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Hydraulics, open channels and fluid power
Equivalent resistance of two parallel quadratic branches
Single resistance that replaces two branches joined at the same two nodes, each obeying h = Ri Qi² with positive flow.
- Ref
- HYD-14
- Rev
- v1
- Field
- Hydraulics, open channels and fluid power
- Result
- Rp [s²/m⁵]
Parallel branches joined at the same two nodes share the same head loss h, and each carries Qᵢ = √(h/Rᵢ). The flows add, so Q = √h Σ Rᵢ^(−1/2), and the equivalent resistance is Rp = [Σ Rᵢ^(−1/2)]⁻². It is smaller than either branch resistance. This calculator takes two branches; flow must be positive and no branch has a pump or a static lift.
Single resistance that replaces two branches joined at the same two nodes, each obeying h = Ri Qi² with positive flow. The branch flows add at a common head, so the resistances combine through their inverse square roots.
Calculator
- Share of the flow in branch 1, √(Rp/R₁)0.6
- Equivalent resistance as a share of the smaller branch0.36
- Head loss at 0.1 m³/s1.44 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.
- Two branches of 400 and 900 s²/m⁵ act as one of 144; the lower-resistance branch carries 60 % of the flow.
- Two identical branches give a quarter of one branch’s resistance, not half, because the flow in each is half and the loss varies with the square.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (s²/m⁵) | Local step | Change in result |
|---|---|---|---|---|---|---|
| R1 Resistance of branch 1 | 1 to 1e8 s²/m⁵ | 0.9365 | 894.6 | 893.7 | +10% | +5.83 % |
| R2 Resistance of branch 2 | 1 to 1e8 s²/m⁵ | 0.907 | 398.4 | 397.5 | +10% | +3.83 % |
Sweep of Resistance of branch 1
| Resistance of branch 1 (s²/m⁵) | Equivalent resistance (s²/m⁵) |
|---|---|
| 1 | 0.9365 |
| 10 | 8.184 |
| 100 | 56.25 |
| 1,000 | 237 |
| 10,000 | 532.5 |
| 100,000 | 750.8 |
| 1,000,000 | 848.3 |
| 1e7 | 883.2 |
| 1e8 | 894.6 |
Sweep one variable
Variables and how each one contributes
Equivalent resistance
What the equation gives you. Everything else feeds it.
Resistance of branch 1
The resistance of branch 1, in s²/m⁵. Raises the equivalent resistance, with less than direct proportion since the other branch shares the flow. Raising it by 10 % (to 440) moves the result by +5.83 %, from 144 to 152.4 s²/m⁵.
Resistance of branch 2
The resistance of branch 2, in s²/m⁵. Raises the equivalent resistance, with less than direct proportion since the other branch shares the flow. Raising it by 10 % (to 990) moves the result by +3.83 %, from 144 to 149.5 s²/m⁵.
Worked example with the values above
- Inverse square roots: 1/√400 + 1/√900 = 0.05 + 0.033333 = 0.083333.
- Square it: 0.0069444.
- Rp = 1 / that = 144 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