7.3 Pipe Flow, Friction, and Pumps

7.3 Pipe Flow, Friction, and Pumps

Quick Answer: Major loss h_f = f (L/D) (v²/(2g)) with f from Re and roughness (Moody). Minor losses h_m = K v²/(2g). Pump fluid power = ρ Q g h_p; shaft power = fluid power / η. NPSH protects against cavitation on the suction side.

Sections 7.1–7.2 classified flow and wrote head balances. This section supplies the loss terms and pump performance ideas that appear throughout plant design and on Domain C of the UPDA/MMUP Chemical exam—cooling-water loops, transfer pumps, and simple ΔP estimates.

Major Losses: Darcy–Weisbach

The standard straight-pipe head loss is:

h_f = f (L / D) (v² / (2 g))

Pressure drop for incompressible flow in a horizontal constant-diameter pipe (no fittings):

ΔP_f = ρ g h_f = f (L/D) (ρ v² / 2)

Laminar circular pipe (Hagen–Poiseuille):

f = 64 / Re (Darcy definition)

Then ΔP = (32 μ L v) / D² , equivalent to the classic laminar formula. Note f falls as 1/Re in laminar flow.

Turbulent flow: f depends on Re and relative roughness ε/D, where ε is absolute roughness of the inner wall (commercial steel, drawn tubing, concrete, etc.). The Moody chart plots f vs Re for curves of constant ε/D. Fully rough turbulent flow makes f nearly independent of Re and controlled by ε/D.

Darcy vs Fanning (Awareness)

Two friction-factor definitions circulate in chemical engineering:

Fanning appears in some ΔP = 2 f_F (L/D) ρ v² forms and in many heat-transfer texts. If you mix definitions, you are off by a factor of 4. On UPDA-style items, read whether the formula shows 64/Re (Darcy laminar) or 16/Re (Fanning laminar).

Minor Losses: Fittings and Valves

Each fitting is modeled as:

h_m = K (v² / (2 g))

sometimes based on the velocity in the pipe approaching the fitting. Total minor loss is Σ K_i v²/(2g) (with care if diameters change).

Exams rarely require memorizing a full K table; they test that fittings add head loss proportional to velocity head and that many elbows + a throttling valve can dominate a short pipe’s major loss.

Equivalent length alternative: represent fittings as extra pipe L_eq so h_f uses (L + L_eq). Same physics, different bookkeeping.

ΔP Estimation Intuition

For quick reasoning without a full Moody read:

1. Compute v and Re (Section 7.1).
2. Laminar → analytical f = 64/Re; ΔP scales with μ, L, v / D².
3. Turbulent rough → f roughly 0.02–0.05 for many plant water lines (order-of-magnitude only—use chart when data given).
4. ΔP scales like v² (hence Q²) in fully turbulent flow → doubling flow roughly quadruples frictional ΔP.
5. Small diameter is expensive: D in the denominator (and v rises as 1/D² for fixed Q, so losses skyrocket).

Worked Example: Laminar Oil Line ΔP

Problem. Oil: μ = 0.40 Pa·s, ρ = 880 kg/m³, D = 0.04 m, L = 50 m, Q = 4.0×10⁻⁴ m³/s. Horizontal pipe, no fittings. Find Re, f (Darcy), h_f, and ΔP.

Velocity: A = π(0.04)²/4 = 1.257×10⁻³ m²
v = Q/A = 4.0×10⁻⁴ / 1.257×10⁻³ ≈ 0.318 m/s

Reynolds: Re = ρvD/μ = (880)(0.318)(0.04)/0.40 ≈ 11.2/0.40 ≈ 28 → laminar.

Friction factor: f = 64/Re ≈ 64/28 ≈ 2.29

Head loss: h_f = f (L/D) v²/(2g) = 2.29 × (50/0.04) × (0.318)² / (2×9.81)
= 2.29 × 1250 × 0.101/19.62
≈ 2.29 × 1250 × 0.00515 ≈ 2.29 × 6.44 ≈ 14.7 m

Pressure drop: ΔP = ρ g h_f ≈ 880 × 9.81 × 14.7 ≈ 1.27×10⁵ Pa ≈ 1.27 bar

Check with Hagen–Poiseuille: ΔP = 32 μ L v / D² = 32(0.40)(50)(0.318)/(0.04)² ≈ 203.5 / 0.0016 ≈ 1.27×10⁵ Pa — matches.

