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 e is 0.05 in. Minor losses can be neglected. Flow leaves the pipe at atmospheric pressure at an average velocity \overline{\nu }=27.5 ft/s. Find the height h if the length L of the pipe is 3,000 ft. Let \mu /\rho =10.76\times 10^{-6}ft^{2}/s. Note, too, that the combination \mu /\rho is called the kinematic viscosity in contradistinction to the absolute viscosity \mu.
Question 10.11: Just because our focus is biomechanics, this should not mean...
For pipe flow,
\left(\frac{p_{1}}{\rho _{1}}+gz_{1}+\frac{1}{2}\alpha _{1}\overline{\nu }^{2}_{1} \right)-\left(\frac{p_{2}}{\rho _{2}}+gz_{2}+\frac{1}{2}\alpha _{2}\overline{\nu }^{2}_{2} \right)=h_{M}+h_{m}.Control Volume and Semi-empirical Methods In the case where we neglect the minor losses, the pipe-flow equation becomes
\left(\frac{p_{1}}{\rho _{1}}+gz_{1}+\frac{1}{2}\alpha _{1}\overline{\nu }^{2}_{1} \right)-\left(\frac{p_{2}}{\rho _{2}}+gz_{2}+\frac{1}{2}\alpha _{2}\overline{\nu }^{2}_{2} \right)=\sum{f\left(Re,\frac{e}{D} \right)\left(\frac{L}{D} \right) }\frac{\overline{\nu }^{2} }{2}.Applying the given conditions, we have (with \overline{\nu }_{1}\ll 1 in lake)
gh-\frac{1}{2}\alpha \overline{\nu }^{2}_{2}=f\left(Re,\frac{e}{D} \right)\left(\frac{L}{D} \right)\frac{\overline{\nu }^{2}_{2} }{2}.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:
Re=\frac{\rho \overline{\nu }D }{\mu }=\frac{\overline{\nu }D }{\mu /\rho }=\frac{(27.5ft/s)(1 ft)}{10.76\times 10^{-6}ft^{2}/s}=2.556\times 10^{6}.Because 2.556\times 10^{6}\gt 2,100, the flow is turbulent and we must use the Moody diagram to find the value of the friction factor f. Also, \alpha \sim 1.08 and e/D=0.05 in./12 in.=0.004. From the Moody diagram (Fig. 10.15), the friction factor f\sim 0.028 (which is simply best approximated by eye). Substituting these values into the pipe-flow equation,
gh-\frac{1}{2}\alpha \overline{\nu }^{2}_{2}=f\left(Re,\frac{e}{D} \right)\left(\frac{L}{D} \right)\left(\frac{\overline{\nu }^{2} }{2} \right)\rightarrow h=\frac{1}{g}\left[f\left(Re,\frac{e}{D} \right)\left(\frac{L}{D} \right)\left(\frac{\overline{\nu }^{2}_{2} }{2} \right)+\frac{\alpha \overline{\nu }^{2}_{2} }{2} \right],we obtain
h=\frac{1}{32.2ft/s^{2}}\left[0.028\left(\frac{3000 ft}{1 ft} \right)\left(\frac{(27.5)^{2}ft^{2}/s^{2}}{2} \right)+\frac{1}{2}(1.08)(27.5)^{2}\frac{ft^{2}}{s^{2}} \right]=999 ft.
Hence, the height h is approximately 1,000 ft.
