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Fluid flow and pumps in process plants

Chemical Engineering (B.E./B.Tech) · Fluid and particle mechanics · free study notes for exams, placements and interviews

In short

Fluid flow and pumps covers how liquids and gases move through process piping, how much pressure is lost doing it, and how pumps and compressors supply that pressure. It rests on three balances — mass (continuity), mechanical energy (Bernoulli with friction) and momentum — plus the Darcy-Weisbach friction equation and pump characteristic curves. Every chemical plant, water utility, refinery and pharma unit sizes pipes, selects pumps and solves cavitation and flow-measurement problems with exactly this material. It is heavily examined in GATE CH (Fluid Mechanics and Mechanical Operations section), in university Fluid Mechanics / Unit Operations-I papers, and in technical rounds at Reliance, IOCL, BPCL, ONGC, L&T Hydrocarbon, Tata Chemicals, UPL, Aarti Industries, Sun Pharma, Thermax, KSB, Kirloskar Brothers and engineering consultants such as Jacobs and Toyo.

Explained

Fluid properties and flow regimes

Continuity, Bernoulli and the mechanical energy balance

Friction losses in pipes and fittings

Pumps: types, selection and characteristics

NPSH and cavitation

Flow measurement and other devices

Indian standards, codes and practice

Key points

Formulas and examples

Continuity: ṁ = ρAu (kg/s); Q = Au (m³/s); A = πD²/4
Reynolds number: Re = ρuD/μ = uD/ν (dimensionless)
Bernoulli with work and friction: p₁/ρg + α u₁²/2g + z₁ + h_pump = p₂/ρg + α u₂²/2g + z₂ + h_f
Darcy-Weisbach: h_f = f (L/D)(u²/2g) [m of fluid]; Δp = ρ g h_f (Pa)
Laminar: f = 64/Re; Hagen-Poiseuille Δp = 32 μ L u / D² ; Q = π D⁴ Δp /(128 μ L)
Blasius (smooth, turbulent): f = 0.316 / Re0.25
Minor loss: h = K u²/2g ; equivalent length h_f = f (L_e/D)(u²/2g)
Pump total head: H = (p₂ − p₁)/ρg + Δz + h_f,total [m]
Hydraulic power: P_h = ρ g Q H (W); Shaft power P_s = ρ g Q H / η ; Motor input = P_s / η_motor
NPSHa = (p_s − p_v)/ρg + z_s − h_f,s [m]; require NPSHa ≥ NPSHr + 0.5–1 m
Affinity laws: Q₂/Q₁ = N₂/N₁ ; H₂/H₁ = (N₂/N₁)² ; P₂/P₁ = (N₂/N₁)³
Specific speed (pumps): Ns = N √Q / H0.75 (units must be stated)
Orifice/venturi: Q = C_d A₂ √[2Δp/(ρ(1 − β⁴))], β = d₂/d₁
Joukowsky surge: Δp = ρ a Δu, a = wave speed (≈1,200–1,400 m/s in water)

Example 1 — Regime, friction loss and pump power for a water line Water (ρ = 1,000 kg/m³, μ = 1×10⁻³ Pa·s) flows at 20 m³/h through 150 m of 50 mm internal-diameter commercial steel pipe (ε = 0.045 mm), with 6 elbows (K = 0.75 each) and 2 gate valves (K = 0.17 each). Pump efficiency 70%, motor efficiency 92%. Static lift 12 m, both ends open to atmosphere.

  1. Q = 20/3,600 = 5.556×10⁻³ m³/s. A = π(0.05)²/4 = 1.963×10⁻³ m². u = 5.556×10⁻³/1.963×10⁻³ = 2.83 m/s (within 1.5–3 m/s design range).
  2. Re = 1,000 × 2.83 × 0.05 / 1×10⁻³ = 1.42×10⁵ → turbulent.
  3. ε/D = 0.045/50 = 0.0009. Moody chart gives f ≈ 0.0215.
  4. u²/2g = 2.83²/(2×9.81) = 8.01/19.62 = 0.408 m.
  5. Straight pipe: h_f = 0.0215 × (150/0.05) × 0.408 = 0.0215 × 3,000 × 0.408 = 26.3 m.
  6. Fittings: ΣK = 6(0.75) + 2(0.17) = 4.84 → h = 4.84 × 0.408 = 1.98 m.
  7. Total head H = 12 + 26.3 + 1.98 = 40.3 m.
  8. Hydraulic power = 1,000 × 9.81 × 5.556×10⁻³ × 40.3 = 2,196 W ≈ 2.20 kW.
  9. Shaft power = 2.20/0.70 = 3.14 kW; motor input = 3.14/0.92 = 3.41 kW → select a 3.7 kW (5 HP) motor.
  10. Annual energy at 8,000 h and ₹8/kWh: 3.41 × 8,000 × 8 = ₹2.18 lakh/year — which is why pipe size and efficiency matter.

