Question 6.6: Compute the loss of head and pressure drop in 200 ft of hori...

• System sketch: See Fig. 6.7 for a horizontal pipe, with \Delta z = 0 and h_f proportional to \Delta p.
• Assumptions: Turbulent flow, asphalted horizontal cast iron pipe, d = 0.5 ft, L = 200 ft.
• Approach: Find Re_d and \epsilon /d; enter the Moody chart, Fig. 6.13; find f, then h_f and \Delta p.
• Property values: From Table A.3 for water, converting to BG units, \rho = 998/515.38 = 1.94 slug/ft^3, \mu = 0.001/47.88 = 2.09E-5 slug/(ft-s).

^*In contact with air.
^{\dagger}The viscosity–temperature variation of these liquids may be fitted to the empirical expression

with accuracy of ±6 percent in the range 0 ≤ T ≤ 100°C.
^{\ddagger}Representative values. The SAE oil classifications allow a viscosity variation of up to ±50 percent, especially at lower temperatures.

\epsilon /d = (0.0004 ft)/(0.5 ft) = 0.0008

• Solution step 2: Find the friction factor from the Moody chart or from Eq. (6.48). If you use the Moody chart, Fig. 6.13, you need practice. Find the line on the right side for \epsilon /d = 0.0008 and follow it back to the left until it hits the vertical line for Re_d \approx 2.79E5. Read, approximately, f \approx 0.02 [or compute f = 0.0198 from Eq. (6.48), perhaps using EES].
\frac{1}{f^{1/2}}=-2.0\log \left(\frac{\epsilon /d}{3.7}+\frac{2.51}{Re_d f^{1/2}}\right)                                     (6.48)
• Solution step 3: Calculate h_f from Eq. (6.10) and \Delta p from Eq. (6.8) for a horizontal pipe:
h_f=(z_1-z_2)+\left(\frac{p_1}{\rho g}-\frac{p_2}{\rho g}\right)=\Delta z+\frac{\Delta p}{\rho g} (6.8)
h_f = f\frac{L}{d}\frac{V^2}{2g} = (0.02)\left(\frac{200  ft}{0.5  ft}\right)\frac{(6  ft/s)^2}{2(32.2  ft/s^2)} \approx 4.5   ft
\Delta p =\rho g h_f = (1.94  slug/ft^3)(32.2  ft/s^2)(4.5   ft) \approx 280  lbf/ft^2
• Comments: In giving this example, Moody [8] stated that this estimate, even for clean new pipe, can be considered accurate only to about \pm10 percent.