# 50 Worked Examples: Units, Conversion Factors & Dimensional Analysis

These worked examples have been fielded as homework problems or exam questions.

Worked Example #1

Determine the base dimensions of each of the following variables:

(a) Plane angle

(b) Specific volume

(c) Force

(d) Stress

(a) Plane angle: A plane angle is defined in terms of the lines from two points meeting at a vertex and is defined by the arc length of a circle subtended by the lines and the radius of that circle. The unit of plane angle is the *radian*. Because it is the ratio of an arc length to the radius, then the plane angle is dimensionless, i.e., a radian is one measurement unit that is already dimensionless, i.e.,

(b) Specific volume: The specific volume is defined as the ratio of volume to mass, i.e., it is the reciprocal of the density. Therefore,

(c) Force: Force is the product of mass times acceleration so

(d) Stress: Stress has units of force per area, but a force is the product of mass times acceleration so

Worked Example #2

*surface tension*, which is given the symbol , and it has dimensions of force per unit length. Write the dimensions of surface tension in terms of its base dimensions.

We are told that the units of surface tension, , have dimensions of force per unit length. We also know that a force is mass times acceleration, which is in base units. Therefore, we can write for the base units of surface tension that

Worked Example #3

Write the primary dimensions of each of the following variables from the field of thermodynamics:

(a) Energy,

(b) Specific energy,

(c) Power,

(a) Energy has units of force times distance, i.e.,

(b) Specific energy has units of energy per unit mass, i.e.,

(c) Power is the rate of doing work, so a force times distance per unit time, i.e.,

Worked Example #4

Determine the primary (base) dimensions of each of the following parameters from thermodynamics:

- Energy,
- Work,
- Power,
- Heat,

1. Energy, , is the ability to do work and is measured in Joules (J) in the SI system and foot-pounds (ft-lb) in the USC system. Notice that “foot-pounds” is the USC unit and not “pounds-foot” or “pounds-feet.” Energy has the same units of work (force times distance), so which are the base units. Notice that the units of force are obtained from the product of mass and acceleration, i.e., [] = = .

2. Work, , is also measured in Joules (J) in the SI system and “foot-pounds” (ft-lb) in the USC system. Work is equivalent to force times distance, so , which are the base units of work.

3. Power, , is the rate of doing work and is measured in Watts (W) in the SI system and foot-pounds per second (ft-lb/s or ft-lb s in the USC system. Power is equivalent to a force times distance per unit time (or force times velocity), so which are the base units of power. In practice, power is measured in terms of kiloWatts (kW) in SI and horsepower (hp) in USC, where one horsepower is equivalent to 550 ft-lb/s.

4. Heat has units of energy (the ability to do work), which have the same units as work, i.e., units of Joules (J) in the SI system and foot-pounds (ft-lb) in the USC system. In base units then, which are the base units of heat.

Worked Example #5

A Watt is a Joule per second. So, one Watt hour is the equivalent of 3,600 Joules per hour. Therefore, one kilo-Watt-hour (kWh) = 3.6 Mega-Joules (MJ). The energy density (in units of kWh/kg) = energy density (in units of MJ/kg)/3.6. Therefore, the energy density of jet fuel in kilowatt-hours per kilogram is approximately 11.9 to 12.5 kWh/kg.

Worked Example #6

The Bernoulli equation can be written as

The first term is the local static pressure. The second term is dynamic pressure. The third term is the local hydrostatic pressure. The sum of the three terms is called total pressure. The Bernoulli equation is a surrogate for the energy equation in a steady, incompressible flow without losses or energy addition.

Each of the terms in the Bernoulli equation has units of pressure. In terms of fundamental dimensions, then

So, we have proved that all terms have the exact fundamental dimensions of M L T.

Worked Example #7

The relationship between and the air properties may be written in a general functional form as

where is the dependent variable. , , and are the independent variables. We will now choose and as repeating variables. Following the Buckingham method, then the products for our problem are:

For the first product

The values of the coefficients , , and must be obtained to make the equation dimensionally homogeneous. In terms of the dimensions of the problem, then

For to be dimensionless, then the powers or exponents of , , and must add to zero, i.e., we must have that

By inspection we get , , and . So, we get the first product as

or

which is non-dimensional, but it is a different grouping to what was obtained with , and as the repeating variables.

This is an interesting outcome because this parameter is a form of Stoke’s Law, which Sir George Stokes determined in the 1840s. He found that the drag force on a sphere of radius moving through a fluid of viscosity at a very low speed is given by

Therefore, in this case, then

For the second product then

so

and setting the sum of the powers to zero gives

In this case , , and and so

i.e.,

which we recognize as the Reynolds number.

