Pipe networks and flow in series and parallel

Why it matters

Real piping is never one straight pipe. Engine cooling circuits split between the radiator and the bypass, fuel and lubrication systems branch to several consumers, and water-supply grids are loops. Engineers need to know how flow divides, what head the pump must supply and how to replace a complicated arrangement with one equivalent pipe. The rules are simple analogies of electric circuits, but the "resistance" is non-linear (h ∝ Q²).

Key ideas

• Pipe resistance. For turbulent flow with a given friction factor, Darcy–Weisbach can be written h_f = k·Q², where k = 8fL/(π²gD⁵). This makes the circuit analogy clear: head loss plays the role of voltage drop and flow rate the role of current, but the drop rises with Q², not Q.
• Pipes in series. The same Q flows through every pipe, and the head losses add (friction plus any minor losses at the joints, such as sudden expansions or contractions).
• Equivalent pipe (Dupuit's equation). A series combination can be replaced by one pipe of uniform diameter D_e and the same total length that gives the same loss for the same Q. With equal f: L/D_e⁵ = Σ(L_i/D_i⁵).
• Pipes in parallel. The head loss between the two junctions is the same in every branch, because each branch connects the same two points with the same total energy. The flows add up to the total. With h = k·Q² for each branch, Q_i ∝ 1/√k_i ∝ √(D_i⁵/(f_i·L_i)). A larger, shorter or smoother branch takes more of the flow.
• Branched pipes (three-reservoir problem). At a junction, continuity requires Σ inflows = Σ outflows. The unknown is the head at the junction: guess it, compute each branch flow from the energy equation, and adjust until continuity balances. The direction of flow in the middle branch depends on whether the junction head is above or below that reservoir's level.
• Looped networks (Hardy Cross method). Two laws apply, as in Kirchhoff's laws: continuity at every node, and the algebraic sum of head losses around every closed loop is zero. Assume flows that satisfy continuity, compute the loop correction ΔQ = −Σ(kQ|Q|)/Σ(2k|Q|), apply it to each pipe in the loop, and repeat until ΔQ is small. Computers now do this, but GATE may test the principle or one iteration.
• Pumps in a network. The pump head appears as an energy gain in the energy equation. The operating point is where the pump's H–Q curve meets the system curve H = static lift + Σk·Q².
• Power transmission through a pipe. Hydraulic power delivered at the outlet is ρgQ(H − h_f). It is maximum when h_f = H/3, giving a maximum transmission efficiency of 2/3 (66.7 %) at that point.
• Empirical alternative. For water distribution, the Hazen–Williams formula (with a roughness coefficient C from tables) is used instead of Darcy–Weisbach. It applies only to water at ordinary temperatures.
• Minor losses are often neglected in long networks (L/D > 1000) but must be included in compact systems such as engine cooling circuits.

Formulas

h_f = 8·f·L·Q² / (π²·g·D⁵) = k·Q² h_f head loss (m), f Darcy friction factor, L length (m), Q flow (m³/s), D diameter (m), g = 9.81 m/s²; k resistance (s²/m⁵).

Q = Q₁ = Q₂ = …, H = h₁ + h₂ + … Pipes in series.

L/D_e⁵ = L₁/D₁⁵ + L₂/D₂⁵ + … Dupuit equivalent pipe (same f; L = L₁ + L₂ + …).

Q = Q₁ + Q₂ + …, h₁ = h₂ = … Pipes in parallel.

Q₁/Q₂ = √[(D₁⁵·f₂·L₂) / (D₂⁵·f₁·L₁)] Flow split between two parallel pipes.

ΔQ = −Σ(k·Q·|Q|) / Σ(2·k·|Q|) Hardy Cross loop correction (n = 2 for Darcy–Weisbach).

P = ρ·g·Q·(H − h_f), η = (H − h_f)/H, P_max at h_f = H/3 Power transmitted by a pipe with supply head H (m).

h_f = 10.67·L·Q^1.852 / (C^1.852·D^4.87) Hazen–Williams (SI; water only; C from tables).

