Sizing a Flow Nozzle for an Ethanol Line: 23-Chem-A2 Unit Operations and Separation Processes, May 2017, Q1

Question A1: Sizing a Flow Nozzle for an Ethanol Line (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.

Quantity	Value
Pipe diameter $D$	0.060 m
Design pressure drop $\Delta p$	4 kPa = 4000 Pa
Volumetric flow $Q$	3×10⁻³ m³/s
Density $\rho$ / viscosity $\mu$	789 kg/m³ / 1.19×10⁻³ Pa·s
Meter type	ISO flow nozzle, $C_n(Re,\beta)$ chart

Find. The throat (nozzle) diameter $d$ that produces exactly 4 kPa at the design flow.

Figure A1 — ISO flow nozzle. The throat diameter $d$ (ratio $\beta=d/D$) sets both the metering area and the discharge coefficient $C_n(Re,\beta)$ read from the chart.

Approach. The pipe-based Reynolds number is fixed by the given flow, so the discharge coefficient depends only on the unknown diameter ratio $\beta=d/D$; iterate the nozzle-meter equation with $C_n$ read from the chart until $d$ is self-consistent.

Pipe velocity and Reynolds number. The approach velocity is $$v=\frac{Q}{\tfrac{\pi}{4}D^2}=\frac{3\times10^{-3}}{\tfrac{\pi}{4}(0.060)^2}=1.061\ \mathrm{m/s},$$ so $$Re=\frac{\rho v D}{\mu}=\frac{(789)(1.061)(0.060)}{1.19\times10^{-3}}=\boxed{4.22\times10^{4}}.$$ This is set by the pipe and the flow — it does not change as we vary the throat.
Nozzle-meter equation. For a flow nozzle the metered flow is $$Q=C_n\,\frac{\pi}{4}d^{2}\,\sqrt{\frac{2\,\Delta p}{\rho\,(1-\beta^{4})}},\qquad \beta=\frac{d}{D}.$$ The velocity-of-approach factor $1/\sqrt{1-\beta^4}$ corrects for the finite upstream velocity. Rearranging for the throat area, $$d^{2}=\frac{Q\,\sqrt{1-\beta^{4}}}{C_n\,\tfrac{\pi}{4}\sqrt{2\Delta p/\rho}}.$$
Read the discharge coefficient. At $Re=4.2\times10^{4}$ the chart curves are still rising. Taking a first guess $\beta\approx0.57$, the interpolated $\beta\!=\!0.6$ curve reads $C_n\approx0.98$. With $\sqrt{2\Delta p/\rho}=\sqrt{2(4000)/789}=3.184\ \mathrm{m/s}$.
Iterate to a consistent diameter. Starting from $\beta=0.57$, $\sqrt{1-\beta^4}=0.946$: $$d^{2}=\frac{(3\times10^{-3})(0.946)}{(0.98)(0.7854)(3.184)}=1.16\times10^{-3}\ \mathrm{m^2}\Rightarrow d=0.0340\ \mathrm{m}.$$ Then $\beta=0.0340/0.060=0.567$, which reproduces the assumed $C_n$ — the loop has converged in one pass. $$\boxed{d\approx34\ \mathrm{mm}\quad(\beta=0.57)}$$

Quantity	Value
Pipe Reynolds number $Re$	4.2×10⁴
Diameter ratio $\beta=d/D$	0.57
Discharge coefficient $C_n$	≈0.98 (chart)
Nozzle throat diameter $d$	≈34 mm

Concept & Theory

Topic. Differential-pressure flow metering with an ISO flow nozzle.

Key relations. $Q=C_n\tfrac{\pi}{4}d^2\sqrt{2\Delta p/[\rho(1-\beta^4)]}$; $\beta=d/D$; $Re=\rho vD/\mu$; $C_n=f(Re,\beta)$ from the ISO chart.

Why this works. The nozzle converts a measurable pressure drop into a flow rate via Bernoulli plus a discharge coefficient that lumps the real-fluid contraction and friction. Because $Re$ here is fixed by the pipe (not the throat), the only unknown feeding $C_n$ is $\beta$, so a single iteration closes the problem.

Common pitfall. Dropping the velocity-of-approach factor $1/\sqrt{1-\beta^4}$ (significant once $\beta>0.3$), or evaluating $Re$ on the throat diameter — the chart's $Re$ is defined on the pipe $D$.

Source. McCabe, Smith & Harriott ch. 8 (flow measurement); Perry's §5; ISO 5167.

Question 1 of 6: Sizing a Flow Nozzle for an Ethanol Line

Part A — Unit Operations

Question A1: Sizing a Flow Nozzle for an Ethanol Line (25 marks)