## 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 Concepts: • 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)

where

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

with

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

and

$\rm Q=Q_1+Q_2$                      (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:

$\mathrm{Re}_{1}=3\times10^{5}\qquad\mathrm{and}\qquad\mathrm{Re}_{2}=8\times10^{5}\,.$

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.