Recall from Chap. 8 that the exit velocity from a simple reservoir is ν=√2gh, which is the same result that one obtains for the velocity of a mass particle that is dropped from a resting position at height h above the surface. If one wishes to determine the latter result (for the first time)

Question

Recall from Chap. 8 that the exit velocity from a simple reservoir is [latex] u =\sqrt{2gh},[/latex] which is the same result that one obtains for the velocity of a mass particle that is dropped from a resting position at height h above the surface. If one wishes to determine the latter result (for the first time) experi-mentally given the general relation [latex] u =f(m,g,h)[/latex] Control Volume and Semi-empirical Methods for a particle of mass m, and neglecting air resistance, one might seek to perform multiple experiments for different masses m and heights h. Let us see what Buckingham Pi would suggest, however.

Accepted Answer

Step 1: Specify the desired functional relationship, [latex] u=f(m, g, h).[/latex]
Step 2: Consider fundamental units such as L, T, and M and identify the unit equations for each variable:

[latex] \left[ u ight]=L^{1}T^{-1}M^{0}, \left[g ight]=L^{1}T^{-2}M^{0},[/latex]

[latex] \left[m ight]=L^{0}T^{0}M^{1}, \left[h ight]=L^{1}T^{0}M^{0}.[/latex]

Step 3: Assign scales. Selections of intrinsic scales are obvious for length and mass. For time, however, we could pick a length divided by the impact velocity or, because gravity is key to the problem and it has a time dimension within, we could also consider the square root of h divided by g; that is, we could let

[latex] L_{s}=h, T_{s}=\sqrt{\frac{h}{g} }, M_{s}=m.[/latex]

Step 4: List the computed Pi groups:

[latex] \pi _{1}=\frac{ u }{(h)^{1}\left(\sqrt{h/g} ight)^{-1}(m)^{0} }=\frac{ u }{\sqrt{gh} }, \pi _{3}=\frac{g}{(h)^{1}\left(\sqrt{h/g} ight)^{-2} }\equiv 1,[/latex]

[latex] \pi _{2}=\frac{m}{(m)^{1}}=1, \pi _{4}=\frac{h}{(h)^{1}}\equiv 1.[/latex]

Step 5: Express our original equation in terms of Pi groups, that is

[latex] \frac{ u }{\sqrt{gh} }=f(1,1,1)[/latex] or [latex]\frac{ u }{\sqrt{gh} }=c.[/latex]

Thus, Buckingham Pi reveals that [latex] u /\sqrt{gh}[/latex] is a constant and all we need to do is to measure multiple values of v for multiple values of h and find the constant, which, of course, we know is [latex]\sqrt{2} .[/latex]

Question 10.6: Recall from Chap. 8 that the exit velocity from a simple res...

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