Example 2 — Laminar flow of a viscous oil Oil (ρ = 900 kg/m³, μ = 0.4 Pa·s) flows at 0.5 m/s in a 40 mm pipe, 60 m long.

  1. Re = 900 × 0.5 × 0.04 / 0.4 = 45 → strongly laminar.
  2. f = 64/45 = 1.422.
  3. u²/2g = 0.25/19.62 = 0.01274 m.
  4. h_f = 1.422 × (60/0.04) × 0.01274 = 1.422 × 1,500 × 0.01274 = 27.2 m of oil.
  5. Check with Hagen-Poiseuille: Δp = 32 × 0.4 × 60 × 0.5 / 0.04² = 384/0.0016 = 2.40×10⁵ Pa; as head = 2.40×10⁵/(900 × 9.81) = 27.2 m — agrees.
  6. Centreline velocity = 2 × 0.5 = 1.0 m/s; a pitot at the axis would read twice the mean.

Example 3 — NPSH check on a hot-water pump Water at 80 °C (p_vap = 47.4 kPa, ρ = 972 kg/m³) is drawn from an atmospheric tank (101.3 kPa) whose level is 2.0 m above the pump centreline. Suction friction loss is 0.8 m. The pump's NPSHr at duty is 3.5 m.

  1. (p_s − p_v)/ρg = (101,300 − 47,400)/(972 × 9.81) = 53,900/9,535 = 5.65 m.
  2. NPSHa = 5.65 + 2.0 − 0.8 = 6.85 m.
  3. Margin = 6.85 − 3.5 = 3.35 m → safe.
  4. If the same water were at 100 °C (p_vap = 101.3 kPa), the pressure term becomes zero and NPSHa = 0 + 2.0 − 0.8 = 1.2 m < 3.5 m → the pump would cavitate; raise the tank or choose a low-NPSHr pump.

Example 4 — Affinity laws and VFD saving A pump delivers 100 m³/h at 50 m head drawing 20 kW at 1,450 rpm. Demand falls to 80 m³/h.

  1. Speed needed: N₂ = 1,450 × (80/100) = 1,160 rpm.
  2. New head = 50 × (1,160/1,450)² = 50 × 0.64 = 32 m.
  3. New power = 20 × (0.8)³ = 20 × 0.512 = 10.24 kW → saving 9.76 kW.
  4. By throttling instead, head would stay near 50 m plus valve loss and power would fall only to about 17–18 kW — the VFD saves roughly 7–8 kW, about ₹4.7 lakh/year at 8,000 h and ₹8/kWh (taking a 7.3 kW average saving). Note the affinity laws assume a purely frictional system curve; with static head the saving is smaller.

Example 5 — Orifice meter An orifice (d₂ = 30 mm) in a 60 mm pipe carries water (ρ = 1,000 kg/m³); manometer Δp = 25 kPa; C_d = 0.61.

  1. β = 30/60 = 0.5; β⁴ = 0.0625; 1 − β⁴ = 0.9375.
  2. A₂ = π(0.03)²/4 = 7.069×10⁻⁴ m².
  3. 2Δp/(ρ(1−β⁴)) = 50,000/937.5 = 53.33 → √ = 7.303 m/s.
  4. Q = 0.61 × 7.069×10⁻⁴ × 7.303 = 3.149×10⁻³ m³/s = 11.3 m³/h.
  5. A venturi of the same bore and Δp with C_d = 0.98 would read 0.98/0.61 = 1.61× this flow, i.e. 18.2 m³/h — the C_d difference reflects the vena contracta and losses in the orifice.
FeatureCentrifugal pumpReciprocating (PD)Rotary (gear/screw)
Flow vs headFalls as head risesAlmost constantAlmost constant
Flow natureSmoothPulsating (needs damper)Fairly smooth
Viscosity limitPoor above ~500 cPHandles highBest for very viscous
Head/pressureLow-moderate (multistage for high)Very highModerate-high
Flow rateLargeSmall-moderateSmall-moderate
Self-primingNo (needs flooded suction)YesYes
Discharge throttlingPermittedForbidden - relief valve neededForbidden
Cost and maintenanceLow, simpleHigh, many wear partsModerate
Typical dutyWater, process liquids, coolingDosing, high-pressure injectionOils, polymers, molasses
MeterC_d / accuracyPermanent lossTurndownNote
Orifice~0.61, ±1–2%High~3:1Cheap, easy to replace
Venturi~0.98, ±0.5–1%Low~3:1Costly, for slurries and big lines
Rotameter±2%Moderate10:1Vertical, visual, local indication
Electromagnetic±0.5%Nil20:1Conductive liquids only
Coriolis±0.1%Low50:1Direct mass flow, expensive

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Related topics

Dimensional analysis and similitude · Flow through packed and fluidised beds · Fluid machinery: turbines, fans and compressors · Heat transfer in process equipment · Process instrumentation and control valves · Piping design and plant utilities

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