Lecture Notes

Dan Curtis

Semester 1

2012/2013

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Contents Lecture 1: Stress, strain-rate and viscosity ............................................................................................. 3

Lecture 2: Models for non-Newtonian fluid behaviourand time dependent viscosity .......................... 9

Lecture 3: The Poiseulle Equation ......................................................................................................... 15

Lecture 4: Velocity profiles and P, V relations for non-Newtonian fluids .......................................... 19

Lectures 5 & 6: Fluid Mixing and Agitation ........................................................................................... 23

Lecture 7: Forces due to fluids in motion ............................................................................................. 28

Lectures 8 & 9 : Forces due to fluids in motion II ................................................................................. 33

Lectures 10 : Forces due to real fluids in motion ................................................................................. 35

Lectures 11 & 12: The Bernoulli Equation ............................................................................................ 43

Lecture 13: Laminar flow in packed beds ............................................................................................. 51

Lecture 14: Turbulent flow in pipes ...................................................................................................... 54

Lecture 15: Head losses in rough and smooth pipes ............................................................................ 59

Lectures 16 & 17: Flow through packed beds II .................................................................................... 62

Lecture 18 & 19: Differential head flow meters ................................................................................... 70

Lecture 20: Summary ............................................................................................................................ 79

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Lecture 1: Stress, strain-rate and viscosity

Rheology is the science of deformation and flow. Before studying topics in fluid mechanics such as pipe flow, mixing and flow in fluidised beds we must first introduce the language of rheology. In this course, we will be concerned mainly with shear deformations, i.e. deformations in which layers of fluid can be thought of as sliding past one another as shown in figure 1. In figure 1a, the block of material initially occupying the region ABCD, is deformed, by the application of a force such that the material moves to occupy the region ABC’D’. The material has been sheared. Figure 1b shows how the same deformation can be thought of as a stack of infinitesimally thin layers of fluid which slide over each other.

Figure 1. i) Shear deformation of a block of material,

ii) representation of shear deformation as sliding of infinitesimally thin layers of fluid.

(i)

(ii)

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Shear Stress In rheological terms, it is convenient to talk in terms of stress rather than force. Shear stress can be defined as the force per unit area either causing of caused by the flow.

Figure 2 shows a block of material being deformed by application of a shearing force, the shear stress is defined as the shearing force divided by the area over which that force is applied (shown in blue). The units of shear stress are Pascals (Pa). The advantage of considering stress instead of force concerns the fact that if the distance L in figure 2 was increased to 2L, the same deformation could be achieved by applying a force double that of the original, however, the stress would remain constant. Hence, when talking about stress we do not need to be concerned with the geometry of the material we are studying.

Figure 2. Stress is defined as the shearing force divided by the area through which that force acts.

Shear Strain Rate (Shear Rate) When a shear stress is applied to a block of material the block will begin to deform as shown in figure 1. We require a quantitative description of this deformation. Consider a volume of fluid, ABCD, of unit length perpendicular to the page).

L

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Now apply a shear stress to through the plane DC. In time , ABCD distorts to ABC’D’ and the change in shape can be expressed by the angle .

Figure 4: Definition of strain rate in terms of the deformation angle. The distance moved by D in time (i.e. the distance DD’) can be expressed as

where is the increment of velocity between AB and CD, hence, figure 4 can be generalised to apply to any small element of fluid in any of the layers shown in figure 1(ii). Now the angle can be calculated from the distances and DD’ (or CC’) using basic trigonometry as

which for small angles simplifies to

This defines a shear strain, which we denote as .

Consider a Hookean solid, in this case there will not be flow but there will be distortion, i.e. a deformation such that

Where G denotes the ‘shear modulus’ of the Hookean material. However, for a fluid, the angle continues to increase for as long as the stress is applied. Therefore, for a fluid we relate the stress to the RATE OF CHANGE OF , i.e. , not .

u

u + du

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Since

then

i.e. the velocity gradient. Shear stress can be related to velocity gradient through the equation

In the above equation, is termed the coefficient of viscosity and is a constant for a Newtonian fluid. The term is called the velocity gradient in the y direction. The equation simply states that applying a stress to a fluid causes a velocity gradient to be set up in the y direction, i.e. each of the fluid layers in figure 1 (ii) moves with a velocity slightly greater than the underlying layer, for a Newtonian fluid the plane at height 2y, will be moving with twice the velocity of the plane at height y. The velocity gradient is also called the shear strain-rate, shear-rate. In practice, the terms shear strain-rate, shear-rate and velocity gradient are used interchangeably. The units of shear rate are reciprocal seconds, s-1. Characteristic shear rates for several processes are given in table 1.

Process Shear Rate / s-1 Applications

Sedimentation 10-6 to 10-4 Paints

Extrusion 100 to 102 Polymers

Eating 101 to 102 Foods

Mixing 101 to 103

Coating 105 to 106 Paper

Lubrication 103 to 107 Engines

Table 1: Typical shear rates associated with some common processes. Newtonian Fluids Newtonian fluids are characterised by a constant shear viscosity, µ, with respect to shear rate, (or equivalently, applied shear stress, σ). However, the viscosity will change with both temperature and pressure. Examples of Newtonian fluids include water, glycerol, simple oils (such as silicone oils), organic liquids and low concentration suspensions. The flow curve for a Newtonian fluid is shown in figure 2.1. Non-Newtonian Fluids Newtonian fluids are characterised by a shear rate independent viscosity, however, many materials do not display such behaviour and the viscosity changes as a function of the applied stress (or resulting shear rate). As a consequence we can no longer assign a single value of viscosity to the fluid and must instead discuss an ‘apparent viscosity’, at a given shear rate, hence .

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Figure 2.1: Flow curve for a Newtonian fluid, the Newtonian viscosity can be calculated from the gradient of the line.

In this course we will consider three types of non-Newtonian fluid, shear thinning, shear thickening and yield stress (or Bingham plastic) materials. The viscosity of a shear thinning fluid (such as ketchup) decreases as a function of applied stress. When a bottle of ketchup is turned upside down, it flows slowly until we squeeze the bottle (increasing the stress) at which point the increased stress causes a decrease in viscosity and the ketchup flows out of the bottle easily. The viscosity of shear thickening fluids increases as a function of applied shear stress, there a fewer examples of shear thickening fluids than the other types of non-Newtonian fluid we will study but a good example is a concentrated corn-flour solution. Many videos of the consequences of shear thickening in these solutions can be found on the internet, for example, it is possible to walk on top of a concentrated corn flour solution provided that one keeps moving, standing still reduces the shear rate, the viscosity decreases, and the subject sinks into the solution. The third class of fluids we will study in this course are materials that flow only after a yield stress has been overcome. These fluids can be described as Bingham plastics. Mud (the pumping of mud is very relevant to drilling and mining operations) and mayonnaise are typical examples of materials that display a yield stress. These material behaviours are summarised in figure 5, note that in figure 5 the axis are both linear.

∆σ

8

Figure 5: Flow curves for Newtonian, shear thinning, shear thickening and Bingham plastic materials.

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Lecture 2: Models for non-Newtonian fluid behaviourand time dependent viscosity

A note on presenting data: Rheologists often plot stress or viscosity as a function of shear rate on log-log plots in order to present the huge range of shear rates that are relevant to the materials behaviour. It is important to note that on such plots, Newtonian, shear thinning and shear thickening fluids can all be represented by straight lines (see figure 1).

Figure 1: Shear stress as a function of shear rate for Newtonian,

shear thinning and shear thickening materials. Note how all three can be represented as straight lines due to the log-log nature of the plot.

We can re-plot the flow curves for Newtonian, shear thinning, and shear thickening behaviour in terms of viscosity as a function of shear rate as shown in figure 2. The viscosity is constant (with respect to shear rate) for the Newtonian fluid, decreases for the shear thinning fluid and increases for the shear thickening fluid. However, the curves shown in figure 2 for the shear thinning and shear thickening fluids are only relevant over a limited range of shear rates, for example, if we were to extrapolate the curve for the shear thinning fluid to very small strain rates, the extrapolated data would suggest an infinitely large viscosity, obviously this is not accurate. Similarly, extrapolating to very large strain rates would result in zero viscosity, again this is obviously inaccurate. The viscosity will actually approach asymptotic values as the shear rate becomes either very small or very large, these asymptotic values of viscosity are termed the zero-shear viscosity and the high shear rate plateau viscosity, respectively. Consequently, a plot of viscosity as a function of shear rate for a real shear thinning fluid would appear as shown in figure 3.

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Figure 2: Apparent viscosity as a function of shear rate for Newtonian,

shear thinning and shear thickening fluids.

Figure 3: Apparent viscosity as a function of

shear rate for a shear thinning material.

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The first model we will consider was proposed by Malcolm Cross, an ICI Rheologist. Whilst working on dye-stuffs and pigment dispersions Cross noted that the viscosity of many suspensions could be described by an equation of the form

This model is known as the Cross model and contains four fitting parameters, the zero shear viscosity, , the high shear plateau viscosity, , K and m. For shear thinning fluids m takes values in the range and indicates the degree of shear thinning. The Cross model describes a Newtonian fluid when and the ‘most shear thinning’ fluids as . One way of measuring viscosity is to use a rheometer. These devices operate in either controlled rate mode, in which a given strain rate is imposed upon a fluid and the resulting stress is measured, or in controlled stress mode in which a given stress is applied to a material and the resulting strain rate measured. Both types of device involve well defined geometries in which uniform strain rates can be achieved. Most commercial rheometers are capable of accurately applying strain rates over a limited range. Whilst this range of shear rates covers several orders of magnitude it is not uncommon for rheological data spanning this range to reveal only a part of the flow curve described by the Cross model, hence it is not always possible to completely define all the parameters in the model. Figure 4 shows viscosity as a function of shear rate for a typical fabric washing liquid, it displays shear thinning behaviour but there appears to be no zero-shear viscosity nor high shear rate plateau viscosity.