Therefore, as a result of our dimensional analysis on the sphere we can write

or

or in explicit form

Worked Example #8

Based on experiments performed in a low-speed wind tunnel, it is determined that the power required at the shaft to drive a propeller forward is a function of the thrust the propeller produces, , the size of the propeller as characterized by its diameter , the rotational speed of the propeller in terms of revolutions per second , and the air density , and the operating free-stream velocity . Find the appropriate non-dimensional groupings that will describe this problem.

The power required for the propeller (the dependent variable), , can be written in functional form as

where is the function to be determined. The power would be given by , where is the torque and is the angular velocity in radians per second.

This foregoing equation can be written in implicit form as

In this case we have 6 variables () and 3 fundamental dimensions () comprising mass (), length (), and time (). This means we have to determine or three products.

The functional dependence can also be written in the form

where , and are the non-dimensional groupings to be determined.

Choose the variables , , and as the repeating variables, which are all linearly independent, and we can confirm this using the dimensional matrix (below). The non-dimensional products can each be written in terms of these repeating variables plus one other variable, that is

where in each case the powers , , and are to be determined in such a way that each of the products must be non-dimensional.

Now we can write down the dimensions of each of the variables. For this problem, we have

and so the dimensional matrix is

By examining the determinants of the submatrix formed by each of the elected repeating variables then we can quickly confirm that they are indeed linearly independent.

Considering the first product then

where , , and are to be determined. We know that

so in this case, in terms of dimensions of the parameters, then

Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

These simultaneous equations have the solution that , and . Therefore, the first product can be written as

which is a form of thrust coefficient, i.e., a non-dimensional measure of thrust.

Considering now the second product then

where new values for , , and are to be determined. In terms of dimensions

Making this equation dimensionally homogeneous gives

These equations have the solution that , and = -5. Therefore, the second product is

which is a form of power coefficient, i.e., non-dimensional measure of power.

Finally, for the third product then

and in terms of dimensions then

Making this final equation dimensionally homogeneous gives

These latter equations have the solution that , and . Therefore, the third product is

which is a non-dimensional airspeed called a tip speed ratio or advance ratio.

Therefore, for this propeller problem we have that

or

or finally as

which allows us to evaluate the performance of the propeller in terms of power coefficient as a function of the thrust coefficient and the tip speed ratio.

Worked Example #9

Consider the internal flow through a rough pipe. The objective is to determine the non-dimensional groupings that will describe this problem. The dependencies include the average flow velocity , the diameter of the pipe, , the density of the fluid flowing through the pipe , the viscosity of the fluid, , the roughness height of the pipe , and the pressure drop along the length of the pipe, .

Proceeding using the Buckingham method, we can write in general functional form that

In this case we have 6 variables () and 3 fundamental dimensions () comprising mass (), length (), and time (). According to the Buckingham Method then we have , so we need to look for three products.

The dimensional matrix is

Choose , and as the repeating variables, a common choice for fluid problems, and these variables are also linearly independent and contain all of the base dimensions.

Considering the first product then

and in terms of dimensions, then

Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

These simultaneous equations have the solution that , and . Therefore, the first product can be written as

which we see is the reciprocal of the Reynolds number, but as discussed before we can just invert this grouping to get the first product as

For the second product then

and in terms of dimensions, then

and we can just reuse the exponents , and for convenience. Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

These simultaneous equations have the solution that , and . Therefore, the second product can be written as

which is a measure of the relative surface roughness.

Finally, for the third product then

and in terms of dimensions, then

Making the equation dimensionally homogeneous by equating the exponents for each of the dimensions in turn gives

These simultaneous equations have the solution that , and . Therefore, the third product can be written as

which is a non-dimensional pressure drop or “head” drop. Normally, this latter grouping is expressed in terms of a friction factor, i.e.,

Therefore, we get to the final result that

which tells us that the frictional pressure drop along the pipe will be a function of the Reynolds number and the effective non-dimensional roughness of the pipe.

Worked Example #10

A solid rocket booster parachutes back to Earth and falls into the sea. The booster bobs around in the sea in an upright position at a frequency of .

Using dimensional analysis, show that the non-dimensional frequency of this motion, , is given by

You may assume that for this problem depends on the diameter of the booster , the mass of the booster, , the density of the water , and acceleration under gravity, .

We are told that

or in implicit form then

Therefore, = 5, = 3 and so we will have 2 products. The repeating variables must collectively include all the units of mass , length and time , so the best choice here is , and . Notice that if we were to choose instead of , then collectively the repeating variables would not include time and so the Buckingham method fails in this case. Of course, the dependent variable, , can never be used as a repeating variable.