Worked examples

Example 1 (standard) — pipes in series and equivalent pipe. Given: Q = 0.08 m³/s of water flows through pipe 1 (L = 300 m, D = 0.3 m) and then pipe 2 (L = 200 m, D = 0.2 m), both with f = 0.02. Neglect minor losses. Find the total head loss and the equivalent diameter for a single 500 m pipe.

1. k₁ = 8 × 0.02 × 300/(π² × 9.81 × 0.3⁵) = 204.0 s²/m⁵ → h₁ = 204.0 × 0.08² = 1.306 m.
2. k₂ = 8 × 0.02 × 200/(π² × 9.81 × 0.2⁵) = 1032.8 s²/m⁵ → h₂ = 1032.8 × 0.08² = 6.610 m.
3. Total H = 1.306 + 6.610 = 7.916 m.
4. Dupuit: 500/D_e⁵ = 300/0.3⁵ + 200/0.2⁵ = 123 457 + 625 000 = 748 457 → D_e = (500/748 457)^(1/5) = 0.2317 m. Answer: H ≈ 7.92 m; equivalent diameter ≈ 0.232 m. The smaller pipe causes most of the loss, because h ∝ 1/D⁵.

Example 2 (GATE level) — two pipes in parallel. Given: a total flow of 0.12 m³/s divides between pipe A (L = 500 m, D = 0.2 m, f = 0.02) and pipe B (L = 400 m, D = 0.15 m, f = 0.025). Find the flow in each and the head loss between the junctions.

1. Equal head loss → Q ∝ √(D⁵/(fL)).
2. a = √(0.2⁵/(0.02 × 500)) = √(3.2 × 10⁻⁵) = 5.657 × 10⁻³; b = √(0.15⁵/(0.025 × 400)) = √(7.594 × 10⁻⁶) = 2.756 × 10⁻³.
3. Q_A = 0.12 × a/(a + b) = 0.12 × 5.657/8.413 = 0.0807 m³/s; Q_B = 0.12 − 0.0807 = 0.0393 m³/s.
4. Check: h_A = k_A·Q_A² = [8 × 0.02 × 500/(π² × 9.81 × 0.2⁵)] × 0.0807² = 2582 × 0.006511 = 16.81 m; h_B = 10 881 × 0.0393² = 16.81 m. ✓ Answer: Q_A ≈ 0.081 m³/s, Q_B ≈ 0.039 m³/s, head loss ≈ 16.8 m.

Common mistakes

• Adding head losses for parallel pipes or adding flows for series pipes — it is the other way round.
• Splitting parallel flow in proportion to area or D², instead of √(D⁵/(fL)).
• Forgetting that h ∝ Q², so doubling the flow quadruples the loss (series) and a branch's share does not scale linearly.
• In the three-reservoir problem, fixing the flow direction in the middle pipe before finding the junction head.
• Applying Hardy Cross corrections with the wrong sign for anticlockwise flows.
• Using Dupuit's equation when the friction factors of the pipes differ.

For GATE ME

Expect: flow split between two parallel pipes, equivalent length or diameter of a series combination, head loss when a pipe is replaced by two parallel pipes (for example, with the total length or diameter changed), condition for maximum power transmission (h_f = H/3, η = 66.7 %), and the principles of Hardy Cross. Practise expressing everything in k·Q² form — it makes ratio questions one-line.

Quick check

1. In parallel pipes, which quantity is common to all branches?
2. Two identical parallel pipes replace one pipe of the same size. If f is the same, by what factor does the head loss fall for the same total Q?
3. Two parallel pipes of equal length and f have D = 0.2 m and 0.1 m. What is Q_large/Q_small?
4. What is the maximum efficiency of power transmission through a pipe?
5. Which two laws does the Hardy Cross method enforce? Answers: 1. Head loss. 2. Each carries Q/2, so h falls to 1/4. 3. 2^2.5 = 5.66. 4. 66.7 % (h_f = H/3). 5. Continuity at nodes and zero net head loss around each loop.