is

1 m s−1 , determine the new time required to fill the vessel.

The volume of liquid required is the volume of a cylinder 2 m in diameter and 1.5 m deep, i.e.

2

the volume is 1.5 π (2.0) or 4.71 m2 . The volumetric flow rate through the pipe V is the product of the

4

superficial or mean velocity and the pipe cross-sectional area, therefore

V = 3.0



π (0.025)2 3 −1

m s

4



or

V = 1.47 × 10−3 m3 s−1

The time to fill the vessel t is then equal to the volume divided by flow rate. Thus

t=



4.71

1.47 × 10−3



s



or

t = 3200 s

Now, because the diameter of the outlet pipe is the same as that of the inlet, the effective net inlet

velocity is reduced to (3.0 – 1.0) = 2.0 m s−1 and the time to fill the vessel will be greater by a factor

of 1.5. Hence

t=



3

× 3200 s

2



or

t = 4800 s



6.2.2 Reynolds’ Experiment

Figure 6.1 illustrates Reynolds’ classic experiment which demonstrated the nature of fluid flow.

A glass tube contains a continuous flow of water and a second very narrow tube enters through the

wall and is positioned along the centre line of the larger tube. This second tube is fed with a dye from

a reservoir such that the flow is negligible relative to the flow of water. Two broad types of flow are

observable: in laminar flow (Fig. 6.1a) the trace of dye remains on the centre line indicating that layers

of water flow over one another and do not mix. Individual elements or packets of fluid move only in

the direction of flow and their components of velocity in other directions are negligible. However,

under certain conditions turbulent flow is observed and the trace moves from side to side and eventually breaks up with the dye being dispersed throughout the water. In turbulent flow (Fig. 6.1b) there



82



6 The Flow of Food Fluids



Fig. 6.1 Reynolds experiment: (a) laminar flow, (b) turbulent flow



is a significant component of velocity in a direction at right angles to the direction of flow and rapid

fluctuations in velocity and pressure. Where temperature gradients exist there will be rapid fluctuations in temperature also. Thus turbulent flow promotes mixing, heat transfer and mass transfer but

there is a disadvantage to this general improvement. The pressure drop in laminar flow along a given

length of pipe is proportional to the flow rate of the fluid, whereas in turbulent flow the pressure drop

increases more rapidly with flow rate and follows an approximately squared relationship.

These differences are reflected in the velocity profiles illustrated in Fig. 6.2. In pipe flow the maximum velocity in the direction of flow is along the centre line of the pipe; at the wall the fluid velocity

is zero. For laminar flow the velocity profile adopts a parabolic shape with the mean velocity equal to

about one half of the maximum; however, when the flow is turbulent a much flatter profile develops

and the mean velocity is approximately 82% of the maximum. The region of more slowly moving

fluid close to the wall is known as a boundary layer. The edge of this layer is sometimes defined as the

point where the velocity is 99% of the maximum velocity. However, the structure of a flowing fluid is

complex and this definition is not absolute.

Reynolds showed that turbulent flow could be obtained by increasing the mean velocity u, the

density ρ and the diameter of the tube d and by decreasing the viscosity μ. Thus the dimensionless

group known as the Reynolds number is used to indicate the degree of turbulence

Re =



ρuL

μ



(6.2)



where L is a length characteristic of the geometry. For pipe flow the characteristic length is the pipe

diameter d and

Re =



Fig. 6.2 Velocity profiles in pipe flow



ρud

μ



(6.3)



6.2



Fundamental Principles



83



In pipe flow the transition between laminar and turbulent flow occurs at a Reynolds number of

about 2000. However, there is not always a sharp boundary between the two and a transition region is

often referred to in the range 2000
Example 6.2

Calculate the Reynolds number for a vegetable oil of viscosity 0.03 Pa s and density 850 kg m−3

flowing in a 50 mm bore pipe at a mean velocity of 0.75 m s−1 .

Taking care to ensure that all quantities are in SI units, the Reynolds number is

Re =



850 × 0.75 × 0.050

0.03



or

Re = 1062.5

The flow is therefore laminar.

