Question 6.18: Assume that the same three pipes in Example 6.17 are now in ...

From Eq. (6.88a) we can solve for each V separately:

\Delta h_{A→B} = \Delta h_1 = \Delta h_2 = \Delta h_3                   (6.88a)

20.3  m = \frac{V^2_1}{2g} 1250f_1 = \frac{V^2_2}{2g} 2500f_2 = \frac{V^2_3}{2g} 2000f_3                         (1)

Guess fully rough flow in pipe 1: f_1 = 0.0262, V_1 = 3.49 m/s; hence Re_1 = V_1d_1/ \nu = 273,000. From the Moody chart read f_1 = 0.0267; recompute V_1 = 3.46 m/s, Q_1 = 62.5  m^3/h. [This problem can also be solved from Eq. (6.51).]

Re_d = -(8 \zeta)^{1/2} \log \left(\frac{\epsilon /d}{3.7} + \frac{1.775}{\sqrt{\zeta}}\right)                  \zeta = \frac{gd^3h_f}{L \nu^2}                    (6.51)

Next guess for pipe 2: f_2 \approx 0.0234,  V_2 \approx 2.61  m/s;  then  Re_2 = 153,000,  and  hence  f_2 = 0.0246,  V_2 = 2.55  m/s,  Q_2 = 25.9  m^3/h.

Finally guess for pipe 3: f_3 \approx 0.0304,  V_3 \approx 2.56  m/s;  then  Re_3 = 100,000,  and  hence  f_3 = 0.0313,  V_3 = 2.52  m/s,  Q_3 = 11.4  m^3/h.

This is satisfactory convergence. The total flow rate is

These three pipes carry 10 times more flow in parallel than they do in series.

EES

This example is ideal for EES. One enters the pipe data (L_i,  d_i,  \epsilon_i); the fluid properties (ρ, μ); the definitions Q_i = (\pi /4)d_i V_i,  Re_i = \rho V_id_i/ \mu,  and  h_f = f_i (L_i/d_i) (V^2_i/2g); plus the Colebrook formula (6.48) for each friction factor f_i. There is no need to use resistance ideas such as Eq. (6.89). Specify that f_i \gt 0 and Re_i \gt 4000. Then, if one enters Q = ΣQ_i = (99.8/3600)  m^3/s, EES quickly solves for h_f = 20.3 m. Conversely, if one enters h_f = 20.3 m, EES solves for Q = 99.8 m^3/h.

\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)

h_f = \frac{Q^2}{(\sum{\sqrt{C_i/f_i}})^2}      where       C_i = \frac{\pi^2 gd_i^5}{8L_i}                      (6.89)