Figure 4: Typical viscosity data for a fabric washing liquid

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In this case we can assume that and , hence the Cross model can be reduced to

or, by a simple change in the parameters to the well know power law model

or

The power law is one of the most commonly used models of non-Newtonian fluid behaviour. Through suitable choice of the power law index n the model is able to describe shear thinning, shear thickening and Newtonian fluids as shown in table 1.

Fluid Behaviour Power Law Index

Shear Thinning 0 < n < 1

Newtonian n = 1

Shear Thickening n > 1

Table 1: Values of the power law index for Newtonian, shear thinning and shear thickening fluids.

However, the power law model must be used carefully as it is only applicable over a limited range of shear rates. Using the power law model, as , but actually , similarly, as , but actually . Hence, extrapolating the data beyond the range of shear rates over which it is known to apply (which will depend of the material, temperature and pressure) can have disastrous consequences. Secondly, the units of (the consistency) depend on the value of n. As such, two materials may have the same value of but the units may differ and hence cannot be used as a basis for comparison between the two materials. In many cases a better fit to the experimental data can be achieved by adapting the power law model to include the high shear rate plateau viscosity,

This is known as the Sisko model and is often used to describe the flow behaviour of suspensions and emulsions. Hence, we have three models to describe the flow behaviour of shear thinning materials with each being applicable over a range of shear rates as summarised in figure 5. Conveniently, the power law model can also be used to describe shear thickening behaviour and all three of the models discussed can be reduced to the Newtonian fluid model as a special case (i.e. for a given combination of parameters).

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Figure 5: Regions of applicability of the Cross, Power law and Sisko models. The dashed line shows the power law model fit to the data and has been extrapolated beyond the limits of applicability to

demonstrate the large deviations from the actual viscosity in the extrapolated region. The Bingham Plastic Bingham plastic fluids exhibit a yield stress below which no flow is observed but above which stress appears to be linear function of strain-rate. Bingham plastic behaviour is shown diagrammatically in figure 6. Mathematically the behaviour can be expressed as

for else

Where represents the yield stress and the plastic viscosity. Toothpaste, mud and many slurries

are examples of materials that display Bingham plastic like flow properties.

Figure 6: Typical flow curve for a Bingham plastic material.

Cross Model

Power Law Model

Sisko Model

∆σ

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Time dependent viscosity In all of the models discussed so far it is important to note that changes in viscosity (at a given temperature and pressure) occur only as the result of a change in shear rate. If a given shear rate is applied to such a material for any length of time the viscosity of the material would remain constant. However, other fluids display a property known as thixotropy, i.e. a time dependent viscosity. The viscosity of a thixotropic material will decrease as a function of the shearing time at a given strain rate. Upon cessation of shear the materials structure will begin to reform and the viscosity will return to its original value. Mayonaisse is a good example of a thixotropic material, stirring the mayonnaise breaks down some of the structure and so the viscosity decreases as a function of the stirring (or shearing) time. When one stops stirring the structure of the mayonnaise will begin to reform. Some fluids display the opposite behaviour in which viscosity is observed to increase as a function of shearing time, such materials are called rheopectic. It is very important to note the difference between shear thinning materials (in which changes in viscosity are associated with a change in shear rate) and thixotropic materials (in which changes in viscosity occur as a function of shearing time). Thixotropic materials are far more common than rheopectic materials.

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Lecture 3: The Poiseulle Equation

In many industrial applications, fluids flow under a laminar conditions through straight pipes, this situation can be modelled quite easily and it is instructive to consider the relationship between pressure drop, volumetric flowrate and pipe radius. The Poiseuille equation relates these variables and can be derived as follows (note: you will not be asked to derive this equation in an exam, however, you may be required to use the result or discuss any assumptions made in its derivation). Consider a balance of forces on a short cylinder of length and radius of fluid in tube flow as shown in figure 1. The flow is slow and steady, is occurring at constant temperature and there is a pressure difference (per unit length). The tube is straight and the flow is fully developed (i.e. the short cylinder we are considering is well away from the entrance/exit of the tube). The equations we will develop are relevant for Reynolds numbers up to approximately 2300, above this the flow will become turbulent and some of the assumptions we will make are no longer valid.

Figure 1: Forces on a short cylinder of fluid in tube flow.

In order to analyse this system we can make several simplifying assumptions:

1) Flow is parallel to the axis of the pipe (i.e. laminar flow) 2) Velocity u of any fluid element is a function of r only 3) Fluid is incompressible 4) Fluid velocity is zero at the wall, i.e. no slip boundary condition. 5) A unique function relates shear rate to stress, i.e.

Each particle of fluid inside a cylinder of length and an (arbitrary) radius moves with a constant velocity, therefore the net force on the cylinder is zero (because there is no net acceleration or deceleration).Hence, a balance of forces is established in which the force driving the flow (i.e. the pressure) is balanced exactly by the forces resisting the flow (i.e. the viscous forces). Mathematically this can be expressed as:

Pressure Force

Viscous Force

(1)

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where is the pressure drop per unit length of tube (i.e. the pressure gradient). If we take the positive direction of shear stress to be opposite to the direction of flow (i.e. in the diagram above, shear stress acts from right to left while the flow is left to right), we can write…

This allows us to have positive values of both and . Rearranging equation 1 allows us to write the following equation for the shear stress at radius …

Hence the shear stress at the wall, W, of the tube can be expressed as

It is convenient for us to express fluid velocity and volumetric flow rate in terms of .

i.e.

Separating the variables ( and ) and integrating from radius to the tube wall gives

This equation tells us that the velocity at any radius r can be calculated if we know , later we will use this equation to derive velocity profiles for Newtonian, Power law and Bingham plastic type fluids. Now, the flow per second through any cross-section, (i.e. the volumetric flow rate, V) is obtained by summing the flows through the annuli between and .

Note:

Note: The no slip criterion gives us the boundary condition u = 0

where

(2)

(3)

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i.e. (by substituting u in equation 3 for equation 2)

Or, by grouping constants and moving outside the integration,

Now, since, , we have

or (see below for simplification steps)

A graph of against W gives a unique line for all radii and pressure gradients. The integral can be simplified using integration by parts for a definite integral,

.

Let

,

therefore

0 0

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The Newtonian Case For the Newtonian fluid, , where is the constant viscosity. Therefore,

Becomes

Substituting

And rearranging gives

This is the Poiseuille Equation. An important consequence of this relationship is that for a Newtonian fluid slowly flowing through a circular pipe at a given volumetric flow rate, the pressure drop per unit length of pipe goes with the inverse of the radius of the pipe to the power 4! In other words, doubling the pipe radius, causes the pressure drop along the pipe to decrease by a factor of 16, (1/24).

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Lecture 4: Velocity profiles and P, V relations for non-Newtonian fluids

In lecture 3 we derived the Poiseuille equation which allows us to predict (subject to the assumptions involved in the derivation) pressure drops for Newtonian fluids flowing slowly in straight cylindrical tubes. In this lecture we will derive the equivalent equation for a Bingham plastic material and look at how we can predict the shape of the velocity profile for slow flow in straight circular pipes. Velocity profile for a Newtonian fluid in tube flow We can use equation 2 (from lecture 3) to calculate the velocity profile within a circular pipe, starting with

we can substitute

to give

or, bringing the constant viscosity outside the integral,

Substituting and (see lecture 3),

Therefore,

(1)

This equation tells us the velocity profile of a fluid (of viscosity ) in a pipe of radius a flowing due to a pressure difference (per unit length) p.

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Figure 2 shows how the position of fluid particles a short time after being in some cross-sectional plane (this is termed the velocity profile). The velocity profile is parabolic.

Figure 2: Velocity profile for a Newtonian fluid in laminar tube flow.

The profile has a parabolic nature. The equation also implies that both the shear rate (i.e. the VELOCITY GRADIENT) and the shear stress vary linearly from a maximum at the tube wall to zero at the pipe axis. Try differentiating equation (1) with respect to r. The Bingham Plastic The Bingham plastic model can be written as

for else

i.e. if the stress is below a yield stress there will be no velocity gradient. Hence, equation 3 (lecture 3), i.e,

the derivation of which assumed no material model, can be written as

or

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This is the Buckingham equation which includes an additional term (when compared to the Poiseuille Equation) that accounts for slippage at the wall. The Power Law Fluid For a power law fluid the equivalent relationship is

Note the effect of pipe variables such as pipe radius, a. For a Newtonain fluid the pressure drop for a given flow rate is proportional to the 4th power of the radius. However, if the material is shear thinning with n = 1/3 (which is typical) then the pressure drop is proportional to the square of the radius. These considerations are very important in the scale up of pipe flow. Velocity profiles Velocity profiles for non-Newtonian flows (slow, incompressible, straight pipe etc) are shown below. The equations governing these profiles can be derived by substituting the relevant fluid model into equation 2 (from lecture 3). Note that the Bingham plastic velocity profile is truncated due to the stress in the central plug of the pipe being below the yield stress. In shear thinning flow the velocity profile is also blunted but this is due to a more gradual increase of viscosity as the pipes centre line is approached. For shear thickening fluids, the viscosity decreases as the pipe centre is approached leading to a ‘sharpening’ of the velocity profile.