Therefore, the products are

and

In this question we are only asked for one of the two products, which is the one involving or the product.

We first set up the dimensional matrix, i.e.,

So, the first product is

where the values of the coefficients , and must be obtained to make the equation dimensionally homogeneous. In terms of the dimensions of the problem then

For to be dimensionless, then the powers or exponents of , and must add to zero, i.e., we must have that

Therefore, = 1/2, = 0, and = -1/2, so the product is

which is what we were asked to prove.

It is straightforward to show (not asked for) that the other product is

For this second product then

In terms of the dimensions then

For to be dimensionless, then

Therefore, = -1, = 0, and , so the product is

which is a buoyancy similarity parameter, i.e., the ratio of the mass (or weight) of the water displaced to the mass (or weight) of the rocket booster.

Worked Example #11

where is the density of the water, is the speed of the ship through the water, and is a length scale associated with the hull.Use the Buckingham method to show that

where is a drag coefficient and the non-dimensional grouping is known as the Froude number.

For this problem we are told that the drag can be written as

In implicit form then

So, we have and so there will be 2 products.

Now we set up the dimensional matrix, i.e.,

We need to choose the repeating variables, for which the standard choice is , and , which will all have a primary influence on drag. They also collectively include all of the fundamental dimensions mass, length and time, and they are clearly linearly independent just by inspection. So, for the first product then

and for the second product then

Continuing with the first product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, = -2, = 0, and , so the product is

or

which is the Froude number, .

For the second product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, = -2, = -1, and , so the product is

or

which is a drag coefficient, .

Therefore, finally we have that

i.e., the drag coefficient on the hull is some function of the Froude number.

Worked Example #12

where is the function to be determined and is the coefficient of viscosity. Using the Buckingham method, find the dimensionless similarity parameter that governs this freely falling behavior.

We are told that

so in implicit form then

We have = 4 and = 3 (by inspection, all of mass, length and time are involved in this problem) so we have just one product.

Setting down the dimensional matrix gives

For the product then we have

and in terms of dimensions then

For to be dimensionless, then

and so = -1, and and = -1.

Therefore, the product is

which is a viscous drag coefficient applicable, in this case, to what is known as a Stokes flow, which is a flow corresponding to Reynolds numbers near unity.

Worked Example #13

Use dimensional analysis to show that the non-dimensional parameter that governs this process, known as a Strouhal number , is given by

where is the flow speed and is the diameter of the cylinder.

In this case, we are told that the frequency of shedding will be a function of the diameter of the cylinder and flow velocity . Working with this information then we have

or in implicit form then

We see , but in this case because only length and time are involved in this group of variables (no mass). So, there is just the one product to determine.

Setting up the dimensional matrix, i.e.,

So, we have

and terms of the dimensions then

For to be dimensionless, then

Therefore, = -1 and , so the product is

which is the Strouhal number , as called for in the question.

We could also have proceeded by recognizing that the frequency of shedding may also be a function of the flow density and its viscosity . In this case, there is an expectation that Reynolds number may be involved. If we do this, then

or in implicit form then

We now have and (mass is now involved) so there are now two products to determine. Of course, there is an expectation that one of these has already been determined.

Setting up the dimensional matrix gives

Proceeding as normal with the selection of the repeating variables (again, the standard choice is , and ) then we have

and

For the first product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, = -1, = 0, and , so the product is

which is the Strouhal number we have derived previously.

For the second product, which we suspect will be a Reynolds number, then

and terms of the dimensions then

For to be dimensionless, then

Therefore, = -1, = -1, and , so the product is

which indeed is the Reynolds number .

Therefore, we have then that the frequency of shedding is expected to be a function of Reynolds number, i.e.,

Worked Example #14

The Reynolds number is a non-dimensional grouping formed in terms of fluid density, , a reference velocity, , a characteristic length scale, , and viscosity, , i.e.,

Reynolds number represents “the ratio of the relative effects of inertial effects to viscous effects,” which can be seen by writing

On the numerator, has units of force so in this case it represents an inertial force. The coefficient of viscosity, , is the shear force per unit area per unit velocity gradient, so the denominator is also a force but a viscous force. Hence, we see the significance of the Reynolds number as a relative measure of inertial effects to viscous effects in a fluid flow.

Worked Example #15

The drag of the golf ball (the dependent variable) , can be written in functional form as

where is a function to be determined. This equation can be written in implicit form as

In this case there are 6 variables () and 3 fundamental dimensions () so there are three products.

The functional dependence can also be written in the form

where , and are the non-dimensional groupings to be determined.