It is often convenient to express the Reynolds number in terms of a mass flow rate m (which is

more easily measurable) than as a function of velocity. Thus the mean velocity is

u=



4m

ρπd 2



(6.4)



4m

μπ d



(6.5)



and the Reynolds number becomes

Re =



Whilst it is clear that the characteristic length should be the diameter of a pipe or sphere, for annular

flow the hydraulic mean diameter de should be used. This is defined as

de =



4 × cross-sectional area of annulus

wetted perimeter



(6.6)



If do is the outer diameter of the annular channel (i.e. the inner diameter of the outer pipe) and di

is the inner diameter of the annular channel (i.e. the outer diameter of the inner pipe), then

2

πdo

4



4

de =



−



πdi2

4



πdo + πdi



(6.7)



which simplifies to

2

do − di2

do + d i



(6.8)



(do − di ) (do + di )

do + d i



(6.9)



de =

and

de =



84



6 The Flow of Food Fluids



and finally to

de = do − di



(6.10)



Example 6.3

Calculate the Reynolds number for water at 50◦ C flowing at a rate of 12 L min−1 through an annulus

between two concentric pipes. The outer diameter of the inner pipe is 10 mm and the inner diameter of

the outer pipe is 20 mm. For water at 50◦ C: density = 988 kg m−3 and viscosity = 5.44 × 10−4 Pa s.

The mass flow rate of water m can be found readily from the volumetric flow rate and the density,

so that

m=



988 × 12 × 10−3

kg s−1

60



or

m = 0.1976 kg s−1

The hydraulic mean diameter [Eq. (6.10)] is the difference between the outer diameter of the inner

cylinder and the inner diameter of the outer cylinder, i.e.

de = 0.02 − 0.01 m

or

de = 0.01 m

Substituting into Eq. (6.5) the Reynolds number becomes

Re =



4 × 0.1976

5.44 × 10−4 π 0.01



or

Re = 46249

Consequently the flow is clearly turbulent.



6.2.3 Principle of Continuity

The analysis of fluid flow is complex. However, considerable understanding can be gained by considering an ideal fluid and applying mass and energy balances, respectively. An ideal, or inviscid, fluid is

one which has no viscosity and for which the velocity is uniform across a given cross-section of the

flow channel. Referring to Fig. 6.3 and equating the mass flow rate at points 1 and 2 gives

ρ1 u 1 A 1 = ρ 2 u 2 A 2 +



d(ρav V)

dt



(6.11)



where A is the cross-section and ρ and u are the density and velocity of the fluid, respectively, ρav being

the average density, and V is the volume of the flow channel between points 1 and 2. The differential



6.2



Fundamental Principles



85

2

1



Fig. 6.3 Principle of continuity



term represents the accumulation of mass. This result is known as the principle of continuity. At steady

state there is no accumulation of mass and Eq. (6.11) becomes

ρ1 u1 A1 = ρ2 u2 A2



(6.12)



A distinction must now be made between liquids and gases. The density of most liquids remains

reasonably constant with pressure, certainly within the pressure ranges encountered in most food

processes. A liquid thus approximates to an incompressible fluid where ρ1 = ρ2 and Eq. (6.12) can

now be written as

u1 A1 = u2 A2



(6.13)



Clearly gases are compressible and the gas laws must be used to find the relationship between

ρ 1 and ρ 2 .

Example 6.4

Two pipelines of 20 and 30 mm internal diameter, respectively, merge into a single line of 50 mm

diameter. The same liquid flows in each branch at mean velocities of 1.2 and 0.8 m s−1 , respectively.

Determine the fluid velocity downstream of the junction.

This problem can be solved using the principle of continuity. Expanding Eq. (6.13), the sum of the

product of velocity and area for the 20 and 30 mm pipelines must equal the product of velocity and

area for the merged pipe, in which the velocity is u. Thus, again taking care to use SI units throughout

π

π

1.2(0.02)2 + 0.80(0.03)2 = u (0.05)2

4

4

which can be solved to give

u = 0.48 m s−1



86



6 The Flow of Food Fluids



6.2.4 Conservation of Energy

The conservation of energy relationship for a flowing fluid is known as Bernoulli’s equation. In

Chapter 4 the steady-flow energy equation was derived for a general process involving the addition

or removal of heat and work to a fluid. If there is no transfer of heat and no work done on or by the

fluid then the temperature and the internal energy of the fluid remain constant. Equation (4.10) then

becomes

mu2

mu2

1

2

+ mZ1 g + mP1υ1 =

+ mZ2 g + mP2υ2

2

2



(6.14)



Dividing through by the constant mass flow rate m and by the acceleration due to gravity and

replacing the specific volume of the fluid υ with density ρ gives

u2

u2

P1

P2

1

+ Z1 +

= 2 + Z2 +

2g

ρ1g

2g

ρ2 g



(6.15)



which is the usual form of Bernoulli’s equation. It should be stressed that the velocity u is the average

velocity and that as written above Bernoulli’s equation is valid only for the flat velocity profile found

in turbulent flow. Each term in Eq. (6.15) has units of metres and represents an energy loss in terms

of a head of the fluid. The terms on each side of the equation are referred to as the velocity head,

the potential head and the pressure head, respectively. For example, water flowing along a pipe with

a mean velocity of 3 m s−1 corresponds to a velocity head of 0.459 m of water. The corresponding

pressure is then

0.459 × 1000 × 9.81 = 4500 Pa.