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Lectures 5 & 6: Fluid Mixing and Agitation A huge variety of agitation equipment is available (See Coulson & Richardson Volume 1), as a guide to impellor selection we can use the fluid viscosity

* Remote clearance, ** Close clearance Typical power requirements for mixing processes are shown below

Low Power Suspending light solids Blending low viscosity fluids

0.2 kW/m3

Moderate Power Gas dispersion Liquid/liquid contacting Some heat transfer problems

0.6 kW/m3

High Power Emulsification Gas dispersion Suspending heavy solids

2.0 kW/m3

Very High Power Blending pastes Dough

4.0 kW/m3

Calculation of the power requirements demands knowledge of many factors such as

o Diameter of agitator o Diameter of tank o Height of agitator from base of tank o Liquid depth o Width of baffles o Speed of agitator o Pitch of agitator o Number of baffles o With of agitator blade

In order to discuss how power input effects problems in fluid mixing scale up, we will use the standard Rushton turbine in which the configuration is completely defined by knowledge of only the tank diameter (DT) as follows:

Impellor Type Fluid Viscosity

Propellor* < 2 mPas

Turbine* < 50 mPas

Paddle* < 1000 mPas

Anchor** Increasing Viscosity Helical Ribbon**

Helical Screw**

Increasing Speed

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Some design data is available that can be used to predict power requirements in mixers of standard configurations such as the Rushton turbine. Figure 1 in the supplementary document to these notes shows the power number as a function of Reynolds number for a standard rushton turbine. Note that the power number is defined as

Whilst the Reynolds number for the mixing unit is defined in terms of the impellor diameter and tip speed

Example: Consider water (at STP), agitated by a standard Rushton turbine where DT = 2m. Calculate the power per unit volume required for a baffle width of 0.1DT. Use a tip speed of 3m/s as a typical design criterion.

As we are told we are dealing with a standard Rushton turbine, we know that if DT = 2m, H = 2m and D = 0.67m. We also know that for water the density is approximately 1000 kg/m3

and the viscosity is 1.0 mPas. Therefore, the impellor speed can be calculated as

Tip speed

Tip patch circumference

4 Baffles

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and the Reynolds number as

We now consult figure 1 in the supplementary information (this would be provided in an examination scenario), we see that for the 10% baffle case at Re =6.4 x 105, the Power number is 6. Hence,

Hence,

Now, we are asked to calculate the power per unit volume, hence,

And the power per unit volume is

Scale up of mixing operations Engineers are frequently called upon to design industrial scale mixing units given processing data obtained in either laboratory or pilot plant scale experimentation. In mixing processes we can use the ‘fully turbulent region’ of figure 1 (in the supplementary information) to guide our scale up calculations. However, it is first necessary to chose a ‘basis for scale up’. Constant mixing time: The initial cost of an industrial size mixer (i.e. impellor + motor + gearbox) is most closely related to torque rather than power and hence we can re-write the power number in terms of torque as

which in the fully turbulent region is a constant. Therefore, if we consider two mixing units (1 and 2) where mixer 2 is a scaled up version of mixer 1 we can write that the power number for the two mixers is equivalent….

Now, approximately, Ntm = constant and hence

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And since we are using a constant mixing time approach, tm1 = tm2 , therefore N1= N2. Also, the density of the liquid doesn’t change on scale up and so

Rearranging gives

and hence

i.e. if , then ! Such a mixer would be extremely expensive as all of the mechanical parts would need to be very heavy and strong to cope with the massive increase in torque. As a result, in scale up we DO NOT expect the mixing times to be constant. Constant tip speed On economic grounds we might be willing to expect to have the same torque per volume ratio on the larger scale as in the smaller scale. This can be written as

In the fully turbulent region

Splitting the D5 into D3D2 allows us to simplify by cancelling the constant T/D3 terms and also the again constant density,

Or by rearranging

i.e.

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Indicating that tip speed decreases with increasing diameter. Since, , we can write:

And so the mixing time is expected to increase on scale up i.e. when . Constant power per unit volume Turbulence theory suggests that for systems such as gas dispersion or liquid-liquid dispersion, the power per unit volume should be the basis for scale up. In this case

in the fully turbulent region

and hence,

or

Example: continuation of previous calculation for power requirement. What speed should the impellor (standard Rushton turbine)be driven at in a geometrically similar 4m diameter tank on the basis of scale up at equal power per unit volume and what power is required from the motor?

For the 2m diameter tank, the power to drive the impellor (of 0.67m diameter) at 1.43 rev/sec was 0.38 kW/m3. Hence for the larger tank, the impellor (now 1.33 m diameter) should be driven at

Since the power per unit volume is to be equal, the power required to run the unit will be

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Lecture 7: Forces due to fluids in motion

Estimating hydrodynamic forces (when acceleration is present) always involves Newton’s second law of motion, for a fixed mass, m, all of which is moving at the same speed u,

Where the arrow above F and u indicate that the force in a particular direction is associated with a movement in that direction, i.e. the force and velocity are vector quantities. Newton’s 2nd law is also applicable to a mass of fluid and can be be found by considering the rate of change of momentum.

Consider a mass of fluid lying within a stream tube ABCD at the beginning of a time interval t,

At the end of the time interval, the fluid has moved to A’B’C’D’

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We can make the following assumptions:

1) The ends of the stream tube is perpendicular to the sides 2) The fluid is in steady motion 3) u1 and u2 are constant across the each end of the stream tube respectively.

After the time interval t, the mass of fluid has moved towards the right and if the fluids

momentum will have changed. Now if we consider a small element of fluid m,

we can write its momentum (in the x-direction) as

where, is the component of velocity in the x-direction. Hence, by summing over many small elements, we can express the total x-momentum of the whole mass of fluid in ABCD (at the

beginning of t) as

And at the end of t, the total x momentum in A’B’C’D’ can be expressed as

Hence, the change in momentum can be expressed as

Or, by expanding

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But since,

The expansion reduces to

Now, let’s make another assumption

4) let dt be very short such that is nearly the same for all m in the volume CDD’C’ and the same (but at a different value) in ABB’A’.This is equivalent to saying that the streamlines do not diverge of converge in short distances.

The mass of fluid that has crossed the boundary CD in time t can be expressed as

And the mass of fluid that has crossed the boundary AB in time t can be expressed as

Hence we can express the momentum leaving and entering the volume ABCD in time t as

and

respectively. So the change in x-momentum in time t of the mass of fluid is the difference between the momentum leaving and the momentum entering the ends of the stream tube in that time.

Now the mass of fluid entering ABCD through the boundary AB in time t can be expressed as

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(i.e the product of the density () and the volumetric flow rate (a1u1) and the time interval t where a1 is the cross sectional area of the streamtube at AB.

Hence the x-momentum passing through AB in time t can be expressed

whilst the x-momentum leaving the ABCD through CD in timet can be expressed as

Hence, we can write the change of x-momentum as

Remembering that Newton’s second law says that the force is equal to the rate of change of momentum we can now write,

i.e.

Which represents the total force (in the x-direction) exerted on the fluid in the stream tube ABCD. The force can be made up of x-components of pressure forces acting normally on the periphery and ends of the stream tube or from the x-components of the tangential forces acting along the periphery of the stream tube. A similar analysis for the y-component leads to

Note that Fx and Fy are the forces exerted on the fluid. Their reactions, -Fx and –Fy are the forces acting by the fluid on the boundaries as a result of the imposed accelerations. It is usually these reaction forces we wish to find. The two force equations

Have been derived for a single stream tube with ‘ends’ of area a1 and a2 over which the fluid is moving with constant velocities u1 and u2 respectively. However, we could equally apply the equations to a ‘bundle’ of adjacent stream tubes not all of which contain fluid moving with the same velocity. The cross-sectional areas of the individual stream tube ends (previously a1 and a2) are then

only small elements (a) of the total cross sectional area for the bundle of stream tubes. In this case

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we simply sum the contributions from each constituent stream tube, this is expressed mathematically as

Where a1 and a2 are the ends of the ‘control volume’. These two equations are the basis for all determinations of the forces exerted by a fluid on its boundaries. It is often possible to make a series of simplifications to these equations. For example, if the fluid is incompressible . If the x-direction is chosen at right angles with one end of the stream tube, and the flow is parallel there, then and hence whilst , therefore

If both u1 and u2 are constant over the areas a1 and a2 then the integrations simplify to

Further, since flow (or discharge) into the control volume is

Then

Finally, it may be possible to make the assumption that changes in direction are small, hence, whilst is finite, therefore…

i.e Fx is the product of the mass flow (Q) and the velocity change (u2-u1).