Choose the standard aerodynamic repeating variables , and , which are all linearly independent. The non-dimensional products can each be written in terms of these repeating variables plus one other variable, that is

where in each case the values of the exponents , , and are to be determined in such a way that each of the products is non-dimensional.

Write down the dimensions of each of the variables. For this problem then

and so the dimensional matrix is

Considering the first product then

and in terms of dimensions then

These simultaneous equations have the solution that , and . Therefore, the first product can be written as

which is a force coefficient.

Considering the second product then

and in terms of dimensions, then

These simultaneous equations have the solution that , and . Therefore, the second product can be written as

which is a non-dimensional length scale, i.e., the ratio of the diameter of the dimples to the diameter of the golf ball.

Considering the third product then

and in terms of dimensions, then

These simultaneous equations have the solution that , and . Therefore, the third product can be written as

or inverting the grouping (it is still non-dimensional)

which is a Reynolds number.

Finally then

or just

Worked Example #16

The relationship between and the properties listed may be written in a general functional form as

or

We are given a hint and told that sound intensity is defined as the acoustic power per unit area, so in SI we could think of this as Watts per unit area. We also know that power is the rate of doing work and so has units of force times displacement per unit time, i.e., when we come to setting up the problem then we will know that

Also, pressure is force per unit area so

In this problem we have and (by inspection mass, length and time are all involved) so there are two products to determine. Setting up the dimensional matrix gives

Proceeding with the selection of the repeating variables, then one choice is , and . Therefore, our two groups are

and

For the first product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, , = -3, and = 0, so the product is

For the second product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, , = -2, and = 0, so the product is

Interestingly, the factor is related to the compressibility modulus of the medium in which the sound propagates. Therefore, we have in this case

Worked Example #17

Consider a liquid in a cylindrical container in which both the container and the liquid are rotating as a rigid body (this is called solid-body rotation).

The elevation difference between the center of the liquid surface and the rim of the liquid surface is a function of the angular velocity , the fluid density , the gravitational acceleration , and the radius . Use the Buckingham method to find the relationship between the height and the other parameters. Show all of your work.

In this problem we are asked to find the effects of the elevation difference between the center of the liquid surface and the rim of the liquid surface, which we are told is a function of the angular velocity , the fluid density , the gravitational acceleration , and the radius , i.e.,

or

We have and (by inspection mass, length and time are all involved) so there are two products to determine. Setting up the dimensional matrix gives

Proceeding with the selection of the repeating variables, then one choice is , and . Therefore, our two groups are

and

For the first product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, , = -1/2, and = 1/2, so the product is

For the second product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, , = 0, and = -1, so the product is

Therefore, we have in this case

Worked Example #18

A liquid of density and viscosity flows by gravity through a hole of diameter in the bottom of a tank of diameter . At the start of the experiment, the liquid surface is at height above the bottom of the tank. The liquid exits the tank as a jet with average velocity straight down. Using the Buckingham method, find a dimensionless relationship for as a function of the other parameters in the problem. Identify any established non-dimensional parameters that appear in your result. Show all of your work. Hint: Notice that there are three length scales in this problem, but for consistency choose as the reference length scale.

In this problem we are asked to find the effects on the exit flow velocity in terms the fluid density , its viscosity , the diameter of the hole . the diameter of the tank , and the height of the liquid surface , i.e.,

or

In this problem we have and (by inspection mass, length and time are all involved) so there are three products to determine. Setting up the dimensional matrix gives

Proceeding with the selection of the repeating variables, the only choice in this case is , and . (Note: Can you explain why?). Therefore, our three groups are

and

and

For the first product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, = -1, = 1, and = 1, so the product is

which we recognize is a Reynolds number.

For the second product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, = 0, = 0, and = -1, so the product is

For the third product then

which quickly follows as per the product as

Therefore, in this case the non-dimensional groupings involved are such that

Worked Example #19

The AIAA Design Build & Fly (DBF) team need to determine the factors that influence the aerodynamic drag on a rectangular banner being towed behind their airplane.

The size of the banner is determined by its length, , and height, . Use the Buckingham method to determine the non-dimensional groupings that will govern this problem. You may also assume that the problem is governed by the airspeed of the airplane, as well as the density and viscosity of the air. Assume further that the banner remains flat and does not flutter in the flow behind the airplane.

The relationship between the drag on the banner and the air properties can be written in the general functional form as

where we are told that the size of the banner is represented by its length, , and height, . Remember that is called the dependent variable and , , , and are are the independent variables. The functional dependence of in implicit form is

Counting up the variables gives and also because there are 3 fundamental dimensions in this problem. Therefore, and we will have three products.

