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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: Π1 = V * m^(1/2) * p^(-1/2) * ρ^(1/2), Π2 = a * V^(-1), Π3 = m * ρ^(-1) * z^3, Π4 = ρ * z * p^(-1). The final dimensionless function could look something like Π1 = f(Π2, Π3, Π4).

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 (ρ) 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’s Pi theorem

Buckingham’s 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 Π) are:\n\n- Π1 = V * m^(1/2) * p^(-1/2) * ρ^(1/2) (all dimensions cancel out).\n- Π2 = a * V^(-1)\n- Π3 = m * ρ^(-1) * z^3\n- Π4 = ρ * 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 Π1 = f(Π2, Π3, Π4). However, the exact form would depend on the specific relationships between the variables, which is not given in this problem.

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Most popular questions from this chapter

The pressure rise, \(\Delta p=p_{2}-p_{1},\) across the abrupt expansion of Fig. \(\mathrm{P} 7.38\) through which a liquid is flowing can be expressed as $$\Delta p=f\left(A_{1}, A_{2}, \rho, V_{1}\right)$$ where \(A_{1}\) and \(A_{2}\) are the upstream and downstream cross-sectional areas, respectively, \(\rho\) is the fluid density, and \(V_{1}\) is the upstream velocity. Some experimental data obtained with \(A_{2}=1.25 \mathrm{ft}^{2}\) \(V_{1}=5.00 \mathrm{ft} / \mathrm{s},\) and using water with \(\rho=1.94\) slugs/ft \(^{3}\) are given in the following table: $$\begin{array}{l|l|l|l|l|r} A_{1}\left(\mathrm{ft}^{2}\right) & 0.10 & 0.25 & 0.37 & 0.52 & 0.61 \\ \hline \Delta p\left(\mathrm{lb} / \mathrm{ft}^{2}\right) & 3.25 & 7.85 & 10.3 & 11.6 & 12.3 \end{array}$$ Plot the results of these tests using suitable dimensionless parameters. With the aid of a standard curve fitting program determine a general equation for \(\Delta p\) and use this equation to predict \(\Delta p\) for water flowing through an abrupt expansion with an area ratio \(A_{1} / A_{2}=0.35\) at a velocity \(V_{1}=3.75 \mathrm{ft} / \mathrm{s}\).

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