Given is a three-pipe series system, as in Fig. 6.24a. The total pressure drop is p_A - p_B = 150,000 Pa, and the elevation drop is z_A - z_B = 5 m. The pipe data are
| Pipe | L, m | d, cm | ε, mm | ε/d |
| 1 | 100 | 8 | 0.24 | 0.003 |
| 2 | 150 | 6 | 0.12 | 0.002 |
| 3 | 80 | 4 | 0.20 | 0.005 |
The fluid is water, ρ = 1000 kg/m^3 and ν = 1.02 × 10^{-6} m^2/s. Calculate the flow rate Q in m^3/h through the system.
The total head loss across the system is
\Delta h_{A→B} = \frac{p_A – p_B}{\rho g} + z_A – z_B = \frac{150,000}{1000(9.81)} + 5 m = 20.3 mFrom the continuity relation (6.84) the velocities are
V_1d^2_1 = V_2d^2_2 = V_3d^2_3 (6.48)
V_2 = \frac{d^2_1}{d^2_2} V_1 = \frac{16}{9} V_1 V_3 = \frac{d^2_1}{d^2_3}V_1 = 4V_1and Re_2 = \frac{V_2d_2}{V_1d_1} Re_1 = \frac{4}{3}Re_1 Re_3 = 2Re_1
Neglecting minor losses and substituting into Eq. (6.86), we obtain
\Delta h_{A→B} = \frac{V^2_1}{2g} \left(\frac{f_1L_1}{d_1} + \sum{K_1}\right) + \frac{V^2_2}{2g} \left(\frac{f_2L_2}{d_2} + \sum{K_2}\right) + \frac{V^2_3}{2g} \left(\frac{f_3L_3}{d_3} + \sum{K_3}\right) (6.86)
\Delta h_{A→B} = \frac{V^2_1}{2g} \left[1250f_1 + 2500 \left(\frac{16}{9}\right)^2 f_2 + 2000(4)^2f_3\right]or 20.3 m = \frac{V^2_1}{2g}(1250f_1 + 7900f_2 + 32,000f_3) (1)
This is the form that was hinted at in Eq. (6.87). It seems to be dominated by the third pipe loss 32,000f_3. Begin by estimating f_1, f_2, and f_3 from the Moody-chart fully rough regime:
\Delta h_{A→B} = \frac{V^2_1}{2g}(\alpha_0 + \alpha_1 f_1 + \alpha_2 f_2 + \alpha_3 f_3) (6.87)
f_1 = 0.0262 f_2 = 0.0234 f_3 = 0.0304Substitute in Eq. (1) to find V_1^2 \approx 2g(20.3)/(33 + 185 + 973). The first estimate thus is V_1 = 0.58 m/s, from which
Re_1 \approx 45,400 Re_2 = 60,500 Re_3 = 90,800Hence, from the Moody chart,
f_1 = 0.0288 f_2 = 0.0260 f_3 = 0.0314Substitution into Eq. (1) gives the better estimate
V_1 = 0.565 m/s Q = \frac{1}{4}\pi d^2_1 V_1 = 2.84 \times 10^{-3} m^3/sor Q = 10.2 m^3/h
A second iteration gives Q = 10.22 m^3/h, a negligible change.