Worked Example: Turbulent Water ΔP (Order of Magnitude)

Water, ρ = 1000 kg/m³, μ = 0.001 Pa·s, D = 0.10 m, L = 200 m, v = 2.0 m/s. Suppose Moody gives f ≈ 0.020.

Re = (1000)(2)(0.10)/0.001 = 2.0×10⁵ (turbulent).

h_f = 0.020 × (200/0.10) × (4)/(19.62) = 0.020 × 2000 × 0.204 ≈ 8.2 m
ΔP ≈ 1000 × 9.81 × 8.2 ≈ 80 kPa.

If an engineer wrongly used f = 64/Re = 64/2e5 = 3.2×10⁻⁴, h_f would be ~0.13 m — catastrophically low. Never apply laminar f in turbulent flow.

Pumps: Power and Efficiency

Fluid (hydraulic) power delivered to the liquid:

P_fluid = ρ Q g h_p = ṁ g h_p

Shaft power (what the driver must supply, ignoring driver losses):

P_shaft = P_fluid / η

where η is the pump efficiency (0 < η < 1), often 50–80% for many centrifugal services at reasonable operating points—use the value given in the stem.

Worked Example: Pump Motor Size Estimate

Q = 0.05 m³/s water, h_p = 40 m, η = 0.70, ρ = 1000 kg/m³.

P_fluid = 1000 × 0.05 × 9.81 × 40 = 19620 W ≈ 19.6 kW
P_shaft = 19.6 / 0.70 ≈ 28.0 kW

Select a motor above 28 kW after service factors—exam items usually stop at shaft power.

System curve vs pump curve (awareness): Required h_p rises with Q² when friction dominates. Centrifugal pumps deliver head that falls as Q rises. The operating point is the intersection. Throttling a discharge valve steepens the system curve (more K) and moves the operating point.

NPSH Concept (Exam Level)

Cavitation occurs when local pressure in the pump falls to vapor pressure: liquid boils, bubbles collapse downstream, damage impellers, and destroy head/capacity.

NPSH = Net Positive Suction Head.

Simplified NPSH_A for a flooded suction from an open tank:

NPSH_A ≈ (P_surface − P_vapor)/(ρ g) + z_surface_to_pump_inlet − h_suction_losses

Exam cues: “Pump is cavitating” → check suction head, temperature/vapor pressure, clogged strainer, valve nearly closed on suction, or undersized suction piping. Fix by lowering the pump, raising liquid level, cooling the fluid, shortening suction line, or selecting a pump with lower NPSH_R.

Centrifugal vs Positive-Displacement (Brief)

Do not dead-head a PD pump without relief; centrifugals at shutoff still have limits (heat, seal issues) but behave differently.

UPDA Chemical Exam Workflow

1. Find v, Re, regime → choose laminar f = 64/Re or turbulent Moody f.
2. Confirm Darcy vs Fanning if a formula is supplied.
3. Compute h_f and h_m; sum into mechanical energy balance.
4. Solve h_p, then P_fluid and P_shaft with η.
5. For suction problems, compare NPSH_A qualitative factors to cavitation risk.

Common Traps

• Using f = 64/Re for Re = 10⁵ water flow.
• Off-by-four error between Darcy and Fanning.
• Reporting fluid power when the question asks for brake/shaft power (missing 1/η).
• Ignoring minor losses when the sketch is all elbows and a globe valve on a short skid.
• Sizing suction lines too small → low NPSH_A and chronic cavitation.
• Confusing pump head (m) with pressure rise (Pa) without ρ g.

With properties, Re, Bernoulli/mechanical energy, friction, and pumps in place, Domain C fluid mechanics connects cleanly to heat transfer (Chapter 8), where Re again appears inside convection correlations.