Question 13.4: Water at 80 F flows through a 0.21-ft-diameter 50-ft-long ca......

The first thing to do is calculate the Reynolds number. The density and viscosity of water at the given temperature can be found in Appendix I. The Reynolds number is

\mathrm{Re}={\frac{\rho v D}{\mu}}={\frac{(62.2\ lb_{\mathrm{m}}/\mathrm{ft}^{3})(53\ \mathrm{ft}/\mathrm{min})(\mathrm{min}/60\ \mathrm{sec})(0.21\ \mathrm{ft})}{0.578\times10^{-3}{\frac{\mathrm{lb}_{\mathrm{m}}}{\mathrm{ft}s}}}}=2\times10^{4}

Figure 13.2 contains roughness parameters for various materials, including cast iron, for which the figure is e = 0.00085:

We next calculate the relative roughness:

{\frac{e}{D}}={\frac{0.00085}{0.21\,{\mathrm{ft}}}}=0.004

We now turn to Figure 13.1. Along the right-side y-axis are values for the relative roughness.We find 0.004 on that axis. Next, follow the line for 0.004 until we reach the value for the Reynolds number from the x-axis. The calculated Reynolds number from above is 2\times10^{4}; where these lines intersect, we read the value for the fanning friction factor on the left-side y-axis to be 0.00825.

Since the flow is turbulent, we use equations (13-3) and (13-16) in combination to calculate the pressure drop:

h_{L}=2f_{f}{\frac{L}{D}}{\frac{v^{2}}{g}}        (13-3)

h_{L}={\frac{\Delta P}{\rho g}}=K{\frac{v^{2}}{2g}}      (13-16)

h_{L}={\frac{\Delta P}{\rho g}}=2f_{f}{\frac{L_{\mathrm{eq}}}{D}}{\frac{v^{2}}{g}}

Rearrange and solve for the pressure change:

\Delta P=2\rho f_{f}\frac{L_{\mathrm{eq}}}{D}v^{2}=2\frac{(62.2\ \mathrm{lb}_{\mathrm{m}}/\mathrm{ft}^{3})}{32.174\ \mathrm{lb}_{\mathrm{m}}\mathrm{ft}/\mathrm{lb}_{\mathrm{f}}s^{2}}(0.00825)\frac{50\,\mathrm{ft}}{0.21\,\mathrm{ft}}(0.88\,\mathrm{ft}/s)^{2}=5.88\,\mathrm{lb}_{\mathrm{f}}/\mathrm{ft}^{2}