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=πd24Q = A v \qquad A = \frac{\pi d^2}{4}

Symbols and SI units:

  • QQ — volumetric flow rate, cubic metres per second (m³/s)
  • AA — internal cross-sectional area, square metres (m²)
  • vv — mean velocity across the section, metres per second (m/s)
  • ddinternal pipe diameter, metres (m)

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

Mass flow rate follows from the density ρ\rho in kg/m³:

m˙=ρQ=ρAv[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 m³/s before substituting.

FromTo m³/sMultiply by
L/sm³/s0.0010.001
L/minm³/s1.667×1051.667 \times 10^{-5}
m³/hm³/s1/36001/3600
US gal/minm³/s6.309×1056.309 \times 10^{-5}

Typical design velocities for water in pipework are roughly 1133 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 — 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 — near zero at the wall, maximum at the centre — 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 belongsA=πd2/4A = \pi d^2/4 and A=πr2A = \pi r^2 are the same thing; using πd2\pi d^2 overstates area fourfold.
  • Leaving diameters in millimetres100100 mm is 0.1000.100 m. In millimetres the area comes out in mm² and the flow rate is off by 10610^6.
  • Using nominal or outside diameter — always the internal bore.
  • Confusing mass and volume flowm˙=ρQ\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 — 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 — flow depends on the pressure drop along the whole system, including fittings and elevation.

示例题目

Step 1: Convert the diameter: d=100 mm=0.100 md = 100\ \text{mm} = 0.100\ \text{m}
Step 2: A=πd2/4=π(0.100 m)2/4=7.854×103 m2A = \pi d^2/4 = \pi (0.100\ \text{m})^2 / 4 = 7.854 \times 10^{-3}\ \text{m}^2
Step 3: Q=Av=(7.854×103 m2)(2.0 m/s)=1.571×102 m3/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: 1.571×102 m3/s×1000=15.7 L/s1.571 \times 10^{-2}\ \text{m}^3/\text{s} \times 1000 = 15.7\ \text{L/s}
Step 5: Per hour: 1.571×102 m3/s×3600 s/h=56.5 m3/h1.571 \times 10^{-2}\ \text{m}^3/\text{s} \times 3600\ \text{s/h} = 56.5\ \text{m}^3/\text{h}
Answer: Q1.57×102Q \approx 1.57 \times 10^{-2} m³/s =15.7= 15.7 L/s =56.5= 56.5 m³/h

Step 1: Convert: Q=10 L/s=0.010 m3/sQ = 10\ \text{L/s} = 0.010\ \text{m}^3/\text{s}, d=0.050 md = 0.050\ \text{m}
Step 2: A=π(0.050 m)2/4=1.963×103 m2A = \pi (0.050\ \text{m})^2 / 4 = 1.963 \times 10^{-3}\ \text{m}^2
Step 3: v=Q/A=(0.010 m3/s)÷(1.963×103 m2)=5.09 m/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 1133 m/s water guidance, so a larger bore should be considered and the limit confirmed against the applicable standard
Answer: v5.09v \approx 5.09 m/s — too fast for typical water pipework

Step 1: m˙=ρQ\dot{m} = \rho Q
Step 2: m˙=(998 kg/m3)(1.571×102 m3/s)\dot{m} = (998\ \text{kg/m}^3)(1.571 \times 10^{-2}\ \text{m}^3/\text{s})
Step 3: m˙=15.7 kg/s\dot{m} = 15.7\ \text{kg/s}
Step 4: The m³ units cancel, leaving kg/s
Answer: m˙15.7\dot{m} \approx 15.7 kg/s

常见问题

Q = Av, where A is the internal cross-sectional area in m² and v is the mean velocity in m/s, giving Q in m³/s. For a round pipe, A = πd²/4 with the internal diameter d in metres.

Rearrange to v = Q/A, converting the flow to m³/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 (m³/s) and mass flow ṁ is the mass passing per second (kg/s). They are linked by density: ṁ = ρQ. For gases the density varies with pressure and temperature, so the two are not interchangeable.

Around 1–3 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 — the binding limit comes from the applicable design code or the pipe manufacturer's data for the specific service.

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