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Hydraulics, open channels and fluid power
Series pipe resistance for quadratic losses
Resistance Rs in hL = Rs Q² for two circular pipes in series, with each pipe’s friction factor held at its duty-point value.
- Ref
- HYD-13
- Rev
- v1
- Field
- Hydraulics, open channels and fluid power
- Result
- Rs [s²/m⁵]
With the quadratic head-loss law hL = f (L/D) V²/(2g) and V = 4Q/(π D²), a pipe’s loss becomes hL = [8 f L/(g π² D⁵)] Q², so its resistance is R = 8 f L/(g π² D⁵). Resistances of pipes in series add. This calculator sums two pipes and includes straight-pipe friction only; add fittings through their equivalent lengths. The friction factors are held at their duty-point values.
Resistance Rs in hL = Rs Q² for two circular pipes in series, with each pipe’s friction factor held at its duty-point value. It includes straight-pipe friction only.
Calculator
- Resistance of pipe 11,653 s²/m⁵
- Resistance of pipe 21,239 s²/m⁵
- Head loss at 0.01 m³/s0.2892 m
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- The two pipes give hL = 2,891.9 Q², so 0.2892 m of head is lost at 0.01 m³/s and four times as much, 1.157 m, at 0.02 m³/s.
- Because of the fifth power, halving a pipe’s diameter multiplies its resistance by 32, so one narrow section can dominate the whole line.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (s²/m⁵) | Local step | Change in result |
|---|---|---|---|---|---|---|
| D1 Diameter of pipe 1 | 0.005 to 3 m | 1,239 | 5.288e9 | 5.288e9 | +10% | −21.7 % |
| D2 Diameter of pipe 2 | 0.005 to 3 m | 1,653 | 3.966e9 | 3.966e9 | +10% | −16.2 % |
| L2 Length of pipe 2 | 0.1 to 10,000 m | 1,677 | 2,480,000 | 2,479,000 | +10% | +4.29 % |
| L1 Length of pipe 1 | 0.1 to 10,000 m | 1,256 | 1,654,000 | 1,653,000 | +10% | +5.71 % |
| g Gravitational acceleration | 1 to 30 m/s² | 945.7 | 28,370 | 27,420 | +10% | −9.09 % |
| f1 Friction factor of pipe 1 | 0.008 to 0.1 | 1,900 | 9,502 | 7,602 | +10% | +5.71 % |
| f2 Friction factor of pipe 2 | 0.008 to 0.1 | 1,983 | 5,784 | 3,801 | +10% | +4.29 % |
Sweep of Diameter of pipe 1
| Diameter of pipe 1 (m) | Series resistance (s²/m⁵) |
|---|---|
| 0.005 | 5.288e9 |
| 0.3794 | 1,242 |
| 0.7538 | 1,239 |
| 1.128 | 1,239 |
| 1.502 | 1,239 |
| 1.877 | 1,239 |
| 2.251 | 1,239 |
| 2.626 | 1,239 |
| 3 | 1,239 |
Sweep one variable
Variables and how each one contributes
Series resistance
What the equation gives you. Everything else feeds it.
Friction factor of pipe 1
The Darcy friction factor of pipe 1. Raises the first pipe’s resistance in direct proportion. Raising it by 10 % (to 0.022) moves the result by +5.71 %, from 2,892 to 3,057 s²/m⁵.
Length of pipe 1
The length of pipe 1, in m. Raises the first pipe’s resistance in direct proportion. Raising it by 10 % (to 11) moves the result by +5.71 %, from 2,892 to 3,057 s²/m⁵.
Diameter of pipe 1
The internal diameter of pipe 1, in m. A larger diameter lowers the first pipe’s resistance with the fifth power. Raising it by 10 % (to 0.11) moves the result by −21.7 %, from 2,892 to 2,265 s²/m⁵.
Friction factor of pipe 2
The Darcy friction factor of pipe 2. Raises the second pipe’s resistance in direct proportion. Raising it by 10 % (to 0.033) moves the result by +4.29 %, from 2,892 to 3,016 s²/m⁵.
Length of pipe 2
The length of pipe 2, in m. Raises the second pipe’s resistance in direct proportion. Raising it by 10 % (to 5.5) moves the result by +4.29 %, from 2,892 to 3,016 s²/m⁵.
Diameter of pipe 2
The internal diameter of pipe 2, in m. A larger diameter lowers the second pipe’s resistance with the fifth power. Raising it by 10 % (to 0.11) moves the result by −16.2 %, from 2,892 to 2,422 s²/m⁵.
Gravitational acceleration
The gravitational acceleration, in m/s². Lowers the resistance in inverse proportion. Raising it by 10 % (to 10.79) moves the result by −9.09 %, from 2,892 to 2,629 s²/m⁵.
Worked example with the values above
- Pipe 1: 8 × 0.02 × 10 / (9.81 π² × 0.1⁵) = 1,652.5 s²/m⁵.
- Pipe 2: 8 × 0.03 × 5 / (9.81 π² × 0.1⁵) = 1,239.4 s²/m⁵.
- Rs = 2,891.9 s²/m⁵.
Assumptions and limits
- Steady incompressible flow, with the friction factor and loss coefficients held at their duty-point values.
- Head losses proportional to the square of the flow, with positive flow in each pipe.
Common mistakes
- Using a loss coefficient that is referred to a different velocity from the one in the formula.
- Mixing the Darcy friction factor with the Fanning factor, which is a quarter of it.
- Adding resistances in parallel as if they were in series.
Statistics and uncertainty
Loss coefficients and friction factors are empirical and are usually known to within 10 % to 20 %, so network results carry that uncertainty. The quadratic forms assume the friction factor stays fixed as the flow changes.
Where it is used
- Building system curves for pumps and checking the effect of fittings.
- Balancing flows in looped water networks, and simplifying branches into one resistance.
Maths you need
- Head loss: energy lost per unit weight, in metres
- Darcy friction factor: f, as in hL = f (L/D) V²/2g
- Continuity: flow into a node equals flow out