Chapter 4

Q. 4.4

Two-pipe Flow with Pump as well as Frictional and Form Losses

Given two water reservoirs (\rm H_A = 5m and \rm H_B =15m) connected by two parallel wrought-iron pipes (\rm D_1 = 0.05m, ε_1 = 0.04mm; \rm L_1 ≈ 15m, and \rm D_2 = 0.10m, ε_2 = 0.06mm; \rm L_2 ≈ 15m with pump (10 kW; η = 0.8) to convey the water (20ºC; ρ = 988 kg/m³ and μ = 10^{−3} kg/(m·s)). From reservoir A to reservoir B. Take \rm K_{bend} ≈ 0.3 and find \rm Q_1 and \rm Q_2 !

\rm h_{P,T}=\frac{P_{P,T}}{\dot mg}                        (4.7b,c)

\rm f^{-1/2}\approx-1.8\ {log}\bigg[{\frac{6.9}{{Re_{D}}}}+\left\lgroup{\frac{\varepsilon /{D}}{3.7}}\right\rgroup ^{1.11}\bigg]                         (4.5e)

Assumptions Sketch
• Steady uniform flow
\rm H_A and \rm H_B are constant
• Minor losses only in parallel pipes
• Turbulent flow, i.e., check that \rm Re=\frac{vD}{\nu}>4000 for all pipes
• Extended Bernoulli with \rm\mathrm{h}_{\mathrm{pump}}=\frac{\dot{W}_{\mathrm{electr}}\cdot{\eta }}{\rho{\mathrm{~g~}}Q}~(\mathrm{see}\,\mathrm{Eq}.\,(4.7\mathrm{b}))
\rm h_L = h_{L,1} + h_{L,2} because Δp is the same for each pipe
• Total losses \rm h_L = h_{friction} + h_{form}
• Friction factors from Eq. (4.5e) or Moody chart
• Flow rate \rm Q = Q_1 + Q_2
• Solve system of n equations with n unknowns simultaneously. Alternatively, guess \rm f_i and use Moody chart or correlation


Verified Solution

With \rm v_A = \rm v_B = 0, \rm p_A = \rm p_B and Δz = \rm H_B − H_A the extended Bernoulli equation yields:

\rm\mathrm{h}_{\mathrm{pump}}=\mathrm{H}_{\mathrm{B}}-\mathrm{H}_{\mathrm{A}}+\mathrm{h}_{\mathrm{L}}={\frac{{\dot{\rm W}}_{\mathrm{el}}\cdot{ \eta }}{\mathrm{\rho g Q}}}                          (E.4.4.1a, b)


\rm \mathrm{h}_{\mathrm{L,i}}=\left[\left\lgroup f\frac{\mathrm{L}}{\mathrm{D}}\right\rgroup _{\mathrm{i}}+\Sigma\mathrm{K}_{\mathrm{i}}\right] \frac{\mathrm{v}_{\mathrm{i}}^{2}}{2{\mathrm{g}}{}};\quad i=1,2                            (E.4.4.2a, b)

Employing Eq. (4.5e):

\mathrm{f}_{\mathrm{i}}\approx\left\{-1.8\log\left[{\frac{6.9}{\mathrm{R}\mathrm{e}_{\mathrm{i}}}}+\left\lgroup{\frac{\varepsilon /{\mathrm{D}_{\mathrm{i}}}}{3.7}}\right\rgroup ^{1.11}\right]\right\}^{-2}                          (E.4.4.3)


\rm\mathrm{Re}_{\mathrm{i}}=\frac{\mathrm{v_{i}}\,\mathrm{D}_{\mathrm{i}}}{\mathrm{\nu}}                        (E.4.4.4)

\mathrm{v_{i}}=\mathrm{Q_{i}}/\left\lgroup{\frac{\pi\,\mathrm{D_{i}}^{2}}{4}}\right\rgroup                    (E.4.4.5a)


\rm Q=Q_1+Q_2                      (E.4.4.5b)

Guess \rm f_1 = 0.02 and \rm f_2 = 0.018 , which leads with

\rm\frac{{\varepsilon }_{1}}{{ D}_{1}}=8\times10^{-4}\qquad\mathrm{and}\qquad\frac{{\varepsilon }_{2}}{{ D}_{2}}=6\times10^{-4}            using (E.4.4.3) to:


Now \rm v_1 and \rm v_2 can be computed as well as Q and \rm h_L (see Eqs. (E.4.4.5a, b and (E.4.4.2a, b)). Clearly, Eq. (E.4.4.1b) has to be fulfilled, which requires \rm h_L and Q have to match! If not, a new f-value has to be assumed and the calculations have to be repeated. The final result is \rm Q_1 = 0.0124 kg/s and \rm Q_2 =0.066 kg/s.