MrsDrPoe: Dimensional Analysis, Part III

Welcome to the final Thesis Tuesday on Dimensional Analysis.  As promised last week, we're going to go through an example problem today so you can see the Buckingham Pi Theorem in action.

Example: Water sloshes back and forth in a tank as shown in the figure.  The frequency of the sloshing, f, is assumed to be a function of the acceleration of gravity, g, the average depth of the water, h, and the length of the tank, l.  Develop a suitable set of dimensionless parameter for this problem using g and l as repeating variables.

Given: f = F(g, h, l)
Find: suitable set of dimensionless parameters for the given problem
Solution:

Step 1: The variables in the problem are: 

frequency, f

tank length, l

gravity, g

water height, h

Step 2: Expressing these variables in basic dimensions:

f = 1/T

l = L

g = L/(T*T)

h = L

Step 3: Determining the number of pi terms:

k = 4 (f, g, l, h) and r = 2 (L, T) so k-r = 2

Step 4: We will choose g and l as our repeating variables because the problem statement told us to; however, we could work the problem using g and h as well.  Note: we cannot use f at all because it is the dependent variable, and we cannot use both h and l because the combination of repeating variables must contain all the basic dimensions.

Step 5: We will find the first pi term:

Pi1 = f*g^(a1)*l^(b1)

(1/T)*(L/T*T)^(a1)*L^(b1) = T^(0)*L^(0)

for L to cancel: a1 + b1 = 0

for T to cancel: -1 -2*a1 = 0

So: a1 = -1/2 and b1 = 1/2

Pi1 = f*sqrt(l/g)

Step 6: We will find the second pi term:

Pi2 = h*g^(a2)*l^(b2)

(L)*(L/T*T)^(a2)*L^(b2) = T^(0)*L^(0)

for L to cancel: 1 + a2 + b2 = 0

for T to cancel: -2*a2 = 0

So: a2 = 0 and b2 =-1

Pi2 =h/l

Step 7: Since it's easy to make a mistake, we need to check the pi terms:

Pi1 = f*sqrt(l/g) = (1/T)*sqrt((L*T*T)/L) = 1

Pi2 = h/l = L/L = 1

Step 8: Next, we will write the pi term containing the dependent variable as a function of the remaining pi terms:

Pi1 = F(Pi2) or f*sqrt(l/g) = F(h/l)

Not too bad, huh?

Additional Considerations

Our variables that we deal with in this process can be classified into three broad categories: those dealing with geometry (lengths, angles), material properties (viscosity, density), or external effects (velocities, external forces).

It is important to note that there is not a unique set of pi terms that arises from a dimensional analysis; however, the required number of pi terms is fixed.

Common Non-dimensional Numbers

Reynolds Number - (rho*V*l)/mu = inertial forces/viscous forces; applicable to all types of fluid dynamics problems

Froude Number - V/sqrt(g*l) = inertial forces/gravitational forces; applicable to flow with a free surface

Euler Number - p/(rho*V*V) = pressure forces/inertial forces; applicable to problems where pressure or pressure difference are important

Mach Number - V/c = inertial forces/compressibility forces; applicable to problems where compressibility of a fluid must be considered

MrsDrPoe: Dimensional Analysis, Part II

Happy Thesis Tuesday!  Today we'll resume our discussion on dimensional analysis by further investigating the Buckingham Pi Theorem.

This theorem is applied to a problem through several simple steps:

Step 1: List all the variables that are involved in the problem.  This step is crucial- without properly identifying all the necessary variables, we will not be able to obtain the correct dimensionless groups.  Each of the properties should be independent.  For example, if density and specific weight are important, we could use rho and gamma, rho and g, or gamma and g, but to use all three would be redundant (since gamma = rho*g).

Step 2: Express each of the variables in terms of basic dimensions.  We will be looking at these quantities in terms of length (L), time (T), mass (M), and temperature (theta), since these are the basic dimensions for SI and BG unit systems.

Step 3: Determine the required number of pi terms.  This number is equal to k-r where k is the number of variables in the problem (from Step 1), and r is the number of reference dimensions required to describe these variables (from Step 2).  Note: if k-r = 1, skip to Step 5.

Step 4: Select a number of repeating variables, where the number required is equal to the number of reference dimensions.  Here we should choose r number of variables from our list of independent variables; there are our "repeating variables."  It is best to choose ones with the simplest representation in basic dimensions (for instance: choose a distance, L, instead of a velocity, L/T).

Step 5: Form a pi term by multiplying one of the non-repeating variables by the product of the repeating variables, each raised to an exponent that will make the combination dimensionless.  For example: Pii = ui*u1^(ai)*u2^(bi)*u3^(ci), where ui is a non-repeating variable, u1, u2, and u3 are repeating variables, and a, b, and c are exponents.  Note: it is acceptable for a, b, or c to equal zero as long as at least one is non-zero.

Step 6: Repeat Step 5 for each of the remaining non-repeating variables.

Step 7: Check all resulting pi terms to make sure they are dimensionless.  It's easy to make mistakes through this process; check your work by writing each of the variables in terms of their basic dimensions (or even units) to see if all the basic dimensions cancel out.

Step 8: Express the final form as a relationship among the pi terms, and think about what it means.  For example: Pi1 = f(Pi2, Pi3, ..., Pik-r), where Pi1 is the pi term containing the dependent variable.


While at this point you may not believe what I said about this process being simple, next week, I'll prove it to you with an example problem.  We'll also look at some common pi terms/dimensionless number in fluid mechanics.  Happy studying!

MrsDrPoe: Dimensional Analysis, Part I

Happy Thesis Tuesday on the blog!  Today we'll look at a brief introduction to the interesting topic of dimensional analysis.

As with any type of experimentation, fluid flow experiments should be performed in a meaningful and systematic manner.  If we consider, for instance, the pressure change across a section of pipe in Poiselle flow, we can see that this change in pressure is affected by fluid properties (density, rho; and viscosity, mu), pipe diameter, and fluid velocity.  By running countless numbers of measurement experiments, each time varying a single parameter, we could determine how the pressure change is affected by each of these items; however, these relationships would only be valid for this pipe with this fluid and this experimental setup.

WHEW!

In order to improve the quality, applicability, and repeatability of our experimental results, we use dimensionless products or dimensionless groups.  These terms are combinations of variables necessary for a problem and often employ basic dimensions (mass, M; length, L; and time, T) or units (pounds, feet, and seconds, etc.).

The method we use to find these dimensionless terms is called the Buckingham Pi Theorem - "If an equation involving k variables is dimensionally homogeneous, it can be reduced to a relationship among k-r independent dimensionless products, where r is the minimum number of reference dimensions required to describe the variables."

Next week, we'll look at an application of this theorem, which is much simpler than it may sound!