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Lectures 8 & 9 : Forces due to fluids in motion II Consider a jet of inviscid fluid of density ρ, cross-sectional area a and uniform speed u striking a large flat surface normal to the x-axis

After the jet hits the surface it eventually leaves the control volume parallel to the surface, i.e. it does not reflect. Let the x direction be along the axis of the jet. The discharge of the jet can be written as and hence the rate at which x momentum arrives in the control volume (i.e. the product of the mass flowrate and the velocity component in the x direction) is . If the plate is large enough, compared with a then the fluid leaves parallel to the surface of the plate (i.e. at right angles to the original flow direction) and since there is no reflection of the jet (i.e. no splash) there is no component of momentum in the x direction leaving the control volume. Hence the rate of change of momentum (which we know to be equal to the force through Newton’s second law) can be written as

Assuming that the jet divides equally on impact such that the mass flowrates of material leaving the control volume in either direction are equal, we can say that the y momentum leaving the control volume to the left is equal to the y momentum leaving to the right and so there is no net momentum in the y direction (where the y axis is perpendicular to the x axis). Further, as the x axis has been defined along the direction of the jet, no y momentum enters the control volume and hence there is no change in momentum in the y-direction. Therfore, no force is exerted on the plate in the y-direction. But what if the plate is inclined?

u

a

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In this case some of the momentum (along the axis of the jet) entering the control volume will be retained by the fluid, however, we can simplify the situation significantly by defining the axis along the XX direction as shown above. In this case the momentum entering the control volume along the XX direction can be written as (again the product of mass flowrate and velocity component in that direction)

Now, after impact none of this momentum remains and the momentum leaving the control volume along the direction XX is zero. Hence the change in momentum in the XX direction (and hence the force) can be written as

Now the component of momentum entering the control volume in the YY direction (defined perpendicular to XX) can be written as

If this component is to change then a force must be exerted on the fluid in the YY direction and this force would be a shear force acting due to the flow of the fluid over the plate. However, we are talking about an inviscid or ideal fluid (i.e. with zero viscosity) and so no shear forces are possible, hence there can be no change in momentum in the YY direction. Hence, the only force caused by the impact of the ideal fluid on the inclined plate is along the XX direction with a magnitude of

35

Lectures 10 : Forces due to real fluids in motion In lectures 7 through 9 we considered forces due to the flow inviscid fluids and found that no forces are exerted when such a fluid flows along a body. However, inviscid fluids do not exist! In this lecture we will consider the forces associated with real fluids flowing along surfaces. When a real fluid flows along a surface the fluid close to the surface is retarded by shear forces due to viscosity. Further away from the surface, outside the retarded, or boundary, layer the fluid is unaffected by the presence of the surface. A common engineering requirement is to find the shear force which has retarded the fluid and produced this boundary layer. If AB is an extensive, flat, solid surface over which fluid flows and A is the leading edge of the plane the boundary layer begins to develop at A and becomes thicker towards B.

Assuming that the pressure is uniform over the surface AB (i.e. there are no pressure forces) and that the only force acting is the shear force F we can say that at some distance Z along the plane the boundary layer thickness is δ and study the velocity profile in the boundary layer as a consequence of the retarding force F.

Outside the boundary layer the velocity is everywhere U but inside the boundary layer the velocity u changes as a function of distance from the surface y and u < U everywhere. (How u varies with y is found through experiment or more advanced theory). The shear force F (between A and Z) can only be found by applying the momentum theorem. The control volume is bounded by AZ, PA and QZ and the curved line PQ (note that QZ= δ).

P

Q

36

Assuming no momentum transfer occurs across PQ (nor AZ) momentum can only be transferred across PA and QZ. Now the momentum entering the control volume (per unit width of surface) can be written as

And the momentum leaving the control volume can be written as

And hence the change in momentum can be written as

Remembering Newton’s second law, this increase in momentum gives us the force exerted in the direction of motion by the surface on the fluid….. But F opposes the direction of motion, i.e. there is a reduction in momentum in direction of motion. Hence,

or

Now, using the continuity equation ( )

Multiplying through by U gives

Hence

Now assuming that is small so that we get

37

We cannot evaluate this integral without knowing u(y) and a fairly accurate approximation is

which leads to

The boundary layer thickness δ depends on: The roughness of the underlying surface Fluid properties Distance AZ The above analysis is valid when the pressure along AB is constant, when the pressure along AB is not constant much more complicated velocity distributions may apply which lead to quite different equations for the force. Drag force on a solid object in a stream Consider relative motion between a solid object and a surrounding fluid. The solid will exert a retarding force on the object and vice versa . One way to measure this force is measure the velocity distribution in the wake of the solid object. Let object ‘O’ be the cross-section of a long cylinder which is subjected to an oncoming velocity U. Behind O, on the cross section XX of the wake the fluid velocity is decreased and varies as a function of distance from the centreline y. At some distance δ from the centre line the fluid is unaffected by the presence of the object and hence the only part of the fluid that looses momentum is that lying in

the region –

38

Assuming that object ‘O’ has the same cross-section in all planes parralell to the diagram then, the mass flow through a slot in the wake wide is:

And hence the change in momentum per unit time is

And the total momentum change per unit time across the whole cross-section XX is given by

Which is the same expression as was derived for the boundary layer. By Newton’s second law this change in momentum is equal to the sum of all forces acting on the fluid between the upstream cross-section (constant velocity U) and the section XX. Those forces are the drag force D and any force caused by the pressure along XX not being the same as the upstream pressure P0. Since the pressure in the wake will differ from P0 but elsewhere the pressure and velocity will be the same upstream and downstream of the object we can write that the total force on the fluid is

And hence the observed momentum change can be expressed as

That is to say that we must measure both the velocity and pressure distribution in the wake of the object if we wish to calculate the drag force.

Only if P = P0 everywhere in the wake will the drag force be expressed as

Drag Coefficient

The flow of fluids past objects such as cylinders and spheres has been studied experimentally by many researchers and it has been found that the parameter of importance in determining the flow behaviour around these objects is the Reynolds Number. This is an extremely important result as it means it is possible to study the flow around wings etc using scaled models in a wind tunnel.

39

Much of the data regarding flow around spheres or cylinders is presented as plots of the drag coefficient as a function of Reynolds number.

In the previous section we defined the drag force on an object, D, however the drag coefficient is defined in terms of the drag force per unit projected area, R’. For a sphere the drag force per unit projected area can be expressed as

And the drag coefficient as

Plots of drag coefficient as a function of Reynolds number can generally be considered as consisting of four regions in which the fluid dynamics display different characteristics. Here we consider the situation for flow around a cylinder.

Region A (Re < 0.2)

In this region we find a linear relationship between drag coefficient and Reynolds number which can be stated as

40

In this region there is laminar flow around the object and the streamlines converge behind it.

Region B 0.2 < Re < 500

As the Reynolds number is increased to around 40 boundary layer separation occurs (see Mechanics of Fluids, B, Massey for an excellent explanation of boundary layer separation) resulting in the appearance of circulating eddies behind the cylinder. These eddies are stable but the wake is limited in length and the main streamlines come together again behind the eddies.

Increasing the Reynolds number further causes the eddies to elongate and eventually lose their stability and the eddies begin to break away from the object and Karman Vortex street is set up.

In this region the point at which the boundary layer separates from the object gradually moves upstream until it reaches a stable position.

In this region the following empirical relation has been employed.

Region C 500 < Re < 2 x 105

In this region, the point at which the boundary layer separates remains constant and hence a constant value of the Drag Coefficient is observed.

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Region D Re > 2 x 105

Until now the flow in the boundary layer has remained laminar, however when the Reynolds number reaches a critical value of 2 x 105 the boundary layer becomes turbulent before separation occurs allowing the boundary layer separation point to move further downstream again thus narrowing the width of the wake. There is a sudden decrease in the drag coefficient which approaches a value that is again independent of Re.

Region A for a sphere

For laminar flow around a sphere it is possible to show that the relation is a consequence of Stokes Law.

First expanding the Reynolds number

Multiplying both sides by gives

Multiplying both sides by the unit projected for a sphere gives

which is Stokes Law.

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Regions B, C and D for flow around a sphere

The drag coefficient for flow around a sphere varies in a similar manner to that discussed above for flow around a cylinder. The main difference is that the symmetrical eddies that are produced and which lead to the Karman Vortex street are replaced by a vortex ring. This ring is initially stable but will become unstable and move downstream as the Reynolds number is increased (the vortex that is flowing downstream is replaced by a new vortex ring immediately behind the object) .

43

Lectures 11 & 12: The Bernoulli Equation

Consider part of a steady, non-uniform flow, a small tube of ideal fluid enclosed by stream lines and which has plane ends – a stream tube.

In a length, δs, the cross sectional area of the stream tube changes from a to a + δa and the velocity from u to u + δu….

The pressure intensity changes from p to p + δp and the elevation from z to z + δz. Further there is a pressure force acting on the sides of the tube, if δs is very short we can assume that the pressure acting on the sides of the tube is equal to the average pressure i.e. p + δp/2

u + δu

u

δs

a+ δa

a

44

All of the flow, in to and out of the stream tube occurs through the ends of the tube and changes in velocity (u) are connected to changes in area (a) since Q = ua. We can derive the Bernoulii equation by considering the forces acting on this element of fluid. There are three forces acting on the fluid element, a force due to the pressure on the sides of the tube, a force due to pressure differences at the ends of the tube and a force due to gravity. Note that the fluid is assumed to be inviscid and so no shear forces are present. Forces due to pressure on the sides of the tube Varying pressure intensities along the sides of the tube cause a force, F1, in the direction of flow. Assuming that δs is small then F1 is the product of the mean pressure along the sides of the tube and the cross-sectional area in the direction of interest. Hence the forces due to pressure along the wall of the tube can be expressed as

But since both δp and δa are small we can neglect second order terms (i.e. we assume that δpδa =0).

u + δu

u

δs

a+ δa

a

δz

z

Datum Level

p + δp

p

p + δp/2

p + δp/2

45

Forces due to the pressure difference between ends of the tube The pressure intensities p and p + δp exert forces on the plane ends of the stream tube in opposite directions. Hence the force due to pressure differences between the ends of the tube, F2, can be written as

If we again neglect second order terms

Force due to the weight of the fluid element The element of the stream tube has a weight force F3 in the direction of motion, it’s mass is

And so the total weight force (which acts vertically downwards) has a component in the direction of flow

Where θ is the angle between the direction of flow and the vertical axis But since we can write

Multiplying out the bracket gives

Again neglecting seconds order terms

Total resulting force on the fluid element By summing F1 F2 and F3 we arrive at the following expression for the total resulting force on the fluid element

46

Now, Newton’s second law can be stated as

But by the chain rule we can write

And hence

Since

We get

And by again neglecting second order terms we can arrive at

Which can be equated with the total force on the fluid element to give

Or by dividing through by δs

Which in the limit of δs becomes

And so integrating with respect to s gives (note that all of the constants of integration can be combined to a single constant, C)

Which rearranges to

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Finally, dividing through by results in the Bernoulli Energy/Head Equation….