We must first find the dimensions of the variables. For each variable the dimensions are

We can now set up the dimensional matrix, i.e.,

Choose , and as the repeating variables, which will all have primary effects on the drag of the banner. These variables also collectively include all the fundamental dimensions of this problem and are linearly independent of each other.

Following the Buckingham method, then the three products are:

For :

The values of the coefficients , and must now be obtained to make the equation dimensionally homogeneous. In terms of the dimensions of the problem then

For to be dimensionless, then the powers or exponents of , and must add to zero, i.e., we must have that

By inspection we get , , and . Therefore, the first product is

or

i.e., a form of drag coefficient. We normally define aerodynamic force coefficients in terms of the dynamic pressure, i.e., so that more conventionally we would write the force coefficient as

It would also be legitimate to write the drag coefficient as

where the banner area is used rather than . Ultimately, how we define is just matter of convenience and/or consistency with established convention.

For :

so

and

Therefore, in this case , , and , so

or

Inverting the grouping gives

which in the latter case is a Reynolds number based on the banner length. Notice that the grouping can be inverted if we want to, such as to follow established convention or just because it is otherwise convenient to do so.

For :

so

giving

Therefore, in this case , , and , i.e.,

so

or we can again invert this grouping (for convenience) giving

which is a length to height ratio or what would be called an aspect ratio .

As a result of our dimensional analysis then

or

Finally, in explicit form the drag coefficient can be written as a function of Reynolds number based on banner length and the aspect ratio of the banner, i.e.,

Worked Example #20

The relationship between the flutter speed and the expected dependencies can be written in general functional form as

or in implicit form as

Hence, in this problem , , , and so we will have three products.

For each variable the dimensions are

Now we can set up the dimensional matrix, i.e.,

Again, as in most aerodynamic problems, we choose , and as the repeating variables. Following the Buckingham method then the products are:

For :

In terms of the dimensions of the problem then

For to be dimensionless we must have

By inspection we get , , and . Therefore, the first product is

or

which is a speed ratio or a non-dimensional flutter speed.

For :

In terms of the dimensions of the problem then

For to be dimensionless we must have

By inspection, we get , , and . Therefore, the second product is

So

or

and, once again, a Reynolds number comes into the problem.

For :

In terms of the dimensions of the problem then

For to be dimensionless we must have

By inspection we get , , and . Therefore, the third product is

or

which is a form of structural non-dimensional frequency or what we might call a structural reduced frequency.

As a result of our dimensional analysis then

or

or in explicit form

Therefore, the dimensional analysis tells us that the non-dimensional flutter speed of the banner will depend on the Reynolds number and its structural reduced frequency.

Worked Example #21

A spherical projectile of diameter is moving supersonically. The drag is assumed to depend on the free-stream velocity , the free-stream density , the free-stream viscosity , and the free-stream temperature , as well as the heat capacities at constant volume and constant pressure, and , respectively. Use the Buckingham method to determine the non-dimensional groupings that will govern this problem.

The relationship between the drag on the sphere and the given variables can be written general functional form as

or in implicit form as

Hence, , , , and so we will have four products. Notice in this case that temperature in explicitly defined so there are 4 fundamental dimensions.

For each variable the units are

Now we can set up the dimensional matrix, i.e.,

Choose , , and as the repeating variables. They are not unique but have primary dependencies on the drag and collectively include all the fundamental dimensions as well as being linearly independent of each other.

Following the Buckingham method then the products are:

For :

In terms of the dimensions of the problem then

For to be dimensionless we must have

By inspection we get , , and . Therefore, the first product is

and

which is a drag coefficient, or we just write it in the conventional way that

For :

In terms of the dimensions of the problem then

For to be dimensionless we must have

By inspection we get , , and . Therefore, the second product is

and

or more conventionally this ratio is written as

Notice that is the internal energy per unit mass of the free-stream flow, so this non-dimensional grouping represents a ratio of kinetic energy to internal energy.

For the process will be identical to that for , which is redundant. But we also see that both and have the same units so we can write immediately that

which is the familiar ratio of specific heats. This ratio would have been a product of the dimensional analysis if either or had been used as a repeating variable.

For :

In terms of the dimensions of the problem then

For to be dimensionless, then we must have

By inspection we get , , and . Therefore, the fourth product is

and

or just

which is the Reynolds number.

As a result of our dimensional analysis then

or in explicit form

But the important grouping that comes out of this problem is

Worked Example #22

The relationship between the force and the tip deflection can be written in the general functional form as

or in implicit form as

where we have used to denote the length of the wing to avoid confusion with the dimensions of length. Hence, , , , and so we will have two products.

