Just because our focus is biomechanics, this should not mean that we cannot look at everyday problems involving continuum mechanics.Indeed, a student should continually try to understand and explain the wide variety of mechanical phenomena that we experience on a daily basis, for in doing so, one

Question

Just because our focus is biomechanics, this should not mean that we cannot look at everyday problems involving continuum mechanics. Indeed, a student should continually try to understand and explain the wide variety of mechanical phenomena that we experience on a daily basis, for in doing so, one is forced to practice the art of formulating and solving problems. For example, let a hydraulic turbine be supplied with water from a mountain lake through a supply pipe. The pipe diameter is 1 ft and the average roughness [latex]e[/latex] is 0.05 in. Minor losses can be neglected. Flow leaves the pipe at atmospheric pressure at an average velocity [latex]\overline{ u }=27.5 ft/s.[/latex] Find the height h if the length L of the pipe is 3,000 ft. Let [latex]\mu / ho =10.76 imes 10^{-6}ft^{2}/s.[/latex] Note, too, that the combination [latex]\mu / ho[/latex] is called the kinematic viscosity in contradistinction to the absolute viscosity [latex]\mu.[/latex]

Accepted Answer

For pipe flow,

[latex] \left(\frac{p_{1}}{ ho _{1}}+gz_{1}+\frac{1}{2}\alpha _{1}\overline{ u }^{2}_{1} ight)-\left(\frac{p_{2}}{ ho _{2}}+gz_{2}+\frac{1}{2}\alpha _{2}\overline{ u }^{2}_{2} ight)=h_{M}+h_{m}.[/latex]

Control Volume and Semi-empirical Methods In the case where we neglect the minor losses, the pipe-flow equation becomes

[latex] \left(\frac{p_{1}}{ ho _{1}}+gz_{1}+\frac{1}{2}\alpha _{1}\overline{ u }^{2}_{1} ight)-\left(\frac{p_{2}}{ ho _{2}}+gz_{2}+\frac{1}{2}\alpha _{2}\overline{ u }^{2}_{2} ight)=\sum{f\left(Re,\frac{e}{D} ight)\left(\frac{L}{D} ight) }\frac{\overline{ u }^{2} }{2}.[/latex]

Applying the given conditions, we have (with [latex]\overline{ u }_{1}\ll 1[/latex] in lake)

[latex] gh-\frac{1}{2}\alpha \overline{ u }^{2}_{2}=f\left(Re,\frac{e}{D} ight)\left(\frac{L}{D} ight)\frac{\overline{ u }^{2}_{2} }{2}.[/latex]

Before we can calculate the height h, we need to determine if the flow is laminar or turbulent. To do this, we need to calculate the Reynolds’ number:

[latex] Re=\frac{ ho \overline{ u }D }{\mu }=\frac{\overline{ u }D }{\mu / ho }=\frac{(27.5ft/s)(1 ft)}{10.76 imes 10^{-6}ft^{2}/s}=2.556 imes 10^{6}.[/latex]

Because [latex]2.556 imes 10^{6}\gt 2,100,[/latex] the flow is turbulent and we must use the Moody diagram to find the value of the friction factor f. Also, [latex]\alpha \sim 1.08[/latex] and [latex]e/D=0.05 in./12 in.=0.004.[/latex] From the Moody diagram (Fig. 10.15), the friction factor [latex]f\sim 0.028[/latex] (which is simply best approximated by eye). Substituting these values into the pipe-flow equation,

[latex] gh-\frac{1}{2}\alpha \overline{ u }^{2}_{2}=f\left(Re,\frac{e}{D} ight)\left(\frac{L}{D} ight)\left(\frac{\overline{ u }^{2} }{2} ight) ightarrow h=\frac{1}{g}\left[f\left(Re,\frac{e}{D} ight)\left(\frac{L}{D} ight)\left(\frac{\overline{ u }^{2}_{2} }{2} ight)+\frac{\alpha \overline{ u }^{2}_{2} }{2} ight],[/latex]

we obtain

[latex] h=\frac{1}{32.2ft/s^{2}}\left[0.028\left(\frac{3000 ft}{1 ft} ight)\left(\frac{(27.5)^{2}ft^{2}/s^{2}}{2} ight)+\frac{1}{2}(1.08)(27.5)^{2}\frac{ft^{2}}{s^{2}} ight][/latex]

[latex]=999 ft.[/latex]

Hence, the height h is approximately 1,000 ft.

Question 10.11: Just because our focus is biomechanics, this should not mean...

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