Unit IV - Dimensional Analysis and Real Flows

BMED3300 Biotransport Lecture Notes C. Ross Ethier 2004, 2013

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Lecture 11: Dimensions. Buckingham Pi theorem. Similitude. Model design.

All (or nearly all) physically measurable quantities have units associated with them. But nature does not use any system of units units are an invention of humans. This means that natural processes should be independent of any system of units. We will see that this can be

formalized by the idea of a -group.

Dimensions: All systems of units are based on fundamental quantities such as mass, length and time. We refer to such quantities as dimensions. Note that dimensions are independent of any system of units, and instead reflect what is being measured or expressed.

We will consider a hierarchy in which the fundamental dimensions are:

Mass, symbol M Length, symbol L Time, symbol T Temperature, symbol

There are other fundamental dimensions, but we will not need them for this class. The dimensions of all other quantities can be derived from these fundamental dimensions. For example, volume is a product of three lengths, and therefore has dimensions of L3. We write this as

[Volume] L3, where the notation [ ] means has dimensions of.

Example: What are the dimensions of force?

Answer: We can determine this by using F = ma to write:

[F] = [m][a] M LT-2

Note that we are not concerned with numerical coefficients when working with dimensions.

Buckingham -Theorem: We argued earlier that nature does not care about systems of units. This implies that mathematical and empirical relationships describing a system should be independent of units, i.e. cannot depend on what are known as dimensional quantities. Instead, such relationships in nature must be formulated in terms of quantities that are independent of units, i.e. that are dimensionless. We call

quantities that are dimensionless -groups, and say that nature talks in

-groups.

Example: The Reynolds number for flow in a pipe is defined by

Re =

=


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where is the diameter of the pipe and is the average fluid velocity in the pipe. Show that the Reynolds number is a -group.

Answer: To show that ReD is a -group, we must show that it has no dimensions. We can do this by listing out the dimensions of the quantities appearing in ReD:

[] 1 [] [] 3

[] []

[/]12

1= 11

That means that the dimensions of are

[] 3 1

11= 1

In other words, has no dimensions and is therefore a -group. Such a combination of quantities is said to be dimensionless.

The Buckingham -theorem provides a formal way to generate -groups

from a set of dimensional quantities. It also tells us how many -groups we can create from such a set. It works are follows. Suppose that there are n variables that are known to play a role in a given problem, 1, 2, . Note that the identification of these n variables can be quite tricky, and is extremely problem-dependent. These n variables will each

include some dimensions 1, 2, , e.g. M, L and T. We create a pn

matrix of the form shown below

1 2 12

[

]

where entry (i,j) is the exponent of in the quantity . We denote the

rank of this matrix as . Note that . (Usually = , but not always.) Then the number of -groups, , is given by

=

To form the -groups: We choose r of the n quantities, and refer to these r quantities as the core group. The r quantities must contain all p

dimensions, and cannot themselves form a -group. Suppose for the purposes of this example we choose 1, 2, as the core group. Then we formulate the first -group by writing:

1 = 12

+1


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The exponents a, b, and c are determined by requiring that 1 be

dimensionless. We then repeat this process to get a second -group

2 = 12

+2

where the exponents a, b, and c are different than for 1. We repeat this

process, each time replacing the last entry in the -group until we have formed = such groups.

Example: The pressure drop per unit length in turbulent pipe flow, /, is known to depend on the pipe diameter, , the pipe wall roughness, , the fluid velocity, , and the fluid viscosity and density, and . What -groups can be formed for this problem?

Answer: It is useful to list all of the relevant variables and their dimensions.

Variable Dimensions

/ ML-2T-2 ML-3 ML-1T-1

L L LT-1

The resulting matrix is:

[1 1 1 0 0 02 3 1 1 1 12 0 1 0 0 1

]

This matrix has rank = 3. Therefore, there will be

= = 6 3 = 3 -groups.

We arbitrarily choose , and as the core group. Then we form the first -group as:

1 = (/)

We can compute the dimensions of 1 from:

[1] (1)(3)22 = 000

where we set the exponents of M, L and T to zero since 1 must be dimensionless. Equating exponents of M, L and T gives 3 equations for a, b, and c:


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M: c +1 = 0 therefore c = -1 L: a + b -3c -2 =0 T: -b -2 = 0 therefore b = -2 and a = 1

This means that the -group is:

1 =

2

=

2

This -group appears so often in pipe flow problems that it is given a

special name. Actually, twice 1 is called , the friction factor.

Now we can do the same procedure to get the next -group, 2.