Particularly for higher velocities, a loss term should be included in Eq. (6.15) as energy is dissipated

between points 1 and 2. This frictional loss is called the head loss due to friction and here is given the

symbol hF . An expression for the frictional loss in straight pipe sections will be derived in Section 6.4

but in addition there are standard values which can be used for the losses through valves, bends and

other pipe fittings. Bernoulli’s equation may now be written as

u2

u2

P1

P2

1

+ Z1 +

= 2 + Z2 +

+ hF

2g

ρ1g

2g

ρ2g



(6.16)



Example 6.5

Water flows through a horizontal pipe. At a particular cross-section X the velocity of the water is

1.5 m s−1 and the pressure is 175 kPa. The pipe tapers gradually from 150 mm at X to 75 mm at a

section Y. Determine the pressure at Y, assuming that the frictional losses are negligible. What must

be the diameter at Y for the pressure there to be reduced to 55 kPa?

This problem concerns a horizontal pipe, there is no change in height above the datum and therefore

Z1 = Z2 . In addition there are no frictional losses and thus Bernoulli’s equation reduces to

u2

u2

P1

P2

1

+

= 2 +

2g ρ1g

2g ρ2g



6.3



Laminar Flow in a Pipeline



87



Using the principle of continuity [Eq. (6.12)], the velocity at cross-section Y, u2 , is

150

75



u2 = 1.5 ×



2



m s−1



or

u2 = 6.0 m s−1

Now multiplying through by the acceleration due to gravity and re-arranging Bernoulli’s equation

explicitly for the pressure at Y gives

P2 = P1 +



ρ 2

u − u2

2

2 1



and therefore

P2 = 1.75 × 105 +

and



1000

1.52 − 6.02 Pa

2



P2 = 158.1 kPa



Now re-arranging Bernoulli’s equation to give the velocity at Y yields

u 2 = u2 +

1



2

(P1 − P2 )

ρ



0.5



Hence, for a pressure at Y of 55 kPa,

u2 = (1.5)2 +



2

(175 − 55) 103

1000



0.5



m s−1



or

u2 = 15.6 m s−1

Again, using the principle of continuity,

2

d2 =



1.5

(0.15)2 m

15.6



where d2 is the diameter at Y and thus

d2 = 0.0465 m



6.3 Laminar Flow in a Pipeline

The most useful relationship concerned with the transport of fluids in pipelines is that between the flow

rate and the pressure gradient responsible for flow. Many food liquids have a high dynamic viscosity

and consequently laminar flow is commonly observed. For the particular case of laminar flow it is

possible to derive an exact relationship between flow rate and pressure.



88



6 The Flow of Food Fluids



dr

r1

r



ΔP, L



Fig. 6.4 Laminar flow in a pipe



Figure 6.4 represents a cylindrical element of fluid, of general radius r, within a circular pipe of

radius r1 , length L and over which the pressure drop due to friction between the fluid and the pipe

wall is ΔP. The pressure gradient along the pipe is assumed to be uniform and therefore equal to LP .

If the momentum of the fluid is constant then the net force acting on the fluid must be zero. In other

words the shear force acting on the curved surface of the element must equal the difference between

the forces acting on the respective ends of the cylinder. This latter force is simply the product of the

pressure difference ΔP and the cross-sectional area π r 2 . Thus

2πrLR = π r2 P



(6.17)



where R is the shear stress at a radius r and 2π rL represents the curved surface of the element.

Consequently

R=



r P

2 L



(6.18)



but from Eq. (5.3) the shear stress equates to the product of viscosity and velocity gradient, in this

case dur where ur is the velocity in the direction of flow, at a radius r. Hence

dr

dur

r P

= −μ

2 L

dr



(6.19)



Separating the variables and integrating between r = r1 (the pipe wall), where the fluid is

stationary, and r = r, where the velocity is ur , gives

ur



μ



r



P

dur = −

2L



0



rdr



(6.20)



r1



and the result

ur =



P 2

(r − r2 )

4Lμ 1



(6.21)