Recall that we began this section by considering experiments to determine the pressure drop Δp associated with flow through a potentially rough circular pipe, which we expect to depend on several factors. The pressure drop may be affected, for example, by the diameter D of the pipe, the length L

Question

Recall that we began this section by considering experiments to determine the pressure drop [latex]\Delta p[/latex] associated with flow through a potentially rough circular pipe, which we expect to depend on several factors. The pressure drop may be affected, for example, by the diameter D of the pipe, the length L of the pipe, the roughness height [latex]e[/latex] of the inner surface of the pipe, the viscosity [latex]\mu[/latex] of the fluid, the density [latex] ho[/latex] of the fluid, and the mean velocity [latex]\overline{ u }[latex] of the fluid. Use the Buckingham Pi Theorem to determine a set of dimensionless groups that can be used to design appropriate experiments and to correlate the associated data.

Accepted Answer

Follow the recipe: Step 1: Specify the relation of interest, namely [latex]\Delta p=f(D,L,e,\mu , ho ,\overline{ u } ).[/latex] Step 2: Consider fundamental units that are appropriate: L, T, and M. Then, for each variable, we have[latex]^{4}[/latex]

[latex]\left[\Delta p ight]=\frac{Force}{Area}=\frac{ML/T^{2}}{L^{2}}=L^{-1}T^{-2}M^{1}, \left[\mu ight]= \frac{Force/Area}{1/Time}=\frac{ML/T^{2}}{L^{2}/T}[/latex]

[latex] =L^{-1}T^{-1}M^{1},[/latex]

[latex]\left[D ight]=L^{1}T^{0}M^{0}, \left[ ho ight]=\frac{Mass}{Volume}=L^{-3}T^{0}M^{1},[/latex]

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

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

Step 3: Assign scales. It is reasonable to let the diameter be the length scale and likewise the ratio of diameter to the mean velocity be the timescale. For mass, it is reasonable to take the density times a volume. Although [latex]\pi D^{2}L/4[/latex] is the fluid volume over the entire length of the pipe, selecting [latex]D^{3}[/latex] as a volume is similarly acceptable. Hence, let

[latex] L_{s}=D, T_{s}=\left(\frac{\overline{ u } }{D} ight)^{-1}=\frac{D}{\overline{ u } }, M_{s}= ho D^{3} .[/latex]

[latex]^{4}[/latex] In some cases, we may not know the dimensions of a parameter directly, such as the viscosity. In such cases, we recall a simple relation that relates the parameter to those having known dimensions (e.g., [latex]\sigma _{xx}=2\mu D_{xy}[/latex]) Control Volume and Semi-empirical Methods Step 4: List the computed Pi groups, namely

[latex]\pi _{1}=\frac{\Delta p}{(D)^{-1}(D/\overline{ u } )^{-2}( ho D^{3})^{1}}=\frac{\Delta p}{ ho \overline{ u }^{2} }, \pi _{5}=\frac{\mu }{(D)^{-1}(D/\overline{ u } )^{-1}( ho D^{3})^{1}}=\frac{\mu }{ ho \overline{ u }D }=\frac{1}{Re},[/latex]

[latex]\pi _{2}=\frac{D}{(D)^{1}(D/\overline{ u } )^{0}( ho D^{3})^{0}}=\frac{D }{D}=1, \pi _{6}=\frac{ ho }{(D)^{-3}(D/\overline{ u } )^{0}( ho D^{3})^{1}}=\frac{ ho }{ ho }=1,[/latex]

[latex]\pi _{3}=\frac{L }{(D)^{1}(D/\overline{ u } )^{0}( ho D^{3})^{0}}=\frac{L }{D}, \pi _{7}=\frac{\overline{ u } }{(D)^{1}(D/\overline{ u } )^{-1}( ho D^{3})^{0}}=\frac{\overline{ u } }{\overline{ u } }=1.[/latex]

[latex]\pi _{4}=\frac{e}{(D)^{1}(D/\overline{ u } )^{0}( ho D^{3})^{0}}=\frac{e }{D},[/latex]

In particular, note that the combination of terms [latex] ho \overline{ u }D/\mu[/latex] appears so com-monly in fluid mechanics that it is given a special symbol Re and is called the Reynolds’ number. It has been mentioned earlier, but its utility will be seen in more detail in Sect. 10.6. Step 5: Express the governing functional equation

[latex] x_{1}=f(x_{2},x_{3},x_{4},x_{5},x_{6},x_{7})\Leftrightarrow \Delta p=f(D,L,e,\mu , ho, \overline{ u } )[/latex]

in terms of Pi-groups, namely

[latex] \pi _{1}=g(\pi _{2},\pi _{3},\pi _{4},\pi _{5},\pi _{6},\pi _{7})\Leftrightarrow \frac{\Delta p}{ ho \overline{ u }^{2} }=g\left(1,\frac{L}{D}.\frac{e}{D},\frac{1}{Re},1,1 ight).[/latex]

Hence, according to the Buckingham Pi Theorem,

[latex] \frac{\Delta p}{ ho \overline{ u }^{2} }=g\left(\frac{L}{D},\frac{e}{D},\frac{1}{Re} ight)=\widetilde{g} \left(\frac{L}{D},\frac{e}{D},Re ight).[/latex]

Note: If we have an equation [latex]y=ax^{2},[/latex] then we say [latex]y=f(x).[/latex] Similarly, if we have an equation [latex]y=a/x^{2},[/latex] then we again say [latex]y=g(x).[/latex] The key here is the functional dependency. As it turns out, extensive experiments have revealed that [latex]\pi _{1}[/latex] depends linearly on [latex]L/D[/latex] and, thus,

[latex] \frac{\Delta p}{\frac{1}{2} ho \overline{ u }^{2} }=\frac{L}{D}f\left(Re,\frac{e}{D} ight),[/latex]

where the function f is called a friction factor. (Note: The functione [latex]\widetilde{g}[/latex] is arbitrary; hence, we can multiply or divide it by a constant such as [latex]\frac{1}{2},[/latex] which we do so as to 562 10. Control Volume and Semi-empirical Methods have a kinetic-energy-type term.) This relation will play a key role in Sect. 10.6. First, however, let us consider a specific example from the lung mechanics literature.

Question 10.7: Recall that we began this section by considering experiments...

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