Flow Rate Calculator

Volumetric flow, pipe velocity and mass flow with step-by-step solutions
Find the flow rate in a 100 mm pipe with a velocity of 2 m/s
Find the water velocity for 10 L/s in a 50 mm bore pipe
Convert 15.7 L/s of water to mass flow rate in kg/s
What pipe diameter carries 20 m^3/h of water at 1.5 m/s?

The Flow Rate Equation Q = Av

Volumetric flow rate is cross-sectional area times average velocity:

Q=AvA=╧Аd24Q = A v \qquad A = \frac{\pi d^2}{4}

Symbols and SI units:

  • QQ тАФ volumetric flow rate, cubic metres per second (m┬│/s)
  • AA тАФ internal cross-sectional area, square metres (m┬▓)
  • vv тАФ mean velocity across the section, metres per second (m/s)
  • dd тАФ internal pipe diameter, metres (m)

Rearranged, v=Q/Av = Q/A тАФ the form used to check whether a chosen pipe size gives an acceptable velocity.

Mass flow rate follows from the density ╧Б\rho in kg/m┬│:

m╦Щ=╧БQ=╧БAv[kg/s]\dot{m} = \rho Q = \rho A v \quad [\text{kg/s}]

The assumption people forget: dd is the bore, not the nominal size or the outside diameter. A DN100 steel pipe is nominally 100 mm but its actual bore depends on wall thickness, and QQ scales with d2d^2, so a 5% diameter error is a 10% flow error.

Units and Sensible Velocities

Flow is quoted in a dozen units; convert to m┬│/s before substituting.

FromTo m┬│/sMultiply by
L/sm┬│/s0.0010.001
L/minm┬│/s1.667├Ч10тИТ51.667 \times 10^{-5}
m┬│/hm┬│/s1/36001/3600
US gal/minm┬│/s6.309├Ч10тИТ56.309 \times 10^{-5}

Typical design velocities for water in pipework are roughly 11тАУ33 m/s: below that, solids settle and the pipe is oversized; above it, noise, erosion and pressure loss climb steeply. Treat those figures as orientation only тАФ the velocity limit for a real installation comes from the governing design standard or the pipe manufacturer's data, not from a rule of thumb.

The assumption people forget: Q=AvQ = Av uses the mean velocity. Real pipe flow has a velocity profile тАФ near zero at the wall, maximum at the centre тАФ and Q=AvQ = Av is correct only because vv is defined as the area-averaged value. It also assumes a full pipe and steady flow; a partly full gravity drain needs open-channel hydraulics instead.

Common Mistakes to Avoid

  • Using the radius where the diameter belongs тАФ A=╧Аd2/4A = \pi d^2/4 and A=╧Аr2A = \pi r^2 are the same thing; using ╧Аd2\pi d^2 overstates area fourfold.
  • Leaving diameters in millimetres тАФ 100100 mm is 0.1000.100 m. In millimetres the area comes out in mm┬▓ and the flow rate is off by 10610^6.
  • Using nominal or outside diameter тАФ always the internal bore.
  • Confusing mass and volume flow тАФ m╦Щ=╧БQ\dot{m} = \rho Q; the two differ by a factor of roughly 10001000 for water in SI units.
  • Applying a liquid density to a gas тАФ gas density changes with pressure and temperature, so a compressible-flow treatment is needed once the pressure drop is significant.
  • Assuming pressure alone sets flow тАФ flow depends on the pressure drop along the whole system, including fittings and elevation.

Examples

Step 1: Convert the diameter: d=100┬аmm=0.100┬аmd = 100\ \text{mm} = 0.100\ \text{m}
Step 2: A=╧Аd2/4=╧А(0.100┬аm)2/4=7.854├Ч10тИТ3┬аm2A = \pi d^2/4 = \pi (0.100\ \text{m})^2 / 4 = 7.854 \times 10^{-3}\ \text{m}^2
Step 3: Q=Av=(7.854├Ч10тИТ3┬аm2)(2.0┬аm/s)=1.571├Ч10тИТ2┬аm3/sQ = Av = (7.854 \times 10^{-3}\ \text{m}^2)(2.0\ \text{m/s}) = 1.571 \times 10^{-2}\ \text{m}^3/\text{s}
Step 4: In litres: 1.571├Ч10тИТ2┬аm3/s├Ч1000=15.7┬аL/s1.571 \times 10^{-2}\ \text{m}^3/\text{s} \times 1000 = 15.7\ \text{L/s}
Step 5: Per hour: 1.571├Ч10тИТ2┬аm3/s├Ч3600┬аs/h=56.5┬аm3/h1.571 \times 10^{-2}\ \text{m}^3/\text{s} \times 3600\ \text{s/h} = 56.5\ \text{m}^3/\text{h}
Answer: QтЙИ1.57├Ч10тИТ2Q \approx 1.57 \times 10^{-2} m┬│/s =15.7= 15.7 L/s =56.5= 56.5 m┬│/h

Step 1: Convert: Q=10┬аL/s=0.010┬аm3/sQ = 10\ \text{L/s} = 0.010\ \text{m}^3/\text{s}, d=0.050┬аmd = 0.050\ \text{m}
Step 2: A=╧А(0.050┬аm)2/4=1.963├Ч10тИТ3┬аm2A = \pi (0.050\ \text{m})^2 / 4 = 1.963 \times 10^{-3}\ \text{m}^2
Step 3: v=Q/A=(0.010┬аm3/s)├╖(1.963├Ч10тИТ3┬аm2)=5.09┬аm/sv = Q/A = (0.010\ \text{m}^3/\text{s}) \div (1.963 \times 10^{-3}\ \text{m}^2) = 5.09\ \text{m/s}
Step 4: That is well above the usual 11тАУ33 m/s water guidance, so a larger bore should be considered and the limit confirmed against the applicable standard
Answer: vтЙИ5.09v \approx 5.09 m/s тАФ too fast for typical water pipework

Step 1: m╦Щ=╧БQ\dot{m} = \rho Q
Step 2: m╦Щ=(998┬аkg/m3)(1.571├Ч10тИТ2┬аm3/s)\dot{m} = (998\ \text{kg/m}^3)(1.571 \times 10^{-2}\ \text{m}^3/\text{s})
Step 3: m╦Щ=15.7┬аkg/s\dot{m} = 15.7\ \text{kg/s}
Step 4: The m┬│ units cancel, leaving kg/s
Answer: m╦ЩтЙИ15.7\dot{m} \approx 15.7 kg/s

Frequently Asked Questions

Q = Av, where A is the internal cross-sectional area in m┬▓ and v is the mean velocity in m/s, giving Q in m┬│/s. For a round pipe, A = ╧Аd┬▓/4 with the internal diameter d in metres.

Rearrange to v = Q/A, converting the flow to m┬│/s and the diameter to metres first. For example, 10 L/s in a 50 mm bore gives 0.010 ├╖ 0.001963 = 5.09 m/s, which is fast enough for water that a larger pipe is usually warranted.

Volumetric flow Q is the volume passing per second (m┬│/s) and mass flow с╣Б is the mass passing per second (kg/s). They are linked by density: с╣Б = ╧БQ. For gases the density varies with pressure and temperature, so the two are not interchangeable.

Around 1тАУ3 m/s is common practice for water in building services and process pipework, balancing pipe cost against noise, erosion and pressure loss. This is orientation, not a specification тАФ the binding limit comes from the applicable design code or the pipe manufacturer's data for the specific service.

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