For each variable the units are

Now we can set up the dimensional matrix, i.e.,

This problem poses a bit of a dilemma because we see that the choice of the repeating variables here is not at all obvious. If we choose , and as the repeating variables they will not be linearly independent, and if we choose , and as the repeating variables (try it!) then the solution to the problem becomes indeterminate, i.e., we cannot uniquely solve for the non-dimensional groupings.

The accepted solution to this type of dilemma is to reduce the number of repeating variables by one and create a third grouping. If we now choose and as repeating variables, which are linearly independent and include all of the fundamental dimensions, then following the Buckingham method then the three products will be:

For :

In terms of the dimensions of the problem then

For to be dimensionless, then

By inspection we get and , so the first product is

so

i.e., a non-dimensional displacement and this seems to be an expected if not obvious grouping.

For :

In terms of the dimensions of the problem then

For to be dimensionless, then

By inspection we get and , so the second product is

and so

which is a non-dimensional form of the second moment of area.

For :

In terms of the dimensions of the problem then

For to be dimensionless, then

By inspection we get and , so the third product is

so

which is a form of non-dimensional force.

Finally, we have all three groupings and so we can write the result in functional form as

or

Worked Example #23

where is the viscosity of the flow. First, proceed to verify that the primary dimensions of are units of time. Second, find a non-dimensional form for the time constant that is based on a characteristic flow velocity, , and a characteristic length, . Comment on your result. Do you see anything interesting?

We are given that

and told that the units of are time. For each variable the units are

so

which confirms that the units of are indeed time.

We are asked to find a non-dimensional form of based on a characteristic flow velocity, , and a characteristic length, . We see that the ratio has units of time so as non-dimensional form could be

which is interesting because it involves a Reynolds number based on particle diameter as well as the ratio of the particle diameter to the length scale. Therefore, the higher the Reynolds number and/or the bigger the particle then the longer it will take for the particle to adjust to any changes in the flow conditions.

Worked Example #24

For this problem we are told that the power output can be written as

In implicit form then

So, we have and again so there will be 2 products.

We first set up the dimensional matrix, i.e.,

Notice the base units of power are M L T.

Now we need to choose the repeating variables. A good choice in this case is , and , which will have a primary influence on power production from the turbine. They also collectively include all of the fundamental dimensions mass, length and time, and they are clearly linearly independent just by inspection.

For the first product then

and for the second product then

Continuing with the first product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, = -3, = -1, and , so the product is

or

which is a form of power coefficient, i.e.,

Considering now the second product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, = -1, = 0, and , so the product is

or

which is a form of advance ratio or tip speed ratio, i.e.,

Therefore, we see based on the information given that the power output of the wind turbine in terms of a power coefficient is related to the wind speed in the form of a tip speed ratio , i.e.,

or

Worked Example #25

1. Write down the functional expression for the drag force in terms of the parameters given above.

2. Write down the dimensional matrix for this problem in terms of base units MLT.

3. Determine the relevant groups for this problem.

4. If the drag force on the sphere with = 0.1 m and = 0.07 m in a specific liquid flowing at an average flow speed of 3 m/s is 600 N, what would the drag force be on a sphere with = 0.4 m and = 0.28 m at 6.7 m/s using the same liquid? Assume that = 900 kg m.

1. In explicit form then

or in implicit form then

2. Setting up the dimensional matrix for this problem gives

3. We see in this (five variables) and by inspection we clearly have all of M, L and T so (i.e., all three fundamental dimensions in terms M, L and T) so there are two products to determine.

4. Use , and as the repeating variables, which the students should hopefully use if they were paying attention. This choice includes all the fundamental dimensions and it is obvious that they are all linearly independent. Following the Buckingham method then the two products are

So, we have for the first product that

and terms of the dimensions then

For to be dimensionless then

Therefore, , , and so the product is

which is a force coefficient.

For the second product then

and terms of the dimensions then

For to be dimensionless, then

Therefore, = 0 and = 0, and = -1 so the product is

which is a non-dimensional length. Therefore, we have

and so finally in explicit form then

5. To examine dimensional similitude then for both cases for dimensional similitude then the force coefficients must be the same so

so by rearrangement then

and also confirming for geometric similarity gives

Worked Example #26

1. Write down the functional relationship for the frequency in terms of the other parameters given, in both implicit and explicit form.

2. How many base dimensions and groupings are involved in this problem

3. Write down the dimensional matrix for the problem.

4. Use the Buckingham method to determine the dimensionless parameter(s) that describes this problem.

5. Rewrite the functional relationship in term of the non-dimensional parameter(s).

1. The frequency of the sound can be written explicitly as

or implicitly as

2. The number of variables is 5 so and the number of base dimensions (mass, length and time are all involved) is 3, so . This means that there are groupings.

