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22-Agric-A5 Principles of Instrumentation · December 2015

Question 7 of 7: Orifice Meter Flow Rate

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 04-Agric-A5 Principles of Instrumentation, National Exams December 2015 — a three-hour open-book exam; any non-communicating calculator is permitted. Questions 1 and 2 are compulsory (20 marks each); candidates then choose any three (3) of Questions 3-7 (20 marks each) for a 100-mark paper. All seven questions are worked here.

Reference texts. E.O. Doebelin, Measurement Systems: Application and Design, 5th ed. (calibration, standards, static/dynamic sensor characteristics, second-order step response); J.P. Bentley, Principles of Measurement Systems, 4th ed. (accuracy vs. precision, error propagation, signal conditioning); P. Horowitz and W. Hill, The Art of Electronics, 3rd ed. (bridge circuits, instrumentation amplifiers, CMRR, ADC architectures); J. Fraden, Handbook of Modern Sensors: Physics, Designs, and Applications, 5th ed. (thermistors, strain gages, photodetectors); R.W. Fox, A.T. McDonald and P.J. Pritchard, Introduction to Fluid Mechanics, 7th ed. (orifice and venturi metering).

Question 7: Orifice Meter Flow Rate (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.

Orifice plateflowD = 10 cmd = 7 cmP1P2ΔP = P1 − P2 = 65 Pa · ρ = 800 kg/m³ · μ = 15 kg/(m·s)
Fig. 4 — Orifice-plate cross-section: pipe ID $D=10$ cm, orifice bore $d=7$ cm, flange pressure taps either side of the plate.

a) Given. Pipe ID $D=0.10\ \text{m}$, orifice bore $d=0.07\ \text{m}$, fluid density $\rho=800\ \text{kg/m}^3$, viscosity $\mu=15\ \text{kg/(m}\cdot\text{s)}$, pressure drop $\Delta P=65\ \text{Pa}$. Find. Volumetric flow rate $Q$.

Approach. Use the standard square-edged orifice-meter equation $Q=C_d A_o\sqrt{2\Delta P/[\rho(1-\beta^4)]}$ with a typical high-Reynolds discharge coefficient $C_d\approx0.61$, then check the resulting Reynolds number against the regime that discharge coefficient assumes.

  1. Geometry. $$\begin{aligned} \beta&=\dfrac{d}{D}=\dfrac{0.07}{0.10}=0.70 \\ A_o&=\dfrac{\pi}{4}d^2=\dfrac{\pi}{4}(0.07)^2=3.848\times10^{-3}\ \text{m}^2 \end{aligned}$$
  2. Velocity-head term. $$\sqrt{\dfrac{2\Delta P}{\rho(1-\beta^4)}}=\sqrt{\dfrac{2(65)}{800(1-0.70^4)}} =\sqrt{\dfrac{130}{800(0.7599)}}=0.4624\ \text{m/s}.$$
  3. Flow rate. With $C_d\approx0.61$ (typical square-edged, corner-tap orifice at high Reynolds number), $$Q=C_d A_o\sqrt{\dfrac{2\Delta P}{\rho(1-\beta^4)}}=(0.61)(3.848\times10^{-3})(0.4624) =1.086\times10^{-3}\ \text{m}^3/\text{s}.$$ $$\boxed{Q\approx1.09\ \text{L/s}\approx3.91\ \text{m}^3/\text{h}\ \ (\dot{m}=\rho Q\approx0.868\ \text{kg/s}).}$$
Check

Checking the pipe Reynolds number at this flow rate, $V_{pipe}=Q/A_{pipe}=1.086\times10^{-3}/7.854\times10^{-3}=0.138\ \text{m/s}$, gives $Re_D=\rho V_{pipe}D/\mu=(800)(0.138)(0.10)/15\approx0.74$ — deep in the creeping (Stokes) flow regime, not the high-Reynolds turbulent regime ($Re_D$ typically above $10^4$) that the $C_d\approx0.6$ correlation used above assumes. The very high stated viscosity (15 kg/(m·s), consistent with a cold, heavy residual fuel oil) drives this. The worked answer follows the exam's own part (b), which explicitly asks for the assumptions made — a constant, high-Reynolds discharge coefficient is the intended simplification for this problem, but a real installation at this Reynolds number would need a laminar/low-Re discharge-coefficient correlation (or a laminar flow element instead of a sharp-edged orifice) rather than the standard $C_d\approx0.6$ value.

QuantityResult
Diameter ratio $\beta$$0.70$
Orifice area $A_o$$3.848\times10^{-3}\ \text{m}^2$
Volumetric flow rate $Q$$1.09\times10^{-3}\ \text{m}^3/\text{s}\approx1.09\ \text{L/s}$
Mass flow rate $\dot{m}$$\approx0.868\ \text{kg/s}$
Pipe Reynolds number $Re_D$$\approx0.74$ (creeping flow — see Verify)

b) The calculation assumes: steady, incompressible, single-phase liquid flow; a standard square-edged, concentric orifice plate with corner (or flange) pressure taps, for which the tabulated $C_d\approx0.6$-$0.61$ applies; fully developed, swirl-free, axisymmetric velocity profile approaching the plate (i.e. adequate straight, unobstructed pipe run both upstream and downstream); negligible fluid expansibility (appropriate for an incompressible liquid, so no expansibility/compressibility correction factor is needed); and — the assumption flagged explicitly in the Verify callout above — that the flow is at a high enough Reynolds number for the discharge coefficient to be treated as the constant $C_d\approx0.61$, which the computed $Re_D\approx0.74$ shows is not actually satisfied here.

c) A venturi meter's gradual converging cone and gentle diverging recovery cone let the flow decelerate smoothly after the throat with minimal separation and turbulent eddy formation, so most of the pressure drop created at the throat is recovered downstream (typical permanent-pressure-loss fraction of only 10-15% of the measured differential). An orifice plate's abrupt, sharp-edged constriction instead forces the flow to separate into a jet with a vena contracta, generating turbulent eddies downstream that dissipate a large fraction of the differential pressure irrecoverably (typically 40-60% permanent loss) — so for the SAME measured $\Delta P$ (i.e. the same metering signal), the venturi wastes far less of the fluid's pumping energy.

d) No — an orifice meter can be installed horizontal, vertical or inclined, PROVIDED the pipe runs completely full of liquid at the plate (no free surface or trapped gas pocket there), since the metering equation assumes the full pipe cross-section is flowing. Orientation matters mainly when the process fluid can carry entrained gas or vapour (or, conversely, entrained liquid in a gas line): pressure-tap placement is then chosen so the taps don't fill with, or get starved of, the entrained phase (e.g. top taps to avoid trapping condensate in a gas line, side/bottom taps to avoid trapping vapour bubbles in a liquid line). Vertical, upward flow is often preferred in practice specifically because it is self-venting and guarantees the line stays completely full at the plate.

e) An upstream elbow introduces secondary swirl and a skewed (asymmetric) velocity profile that has not had the straight-pipe run needed to redevelop into the axisymmetric, fully developed profile that the tabulated discharge coefficient $C_d(\beta,Re)$ was calibrated against. That distorted approach flow produces an incorrect (biased) differential-pressure reading at the plate and hence an erroneous inferred flow rate, even though nothing about the orifice plate itself has changed. The standard remedy is a minimum straight, unobstructed upstream pipe run — a function of $\beta$ and the type of upstream fitting, typically tens of pipe diameters for a close-coupled elbow — sufficient to let the profile redevelop, or a flow straightener when the available straight run is too short.

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