Orifice Flow Calculator

Orifice plate flow, bore sizing and pressure drop with step-by-step solutions
Find the water flow through a 50 mm orifice in a 100 mm pipe with a 20 kPa differential
Find the discharge from a 25 mm orifice in a tank under 3 m of head
What differential pressure gives 5 L/s through a 50 mm orifice in a 100 mm pipe?
Find the beta ratio of a 60 mm orifice in a 150 mm pipe

The Orifice Equation

An orifice plate creates a measurable pressure drop that maps to flow:

Q=CdтАЙEтАЙA02тАЙ╬ФP╧Б,E=11тИТ╬▓4,╬▓=dDQ = C_d\,E\,A_0\sqrt{\frac{2\,\Delta P}{\rho}}, \qquad E = \frac{1}{\sqrt{1-\beta^4}}, \qquad \beta = \frac{d}{D}

Symbols and SI units:

  • QQ тАФ volumetric flow, m┬│/s
  • CdC_d тАФ discharge coefficient, dimensionless; typically 0.600.60тАУ0.620.62 for a sharp-edged concentric plate
  • EE тАФ velocity-of-approach factor, dimensionless
  • A0=╧Аd2/4A_0 = \pi d^2/4 тАФ orifice bore area, m┬▓; dd bore, DD pipe internal diameter, m
  • ╬ФP\Delta P тАФ differential pressure across the taps, pascals (Pa)
  • ╧Б\rho тАФ fluid density, kg/m┬│

For free discharge from a tank under a head hh (m), ╬ФP/╧Б=gh\Delta P/\rho = gh and the equation reduces to Q=CdA02ghQ = C_d A_0 \sqrt{2gh}.

The assumption people forget: the flow is treated as incompressible. For gases you must add the expansibility factor ╬╡\varepsilon, and CdC_d itself depends on Reynolds number, ╬▓\beta and tap arrangement.

Sizing, Pressure Loss and the Standard

Because QтИЭ╬ФPQ \propto \sqrt{\Delta P}, the relationship is strongly non-linear: halving the flow quarters the differential, which is why an orifice meter has a usable turndown of only about 3:1.

To size a bore, fix a target differential at maximum flow and solve for A0A_0:

A0=QCdE2╬ФP/╧Б,d=4A0╧АA_0 = \frac{Q}{C_d E \sqrt{2\Delta P/\rho}}, \qquad d = \sqrt{\frac{4A_0}{\pi}}

Keep ╬▓\beta within roughly 0.20.2тАУ0.750.75; outside that band the coefficients are not well characterised.

The measured ╬ФP\Delta P is not the permanent loss. Some pressure recovers downstream, and the unrecovered loss is roughly (1тИТ╬▓1.9)╬ФP(1-\beta^{1.9})\Delta P тАФ around 70%70\% of the differential at ╬▓=0.5\beta = 0.5. That loss is a permanent pumping cost.

The rule that governs the real installation: a metering orifice must be designed and installed to ISO 5167 (or an equivalent standard such as AGA 3 for gas), which fixes the plate geometry, tap positions, straight-run lengths and the CdC_d correlation. The hand calculation here shows the method; it is not a substitute for the standard.

Common Mistakes to Avoid

  • Dropping the velocity-of-approach factor тАФ at ╬▓=0.5\beta = 0.5, E=1.033E = 1.033, and it grows quickly for larger ╬▓\beta.
  • Mixing pressure units тАФ ╬ФP\Delta P must be in pascals. 2020 kPa is 20,00020{,}000 Pa; a bar is 100,000100{,}000 Pa.
  • Assuming flow is proportional to ╬ФP\Delta P тАФ it goes with the square root, so a differential-pressure transmitter output must be square-rooted before it means flow.
  • Treating the measured differential as the permanent loss тАФ much of it recovers downstream.
  • Using a liquid CdC_d for compressible gas flow тАФ the expansibility factor is required, and choked flow needs a different treatment entirely.
  • Ignoring upstream straight run тАФ an elbow too close to the plate biases the reading regardless of how good the arithmetic is.

Examples

Step 1: ╬▓=d/D=50/100=0.50\beta = d/D = 50/100 = 0.50, so ╬▓4=0.0625\beta^4 = 0.0625 and E=1/1тИТ0.0625=1.0328E = 1/\sqrt{1-0.0625} = 1.0328
Step 2: A0=╧А(0.050┬аm)2/4=1.963├Ч10тИТ3┬аm2A_0 = \pi (0.050\ \text{m})^2/4 = 1.963 \times 10^{-3}\ \text{m}^2
Step 3: 2╬ФP/╧Б=2(20,000┬аPa)/(998┬аkg/m3)=40.08┬аm2/s2=6.331┬аm/s\sqrt{2\Delta P/\rho} = \sqrt{2(20{,}000\ \text{Pa})/(998\ \text{kg/m}^3)} = \sqrt{40.08\ \text{m}^2/\text{s}^2} = 6.331\ \text{m/s}
Step 4: Q=(0.62)(1.0328)(1.963├Ч10тИТ3┬аm2)(6.331┬аm/s)=7.96├Ч10тИТ3┬аm3/sQ = (0.62)(1.0328)(1.963 \times 10^{-3}\ \text{m}^2)(6.331\ \text{m/s}) = 7.96 \times 10^{-3}\ \text{m}^3/\text{s}
Step 5: In litres: 7.96┬аL/s7.96\ \text{L/s}, or 28.7┬аm3/h28.7\ \text{m}^3/\text{h}
Step 6: m╦Щ=╧БQ=(998┬аkg/m3)(7.96├Ч10тИТ3┬аm3/s)=7.94┬аkg/s\dot{m} = \rho Q = (998\ \text{kg/m}^3)(7.96 \times 10^{-3}\ \text{m}^3/\text{s}) = 7.94\ \text{kg/s}
Answer: QтЙИ7.96Q \approx 7.96 L/s (28.728.7 m┬│/h), m╦ЩтЙИ7.94\dot{m} \approx 7.94 kg/s