3. The dimensional matrix is

4. Choose the variables as , , and , which are a standard choice for aerodynamic problems. We can now proceed to find the two groups, i.e.,

and

For then

Raising the repeating variables to unknown powers gives

In terms of dimensions then

For to be dimensionless, the powers of each base dimension must add to zero, i.e.

Solving the equations gives , , and so

which is the inverse of the well known Reynolds number so we can invert (still having a dimensionless grouping) giving

Solving for gives

so

In terms of dimensions then

For to be dimensionless, the powers of each base dimension must add to zero, i.e.

Solving the equations gives , , and so

which is a Strouhal number, i.e.,

5. Therefore, based on the foregoing analysis, we have that

or

or

so the Strouhal number is a function of the Reynolds number.

Worked Example #27

Based on the previous problem then the two relevant similarity parameters in this case are the Reynolds number and the Strouhal number, and the Strouhal number is a function of the Reynolds number. The actual power wires the Reynolds number based on diameter = 2.2 cm will be

using the highest wind speed of 70 mph and MSL ISA values for air. Notice that 70 mph is 102.67 ft/s.

In the wind tunnel, the wire available is only 1.1 cm in diameter, i.e., . So, to get the same Reynolds number the flow speed will need to be twice, i.e., 140 mph or 205.3 ft/s, but this is significantly less than the maximum flow speed of the wind tunnel. Even if the wire to be used in the tunnel was 2.2 cm in diameter, the required flow speed to match the Reynolds number would be higher than is attainable.

One solution would be to use a wire of diameter of say 3.3 cm in the wind tunnel, which would need a flow speed of

to match the Reynolds numbers and this is easily achievable, and it is a factor of 0.67 of the actual wind speed.

Therefore, if the Reynolds number is matched by increasing the wire diameter the question is can we also match the Strouhal number? In this case we would get the same sound frequency if

but in this case we have

So even though the Reynolds number could be matched we would not get the same frequency of the Aeolian sounds. Nevertheless, by matching Reynolds number in the wind tunnel we would get the same Strouhal number so the frequency obtained in this case would be higher by a factor of 2.24. In principle then, it would be possible to study the singing sound behavior of the wires in the wind tunnel it is just that the frequencies so obtained would be higher.

This is another example of the challenges in sub-scale testing to study real problems. But it can be seen that with a little ingenuity the problem can be studied by matching, or by matching as closely as possible, the similarity parameters that govern the physics.

Worked Example #28

A Covid-19 particle has a density and characteristic size . It is carried along in the air of density and viscosity . In still air, the particles slowly settle out only very slowly under the action of gravity and reach a terminal settling speed . It can be assumed that depends only on , , , and the density difference .

- Write down the functional dependency of and the other variables , , . and in both explicit and implicit forms.
- How many base dimensions and groupings are involved in this problem?
- Write down the dimensions of each of the variables involved.
- Form the dimensional matrix for this problem.
- Choose , and as the repeating variables, so then proceed to find the grouping involving .

1. We can let . The relationship between the settling velocity and other properties can be written in the general functional form as

and in implicit form

2. and ; there are 3 fundamental dimensions in this problem. Therefore, , so we will have two products.

3. For each variable the dimensions are

4. We can now set up the dimensional matrix, i.e.,

5. We are told to choose , , and as the repeating variables, which will all have primary effects on the settling velocity. They also collectively include all the fundamental dimensions and are linearly independent of each other. We are asked just to find the grouping that involves so

In terms of the dimensions of the problem then

For to be dimensionless, then the powers or exponents of , and must add to zero, i.e., we must have that

By inspection we get , and so

or

Worked Example #29

A weir is an obstruction in an open channel water flow. The volume flow rate over the weir depends on acceleration under gravity , the width of weir (into the screen), and the water height above the weir.

- Write out the functional relationship in explicit and implicit form.
- How many fundamental (base) dimensions are involved?
- Write out the dimensional matrix for this problem.
- Determine the non-dimensional groupings that govern this problem.

1. The explicit form of the relationship is

and in implicit form then

2. In this problem we see that there are actually only two base dimensions, length and time , so the procedure should be easier.

The units involved are

3. We can now set up the dimensional matrix, i.e.,

4. We will have two products as the number of the basic quantities and number of variables in this problem . We will use and as repeating variables, which collectively include all the fundamental dimensions and are linearly independent of each other. Note: We cannot choose both and repeating variables because they have the same units and, therefore, they are not linearly independent of each other.

