Calculate flow rate from pressure and pipe diameter

Primary Tool

Pipe Head Loss Calculator

The solved flow from the third case is loaded into the Pipe Head Loss Calculator below, which runs the forward calculation. If the pressure drop it reports matches the $50$ kPa you started with, the iteration has converged. The Flow Rate Calculator and Orifice Flow Calculator cover the first two cases.

Which case are you in?

A pressure reading becomes a flow rate only through a relationship between the two, and which relationship applies depends on where the pressure is being used up. A pressure of $50$ kPa across a short sharp-edged restriction, across a long straight pipe, and at a closed valve gives three entirely different flows. Start by asking what the pressure difference is acting against.

•The velocity is known or measured: use $Q = A v$ . Pressure is not needed.
•The pressure difference is across a short restriction such as an orifice, nozzle, or venturi: use the orifice equation.
•The pressure difference is across a length of pipe, so friction uses it up: use Darcy-Weisbach, which needs iteration.

Case 1: velocity is known

For water at $v = 2$ m/s in a $D = 50$ mm pipe, the flow area is $A = π D^{2} /4 = 1.963 \times 1 0^{- 3}$ m $^{2}$ and $Q = A v = 3.927 \times 1 0^{- 3}$ m $^{3}$ /s, or $3.927$ L/s. Pressure played no part. If a pressure gauge is the only measurement you have, this case does not apply.

Case 2: flow through an orifice or nozzle

A $d = 25$ mm sharp-edged orifice sits in the same $D = 50$ mm line, with water at $ρ = 998$ kg/m $^{3}$ and a pressure difference of $Δ P = 50$ kPa across it. The ratio $β = d / D = 0.5$ , and the discharge coefficient is $C_{d} = 0.61$ .

1Orifice area: $A_{o} = π (0.025)^{2} /4 = 4.909 \times 1 0^{- 4}$ m $^{2}$ .
2Velocity term: $\frac{2 Δ P}{ρ ( 1 - β ^{4} )} = \frac{2 ( 50 , 000 )}{998 ( 0.9375 )} = 10.34$ m/s.
3Flow: $Q = C_{d} A_{o} \times 10.34 = 0.61 (4.909 \times 1 0^{- 4}) (10.34) = 3.096 \times 1 0^{- 3}$ m $^{3}$ /s, which is $3.10$ L/s.

Case 3: pressure lost to friction along a pipe

Now there is no restriction, only $L = 30$ m of $D = 50$ mm commercial steel pipe ( $ε = 0.046$ mm) with $Δ P = 50$ kPa along it. Darcy-Weisbach gives $Δ P = f \frac{L}{D} \frac{ρ v ^{2}}{2}$ . Solving for velocity gives $v = \frac{2 Δ P D}{f L ρ}$ . The trouble is that $f$ depends on the Reynolds number $R e = v D / ν$ , and so on $v$ . The unknown sits on both sides, so iterate.

1Guess $f = 0.02$ . Then $v = 2 (50, 000) (0.05) / (0.02 \cdot 30 \cdot 998) = 2.890$ m/s and $R e = 144, 000$ .
2At that $R e$ with $ε / D = 0.00092$ , the Colebrook-White equation gives $f = 0.02120$ . Recompute: $v = 2.807$ m/s and $R e = 140, 300$ .
3Colebrook at $R e = 140, 300$ gives $f = 0.02125$ , so $v = 2.8036$ m/s. One more pass changes $f$ in the sixth decimal place, so the iteration has converged.
4Result: $v = 2.803$ m/s, $f = 0.02125$ , and $Q = A v = 1.963 \times 1 0^{- 3} (2.803) = 5.50 \times 1 0^{- 3}$ m $^{3}$ /s, or $5.50$ L/s.

Why the iteration settles so fast

$f$ depends on $R e$ only through a logarithm, so a $3%$ change in velocity moves $f$ by well under $1%$ . Each pass shrinks the error by a large factor. Starting from a typical turbulent guess of $f = 0.02$ is almost always enough, and three passes reach four significant figures.

The embedded Pipe Head Loss Calculator runs the forward calculation with $Q = 5.5045$ L/s and reports a pressure drop of $50.0$ kPa, which confirms the result. For comparison, the assumed $2$ m/s in Case 1 gave $3.93$ L/s, the orifice passed $3.10$ L/s at $50$ kPa, and the open $30$ m pipe passed $5.50$ L/s at the same $50$ kPa.

Common Pitfall

The pressure you plug in is the difference across the part of the system that is using it up, not the supply or gauge pressure. A gauge reading $50$ kPa on a line that discharges to atmosphere through a long pipe and a throttled valve is spent across both, and putting all of it into the pipe equation overestimates the flow. With a closed valve the gauge reads the same $50$ kPa and the flow is zero.

Try a Variation

Keep $Δ P = 50$ kPa and $L = 30$ m but change the pipe to $D = 100$ mm. Before running the iteration, predict whether the flow doubles, quadruples, or does something else. The velocity also rises when the diameter grows, so the area ratio of $4$ is not the whole story. Check your prediction with the Pipe Head Loss Calculator or by repeating the iteration.

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Worked ExampleEngineeringPipe Head Loss Calculator

Calculate pressure drop from flow rate

With the flow known, pressure drop in a pipe is a direct calculation: find the velocity, the Reynolds number, the friction factor, then apply Darcy-Weisbach and add fitting losses. For $4$ L/s through $80$ m of $50$ mm pipe with two elbows it comes to about $78.7$ kPa.

Open worked example →