Note that the Bernoulli equation (as derived here) only applies to incompressible fluids in steady flow. Each term has units of length and can be regarded as a quantitiy of energy in a volume of fluid of unit weight (note that this mean Joules/Newton and that in the SI system I J = 1 Nm). The first term in the Bernoulli equation (as written above) is the ‘velocity head’ and represents kinetic energy. The second term, the pressure head, represents energy due to pressure within the fluid and the final term represents the potential energy due to gravity. The Bernoulli equation is often called the total energy of head equation. Example Water flows through a pipe of inside diameter 200 mm at a rate of 100 m3/hr. The flow abruptly enters a section reducing the pipe diameter to 150 mm for which the head loss is equivalent to 0.2 velocity heads based in the smaller pipe. Determine the force required to hold the pipe in position. Upstream of the reducing section the gauge pressure is 80 kN/m2.

In order to find the force Fx we must first find the velocity in the smaller pipe and then calculate the change in momentum occurring in the control volume (marked by the dashed lines in the diagram) [Remembering that there is also a pressure force being exerted on each end of the control volume]. The cross sectional area of the larger pipe, a1, is

Fx

Fx

48

And hence the velocity in the larger pipe is

Whilst for the smaller pipe (through continuity)

Now we are told that the head loss in the reducing section is 0.2 velocity heads based on the smaller pipe

We are told that the pressure upstream of the reducing section (i.e. in the larger pipe) is 80 kN/m2 but need to calculate the pressure in the smaller pipe. We do this by applying the Bernoulli equation…

Now since z1 = z2

Rearranging for p2

Hence

We can now work out the force driving the flow due to upstream pressure and the force resisting flow due to downstream back-pressure. The upstream and downstream forces are…..

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and

respectively. We can write for the force in the x direction

And hence

i.e. a force of 1096 N is required to hold the reducing section in place (in the opposite direction to flow). Example 2 A pipe carrying water tapers from a cross section of 0.3 m2 at A to 0.15 m2 at B. At A the velocity, assumed uniform is 1.8 m/s and the pressure 117 kPa gauge. If frictional effects are negligible, determine the pressure at B which is 6 m above the level of A.

Firstly, we need to know the velocity at B. Hence we calculate the volumetric flowrate using Q = ua at A and then use Q to calculate u2.

Hence,

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Through continuity,

We now have enough information to apply the Bernoulli equation

We are told that frictional losses are negligible hence HL by taking Iz1 as the datum level we get

Which rearranges to give

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Lecture 13: Laminar flow in packed beds

Consider the following arrangement of packed solid particles.

Flow occurs in the spaces between the particles, we call these spaces the voids of the bed and we term the volume fraction of the bed of particles that is void space the voidage, , of the bed. Hence, the volume fraction of the bed occupied by solid particles can be expressed as 1 – . For a bed of spheres in a cubic packing . We can also define SB as the surface area of the particles that is contacted by fluid per unit volume of bed and S as the surface area of the particles that is contacted by fluid per unit volume of particles. For a bed of n spherical particles, each of diameter, d, the total surface area can be written as

And the total volume of particles can be written as

Hence, S can be written

Since the particles only occupy 1 – volume fraction of bed then in general

We also need to define an effective diameter of the pore spaces and we chose to do this, in a similar manner to the ‘hydraulic mean diameter’, as

which reduces to

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Now the fluid must flow around the particles and as such follows a tortuous path through the bed. If the length of the path actually travelled by the fluid elements through a bed of depth L is L’, then the totuosity factor, τ, can be defined through the equation

Usually is not calculated but is introduced into equation in experimentally determined constants.

We now need to define two differenct velocities, the superficial and interstitial velocities. The superficial velocity, uc, is the velocity that the fluid would travel at through the tube containing the bed if the bed was not present, i.e, if the tube were empty of particles. This can be calculated in the standard way by dividing the volumetric flowrate (Q) by the area available for flow (a),

The interstitial velocity is the average velocity of the fluid as it travels through the pores of the voids in the bed. To calculate the interstitial velocity we must take into account the ‘effective cross-sectional area’ which is simply the area available for flow when the bed is present (fluid cannot flow through the particles). The interstitial velocity can be expressed in terms of the superficial velocity and the voidage of the bed as

The Carmen-Kozeny Equation Just as in previous work (see flow through pipes), engineers require (or would like) general expressions for pressure drop and mean flow velocity. For fluid flow through beds these should ideally be in terms of voidage and specific surface as these are known or measurable quantities. An equation for stream line flow in circular tubes can be written in the following form:

where u is the mean velocity of fluid, µ the coefficient of viscosity dt the tube diameter, lt the tube length and the pressure drop along the tube (see Poiseuille Equation). By arguing that the void space of the bed (where flow occurs) consists of a network of many interconnected tubes we can re-write the above equation to apply to flow through a bed as

where ui is the mean velocity of fluid in the voids (the interstitical velocity), µ the coefficient of viscosity, dm the equivalent tube diameter, l’ the mean channel length length, the pressure drop along the bed and K’ is some constant that replaces the constant ‘32’ that appears in the original equation.

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Remember that , hence,

Now, in order to make progress we need an expression for dm and Kozeny propsed that

[NOTE: This is related to the previously defined effective diameter dm by a constant factor of 4, as this is a constant it will simply get combined with the constant K’ in the next step to give the new constant K’’] Given that we can now write

[NOTE: the constant τ has also been combined with K’ in defining the new constant K’’] Experimentally Kozeny reported that K’’ = 5 and hence

or if we gather the voidage terms together for neatness

Now for spheres we know that (see previous) S = 6/d, hence

Or

This is the Carmen-Kozeny Equation for stream line flow which can be applied in the design of packed beds operating under laminar flow conditions. We will see how the equation is used in lectures 16/17.

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Lecture 14: Turbulent flow in pipes

Smoke rising from a lit cigarette in an ash tray is a typical example of the two types of fluid flow, initially the smoke flows in a laminar fashion but as the smoke rises the pattern becomes more disordered and chaotic, here the flow regime may be termed turbulent. Although these different fluid behaviours had been observed previously, it was not until the pioneering work of Osborne Reynolds, Professor of Engineering at Manchester University around 1880 that the nature of the two types of flow was described. Reynolds’ Experiment Reynolds fitted a small bore glass tube to a large tank of quiescent water and allowed the water to drain out of the tank at a rate set by a valve at the end of the glass tube. A stream of dye was then introduced at the entrance to the tube (see figure 1). By studying the behaviour of the dye stream at different flowrates Reynolds was able to determine the factors that affect the transition from laminar to turbulent flow.

At low flowrates the dye appeared to move along the central axis of the tube as a single stream tube with no transfer of material across the stream tubes. That is to say that the fluid particles move along smooth parallel layers in the direction of flow and the only velocity component is in the direction of flow. Any movement of material that does occur across stream lines is a result of molecular diffusion only. As the flowrate was increased, the dye stream became unstable before eventually becoming dispersed throughout the width of the tube, i.e. the dye has been mixed with the water due to random chaotic motion of the fluid particles, turbulence now dominates the system. As the flowrate was increased still further, the length along the tube at which the turbulent flow set in became smaller until it reached a very small distance in front of the nozzle. By carrying out the above experiment for different tube diameters and fluids Reynolds was able to define the set of variables that affect the transition to turbulent flow, these parameters are the diameter of the tube, the average fluid velocity, the viscosity of the fluid and the density of the fluid. These parameters can be arranged into a dimensionless group called the Reynolds Number

which represents the ratio of inertial to viscous forces. (This is the Reynolds number for pipe flow, other Reynolds numbers exist for other situations such as flow around a sphere or flow in a packed bed).

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There are 3 regimes of fluid behaviour that can be characterised by the Reynolds Number of the flow. For Re < 2000 the flow is laminar, as the Reynolds exceeds 2000 the flow begins to ‘break up’ Until for Re > 4000 the flow is fully turbulent. The region 2000 < Re < 4000 is termed the transition region and here a complex interplay between laminar and turbulent flows exists. In previous lectures we have used the Poiseuille Equation to look at a pressure drops under laminar flow conditions. In this lecture, we will look at how we can predict pressure drops in pipes in which turbulent flow exists. Pressure drop due to friction Consider flow along a long straight pipe of length L (the nature of the flow, i.e. laminar or turbulent, is unimportant at this point), see figure 2. We have a pressure, p1 acting at one end of the pipe and a pressure p2 acting at the other end of the pipe but p1 > p2 and a result we have a pressure difference and hence a force driving the flow of fluid. However, as a result of the flow (and friction) a stress, σ, will exist at the wall of the pipe acting in the opposite direction to the flow.