For the first product then

In terms of the dimensions of the problem then

For to be dimensionless, then the powers or exponents of and must add to zero, i.e., we must have that

By inspection we get and so the grouping is

or

For the second product then

In terms of the dimensions of the problem then

For to be dimensionless, then the powers or exponents of and must add to zero, i.e., we must have that

By inspection we get and so the grouping is

Therefore, we have that

or

Worked Example #30

The relationship may be written in a general functional form as

or in implicit form then

We have and because the problem clearly includes the dimensions of mass, length and time, and so we have just one product. Energy is the ability to to work and so the units of energy will be the same as the units of work, which is equivalent to a force times a distance, i.e., . Setting up the dimensional matrix gives

Following the Buckingham method, then the product is

where the specific values of the coefficients , and must be obtained to make the equation dimensionally homogeneous. In terms of the dimensions of the problem then

In this case we see that so from the second equation we get that . This further gives that , and . Therefore, we get that

or

In final form we can write

In the second part of the question we are asked about how the radius of the blast wave changes by a doubling of . According to the relationship we have derived then the radius would increase by a factor of or about 1.15. By doubling the time then the radius of the wave would increase by a factor of or about 1.32.

Worked Example #31

An ocean surface wave is a sinusoidal-like disturbance propagating along the surface of the ocean, as shown in the figure below. The speed of the wave, , is found to be a function of the surface tension of the sea water, , the density of the sea water, , and the wavelength, , of the wave.

1. Write down the functional relationship for the speed of the wave in terms of the given parameters, in both implicit and explicit form.

2. How many base dimensions and groupings are involved in this problem?

3. Write down the dimensions of all the parameters in base units. **Notice: The units of surface tension are force per unit length.**

4. Create the dimensional matrix for this problem.

5. Determine the dimensionless parameter(s) that describe this problem.

1. In explicit form, the speed of the wave is

where is some function. In implicit form then

where is some other function.

2. The base dimensions and groupings involved in this problem are:

3. Below are the dimensions of all the parameters in this problem in terms of base units:

The speed of the wave, , has base units of

Surface tension, , is given in the question as a “force per unit length” so it has base units of

Density, , has base units of

Wavelength, , has base units of

4. From the previous results, then the dimensional matrix is

5. The wave speed cannot be a repeating variable so the only possible choice of the repeating variables is , , and . The group will be

Therefore, we have that

In terms of dimensions then

or alternatively the foregoing can be written as

For this equation to be mathematically balanced on the left and right sides, i.e., to be dimensionally homogeneous, then

Solving the foregoing equations gives , , so that the resulting grouping is

or

As a final check, it is easy to show that this grouping is non-dimensional because

Worked Example #32

The flow rate in a pipe is to be measured with an orifice plate, as shown in the figure below. The static pressure before and after the plate is measured using two pressure gauges. The volumetric flow rate is found to be a function of the measured pressure drop across the plate, , the fluid density, , the pipe diameter, , and the orifice diameter, .

- Write down the functional relationship for the volumetric flow rate in terms of the other parameters given, in both implicit and explicit form.
- Write down the base units of the parameters involved in this problem.
- How many base dimensions and dimensionless groupings are involved in this problem?
- Write out the dimensional matrix for this problem.
- Choose the repeating variables and explain your choice.
- Determine the non-dimensional grouping(s) for the parameters involved.
- Write down the final dimensionless functional relationship(s).

where is some function, and implicit form as

where is some other function.

2. The dimensions of the parameters involved are:

3. The number of base dimensions and groupings are involved in this problem:

- Number of variables: .
- Number of base dimensions (mass, length, and time are all involved): .
- Number of groups: .

4. The dimensional matrix is

5. The only possible choice of repeating variables are: , , and . Obviously, we cannot choose both and as repeating variables because they are not linearly independent. But we should choose one or the other, so we will choose and so the repeating variables will be , , and .

6. The groups will be formed from:

and

We now have to solve for the dimensionless groupings.

For then:

so

and inserting the dimensions for each parameter gives

or

For to be dimensionless the powers must add to zero, i.e.,

Solving the equations we get: , , and . This means, therefore, that is

or

As a check to see if this is a non-dimensional grouping, we can substitute the units of the parameters so that

so confirming that the grouping is indeed nondimensional.

For then:

so

and inserting the dimensions for each parameter gives

or

For to be dimensionless the powers must add to zero, i.e.,

Solving the equations we get: , , and . This means, therefore, that is

In this case, it is obvious that the ratio of one length to another is non-dimensional.

7. Finally then, the non-dimensional relationship between the volumetric flow rate and the other parameters is