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Introduction to Food Processing

Fluid Flow

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Introduction

The processing of many foods involves the transport of fluids (liquids and gases). It is therefore useful for food scientists to have some understanding of the principles of fluid flow.

Fluid Mechanics is concerned with the effects of forces on fluids. These forces usually arise as a result of a pressure difference in the fluid system. There are two major types of fluid, compressible and incompressible. Compressible fluids undergo a reduction of volume and an increase in density as pressure is applied whereas the density and volume of incompressible fluids remains essentially constant under varying pressure. Compressible fluids are usually gases and incompressible fluids are usually liquids. However gases operating at low pressure differences can be treated as if they are incompressible fluids.

This part of your course will be concerned with the steady flow of incompressible fluids (liquids and gases at low ΔP)

The principles of fluid flow are largely based on the conservation of mass and the conservation of energy.


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Flow of fluids in pipes velocity volume and mass flow

The volume and mass flows are measures of the total quantity of fluid flowing, the velocity is a measure of how fast the fluid is flowing. Volume and mass flows are related by density from the relationship

Volume flow = Mass flow x density, or Q=m

For a given volume flow, the velocity is dependant on the cross sectional area of the pipe. The relationship is

Volume flow = Velocity x Area, or Q=vA

For a circular pipe, the velocity is the area of a circle, i.e. A=π r2. However, it is usually more convenient to use an alternative form, i.e. π d2/4

For a fluid flowing at constant rate in a pipe, the mass flow will be constant throughout the pipe, this is an expression of the conservation of mass. If the fluid is incompressible, the density of the fluid will be constant, so the volume flow will also be constant throughout the pipe. This applies to all liquids and (approximately) to gases at low pressure. A consequence of this is that velocity x area is also constant throughout the pipe. This is known as the continuity equation and is normally expressed in the form

v1A1=v2A2

The subscripts 1 and 2 refer to two different points in the pipe, for example;

fluid continuity
Fig. 1 Fluid continuity

If flows are split or combined, the combined volume flow will be the sum of the two smaller volume flows.

This may be written in the form of an equation

v1A1+v2A2=v3A3

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Pressure in fluids

Pressure, or more strictly pressure difference is the driving force determines the rate at which fluids flow. The pressure in a fluid may arise from two main sources

In the diagram below, the pressure to move the fluid to the upper tank will be a combination of these.


Fig 2 Pumping fluids

There will be pressure in the fluid at the base of the storage tank (1) due to the weight of fluid in the tank. This will not be enough to lift the fluid to the upper tank (3), so a pump (2) is needed to add extra pressure to move the fluid. The total pressure at the discharge of the pump must be sufficient to

  1. Raise the fluid above the level of the fluid in the storage tank
  2. Overcome any pressure in the upper tank
  3. Move the fluid along the pipe at the desired rate
  4. Overcome any frictional losses in the pipes.

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Bernoulli's Equation

Bernoulli's equation is based on an energy balance in the fluid. The total energy in the fluid at any point in the system is constant (conservation of energy), but, it is redistributed as the fluid flows through the system

The total energy is the sum of
Pressure energy + kinetic energy + potential energy

Pressure energy is the pressure in the fluid at any point. This is based on any externally applied pressure plus the pressure due to the head of liquid which is found from head x density x acceleration due to gravity. Hence;

P = Pext + hρg

By convention, pressure is often related to atmospheric pressure, so if Pext is atmospheric pressure, then it is treated as zero. Such a measure of pressure is called gauge pressure.

The kinetic energy is ½ mass x velocity2 expressed in pressure terms as ρv2/2

The potential energy is found from head x density x acceleration due to gravity, hρg

Thus the Bernoulli equation is expressed as;

Bernoulli equation: P-one plus rho V-one squared over two plus H-one rho G equals P-two plus rho V-two squared over two plus H-two rho G

The equation needs to be modified for practical systems by including two extra terms,

The equation, thus becomes

modified Bernoulli equation: P-one plus rho V-one squared over two plus H-one rho G plus Delat P sub E equals P-two plus rho V-two squared over two plus H-two rho G plus Delta P sub f

Example

Let us suppose that the flow of the fluid is to be 15 m3 h-1, depth of liquid in the storage tank is 3 m, the pipe is 50 mm i.d. (inside diameter) and the upper tank is 10 m above the pump. What pressure must be added at the pump? The liquid is water, density = 1000 kg m-3. The frictional losses are 5 kPa

Pressure available at the base of the storage tank. This is dependant on the weight of fluid (= mass x g, acceleration due to gravity) and is found from height x density x g

or; P=h ρ g = 3 x 1000 x 9.81 = 29430 Pa (say 29400 Pa)

Pressure required to raise the fluid 10 m

P = h ρ g = 10 x 1000 x 9.81 = 98100 Pa

Pressure required to move fluid along the pipe. This is dependant on the kinetic energy of the fluid and is found from ½ density x velocity-squared.