We can construct a force balance for this situation, for steady state conditions, the force driving the flow can be expressed as

Whilst the force resisting the flow can be expressed as

(i.e. the stress, the force per unit area multiplied by the surface area of the pipe in contact with the fluid) As there is no net force on the fluid (which would result in acceleration) we can equate the above equations to give

which reduces to

56

and rearranges to

Eqn (1) Now, as the stress is related to the kinetic energy per unit volume we can write,

and hence,

Eqn (2) Where f is some dimensionless coefficient that is termed the “Fanning friction factor”. Introducing equation 2 into equation 1 allows us to write

This is called the “Darcy Equation” and is equally applicable to laminar and turbulent flows with appropriate choice of the coefficient f. Estimating the Fanning friction factor There have been huge efforts over the last 150 years to establish relationships to predict the nature of fluid flow in smooth and rough walled pipes but still no theory exists which is testament to the complexity of the situation. However, many very useful empirical relationships have been proposed that allow estimation of the Fanning friction factor and hence use of the Darcy equation in the design of pipelines. Blasius (1911): Studied the flow in smooth walled pipes and found that f was a function of Re only. He proposed that f could be predicted using

For re in the range 4000 < Re < 10 000 Stanton & Pannel (1914): Studied the flow of various fluids in a variety of pipes (of different material), they suggested that f could be predicted using

57

Where the constant, a, b, and c depend on the construction material of the pipe. They presented their results in the form of a plot of f as a function of Re, a form that would later be adopted by Moody. Nikuradse (1936) and von Karmen: One of the most important studies was made by the German Engineer Johann Nikuradse. Nikuradse glued finely graded sand to the inside of smooth pipes to create a series of pipes with defined roughness. The relative roughness of the pipe was defined as the diameter of the sand particles, , divided by the pipe diameter, d. He then studied the flow of various fluids through these well defined pipes. Nikuradse’s data was fitted by von Karmen to the empirical relationship

Colebrook and White Colebrook and White extended the work of NIkuradse, von Karmen and using Prandtl’s Boundary layer theory propsed the following relationship

Moody (1944) In 1944, Moody produced a composite plot of f as a function of Re which included all of the work described above and assigned a value k to the roughness (comparable to the of Nikuradse) to several pipe construction materials. The Moody plot is still used today to estimate values of f for the design of pipelines. The Moody plot contains:

i) A straight line relationship for laminar friction factor ii) Smooth pipe turbulent friction factor iii) Rough pipe turbulent friction factor iv) The concept of relative roughness

The Special Case for Laminar Flow For laminar flow in a pipe, the Poiseuille equation tells us that the average velocity of the flow can be expressed as:

Hence

But the Darcy Equation also tells us that

58

Hence, by equating the above we get

Which simplifies to

And rearranging gives

Or

i.e. For laminar flow in a pipe (smooth or rough), f is inversely proportional to Re.

59

Lecture 15: Head losses in rough and smooth pipes

The Darcy-Weisbach Equation The Darcy equation can be written such that it expresses a frictional head loss suitable for use with flow calculation involving the Bernoulli equation. In this form the equation is known as the Darcy-Weisbach equation:

where Hf Head loss due to friction (m) f Fanning friction factor u is the mean fluid velocity (m/s) L is the length of pipe (m) d is the pipe diameter (m) g is the acceleration due to gravity (9.81 m/s2) The Darcy-Weisbach equation is sometime expressed as

where . Example Calculations

1. Determine the head lost to friction when water flows through 300 m of 150 mm i.d.galvanised steel pipe at 50 litres/s.

The first thing we need to do is calculate the mean velocity in the pipe, this can be calculated by dividing the volumetric flowrate (m3/s) by the area availbel for flow (m2).

Next we calculate the Reynolds number (assuming that the density is 1000 kg/m3 and the viscosity is 0.001 Pa.s)

We also need to know the relative roughness of the pipe, hence, from the Moody Chart we find that the absolute roughness of galvanised steel pipes (k) is 0.15 mm. So the relative roughness (k/d) is

We can now use the Moody Chart to estimate the Friction factor

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And finally we can use the Darcy-Weisbach equation to calculate the head loss

2. Calculate the steady rate at which oil (kinematic viscosity = 10-5 m2/s) will flow through a cast iron pipe 100 mm in diameter and 120 m long under a head difference of 5 m.

As yet Re is unknown (as we can’t calculate u), hence we must use an iterative approach. Firstly, we calculate the relative roughness (from the Moody Chart, for cast iron pipes k = 0.25 mm).

We now assume that we are working in the fully turbulent region hence, from the Moody Plot

We now rearrange the Darcy-Wiesbach equation to make u the subject…..

Now we can calculate Re

We now refer back to the Moody Chart and check the assumed value of f. At Re = 17 700, f = 0.0079 which is significantly different to the assumed value, hence we make another iteration…. Recalculate u using the new value of f

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Recalculate Re

We now use the Moody Chart once again to check our value of f . At Re = 16 000, f = 0.008. Although this value is close to the previous iteration it is best to check what effect if has on the velocity Recalculate u

This is close enough to the previous iteration to be accepted, hence we now calculate the flowrate

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Lectures 16 & 17: Flow through packed beds II

In lecture 13 we considered flow through packed beds and derived the Carman-Kozeny equation which relates the superficial velocity to the pressure drop across a bed of spherical particles provided that the flow rate is small enough for the flow within the void space of the bed to be described as laminar. In this lecture, we will first define a criterion (a modified Reynolds Number) to determine the nature of the flow within the pore space of the bed before studying how we can use the friction factor to estimate the pressure drop across beds where the flow regime is not laminar. A modified Reynolds number, Rei. We know that for flow in a pipe the Reynolds number (i.e. the ratio of inertial to viscous forces) is defined as

where ρ and µ are the density and viscosity, respectively, of a fluid flowing with mean velocity u in a pipe of inside diameter d. If we are considering a packed bed to be composed of a network of interconnected channels with equivalent diameter dm in which the fluid is moving with an average velocity equal to the interstitial velocity, ui, we can write an equivalent Reynolds number as

But we know from lecture 13 that dm can be expressed in terms of the bed voidage, , and the surface area of particles per unit volume of bed, SB, as

Further, we know that it is possible to express the interstitial velocity (which is difficult to measure) in terms of the voidage and superficial velocity, uc, (which are both relatively easy to determine) as

Hence, we can rewrite the modified Reynolds number as

Since the factor 4 serves only as a multiplier it can be dropped and the modified Reynolds number can be written

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A modified Friction Factor In the same manner as we have defined a modified Reynolds number for flow in a packed bed we can also define a modified friction factor which we define in terms the resistive force in direction of motion per unit area of particle surface,R1, as

Note how this is different to the definition of drag coefficient for flow around a single sphere/cylinder (see lecture 10) in which R’ was defined as the drag force per unit projected area. Now consider a packed bed of unit cross sectional area and length L The volume of particles in the bed can be expressed as

Or, since the bed is of unit cross sectional area

Then the total surface are in the bed can be written as

Where S is the surface area per unit volume of particles. Hence, since we previously defined R1 in terms of the resistive force per unit area of particle surface, we can write that the total resistive force in the direction of motion is

Now, if we consider a system in steady state there can be no net acceleration or deceleration of the fluid and hence we must have a balance of forces within the system, i.e. the force driving the flow must equal this total resistive force due to the presence of the bed. The force driving the flow is (as usual) the pressure difference across the bed, but what area should we multiply this pressure by in order to obtain the force driving the flow? Only a proportion of the cross sectional area of the bed is actually fluid, the remaining portion is the particle bed. If we consider all of the particles in the bed compressed into a single continuous solid mass on one side of the bed we can see that the pressure actually acts (such that it causes flow) over an area equal to a, or since we have specified unit cross-sectional area, . Hence we can write the force balance as

or, by rearranging,

And by dividing through both sides by we obtain

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But we would prefer to have an expression in terms of the superficial velocity rather than the interstitial velocity and hence write,

Or

Where S is the surface area contacted by the fluid per unit volume of particles is the voidage of the bed uc is the superficial velocity ui is the interstitial velocity

Carman plotted as a function of Re’ (using log coordinates) and found that his data for flow

through randomly packed beds of solid particles could be approximated by a single curve of the form

which has been plotted as figure 1. For Re1 < 2 the second term is small and so we can approximate the above equation to

i.e. that is a function of Reynolds number only. (We have seen this behaviour for laminar

flow before!). The second term can be related to turbulent eddies and its contribution to

gradually grows in significance as Re1 is increased from 2 towards 100, i.e. the slope of the plot gradually changes from -1 to – ¼ . At higher Reynolds numbers the similarity with pipe flow is lost and entirely empirical relationships are used. Ergun proposed the following relationship for a sphere which can be used over a wider range of Re.

or

Using the relationships proposed by both Carman and Ergun the second term gradually becomes more significant when Re1 > 2 (see figure 2). Hence for our modified Reynolds number we have a transition from the laminar to turbulent flow regimes occurring very gradually at Re1 > 2. Why might we expect a gradual change in the overall flow pattern?

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Figure 1:

as a function of modified Re as described by the Carman correlation.

Figure 2: Comparison on Carman correlation and the Ergun equation.

10-2

10-1

100

101

102

10-1

100

101

102

103

Re1

R1/

ui2

10-2

10-1

100

101

102

103

100

101

102

103

Re1

R1/

u2 c

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Example A packed bed is composed of cylinders having diameter and length 0.02 m. The density of the packing material is 1600 kg/m3 whilst the density of the bed has been measured as 962 kg/m3. Calculate the void fraction of the bed.

We can assume 1 m3 of bed hence the mass of particles within the bed is 962 kg/m3. Hence the volume of the particles in the bed can be calculated using material density

And so

Example A water softener consists of a vertical cylindrical pipe 0.5 m long and 50 mm internal diameter. The unit is packed with spherical particles whose diameter is 1 mm. The bed porosity is 0.33. The column runs full of water under a head of 0.2 m. Treated water trickles out of the bottom of the unit and is collected for later use. Using the Carmen-Kozeny Equation, estimate the flow-rate of treated water leaving the unit and the mean residence time of the water within the bed. Verify that the Carmen-Kozeny equation is applicable. The pressure at the exit plane is atmospheric (i.e. gauge pressure is zero) so the hydrostatic head driving the flow is the 0.2 m + 0.5 m = 0.7 m, this must be converted to Pa in order to use the Carman-Kozeny equation…..

