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Chapter 7: Problem 26

A coach has been trying to evaluate the accuracy of a baseball pitcher. After two years of studying, he proposes a function that can be presented as the accuracy of any pitcher: \\[ \mathrm{Acc}=f(V, a, m, \rho, p, z) \\] where Acc is the dimensionless accuracy, \(V\) is the velocity of the ball, \(a\) is the age of the pitcher, \(m\) is the mass of the pitcher, \(\rho\) is the density of air where the game is played, \(p\) is the pressure where the game is played (which varies with elevation), and \(z\) is the elevation above sea level. Find the dimensionless groups for this function.

Short Answer

Expert verified
The dimensionless groups for the given function are: 螤1 = V * m^(1/2) * p^(-1/2) * 蟻^(1/2), 螤2 = a * V^(-1), 螤3 = m * 蟻^(-1) * z^3, 螤4 = 蟻 * z * p^(-1). The final dimensionless function could look something like 螤1 = f(螤2, 螤3, 螤4).

Step by step solution

01

Identify the physical dimensions of all the variables

Firstly, identify what physical dimension each variable in the problem represents. Here, we have: \n\n- Velocity (V) with dimensions of Length/Time (LT^-1).\n- Age (a) with dimensions of Time (T).\n- Mass (m) with dimensions of Mass (M).\n- Density (蟻) with dimensions of Mass/Length^3 (ML^-3).\n- Pressure (p) with dimensions of Force/Area = Mass*Length/Time^2/Area = Mass/Length*Time^2 (ML^-1T^-2).\n- Elevation (z) with dimensions of Length (L).
02

Applying Buckingham鈥檚 Pi theorem

Buckingham鈥檚 Pi theorem states that the number of dimensionless groups is equal to the difference between the number of variables and the number of fundamental dimensions. Here, the total number of variables is 7 (including 'Acc') and the fundamental dimensions are M, L and T; hence, 3. So, we have 7 - 3 = 4 dimensionless groups.
03

Formulating the dimensionless groups

To form the dimensionless groups, we combine variables such that their dimensions cancel out. For example, we choose velocity (V), mass (m), pressure (p), and elevation (z) for the four groups. \n\nSo, the dimensionless groups (represented as 螤) are:\n\n- 螤1 = V * m^(1/2) * p^(-1/2) * 蟻^(1/2) (all dimensions cancel out).\n- 螤2 = a * V^(-1)\n- 螤3 = m * 蟻^(-1) * z^3\n- 螤4 = 蟻 * z * p^(-1)
04

Final dimensionless function

Finally, the dimensionless function representing the accuracy of the pitcher can be represented using these dimensionless groups.\n\nIt could be something like 螤1 = f(螤2, 螤3, 螤4). However, the exact form would depend on the specific relationships between the variables, which is not given in this problem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dimensionless Groups
In fluid mechanics and many other branches of physics, we often convert equations to dimensionless form. This simplifies analysis and compares systems without worrying about unit differences. Dimensionless groups are crucial here. They are combinations of variables where all units cancel out, meaning these groups have no dimensions. This can help identify similarities between different physical scenarios.

Using the Buckingham Pi Theorem is a common way to find these groups. This theorem provides a method to reduce a complex problem with various variables to a simpler one. The number of dimensionless groups is derived by subtracting the number of fundamental dimensions from the total number of variables. In our example, there were seven variables and three fundamental dimensions, resulting in four dimensionless groups.

The process involves combining variables strategically until all units cancel out. This generates "Pi terms," which are the dimensionless groups that allow easier manipulation of equations without losing any physical significance. These terms can often reveal underlying relationships between variables that are not immediately obvious.
Physical Dimensions
Every physical quantity can be expressed in terms of fundamental dimensions. These are basic units like mass (M), length (L), and time (T). Understanding these helps us analyze and manipulate the equations and eventually derive dimensionless groups.

