A1 — Three-reservoir branching pipe network: 23-Chem-B1 Transport Phenomena, December 2017, Q1

Question 1: A1 — Three-reservoir branching pipe network (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Pipe / quantity	Symbol	Value
Flow, A→junction	$Q_A$	$8.64\ \text{L/s}=8.64\times10^{-3}\ \text{m}^3/\text{s}$
Flow, junction→B / →C	$Q_B=Q_C$	$4.32\ \text{L/s}=4.32\times10^{-3}\ \text{m}^3/\text{s}$
Pipe A: length, diameter	$L_A,\ D_A$	$61\ \text{m},\ 0.078\ \text{m}$
Pipe B / C length	$L_B,\ L_C$	$30.5\ \text{m},\ 45.75\ \text{m}$
Reservoir levels (vs. datum)	$z_A,z_B,z_C$	$+15.25,\ +3.05,\ -15.25\ \text{m}$
Darcy friction factor	$f$	$0.008$

Find. The pipe diameters $D_B$ and $D_C$ that carry the stated flows to reservoirs B and C.

Fig. A1: Reservoir A (highest) drives flow to the junction J; continuity ($Q_A=Q_B+Q_C$) splits it to B and C. Pipe A is fully specified, so it fixes the piezometric head at J, which then sizes pipes B and C.

Approach. Confirm continuity at the junction, use the fully-specified pipe A to compute its friction loss and hence the piezometric head at the junction, then apply the head balance on each of pipes B and C — each written as Darcy–Weisbach in terms of flow $Q$ — and solve for the unknown diameters.

Continuity fixes the flow directions. The paper says water flows “out of” all three reservoirs, but that cannot hold literally: with no other inflow, 8.64 + 4.32 + 4.32 = 17.28 L/s would enter the junction and nothing would leave it. The only reading that satisfies continuity is that the 8.64 L/s leaving A splits at J into the 4.32 L/s in each of the pipes from B and C, i.e. flow runs A→J, J→B and J→C: $Q_A=Q_B+Q_C=4.32+4.32=8.64\ \text{L/s}$. Step 4 confirms it, because the junction head exceeds both B’s and C’s levels (this assumption is stated per Note 1 of the paper).
Velocity and friction loss in pipe A. The area is $A_A=\tfrac{\pi}{4}D_A^2=\tfrac{\pi}{4}(0.078)^2=4.778\times10^{-3}\ \text{m}^2$, so $V_A=Q_A/A_A=8.64\times10^{-3}/4.778\times10^{-3}=1.808\ \text{m/s}$. The Darcy–Weisbach loss is $$h_{f,A}=f\frac{L_A}{D_A}\frac{V_A^2}{2g}=0.008\cdot\frac{61}{0.078}\cdot\frac{1.808^2}{2(9.81)}\ \Longrightarrow\ \boxed{h_{f,A}\approx1.04\ \text{m}.}$$
Piezometric head at the junction. Following the energy line from A’s surface (velocity head negligible in the reservoir) down to J, $$H_J=z_A-h_{f,A}=15.25-1.04\ \Longrightarrow\ \boxed{H_J\approx14.21\ \text{m}.}$$ This single head now drives both downstream branches.
Head available to each branch. The loss in each downstream pipe equals the head drop from J to that reservoir surface: $$h_{f,B}=H_J-z_B=14.21-3.05=11.16\ \text{m},\qquad h_{f,C}=H_J-z_C=14.21-(-15.25)=29.46\ \text{m}.$$ Both are positive, confirming flow leaves J toward B and C.
Invert Darcy–Weisbach for the diameter. Writing the loss in terms of flow, $h_f=f\dfrac{L}{D}\dfrac{V^2}{2g}$ with $V=\dfrac{4Q}{\pi D^2}$ gives $h_f=\dfrac{8fLQ^2}{\pi^2 g D^5}$, hence $$D=\left(\frac{8fLQ^2}{\pi^2 g\,h_f}\right)^{1/5}.$$
Pipe B diameter. $$D_B=\left(\frac{8(0.008)(30.5)(4.32\times10^{-3})^2}{\pi^2(9.81)(11.16)}\right)^{1/5}\ \Longrightarrow\ \boxed{D_B\approx0.0320\ \text{m}=3.20\ \text{cm}.}$$
Pipe C diameter. The larger driving head on C (its reservoir sits 15.25 m below datum) allows a slightly smaller bore for the same flow: $$D_C=\left(\frac{8(0.008)(45.75)(4.32\times10^{-3})^2}{\pi^2(9.81)(29.46)}\right)^{1/5}\ \Longrightarrow\ \boxed{D_C\approx0.0286\ \text{m}=2.86\ \text{cm}.}$$

Quantity	Result
Friction loss in pipe A	$h_{f,A}\approx1.04\ \text{m}$
Junction piezometric head	$H_J\approx14.21\ \text{m}$
Diameter of pipe B	$D_B\approx3.20\ \text{cm}$ ($h_{f,B}=11.16\ \text{m}$)
Diameter of pipe C	$D_C\approx2.86\ \text{cm}$ ($h_{f,C}=29.46\ \text{m}$)

Check — constant friction factor and reservoir velocity heads

The problem fixes a single Darcy $f=0.008$ for every pipe; in reality $f$ depends on Reynolds number and roughness, so a rigorous design would iterate $f$ with a Colebrook/Moody update on the computed $D_B,D_C$. The reservoir-surface velocity heads and any entrance/exit minor losses are neglected (standard for this branching-pipe idealisation); including them would slightly increase the required diameters.

Concept & Theory

Topic. Steady incompressible flow in a branching (three-reservoir) pipe network; Darcy–Weisbach friction and the junction-head method.

Key relations. Continuity at a node $\sum Q_{in}=\sum Q_{out}$; energy line $H=z+p/\rho g+V^2/2g$; Darcy–Weisbach $h_f=f\frac{L}{D}\frac{V^2}{2g}=\frac{8fLQ^2}{\pi^2 g D^5}$.

Why this works. All three pipes meet at one point, so they share a single junction head $H_J$. Because pipe A is fully specified, it fixes $H_J$ directly; the head drop from $H_J$ to each downstream reservoir surface then equals that pipe’s friction loss, and Darcy–Weisbach solved for $D$ gives the required bore. No iteration on $H_J$ is needed here since the flow split is given.

Common pitfall. Mis-signing reservoir C’s level (it is below datum, so its driving head is $H_J-(-15.25)$, the largest of the three), or confusing Darcy $f$ with the Fanning factor $f_F=f/4$ — using the wrong one changes $h_f$ (and hence $D^5$) by a factor of four.

Source. White, Fluid Mechanics, Ch. 6 (three-reservoir problem); Welty et al., Momentum, Heat and Mass Transfer, Ch. 13 (pipe-flow friction).

Question 1 of 6: A1 — Three-reservoir branching pipe network

Question 1: A1 — Three-reservoir branching pipe network (25 marks)