Now the Carman-Kozeny equation for streamline flow can be written as

i.e. 6.1 mm/s

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Hence the volumetric flowrate (remembering that we have just calculated the superficial velocity) is given by

But the Carman-Kozeny equation is only valid in the laminar flow regime so we need to check that the column is operating under these conditions by calculating the modified Reynolds number….

Now S is the surface area per unit volume of particles and as we have previously seen for n spherical particles the surface area per unit volume of particles can be expressed as

and hence the modified Reynolds number is

i.e. laminar flow conditions (Re1 < 2) and so we can be confident about using the Carman-Kozeny equation. Finally the question asks us to calculate the mean residence time of water within the bed, as usual, the residence time can be calculated by dividing the volumetric flowrate by the vessel volume, however, as the water can only occupy the void space we must use only this volume in calculating the residence time….

Example A packed bed tubular reactor has been designed in which catalyst is bound to the surface of 2 mm diameter particles. The bed voidage is 0.4 and reactants are to be fed to the system at a flowrate of 500 L/min, calculate the pressure drop along the reactor if the proposed tubular reactor is 2 m long and 100 mm in diameter. The density of the reactant is 1 kg/m3 and the viscosity is 10-5 Pa.s. First we need to calculate the modified Reynolds number and hence need the superficial velocity.

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and so the Reynolds number can be calculated as

This Ergun equation is most applicable (not that we cannot use the Carman-Kozeny Equation)

And so we can estimate the pressure drop by rearranging the following

Example 4 Calculate the pressure drop of air flowing at 30oC and 1 bar pressure through a bed of 1.25 cm diameter spheres, at a rate of 60 kg/min. The bed is 125 cm diameter and 250 cm height. The porosity of the bed is 0.38. The viscosity of air is 0.0182 mPa.s and the density is 0.001156 g/cm3. In order to calculate the Reynolds number we first need to change the mass flowrate into a volumetric flowrate and then calculate the superficial velocity

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and so

In this case the Reynolds number is too high for us to use the Carman correlation but we can use the Ergun equation.

Now using the general equation for the friction factor

And rearranging for ∆p

Hence,

Hence

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Lecture 18 & 19: Differential head flow meters

Differential head flowmeters indirectly measure velocity and therefore ‘flow’ by measuring a differential head and are sometimes referred to as “head meters” of “rate meters”. Many types of differential head flow meters exist but the main group are Venturi meters, orifice plates, nozzles and Pitot tubes. All of these work on the same principle which is based on the Bernoulli equation, when a fluid flows through some restriction its velocity is forced to increase (by continuity), hence, there is an increase in kinetic energy which evolves from a reduction in pressure through the restriction. By measuring the decrease in pressure and applying the Bernoulli equation we can calculate the flow velocity and hence the flow rate through the restriction. Before considering specific devices let’s consider a general contraction, consider a circular pipe in which a constriction is present such that the pipe cross-sectional area reduces from a1 to a2 in a distance of about

, which causes the flow velocity to increase from u1 to u2.

Through continuity we can write

Hence

And if we assume that the velocity is uniform across each cross-section we can write that the Bernoulli equation applies for every stream line passing through every cross-section

And since the total head is assumed to be constant along every stream-line we can write that

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which rearranges as

but, from continuity we have

and so

or

and by rearranging for u2 we get

which can be written as

hence, assuming there is no degradation of energy, the flowrate can be expressed as

however, since there will be some degradation of energy we must introduce a coefficient of discharge, Cd such that

For a well shaped convergence, Cd is approximately 0.98 meaning that the measured flowrate is approximately 98% of the Bernoulli prediction. So Engineers can use contractions as flowmeters provided that the pressure difference between two cross-sections of the convergence are known (i.e. are measureable by manometers) along with the constants a1, a2, z1 and z2.

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The Venturi Meter Venturi meters are shaped to allow as much pressure recovery as possible after the constriction and consist of a short constriction followed by a short straight throat before the pipe diameter is gradually returned to its original size. Usually, the throat diameter is around half of the initial pipe diameter and the diverging part of the Venturi meter is angled at about 6° which has found to be optimum in terms of allowing sufficient pressure recovery in an acceptable length of pipe (and hence cost).

Example 1: Horizontal Venturi Meter A Horizontal Venturi meter has a Cd = 0.96, it is to be used to measure the flowrate of water up to 0.025 m3/s in a pipeline of inside diameter 100 mm. The meter is connected to a differential manometer containing mercury (specific gravity 13.6). If the maximum allowable difference in mercury levels is 80 cm what is the diameter of the throat? First we apply Bernoulli’s equation at some point upstream of the Venturi meter and at the throat, since the Venturi meter is horizontal z1 = z2 Bernoulli’s equation can be written

Assuming continuity,

Hence,

And

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Introducing Cd as Q = Cda1u1

We now need to rearrange for a2 as follows

Now, we can calculate all of the terms on the right hand side of this equation and hence calculate a2

The pressure difference p1 –p2 needs to be calculated using the density of mercury (which we are told through the specific gravity)

Hence,

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Hence, the throat diameter is

Example 2: Vertical Venturi Meter A vertical Venturi meter carries liquid of relative density 0.8 and has inlet and throat diameters of 150 mm and 75 mm respectively. The pressure connection at the throat is 150 mm above that at the

inlet. If the actual rate of flow is 40 litres/s and the coefficient of discharge is 0.96 calculate a) the pressure difference between the inlet and throat, and b) the difference of level in a vertical U-tube mercury manometer connected between thee points, the tubes above the mercury being full of the liquid. Relative density of mercury is 13.56. To tackle this question we again apply the Bernoulli equation between the cross-sections 1 and 2. However, here we must include the z terms.

Again, we can apply the continuity equation (assuming that the fluid is incompressible) so that

Hence, we can rearrange the Bernoulli equation and write

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or

Hence,

We know the diameter of the pipe and the throat and can therefore calculate the cross-sectional area at points 1 and 2…

As we know the flowrate of the fluid 40 litres/s and the diameter of the pipe we can calculate the mean flow velocity, however we must remember to include the coefficient of discharge in this equation!

Hence,

For the vertical monometer, not only do we need to take into account the difference in the level of the manometric fluid in the two limbs of the manometer but we also need to take into account the additional mass of liquid (in this case oil) between z1 and z2 in the right hand limb in the diagram. Hence,

Therefore

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Orifice Meter One of the simplest and cheapest arrangement for flow measurement is sharp edged orifice plate inserted in flow. Like the Venturi meter this technique relies on the fact that in order to maintain the flowrate through the orifice the fluid is forced to increase its velocity resulting in a decrease in pressure across the plate. In this case the difference in pressure is measured between the vena-contracta and upstream of the orifice plate. The orifice plate suffers are far greater degradation of energy (downstream eddies) and downstream pressures will generally only recover to about 60% of the upstream pressure.

Example 3: Orifice meter in a vertical pipe Oil of density 860 kg/m3 flows upwards in a vertical pipe section of diameter 225 mm. A manometer filled with a mnometric fluid (shown in green) of density 1075 kg/m3 is used to measure the pressure drop across a pipe orifice plate with a throat diameter of 75 mm. What is the flowrate of the oil if the deflection of the manometer fluid is 0.5 m and Cd = 0.659 for the orifice.

As for the Venturi meter we start to tackle this question by applying the Bernoulli equation between points 1 and 2.

Applying the continuity equation (hence assuming incompressibility)

Substituting for u2 in the Bernoulli equation we get

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or

But how do we asses for the vertical manometer? Not only do we need to take into account the difference in the level of the manometric fluid in the two limbs of the manometer but we also need to take into account the additional mass of liquid (in this case oil) between z1 and z2 in the right hand limb in the diagram. Hence,

Substituting into the rearranged Bernoulli equation we get

In which the terms involving elevation cancel,

Now, we simply rearrange to get u1

It is possible to simplify the term a1/a2 so that it is expressed in terms of pipe and constriction diameters

and the ratio of pipe diameter to constricgtion diameter is often given the symbol so that

and

Hence,

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which solves to give

And hence the flowrate, Q, is

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Lecture 20: Summary

WARNING: The following table is summary of the topics we have covered in the course. Whilst it can be used to help revision it should be used in conjunction with the

course notes, class examples, tutorial sheets and past papers.

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Lecture Concepts/Definitions Important Equations

Lecture 1: Stress, Strain Rate & Viscosity

Shear Stress: Shear force either causing or caused by flow divided by the surface area through which that force acts. The units of shear stress are Pascals, Pa.

Shear Strain Rate: Velocity gradient perpendicular to the direction of flow. If in time dt the element of fluid bounded by ABCD moves such that it now occupied ABC’D’ the shear rate can be defined in terms of the rate of change of the angle θ. The units of shear rate are reciprocal seconds.

Viscosity For a Newtonian fluid the viscosity (µ) is defined as the constant of proportionality between shear stress and shear strain rate, it has units of Pa.s and changes with both temperature and pressure but is constant with respect to shear rate. Also sometimes called the dynamics viscosity. The kinematic viscosity is the ratio of dynamic viscosity and fluid density.

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Lecture Concepts / Definitions Important Equations

Lecture 1: Stress, Strain Rate & Viscosity

Apparent Viscosity Non-Newtonian fluids are not characterised by a constant viscosity and hence we define an apparent viscosity (η) at a given strain rate. Units of Pa.s.

Stress vs Strain Rate Plots

Lecture 2: Models for non-Newtonian fluid behaviour and time dependent viscosity

Cross Model Contains 4 fitting parameters. Can model viscosity in zero shear and high shear plateau viscosity regions. M describes the degree of shear thinning.

Power Law Contains 2 fitting parameters, n and k2. Can only be used in the ‘power law region’ of material behaviour. With appropriate choice of n the power law model can be applied to shear thinning (n < 1), shear thickening (n>1) or Newtonian behaviour (n = 1) .

