Question 1 of 4: Three Cast-Iron Pipes in Parallel
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-Agric-A4, Fluid Flow — National Exams, December 2017. Open-book, 3-hour exam; four questions of equal value, all requiring calculation.
Reference texts. White, Fluid Mechanics (pipe friction/Colebrook, cavitation, open-channel hydraulic jumps and bumps, pumps & the energy equation) — the standard undergraduate text for this subject.
Problem 1: Three Cast-Iron Pipes in Parallel (25 points)
Given. Three cast-iron pipes connect the same two junction points (so each sees the identical head loss $h_f$), carrying a combined flow of $200\ \text{m}^3/\text{h}$ of 20°C water.
Find. (a) The flow split $Q_1,Q_2,Q_3$ between the three pipes; (b) the common pressure drop $\Delta p$ across the parallel system.
Figure 1 — Three pipes connect the same two junctions A and B, so every pipe carries the same head loss $h_f$; the flow $Q=200\ \text{m}^3/\text{h}$ splits among them accordingly.
Approach. All three pipes share the same head loss $h_f=f_i(L_i/D_i)(V_i^2/2g)$ because they connect the same two nodes. Guess a common trial $h_f$, get each $V_i$ from Darcy–Weisbach with Colebrook friction factors (iterating $f_i$ on the resulting Reynolds number), sum the three flows, and adjust $h_f$ until $Q_1+Q_2+Q_3$ matches the given 200 m³/h.
Set up the shared-head-loss condition. For pipe $i$,
$$h_f = f_i\frac{L_i}{D_i}\frac{V_i^2}{2g} \;\Rightarrow\; V_i = \sqrt{\frac{2g h_f D_i}{f_i L_i}},$$
with $f_i$ from Colebrook using $\text{Re}_i=\rho V_i D_i/\mu$ and $\epsilon/D_i$. Starting from the hinted guess $f_1=f_2=f_3\approx0.025$ gives a first estimate of the flow split, which is then refined by iterating $f_i$ to convergence at the trial $h_f$.
Search $h_f$ for $\sum Q_i=Q_{\text{tot}}$. Bisecting on $h_f$ (recomputing every $f_i$ at each trial) converges to
$$\boxed{h_f \approx 51.38\ \text{m}}.$$
(a) Resulting flow split. At this $h_f$:
$$\begin{aligned}
\text{Pipe 1: } V_1&=2.484\ \text{m/s},\ f_1=0.0245,\ Q_1=101.1\ \text{m}^3/\text{h}\\
\text{Pipe 2: } V_2&=2.211\ \text{m/s},\ f_2=0.0275,\ Q_2=40.0\ \text{m}^3/\text{h}\\
\text{Pipe 3: } V_3&=2.081\ \text{m/s},\ f_3=0.0259,\ Q_3=58.8\ \text{m}^3/\text{h}
\end{aligned}$$
$$\boxed{Q_1\approx101\ \text{m}^3/\text{h},\quad Q_2\approx40.0\ \text{m}^3/\text{h},\quad Q_3\approx58.8\ \text{m}^3/\text{h}}$$
(sum $=200.0\ \text{m}^3/\text{h}$, as required). The widest, shortest pipe (1) takes about half the total flow, while the narrowest pipe (2) — despite the shortest run being only moderately shorter — carries the least, since diameter dominates the resistance more strongly than length.
(b) Pressure drop across the system.
$$\Delta p = \rho g h_f = (998)(9.81)(51.38) = \boxed{\Delta p \approx 503\ \text{kPa}}.$$
