Question 6.14: What should the reservoir level h be to maintain a flow of 0...

• Assumptions: Fully developed annulus flow, minor losses neglected.
• Approach: Determine the Reynolds number, then find f and h_f and thence h.
• Property values: Given ρ = 1000 kg/m^3 and ν = 1.02E-6 m^2/s.

• Solution step 1: Calculate the velocity, hydraulic diameter, and Reynolds number:

Re_{D_h} = \frac{VD_h}{\nu} = \frac{(1.99  m/s)(0.04  m)}{1.02E-6  m^2/s} = 78,000                        (turbulent flow)

• Solution step 2: Apply the steady flow energy equation between sections 1 and 2:

or              h = \frac{\alpha_2 V_2^2}{2g} + h_f = \frac{V_2^2}{2g} \left(\alpha_2 + f\frac{L}{D_h}\right)                       (1)

Note that z_1 = h. For turbulent flow, from Eq. (3.43c), we estimate \alpha_2 ≈ 1.03

\beta = \frac{(1 + m)^2(2 + m)^2}{2(1 + 2m)(2 + 2m)}                 (3.43c)

• Solution step 3: Determine the roughness ratio and the friction factor. From Table 6.1, for (new) commercial steel pipe, ε = 0.046 mm. Then

For a reasonable estimate, use Re_{D_h} to estimate the friction factor from Eq. (6.48):

\frac{1}{\sqrt{f}} = -2.0 \log_{10} \left(\frac{0.00115}{3.7} + \frac{2.51}{78,000 \sqrt{f}}\right)                            solve for f ≈ 0.0232

For slightly better accuracy, we could use D_{eff} = D_h / \zeta. From Table 6.3, for b/a = 3/5, 1/ζ = 0.67. Then D_{eff} = 0.67(40 mm) = 26.8 mm, whence Re_{D_{eff}} = 52,300, \epsilon /D_{eff} = 0.00172, and f_{eff} ≈ 0.0257. Using the latter estimate, we find the required reservoir level from Eq. (1):

• Comments: Note that we do not replace D_h  by  D_{eff} in the head loss term fL/D_h, which comes from a momentum balance and requires hydraulic diameter. If we used the simpler friction estimate, f ≈ 0.0232, we would obtain h ≈ 3.72 m, or about 9 percent lower.