Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 04-Agric-A4, Fluid Flow. Three-hour, open-book exam; a Casio or Sharp approved calculator is permitted. Format: four questions constitute a complete paper, with a choice between 1a/1b and between 4a/4b; all questions require calculation. Both alternatives are solved below for completeness.
Reference texts: White, Fluid Mechanics (7th/8th ed., McGraw-Hill) — open-channel hydraulics (hydraulic jump, gradually varied flow over a bump), pipe friction (Colebrook–White), rotating control volumes (sprinkler reaction), and turbomachinery (pump performance curves, affinity laws).
Question 3: Series-Parallel Pipe System (equal value)
Given. Two parallel branches (A and B) between nodes 1–2 rejoin into a single series pipe C; all three pipes share the same 8 cm diameter and asphalted-cast-iron roughness; total drop across the whole system is 750 kPa; minor losses neglected.
Find. The total flow rate $Q$ (m³/h) delivered through the system.
Figure 3 — Series-parallel network: branches A and B share the same head loss between nodes 1 and the junction; their combined flow then passes through series pipe C to node 2.
Approach. Branches A and B share the same pressure drop $\Delta p_{AB}$ (same end nodes); pipe C then carries the combined flow $Q_C=Q_A+Q_B$ under $\Delta p_C = 750\ \text{kPa}-\Delta p_{AB}$. Guess $\Delta p_{AB}$, solve each branch's Darcy–Weisbach/Colebrook equation for its flow, sum for $Q_C$, check $\Delta p_C$, and iterate to convergence.
Governing equations (identical for every pipe, same $D$, $\epsilon/D=0.0015$).
$$\Delta p = f\frac{L}{D}\frac{\rho V^2}{2}, \qquad \frac{1}{\sqrt f} = -2\log_{10}\!\left(\frac{\epsilon/D}{3.7}+\frac{2.51}{Re\sqrt f}\right), \qquad Re=\frac{\rho V D}{\mu}.$$
Iterate on the shared parallel-branch pressure drop $\Delta p_{AB}$. For a trial $\Delta p_{AB}$, solve the Colebrook/Darcy pair for $V_A$ (using $L_A=250\,$m) and $V_B$ (using $L_B=100\,$m), form $Q_A=V_A A$, $Q_B=V_B A$ ($A=\pi D^2/4$), then compute $\Delta p_C$ for $Q_C=Q_A+Q_B$ through $L_C=150\,$m and check whether $\Delta p_{AB}+\Delta p_C=750\,$kPa. Bisecting on $\Delta p_{AB}$ converges to
$$\boxed{\Delta p_{AB} \approx 152.7\ \text{kPa}}, \qquad \Delta p_C \approx 597.3\ \text{kPa} \quad(\text{sum } = 750.0\ \text{kPa}).$$
Branch flows at the converged $\Delta p_{AB}$.
$$V_A = 2.06\ \text{m/s}\ (f_A=0.0230,\ Re_A\approx1.65\times10^5) \;\Rightarrow\; Q_A = V_A A = \boxed{Q_A \approx 37.3\ \text{m}^3/\text{h}},$$
$$V_B = 3.29\ \text{m/s}\ (f_B=0.0226,\ Re_B\approx2.63\times10^5) \;\Rightarrow\; Q_B = V_B A = \boxed{Q_B \approx 59.6\ \text{m}^3/\text{h}}.$$
Branch B (shorter, less resistance) naturally carries more flow than the longer branch A for the same $\Delta p_{AB}$.
Total system flow.
$$Q = Q_A+Q_B = 37.3+59.6 = \boxed{Q \approx 96.9\ \text{m}^3/\text{h}}$$
(check: through pipe C, $V_C=Q/A=5.36\ \text{m/s}$, $f_C=0.0223$, $Re_C\approx4.28\times10^5$, giving $\Delta p_C=597\,$kPa as required).