Step 1: Free discharge, so use Q=CdA02ghQ = C_d A_0 \sqrt{2gh}
Step 2: A0=╧А(0.025┬аm)2/4=4.909├Ч10тИТ4┬аm2A_0 = \pi (0.025\ \text{m})^2/4 = 4.909 \times 10^{-4}\ \text{m}^2
Step 3: 2gh=2(9.81┬аm/s2)(3.0┬аm)=58.86┬аm2/s2=7.672┬аm/s\sqrt{2gh} = \sqrt{2(9.81\ \text{m/s}^2)(3.0\ \text{m})} = \sqrt{58.86\ \text{m}^2/\text{s}^2} = 7.672\ \text{m/s}
Step 4: Q=(0.61)(4.909├Ч10тИТ4┬аm2)(7.672┬аm/s)=2.30├Ч10тИТ3┬аm3/sQ = (0.61)(4.909 \times 10^{-4}\ \text{m}^2)(7.672\ \text{m/s}) = 2.30 \times 10^{-3}\ \text{m}^3/\text{s}
Step 5: The pipe area is large compared with the orifice, so EтЙИ1E \approx 1 and is omitted
Answer: QтЙИ2.30├Ч10тИТ3Q \approx 2.30 \times 10^{-3} m┬│/s =2.30= 2.30 L/s

Step 1: From the first example, CdEA0=(0.62)(1.0328)(1.963├Ч10тИТ3┬аm2)=1.257├Ч10тИТ3┬аm2C_d E A_0 = (0.62)(1.0328)(1.963 \times 10^{-3}\ \text{m}^2) = 1.257 \times 10^{-3}\ \text{m}^2
Step 2: 2╬ФP/╧Б=Q/(CdEA0)=(0.0050┬аm3/s)├╖(1.257├Ч10тИТ3┬аm2)=3.977┬аm/s\sqrt{2\Delta P/\rho} = Q/(C_d E A_0) = (0.0050\ \text{m}^3/\text{s}) \div (1.257 \times 10^{-3}\ \text{m}^2) = 3.977\ \text{m/s}
Step 3: Square it: 2╬ФP/╧Б=15.82┬аm2/s22\Delta P/\rho = 15.82\ \text{m}^2/\text{s}^2
Step 4: ╬ФP=(15.82)(998┬аkg/m3)/2=7890┬аPa=7.89┬аkPa\Delta P = (15.82)(998\ \text{kg/m}^3)/2 = 7890\ \text{Pa} = 7.89\ \text{kPa}
Step 5: Check the square-root law: (5.0/7.96)2├Ч20┬аkPa=7.89┬аkPa(5.0/7.96)^2 \times 20\ \text{kPa} = 7.89\ \text{kPa}
Answer: ╬ФPтЙИ7.89\Delta P \approx 7.89 kPa

Frequently Asked Questions

Q = Cd ┬╖ E ┬╖ AтВА ┬╖ тИЪ(2╬ФP/╧Б), where Cd is the discharge coefficient, E = 1/тИЪ(1тИТ╬▓тБ┤) is the velocity-of-approach factor, AтВА is the bore area in m┬▓, ╬ФP is the differential in pascals and ╧Б is the density in kg/m┬│. For free discharge from a tank it simplifies to Q = Cd┬╖AтВА┬╖тИЪ(2gh).

Around 0.60 to 0.62 for a sharp-edged concentric plate in turbulent liquid flow. The exact value depends on the beta ratio, the Reynolds number and the tap arrangement, and for a metering installation it should come from the ISO 5167 correlation rather than a single assumed figure.

No. Part of the differential recovers downstream of the plate as the jet re-expands. The permanent, unrecovered loss is roughly (1 тИТ ╬▓^1.9) times the measured differential тАФ about 70% of it at ╬▓ = 0.5 тАФ and that is what costs pumping energy.

Use it to understand the method and get a first estimate, but a metering orifice must be designed and installed to ISO 5167 (or AGA 3 for gas), which specifies the plate geometry, tap locations, straight-run requirements and the coefficient correlation. The final bore and installation must follow that standard.

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