Optimum design: economic pipe diameter

Why it matters

Piping can be a large share of a plant's fixed capital, and pumping power runs every hour of the plant's life. A pipe that is too small is cheap to buy but expensive to pump through; one that is too large wastes capital. The economic pipe diameter is the classic example of design optimisation: write the total annual cost as a function of one variable, and find its minimum.

Key ideas

The trade-off. For a fixed flow rate:

• Fixed charges on the pipe (depreciation, maintenance, interest, taxes on the installed cost of pipe, fittings and supports) rise with diameter. Installed cost per metre is often fitted as a power law in D with an exponent n of roughly 1 to 1.5.
• Pumping cost falls very steeply with diameter. At a fixed volumetric flow, velocity varies as 1/D², and the Darcy–Weisbach pressure drop varies as f/D⁵. With f constant (fully rough flow) pumping cost ∝ D⁻⁵; with f ∝ Re^(−0.2) (smooth turbulent flow) it is ∝ D^(−4.8). The sum of a rising term and a falling term has a minimum — the economic (optimum) diameter.

The general result. If the annual cost per metre is C_T = A·Dⁿ + B·D^(−m), setting dC_T/dD = 0 gives the optimum. At the optimum the ratio of pumping cost to fixed charges equals n/m — a quick check on any answer. Because the curve is flat near the minimum, choosing the next standard pipe size above the calculated diameter costs little and adds margin for future capacity.

Rule-of-thumb correlations. Substituting typical values of pipe cost, power price, fixed-charge rate and pump efficiency gives correlations of the form D_opt = K·Q^0.45·ρ^0.13 for turbulent flow (viscosity has only a tiny effect, exponent about 0.02–0.03), and a different form with viscosity for laminar flow. Peters and Timmerhaus give K ≈ 0.363 in SI units for steel pipe; other sources differ slightly because their cost bases differ. Use them for quick estimates and take the constant from your data book. The exponents follow from the D-dependence above: with n = 1.5 and m = 4.8, Q appears as Q^(2.8/6.3) = Q^0.44 and ρ as ρ^(0.8/6.3) = ρ^0.13.

Economic velocity. The optimum diameter corresponds to typical velocities: roughly 1–3 m/s for water-like liquids in process lines, lower for viscous liquids and for pump suctions, and roughly 10–30 m/s for gases and vapours at moderate pressure. These ranges are guides; erosion, noise, pressure-drop limits and NPSH can override economics.

What the analysis assumes. Steady flow at the design rate for a known number of hours per year; a fixed electricity price; pump and motor efficiency independent of size; no change in pump capital cost with pressure drop (it can be added as another term). Pipe must also meet pressure rating and corrosion allowance, which set wall thickness independently.

Formulas

Pressure drop: ΔP = f·(L/D)·(ρ·v²/2), with v = 4Q/(π·D²), so ΔP = 8·f·L·ρ·Q² / (π²·D⁵) Pumping power: P = Q·ΔP / η Annual pumping cost: C_p = P·H·c_e Total annual cost per metre: C_T = A·Dⁿ + B·D^(−m) Optimum: D_opt = [m·B / (n·A)]^(1/(n + m)) At the optimum: (B·D^(−m)) / (A·Dⁿ) = n/m Turbulent-flow correlation (steel pipe, SI): D_opt ≈ 0.363·Q^0.45·ρ^0.13

Symbols: D inside diameter (m); L length (m); f Darcy friction factor (–); ρ density (kg/m³); v velocity (m/s); Q volumetric flow (m³/s); ΔP pressure drop (Pa); η pump-and-motor efficiency (–); P power (W); H operating hours per year (h/yr); c_e electricity price (₹/kWh, with P in kW); A coefficient of the fixed-charge term (₹ per m length per yr per mⁿ); B coefficient of the pumping-cost term (₹·m^m per m length per yr); n, m exponents (–).

Worked examples

Example 1 (standard). Annual cost per metre of pipe is C_T = 1200·D + 0.002/D⁵ ₹/yr (D in m). Find the optimum diameter and the costs there.

1. n = 1, m = 5. D_opt = [5 × 0.002/(1 × 1200)]^(1/6) = (8.333 × 10⁻⁶)^(1/6).
2. ln(8.333 × 10⁻⁶) = −11.695; ÷ 6 = −1.949; e^(−1.949) = 0.1424.
3. Fixed charges = 1200 × 0.1424 = ₹170.9/yr; pumping = 0.002/0.1424⁵ = ₹34.2/yr; total ₹205.0 per metre per year.
4. Check: 34.2/170.9 = 0.20 = n/m = 1/5. ✓
5. D_opt ≈ 0.142 m (142 mm); minimum cost ≈ ₹205 per metre per year.

Example 2 (GATE level). Water (ρ = 1000 kg/m³) flows at Q = 0.05 m³/s. Take f = 0.02 (constant), pump-and-motor efficiency 0.70, 8000 h/yr, electricity ₹8/kWh. Installed pipe cost is ₹60 000·D^1.5 per metre (D in m), and annual fixed charges are 20% of installed cost. Find the economic diameter and velocity.

1. Fixed-charge term: A·D^1.5 with A = 0.20 × 60 000 = 12 000.
2. Power per metre: P = 8·f·ρ·Q³/(π²·η·D⁵) = 8 × 0.02 × 1000 × 1.25 × 10⁻⁴ /(9.8696 × 0.70 × D⁵) = 0.002895/D⁵ W per m.
3. Annual pumping cost per metre = (0.002895/1000 kW) × 8000 × 8 /D⁵ = 0.18527/D⁵ ₹/yr, so B = 0.18527, m = 5.
4. D_opt = [5 × 0.18527/(1.5 × 12 000)]^(1/6.5) = (5.146 × 10⁻⁵)^(0.1538) = 0.219 m.
5. v = Q/(πD²/4) = 0.05/(0.7854 × 0.0479) = 1.33 m/s — inside the usual range for water.
6. Check: pumping ₹368.7, fixed ₹1228.9 per m per yr; ratio 0.30 = 1.5/5. ✓
7. D_opt ≈ 0.22 m, v ≈ 1.3 m/s. (The correlation 0.363 × 0.05^0.45 × 1000^0.13 gives 0.23 m, close.)

Common mistakes

• Writing pressure drop as ∝ 1/D instead of ∝ 1/D⁵ at constant flow (velocity also depends on D).
• Treating ΔP × cost-per-pascal as an annual cost without power, hours and efficiency.
• Forgetting pump efficiency, or using hours per year inconsistently with kW.
• Using the total installed cost instead of the annual fixed charges in the cost equation.
• Not checking the answer against typical economic velocities.

For GATE CH

Expect optimisation of a two-term cost function (one rising, one falling with D) by differentiation, the scaling of pumping power with diameter at fixed flow (∝ D⁻⁵), and the effect on optimum diameter of a change in flow rate or energy price. Practise fractional powers and the n/m ratio check.

Quick check

1. At fixed flow and constant f, by what factor does pumping power change if D is halved?
2. C_T = 500·D + 0.01/D⁴. D_opt?
3. If the flow rate doubles, roughly how does the economic diameter change (D ∝ Q^0.45)?
4. Why is choosing the next larger standard size usually acceptable?

Answers: 1. It increases 2⁵ = 32 times. 2. D = (4 × 0.01/500)^(1/5) = (8 × 10⁻⁵)^(0.2) = 0.152 m. 3. It increases by 2^0.45 = 1.37 times. 4. The total-cost curve is flat near the minimum, and the larger pipe gives capacity margin.