2 =

Once again substitute in dimensions:

[2] (1)(3) = 000

Solving, we get a = -1; b = c = 0, so that 2 = / , the relative roughness. Finally, doing the same thing for the last -group we get:

3 =

Once again substitute in dimensions:

[3] (1)(3)11 = 000

Solving, we get a = b = c = -1, so that 3 =

=

1

Re.

Notes:

Any product of -groups is also a -group.

A -group to any power is also a -group.

A -group multiplied by a numerical factor is also a -group.

From the above it should be clear that there are infinitely many possible

-groups. However, only = of these are independent. Different choices for the core group of variables will lead to different -groups. To

keep things manageable, certain -groups are standard and are always used. In this problem they are , / and .

The Buckingham -theorem is useful in two different ways.

Efficiency in Data Correlation and Relationships: Think about our

example of the pressure drop in pipe flow. Without -groups, the most that we can say about the problem is:


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/ = 1(, , , , )

which means that the unknown function 1 depends on 5 variables. Discovering this function is very hard from experimental measurements, and even once it is determined, presenting it in charts or tables is almost impossible.

It is much simpler to use -groups. Knowing that the fundamental

relationships about pressure drop in pipe flow must involve only -groups, we can write:

= 2 (

, )

Notice that f is the -group that includes the pressure drop, and depends on only two other quantities. The form of 2 is much easier to determine from experiments than the form of 1, and it is also easier to express. In fact, it all fits on one page (the Moody chart). Notice that -group analysis does not tell us anything about the form of 2. To determine this we still need to do experiments or analysis, but the -groups make this easier.

Example: The pulse wave speed in an artery, , is a function of the fluid density, , and the distensibility of the artery, , defined by:

=1

where is the cross-sectional area of the artery. Find an expression for in terms of and .

Answer: We first determine the dimensions of the various quantities involved in the problem, as follows:

Variable Dimensions

LT-1 ML-3 LT2M-1

Notice from the definition of , it has the same units as 1/p. We can then form our matrix of coefficients as:

[0 1 11 3 11 0 2

]

This matrix is degenerate, and has rank = 2. Therefore, we can form = 3 2 = 1 -group. It is easy to compute this -group as


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1 = 2

This does not seem that useful, but if we think about it a bit more we can actually get something very useful from it. In general, the behaviour of a

physical system must be governed by a relationship of the form 1 =

function(2, 3, ). However, for the case of this example there is only

one -group, so that the function in the above expression takes no arguments. The only such function is a constant, so that we immediately

have that 1 = const. Therefore we can write the wave speed as:

=

Dimensional analysis does not tell us what the constant is in the above expression, but it does tell us about the form of this expression. This is quite handy.

Similitude: The second way in which -groups are useful is in similitude, which is the scaling of transport phenomena from one situation to another. This usually involves trying to extrapolate the performance of a model to that of a full-scale device. To understand how this works we define 3 different types of similitude.

1. Geometric Similitude: Two objects are said to be geometrically similar if one is a simple enlargement of the other, i.e. all lengths are scaled up by some constant factor. This scaling can be quite tricky, including, for example, surface roughness, etc.

2. Kinematic similitude: The velocity fields around two geometrically similar objects are kinematically similar if all velocities around one object are identical to the velocities around the second object to within a scaling factor.

In other words, vA = C vA for all pairs of points AA. Note that if model 2 is k times larger than model 1, then the distance from A to the origin must be k times larger than the distance from A to the origin. In other words, we must compare velocities at scaled locations.

Model 1 Model 2

A A

vA vA


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3. Dynamic similitude: Each of the terms in the Navier-Stokes equations can be interpreted as a force per unit volume (or inertia per unit volume). As is the case for all equations, each term must have the

same dimensions as the other terms. We can therefore form -groups by taking ratios of the terms in the Navier-Stokes equations.

There are three common -groups that result when this is done.

The Reynolds Number: Re =

=

inertia

viscous forces

The Froude Number: =2

=

inertia

gravitational forces

The Euler Number: =

2=pressure

inertia

where , and are suitable length, velocity and pressure scales. If these -groups match between two flows, then the flows are said to exhibit dynamic similitude. In such cases, all force-related quantities in the flows can be scaled using similitude.

Aside: Show that the Reynolds number is the ratio of inertial to viscous effects. We can do this by considering the relevant terms in the governing equations:

Inertia scales as

2

Viscous forces scale as 2

2

The ratio is 2/

/2=

= Re

In addition to the -groups arising from the governing equation,

there can also be -groups arising from the boundary conditions. This is very problem-dependent. Note that in real problems, one can

often argue on physical grounds that one or more of the above -groups do not play a role in the problem, e.g. the Froude number is usually only important when there are free-surface flows. In that case

we do not need to include the unimportant -groups in the analysis. Knowing what groups are/are not important can be very TRICKY!