To find velocity, we need the X-sectional area of the pipe and the volume flow in m3 s-1.

X-sectional area of pipe = π d2/4 = π x (50/1000)2/4 = 1.96x10-3 m2
(Note; the "50/1000" is to convert mm to m)

velocity = volume flow/area, i.e. v=Q/A = (15/3600)/ 1.96x10-3 = 2.1 m s-1

This may now be used to find the pressure required to move the fluid along the pipe.

P = ρ v2/2 = 1000 x 2.12/2 = 2205 Pa (say 2200 Pa)

Pressure loss due to friction = 5000 Pa

In the absence of other information we will assume the storage tank and the upper tank are both at atmospheric pressure. This means there is no net pressure to overcome in the upper tank.

Total pressure needed to move the fluid = 98100+2200+5000 = 105300 Pa

There is, however a pressure of 29400 Pa at the base of the tank, so the pump must add

105300-29400 = 75900 Pa = 75.9 kPa

Usually, the information really needed is the power of the pump. This can be found since

Energy due to pressure = pressure x volume, hence;


Power = pressure x volume flow
∴ Power required by pump = 75900 x (15/3600) = 316 W.

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Laminar and turbulent flow

There are two basic patterns of fluid flow called laminar and turbulent flow. These differ in the degree of cross mixing occurring when the fluid is flowing.

Laminar flow

At low flow velocities, no mixing of the fluid takes place in directions perpendicular to the direction of flow. This is called laminar or streamline flow. Fig 6 illustrates laminar flow in a pipe.


Fig 3 Laminar flow

Turbulent flow

As the flow velocity increases, the fluid in the centre of the pipe starts to behave erratically and to break away and form small eddies. This causes some cross mixing of central core of the fluid. This flow behaviour is known as turbulence. You may have seen it in a river in flood or in a fast flowing stream. As the flow velocity increases, this turbulent core increases until it leaves only a very small layer of fluid next to the pipe wall in laminar flow. This thin layer is called the boundary layer. Fully developed turbulent flow is illustrated in Fig 7


Fig 4. Turbulent flow

The onset of turbulence does not only depend on the velocity, u of the fluid, but also on the pipe diameter, d, the viscosity, μ and the density, ρ.

These properties are related together by Reynold's Number, Re and it is the value of Reynold's number which determines whether the flow is streamline or turbulent.

Reynold's number is defined as

Reynold's number equation: Re equals D rho U over mu

The relationship between the type of flow and Reynold's number is:


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Estimating Friction Losses

Pressure loss due to friction is dependant on;

The pressure loss due to friction is found from

pressure loss due to friction equation: Delta P sub f equals 4 f L over D time rho V-squared over 2

The quantity "f" is known as the friction factor

For laminar flow, surface roughness is not important and the friction factor is found from

f = 16/Re

In turbulent flow friction factor must be found from charts. However, Over a typical range of flows for liquids with "typical" piping, the value of f is approximately 0.000625. (i.e. 4f = 0.0025)

Thus the pressure loss due to friction may be approximated from

simplified pressure loss due to friction equation: Delta P sub f equals 0.0025 L over D time rho V-squared over 2

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Problems

  1. A liquid food is being transported through a pipe 60 mm i.d. at a rate of 18 m3 h-1. The pipe contains a reducing section at one point which reduces the diameter to 50 mm. What is the velocity of the fluid in each section of the pipe?


  2. A 50 mm diameter pipe containing sucrose solution flowing at a velocity of 1.5 m s-1 divides into two. The larger, 30 mm diameter branch takes 60% of the volume flow and the smaller branch, of 20 mm diameter takes the remaining 40% of the flow. What is the velocity in each branch?


  3. Milk is to be pumped from a storage tank to a pasteuriser. The tank has 2.5 m depth of milk in it and the pasteuriser is on an upper floor at an elevation of 4 m. The pressure pasteuriser is 2.5 kPa and the milk tank is at atmospheric pressure. How much pressure must be added to the milk for a flowrate of 20 m3 h-1 through a 60 mm i.d. pipe? Density of milk = 1032 kg m-3.

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