Identifying the physical dimensions of each variable is the first step in using the Buckingham Pi Theorem. For instance:
  • Velocity (V) has dimensions of length over time, expressed as \(LT^{-1}\).
  • Age (a) is simply time, so \(T\).
  • Mass (m) has the dimension of mass, \(M\).
  • Density (\(\rho\)) combines mass and volume, \(ML^{-3}\).
  • Pressure (p) mixes force, area, and time, leading to \(ML^{-1}T^{-2}\).
  • Elevation (z) is just length, \(L\).
Thinking in terms of these fundamental dimensions helps streamline the process of forming dimensionless groups, which simplifies complex analyses.
Variables in Fluid Mechanics
In fluid mechanics, we often deal with multiple variables like velocity, pressure, and density, each affecting flow behavior. These variables have specific physical dimensions and play a crucial role in categorizing the types of flow and the behavior of systems under different conditions.

To understand their interactions, converting them into dimensionless groups often offers deeper insight. In our example, variables like velocity and pressure interact to form a system's dynamics. By examining these in dimensionless form, one can explore relationships between speed, pressure changes, and other factors like elevation.

For example, you might discover how a change in elevation affects pressure and density, affecting a baseball pitcher's accuracy. Understanding these interactions is essential in predicting real-world behavior and optimizing systems in engineering, sports analytics, weather predictions, and various other fields.

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Most popular questions from this chapter

The pressure rise, \(\Delta p,\) across a pump can be expressed as \\[ \Delta p=f(D, \rho, \omega, Q) \\] where \(D\) is the impeller diameter, \(\rho\) the fluid density, \(\omega\) the rotational speed, and \(Q\) the flowrate. Determine a suitable set of dimensionless parameters.

At a sudden contraction in a pipe the diameter changes from \(D_{1}\) to \(D_{2} .\) The pressure drop, \(\Delta p,\) which develops across the contraction is a function of \(D_{1}\) and \(D_{2},\) as well as the velocity, \(V\), in the larger pipe, and the fluid density, \(\rho,\) and viscosity, \(\mu .\) Use \(D_{1}, V,\) and \(\mu\) as repeating variables to determine a suitable set of dimensionless parameters. Why would it be incorrect to include the velocity in the smaller pipe as an additional variable?

For a certain fluid flow problem it is known that both the Froude number and the Weber number are important dimensionless parameters. If the problem is to be studied by using a 1: 15 scale model, determine the required surface tension scale if the density scale is equal to \(1 .\) The model and prototype operate in the same gravitational field.

As winds blow past buildings, complex flow patterns can develop due to various factors such as flow separation and interactions between adjacent buildings. (See Video \(\vee 7.13\).) Assume that the local gage pressure, \(p\), at a particular location on a building is a function of the air density, \(\rho,\) the wind speed, \(V\), some characteristic length, \(\ell,\) and all other pertinent lengths, \(\ell_{i},\) needed to characterize the geometry of the building or building complex. (a) Determine a suitable set of dimensionless parameters that can be used to study the pressure distribution. (b) An eight-story building that is \(100 \mathrm{ft}\) tall is to be modeled in a wind tunnel. If a length scale of 1: 300 is to be used, how tall should the model building be? (c) How will a measured pressure in the model be related to the corresponding prototype pressure? Assume the same air density in model and prototype. Based on the assumed variables, does the model wind speed have to be equal to the prototype wind speed? Explain.

A very small needle valve is used to control the flow of air in a \(\frac{1}{8}-\) in. air line. The valve has a pressure drop of 4.0 psi at a flow rate of \(0.005 \mathrm{ft}^{3} / \mathrm{s}\) of \(60^{\circ} \mathrm{F}\) air. Tests are performed on a large, geometrically similar valve and the results are used to predict the performance of the smaller valve. How many times larger can the model valve be if \(60^{\circ} \mathrm{F}\) water is used in the test and the water flow rate is limited to 7.0 gal/min?
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