Two reservoirs, two parallel laminar oil pipes: 20-Bio-A7 Bioinstrumentation, May 2015, Q3

Question 3: Two reservoirs, two parallel laminar oil pipes (20 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.

Check — driving head. Each pipe connects tank A's free surface ($z_A=22$ m) directly to tank B's free surface ($z_B=15$ m); both surfaces are open to atmosphere with negligible velocity head in the large tanks, so the energy equation gives the SAME total head loss, $h_f = z_A - z_B = 7$ m, for each pipe independently — the 6 m vertical offset between the two pipes is a layout detail (needed to draw the figure) and does not itself enter the head-loss calculation.

Given.

Quantity	Value
Oil density ρ	891 kg/m³
Oil viscosity μ	0.29 Pa·s
Pipe length L (each)	9 m
Pipe 1 diameter D₁	5 cm
Tank surface elevations	$z_A=22$ m, $z_B=15$ m
Flow ratio	$Q_2 = 2Q_1$

Find. (a) D₂; (b) whether both flows are laminar; (c) Q₂ in m³/s.

Two reservoirs connected by two independent, equal-length pipes at different elevations.

Approach. Each pipe independently sees the same 7 m free-surface-to-free-surface head loss. Given the very viscous oil and small pipe diameters, assume laminar flow, apply the Hagen–Poiseuille relation to pipe 1 (fully specified), use the fact that $Q\propto D^4$ at fixed head/length/viscosity to back out $D_2$ from the given flow ratio, then verify the laminar assumption with the Reynolds number.

Establish the common driving head. $$\boxed{h_f = z_A - z_B = 22 - 15 = 7\ \text{m}}$$ applies to both pipes independently, since each connects the same two atmospheric free surfaces.
Solve pipe 1 by Hagen–Poiseuille (assuming laminar flow). For laminar flow, $Q = \dfrac{\pi \rho g h_f D^4}{128\mu L}$ (equivalent to $h_f = 32\mu L V/(\rho g D^2)$). With $D_1=0.05$ m: $$Q_1 = \frac{\pi(891)(9.81)(7)(0.05)^4}{128(0.29)(9)} = \frac{1.201}{334.1} = 3.596\times10^{-3}\ \text{m}^3/\text{s} = 3.60\ \text{L/s}$$
Find D₂ from the flow ratio (part a). With $h_f$, $L$, $\rho$, $\mu$ identical for both pipes, Hagen–Poiseuille gives $Q\propto D^4$, so $$\frac{Q_2}{Q_1} = \left(\frac{D_2}{D_1}\right)^4 = 2 \quad\Longrightarrow\quad \boxed{D_2 = D_1\,2^{1/4} = (5\ \text{cm})(1.1892) = 5.95\ \text{cm}}$$
Check the laminar assumption via Reynolds number (part b). $V_1 = Q_1/A_1 = (3.596\times10^{-3})/(1.963\times10^{-3}) = 1.831$ m/s, so $$Re_1 = \frac{\rho V_1 D_1}{\mu} = \frac{(891)(1.831)(0.05)}{0.29} = 281$$ For pipe 2, $Q_2=2Q_1=7.192\times10^{-3}$ m$^3$/s, $A_2=\pi(0.0595)^2/4=2.776\times10^{-3}$ m$^2$, $V_2=2.591$ m/s, so $$Re_2 = \frac{(891)(2.591)(0.0595)}{0.29} = 473$$ $$\boxed{Re_1 = 281 < 2300, \quad Re_2 = 473 < 2300 \implies \text{both flows are laminar}}$$ confirming the Step 2 assumption was valid.
State the flow rate in pipe 2 (part c). $$\boxed{Q_2 = 2Q_1 = 7.19\times10^{-3}\ \text{m}^3/\text{s} = 7.19\ \text{L/s}}$$

Quantity	Value
(a) Pipe 2 diameter D₂	5.95 cm
(b) Reynolds numbers	Re₁≈281, Re₂≈473 — both laminar
(c) Flow rate in pipe 2 Q₂	7.19×10⁻³ m³/s

Topic. Laminar pipe flow between two open reservoirs (Hagen–Poiseuille).

Key relations. $Q = \pi\Delta p\,D^4/(128\mu L)$ with $\Delta p=\rho g h_f$; equivalently $h_f=32\mu LV/(\rho g D^2)$; $Re=\rho VD/\mu$, laminar if $Re<2300$.

Why this works. Any pipe that directly joins two open reservoirs sees a head loss equal to the difference in the reservoirs' free-surface elevations, because both ends are at atmospheric pressure with negligible velocity in the large tanks — the pipe's own routing or elevation between the reservoirs is irrelevant to that total. For laminar (Poiseuille) flow, $Q$ is extremely sensitive to diameter (fourth power), which is why doubling the flow rate only requires a modest 19% increase in diameter.

Common pitfall. Assuming turbulent flow and reaching for the Darcy–Weisbach/Colebrook machinery by default — always check $Re$ first; here the oil's viscosity (about 300× water's) keeps both pipes comfortably laminar despite a 7 m driving head. A second trap is treating the stated 6 m pipe-elevation offset as part of the head-loss calculation; it only matters for confirming the pipe stays below both free surfaces (so it runs full), not for the magnitude of $h_f$.

Source. White, Fluid Mechanics, 8th ed., §6.4 (laminar fully developed pipe flow, Hagen–Poiseuille).

Question 3 of 4: Two reservoirs, two parallel laminar oil pipes

Question 3: Two reservoirs, two parallel laminar oil pipes (20 marks)