The important point from all of this is that if both geometric similitude and kinematic similitude exist for flow over two different objects, then dynamic similitude also exists for these flows. Alternatively, if geometric similitude and dynamic similitude exist, then kinematic similitude does also. In practice, we usually try to enforce geometric and dynamic similitude in experimental testing of fluid mechanical performance using scaled models. More specifically, when we make a model, we must


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ensure that it has geometric similitude to the actual device. Then we

determine what -groups are important based on the details of the

problem, and enforce dynamic similitude using those -groups. Note

that if we have -groups in a problem, we must only match 1 of them, since we know that i = (1,2, i1). In other words, matching 1 of the -groups is sufficient to fully constrain the system and ensure matching of all the -groups.

Example: A blood oxygenator is being tested to ensure that the pressure drop on the blood side is not too large. A 3 scale model is made, and a blood analogue fluid with viscosity 19 cP and density 1.8 g/cm3 is pumped through it. If the design flow rate is 5 L/min, what should be the blood analogue fluid flow rate? If a 32 mmHg pressure drop is measured across the 3 scale model, what will the corresponding pressure drop be for the real system?

Answer: It is convenient to summarize everything in a table:

Quantity Configuration 1 (Blood) Configuration 2 (3 larger)

3.5 cP 19 cP

1.05 g/cm3 1.8 g/cm3 Q 5 L/min ?

p ? 32 mmHg

Treating the oxygenator as a series of pipes (tubes), we expect that the

pressure drop can be computed from f = fcn(/D, ReD). To ensure similitude we need 1/1 = 2/2, which is part of the requirement for geometric similitude. We also need Re1 = Re2, which we can write as:

41

11=

42

22 (*)

where we have rewritten the Reynolds number in terms of the flow rate for a pipe of diameter D. By manipulating equation (*) we obtain a constraint on the flow rate in Configuration 2, namely

2 = 1 (21)(21) (12)

= (5

) (3) (

19

3.5) (1.05

1.8)

= 47.5

min

This is the flow rate which will ensure that Reynolds numbers match between the two configurations.

In this example there are only three relevant -groups, so that if we specify two of them, then the third must be determined uniquely. This implies that by matching Reynolds number and relative roughness


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between the two configurations, the friction factor must match between the two configurations. We can therefore write:

1 = 2

1

112

1

1=

2

222

2

2 (**)

From the geometry of the model, we have that 2 = 31 and 2 = 31, so that the / terms cancel out. We can therefore write:

1 = 2 (12)(12

22)

It is convenient to eliminate the velocities in favour of flow rates ,

which we can do by noting that =4

2, so that equation (**) becomes:

1 = 2 (12) (12)2

(21)5

(L1L2)

= 32 (1.05

1.8) (

5

47.5)2

(3)4

= 16.8

Lecture 12: What is a boundary layer? Laminar boundary layer equations & Blasius solution.

Consider what happens when we immerse an object, say a sphere, into a stream of moving fluid. We will suppose that the Reynolds number is large, which means that viscous effects are not very important. Yet, at the surface of the sphere we must still satisfy the no-slip condition, which comes about due to viscosity. This presents an apparent contradiction: it seems that viscosity is both important and not important.

The resolution to this apparent paradox is the boundary layer, or more specifically, the momentum boundary layer. This is a thin region next to a solid surface where viscous effects are important. Note that it only makes sense to talk about a boundary layer in the context of a high Reynolds number flow, since by definition the boundary layer is a small region of the flow. As the Reynolds number increases, we expect that the thickness of the boundary layer will decrease.

To analyze the boundary layer in more detail, it is best to consider the simpler case of a thin flat plate in a uniform flow. This is shown schematically below. Outside the boundary layer, viscous effects are unimportant. For the case of the flat plate, the streamlines outside the boundary layer are essentially parallel, and the velocity (the free-stream velocity) is constant and equal to .


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Inside the boundary layer, viscous effects are important and there is an appreciable velocity gradient. We arbitrarily define the edge of the boundary later to be the location where is 99% of its free-stream value. This defines a boundary layer thickness, (), which depends on axial position .

The boundary layer has an initial laminar region, followed by a turbulent region. The transition between the laminar and the turbulent parts depends on the Reynolds number based on x:

=

The transitional Reynolds number is somewhere between 2 105 and

3 106, depending on the roughness of the plate and the level of free-stream turbulence.

Let us now carry out a more quantitative analysis of the boundary layer. In the y-direction, all changes take place over the length scale . In the x-direction, the length scale for changes to take place is x. Notice that

Unit IV - Dimensional Analysis and Real Flows

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