Limitations of the Power Law model Only applicable in power law region, extrapolation outside this region leads to large errors. Units of k2depend on n and so cannot (in general) be used as a basis of comparison between two fluids.

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Lecture Concepts / Definitions Important Equations

Lecture 2: Models for non-Newtonian fluid behaviour.

Sisko Model Extends the power law model to include the high shear plateau region

Applicablity of Cross, Power Law and Sisko Models to shear thinning fluid.

Bingham Plastic Models fluids that display a yield stress (toothpaste)

Thixotropic/Rheopectic Materials Viscosity is function of duration of shear (mayonnaise).

Lecture 3: The Poiseuille Equation

The Poiseuille Equation Relates volumetric flow rate (V) to the fluid viscosity (µ) pipe radius (r) and pressure drop per unit length(p) along a straight pipe. The equation is applicable to Newtonian fluids. Assumptions Laminar Flow Velocity of any element of fluid is a function of r only Incompressible fluid No slip boundary condition

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Lecture Concepts / Definitions Important Equations

Lecture 4: Velocity profiles and ∆p – V relationships for non-Newtonian fluids

Velocity Profile for Newtonian fluid under laminar flow

Parabolic shape, shear rate is maximum at the wall and zero at the centre of the pipe, stress is hence maximum at the wall and zero at the centre of the pipe.

Velocity Profile for a Bingham Plastic

Velocity profile is truncated in the central core of the pipe because the stress drops below the yield stress of the fluid.

No shear rate in the central core. Shear occurs in the Newtonian slip layer between the pipe wall and the core. Note that the central core does move.

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Lecture Concepts / Definitions Important Equations

Lecture 4: Velocity profiles and ∆p – V relationships for non-Newtonian fluids

Velocity Profile for Power Law fluids

Velocity profile for a shear thinning material is truncated whilst the velocity profile for a shear thickening fluid becomes more pointed. The truncation/pointedness increases as the fluid become more shear thinning/shear thickening.

Buckingham Equation The Bingham plastic version of the Poiseuille equation. Reduces to the Poiseuille Equation where .

∆P-V relation for Power Law Fluid For power law fluids the pressure drop no longer goes with the inverse of the 4th power of radius. This is important in pipe scale up calculations.

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Lecture Concepts / Definitions Important Equations

Lecture 5: Fluid Mixing and Agitation

The Standard Rushton Turbine

4 Baffles

Factors that affect power requirements In addition to the above (DT, D, ZA, H and WB), the following are also important:

Agitator Speed Agitator Pitch Width of blades of agitator paddle Fluid viscosity Type of agitator

The Power Number Dimensionless group used in calculations of power requirement and scale up of mixing equipment. In the fully turbulent region it is acceptable to assume that (for a 10% baffle condition / standard Rushton Turbine) the power number is 6.

Reynolds number for Mixing Vessels For mixing vessels the Reynolds number is defined in terms of the impellor diameter, D.

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Lecture Concepts / Definitions Important Equations

Scale Up of Mixing Vessels Scale up can be carried out on the basis of constant tip speed (equivalent to constant torque per unit volume) or constant power per unit volume by assuming that both small and large scale tanks are being operated in the fully turbulent region. Constant Mixing Time Scale up is not carried out using a constant mixing time criterion due to the very high torque required. Constant tip speed Equivalent to constant torque per unit volume. Not generally used for fully turbulent operations. Constant power per unit volume Most widely used scale up criterion (of those studied)

If both tanks are operated in the fully turbulent region

Mixing times can be related by

Lecture 7-9: Forces due to fluids in motion

Newtons 2nd Law can be used to determine the forces associated with fluids in motion. The force can be expressed as the rate of change of momentum occurring within a given control volume.

For a fluid jet striking a plate normally, if the xx axis is set along the direction of flow then no momentum leaves the control volume in the xx direction whilst the momentum leaving the control volume in the +yy direction is equal to the momentum leaving the control volume in the –yy direction. Hence the force on the fluid can be expressed as the product of the mass flow rate and the component of velocity along the xx axis. (note the xx axis is here defined along the direction of flow of the jet)

Force = Mass Flowrate x component of velocity in the xx direction

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Lecture Concepts / Definitions Important Equations

For a fluid jet striking a plate at an angle, if the xx axis is set along normal to the plate then no momentum leaves the control volume in the xx direction. Hence the force on the fluid can be expressed as the product of the mass flow rate and the component of velocity along the xx axis. If the fluid is assumed to be inviscid then no forces are possible in the yy direction (perpendicular to xx).

Force = Mass Flowrate x component of velocity in the xx direction

Lecture 10: Forces due to real fluids in motion

Boundary Layer When a real fluid flows along a surface the fluid next to the surface is retarded by the

presence of the surface whilst fluid elements further from the surface will be unaffected. The force generated (in the direction of motion) by the fluid flowing in this boundary layer can be calculated by applying the momentum theorem and can be expressed as….(see eqn)

Drag Force The drag force on an object can be determined by measuring the velocity and pressure profile in the wake of the object.

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Lecture Concepts / Definitions Important Equations

Lecture 10: Forces due to real fluids in motion

Drag Force per unit projected area For a sphere……

Drag Coefficient Data for flow around standard objects is available and is often presented in the form of drag coefficient as a function of Reynolds number

Flow around a cylinder

Region A (Re < 0.2): Linear relationship between drag coefficient and Re. Laminar flow around the object streamlines converge behind it.

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Lecture Concepts / Definitions Important Equations

Lecture 10: Forces due to real fluids in motion

Region B (0.2 < Re < 500): Flow separation occurs. Eddies in the wake of the object. Carman vortex street may be set up. Point at which separation occurs gradually moves forward until it reaches a stable position.

Region C (500 < Re < 2 x 105): Separation point remains at a stable position and drag coefficient is hence constant

Region D (Re > 2 x 105): Transition to turbulent boundary layer before separation. Separation point moves to a new stable position towards the rear of the cylinder. Drag coefficient approaches a second constant value.

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Lecture Concepts / Definitions Important Equations

Lecture 10: Forces due to real fluids in motion

Flow around a sphere Drag coefficient varies in a similar manner to flow around a cylinder. However, symmetrical eddies (which occur for flow around a cylinder and lead to the Vortex street) are replaced with a vortex ring.

Lecture 11 and 12: The Bernoulli Equation

The Bernoulli equation can be derived from Newton’s second law. Contains 3 terms, the pressure head, the velocity head and the elevation.

Lecture 13: Laminar flow in packed beds

Voidage: Fraction of the bed not occupied by solids (i.e. the portion through which flow CAN occur)

Hydraulic mean diameter: Effective diameter of pore spaces in which flow occurs SB (Surface area of particles per unit volume of bed)

Tortuosity: Flow occurs along a tortuous path within the bed. The length of the actual flow path (L’) is related to the height of the bed (L) through the tortuosity factor (τ)

Superficial velocity: Flow velocity in the tube containing the bed if the bed were not present

Interstitical velocity: Actual flow velocity in the pore of the bed.

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Lecture Concepts / Definitions Important Equations

Lecture 13: Laminar flow in packed beds

Karmen-Cozeny Equation Derived for laminar flow in a packed bed by assuming that the pore spaces can be represented by an interconnected network of small circular tubes. (d is particle diameter and l the bed height.

Lecture 14: Tubrulent flow in pipes

Reynolds experiment: Reynolds studied flow in long straight circular tubes and was able to define the transition between laminar and turbulent flows in terms of the Reynolds number Re < 2000 – Laminar Flow 2000 < Re < 4000 – Transition region Re > 4000 – Fully turbulent flow

The Darcy Equation Derived from a force balance assuming steady flow conditions Equally applicable to laminar and turbulent flows with appropriate choice of the “Fanning friction factor, f”.

The Fanning Friction Factor For laminar flow it can be shown that f is inversely related to Re by equating the Darcy equation and the Poiseuille equation to give:

For turbulent flow in rough or smooth pipes we can use the Moody Chart to estimate f The Moody Chart contains several important features

1) A straight line relationship for laminar friction factor 2) Smooth pipe turbulent friction factor 3) Rough pipe turbulent friction factor 4) The concept of relative roughness defined as k (the absolute roughness)

divided by the pipe diameter

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Lecture Concepts / Definitions Important Equations

Lecture 15: Head losses in rough and smooth pipes

The Darcy Weisbach equaton A form of the Darcy equation in which the pressure drop is expressed in terms of head loss (it can thus be used in conjunction with the Bernoulli equation to estimate frictional losses)

Lecture 16 and 17: Flow through packed beds II

Modified Reynolds Number We define a modified Reynolds number that we can use to assess the nature of flow within a packed bed.

Modified friction factor We define a modified friction factor that can be used to determine resistance to flow in a packed bed. The modified friction factor is expressed in terms of the resistive force in the direction of motion per unit area of particle surface, R1

Pressure drop can be related to modified friction factor through

Estimating the modified friction factor Several relationships between the modified friction factor and modified Reynolds number have been proposed. Carman proposed that for laminar flow (where Re1 < 2)…..(see eqn). Other relationships include the Ergun relation (see notes).

Lectures 18 & 19: Differential head flow

meters

Venturi Meter Shaped to allow maximum pressure recovery. Calculations based on Bernoulli equation. Coefficient of discharge introduced to continuity equation in order to account for some degradation of energy. Cd approximately 0.98 for a well shaped Venturi meter.

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Lecture Concepts / Definitions Important Equations

Lectures 18 & 19: Differential head flow

meters

Orifice Plate Significant pressure loss due to downstream turbulence. Pressure difference measured between upstream and vena contracta. Cd approximately 0.65

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