12.8 Fluid Transport Systems and Flow Measurement

12.8 Fluid Transport Systems and Flow Measurement

Two NCEES Fluid Mechanics sub-topics are covered here: "Fluid transport systems (e.g., series and parallel operations)" and "Flow measurement (e.g., pitot tube, venturi meter, weir)." Both reward recognizing a rule rather than deriving anything.

Pipes in Series and Parallel

The two configurations invert which quantity is shared, and confusing them is the standard error:

The memory hook: it is the reverse of electrical resistors. In a series pipe run the flow is common and losses accumulate; in a parallel run the head loss is common (both branches connect the same two nodes, so they must experience the same pressure drop) and the flows accumulate.

The parallel condition is powerful: because both branches span the same two junctions, physics forces equal head loss, and you solve for the flow split that makes it so.

$\frac{f_1 L_1}{D_1}\frac{V_1^2}{2g} = \frac{f_2 L_2}{D_2}\frac{V_2^2}{2g}$

Pumps in Series and Parallel

Use series pumping (or a multistage pump) when you need more head; use parallel pumping when you need more flow. Note this is the opposite pattern from pipes, because a pump adds energy while a pipe removes it.

Worked Example: Parallel Pipe Flow Split

Two parallel pipes connect the same two reservoirs. Pipe 1: $L = 400$ m, $D = 200$ mm. Pipe 2: $L = 250$ m, $D = 150$ mm. Assume the same friction factor $f = 0.020$ in both. Total flow is 0.10 m³/s. Find the split.

Head loss must be equal. Writing $h_L$ in terms of $Q$ using $V = Q/A = 4Q/(\pi D^2)$:

$h_L = \frac{fL}{D}\frac{V^2}{2g} = \frac{fL}{D}\frac{16Q^2}{2g\pi^2D^4} = \frac{8fLQ^2}{g\pi^2D^5}$

Setting $h_{L1} = h_{L2}$ with equal $f$:

$\frac{L_1Q_1^2}{D_1^5} = \frac{L_2Q_2^2}{D_2^5} \;\Rightarrow\; \frac{Q_1^2}{Q_2^2} = \frac{L_2}{L_1}\cdot\frac{D_1^5}{D_2^5}$

$\frac{Q_1}{Q_2} = \sqrt{\frac{250}{400}\left(\frac{200}{150}\right)^5} = \sqrt{0.625(4.214)} = \sqrt{2.634} = 1.623$

With $Q_1 + Q_2 = 0.10$:

$Q_2 = \frac{0.10}{1+1.623} = 0.0381\ \text{m}^3/\text{s}, \qquad Q_1 = \boxed{0.0619\ \text{m}^3/\text{s}}$

Note the $D^5$ sensitivity. Pipe 1 is only 33% larger in diameter but carries 62% more flow, despite being 60% longer. Diameter dominates every pipe-network question — a consequence of area scaling with $D^2$ and velocity head with the square of that.

Flow Measurement

Pitot and Pitot-Static Tubes

A pitot tube faces the flow and brings it to rest, reading stagnation pressure. Combined with a static port, the difference gives velocity:

$p_0 = p + \frac{\rho V^2}{2} \qquad\Longrightarrow\qquad V = \sqrt{\frac{2(p_0-p)}{\rho}} = \sqrt{2g\,\Delta h}$

where $\Delta h$ is the differential head in the flowing fluid.

Two cautions. A pitot tube measures local point velocity, not average velocity — a centerline reading must be corrected ($V_{\text{avg}} \approx 0.5V_{\max}$ laminar, $\approx 0.82V_{\max}$ turbulent). And it must be aligned with the flow; a few degrees of yaw introduces measurable error.

Venturi and Orifice Meters

Both create a constriction and infer flow from the pressure drop, using continuity plus Bernoulli:

$Q = C_d A_2\sqrt{\frac{2(p_1-p_2)/\rho}{1-(A_2/A_1)^2}}$

The venturi's gradual expansion recovers most of the pressure, while the orifice's abrupt expansion dissipates it in turbulence. The engineering trade-off is capital cost versus lifetime pumping cost.

The square-root problem, common to all differential-pressure meters: since $Q \propto \sqrt{\Delta p}$, halving the flow quarters the signal. At 25% of full flow the differential pressure is only 6.25% of full scale, so turndown ratio is poor — typically 3:1 or 4:1. This is why a plant needing wide-range measurement uses a magnetic, ultrasonic, or Coriolis meter rather than an orifice plate.

Rotameter (variable-area meter): a float rises in a tapered tube until drag balances its weight. Here the area varies and the pressure drop stays roughly constant, giving a nearly linear scale and much better turndown (10:1) — the complementary trade to the differential-pressure meters.

Weirs

A weir is an obstruction over which open-channel flow spills; the head over the crest determines discharge.

Rectangular weir:

$Q = C_w\,\frac{2}{3}\sqrt{2g}\;L H^{3/2} \approx 1.84\,L H^{3/2} \quad\text{(SI, sharp-crested, } C_w \approx 0.62)$

V-notch (triangular) weir, notch angle $\theta$:

$Q = C_w\,\frac{8}{15}\sqrt{2g}\;\tan\!\left(\frac{\theta}{2}\right)H^{5/2} \approx 1.38\,H^{5/2} \quad\text{(90° notch, SI)}$

Why the V-notch is the low-flow instrument. With a 5/2 exponent, halving the flow reduces the head by only $0.5^{2/5} = 0.76$ — a 24% change in head. A rectangular weir at the same low flow would produce a head so small that surface tension and approach-velocity errors dominate. The steeper exponent spreads small discharges over a measurable head range.

Worked Example: V-Notch Weir Discharge

A 90° V-notch weir operates under a head of 0.25 m. Find the discharge, then the head required to double it.

$Q = 1.38\,H^{5/2} = 1.38(0.25)^{2.5}$

$(0.25)^{2.5} = e^{2.5\ln 0.25} = e^{2.5(-1.3863)} = e^{-3.4657} = 0.03125$

$Q = 1.38(0.03125) = \boxed{0.0431\ \text{m}^3/\text{s}}$

To double the discharge to 0.0862 m³/s:

$H = \left(\frac{0.0862}{1.38}\right)^{2/5} = (0.06246)^{0.4} = e^{0.4\ln 0.06246} = e^{0.4(-2.7734)} = e^{-1.1094} = 0.330\ \text{m}$

Doubling the flow requires only a 32% rise in head ($0.25 \to 0.33$ m) — the flip side of the 5/2 exponent: excellent resolution at low flow, but the weir does not need much more depth to pass much more water. Note that $2^{2/5} = 1.32$ gives the ratio directly, without recomputing.