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Air compressors

Mechanical Engineering (B.E./B.Tech) · Thermodynamics and thermal engineering · free study notes for exams, placements and interviews

In short

An air compressor raises the pressure of atmospheric air by doing mechanical work on it, so that the stored energy can later drive tools, actuators, valves, process nozzles or a gas-turbine combustor. Compressed air is the costliest common utility in an Indian factory — roughly 5.5 to 7 kW of shaft power for every m³/min of free air at 7 bar — so compressor selection, staging and leak control are directly money questions. In university exams and GATE the topic is examined as work-done integrals on the p–V diagram, volumetric efficiency with clearance, optimum multistage pressure and intercooling, and the surging/choking behaviour of dynamic machines. Interviews at Elgi, Kirloskar Pneumatic, Atlas Copco, Ingersoll Rand, Siemens Energy, Thermax, cement and steel plants, and railway workshops lean on the same material plus compressed-air system practice.

Explained

Classification and basic terms

Reciprocating compressor: indicator diagram and work

Volumetric efficiency, other efficiencies and capacity control

Multistage compression and intercooling

Rotary positive-displacement compressors

Dynamic compressors and their instabilities

System accessories, Indian codes and energy practice

Key points

Formulas and examples

Swept volume (single-acting, one cylinder): Vs = (π/4)·d²·L — d = bore (m), L = stroke (m), Vs in m³/cycle. Cycles per minute = N (single-acting) or 2N (double-acting), N in rpm.
Clearance ratio: C = Vc/Vs. Volumetric efficiency: η_vol = 1 + C − C·(p₂/p₁)1/n
Polytropic work per cycle: W = [n/(n−1)]·p₁V₁·[(p₂/p₁)(n−1)/n − 1] — p in Pa, V₁ = induced volume (m³), W in J
Isothermal work: W_iso = p₁V₁·ln(p₂/p₁); Isentropic: replace n by γ = 1.4
Indicated power = W × (cycles per minute)/60, in W. Mean effective pressure p_m = W / Vs (Pa)
Delivery temperature: T₂ = T₁·(p₂/p₁)(n−1)/n, T in K
Isothermal efficiency η_iso = W_iso / W_indicated; Mechanical efficiency = indicated power / shaft power
Minimum work, z stages, equal ratios, perfect intercooling: W = z·[n/(n−1)]·p₁V₁·[(p_f/p₁)(n−1)/(z·n) − 1]; ideal stage ratio = (p_f/p₁)1/z
Mass flow: ṁ = p₁V̇₁/(R T₁), R = 287 J/kg·K for air. Intercooler heat = ṁ·c_p·(T₂ − T₁), c_p = 1.005 kJ/kg·K
Centrifugal work input: W = ψ·σ·u₂² (J/kg), u₂ = πD₂N/60; pressure ratio = [1 + η_c·ΔT₀/T₁]γ/(γ−1)
Receiver pump-up test: FAD = V_receiver·(p₂ − p₁)/(p_atm·t) — all pressures absolute, same units; V in m³, t in min
Receiver size for a peak draw: V = Q·t·p_atm/(p_start − p_stop), Q = demand (m³/min FAD), t = duration (min)

Example 1 — single-stage reciprocating compressor with clearance. Single-acting, bore 150 mm, stroke 200 mm, N = 300 rpm, p₁ = 1 bar, T₁ = 300 K, p₂ = 7 bar, n = 1.3, C = 5%.

  1. Vs = 0.7854 × 0.15² × 0.20 = 0.003534 m³ (3.534 litres).
  2. 71/1.3 = 70.7692 = 4.468 → η_vol = 1 + 0.05 − 0.05 × 4.468 = 0.827 (82.7%).
  3. Induced volume V₁ = 0.827 × 0.003534 = 0.002922 m³/cycle → FAD = 0.002922 × 300 = 0.877 m³/min (at 1 bar, 300 K).
  4. 70.3/1.3 = 70.2308 = 1.567 → W = (1.3/0.3) × 10⁵ × 0.002922 × (1.567 − 1) = 4.333 × 10⁵ × 0.002922 × 0.567 = 717 J/cycle.
  5. Indicated power = 717 × 300/60 = 3.59 kW. MEP = 717/0.003534 = 2.03 × 10⁵ Pa = 2.03 bar.
  6. W_iso = 10⁵ × 0.002922 × ln 7 = 292.2 × 1.946 = 569 J → 2.84 kW → η_iso = 2.84/3.59 = 79.2%.
  7. T₂ = 300 × 1.567 = 470 K = 197 °C — already high enough to justify two stages above this ratio.

Example 2 — two-stage with perfect intercooling versus single-stage. FAD 10 m³/min at 1 bar, 300 K, delivered at 36 bar, n = 1.3.

  1. Ideal intermediate pressure = √(1 × 36) = 6 bar, so each stage ratio = 6. 60.2308 = 1.512.
  2. p₁V̇₁ = 10⁵ × 10 = 10⁶ J/min. W = 2 × 4.333 × 10⁶ × 0.512 = 4.438 × 10⁶ J/min → 74.0 kW.
  3. Single-stage: 360.2308 = 2.286 → W = 4.333 × 10⁶ × 1.286 = 5.574 × 10⁶ J/min → 92.9 kW.
  4. Saving = 18.9 kW, i.e. 20.4% — and the delivery temperature per stage drops from 300 × 2.286 = 686 K (413 °C, which would char the oil) to 300 × 1.512 = 454 K (181 °C).
  5. Intercooler duty: ṁ = 10⁵ × 10/(287 × 300) = 11.61 kg/min; Q = 11.61 × 1.005 × (454 − 300) = 1,797 kJ/min = 29.9 kW of heat to be removed.

Example 3 — measuring FAD by a receiver pump-up test. A 1.5 m³ receiver, initially at atmospheric 1 bar abs, is pumped to 7 bar abs in 2 minutes; temperature is allowed to settle back to ambient before reading. FAD = 1.5 × (7 − 1)/(1 × 2) = 4.5 m³/min. If the nameplate says 5.2 m³/min, the machine has lost about 13% of capacity — look for worn valves, rings, a choked intake filter or a leaking intercooler. (The test assumes isothermal conditions, so take the final pressure after the air has cooled, not at cut-off.)

Example 4 — what leaks and over-pressure cost, in rupees. A plant runs a 75 kW screw compressor 6,000 h a year at an average 70% load; electricity ₹8/kWh.

  1. Energy = 75 × 0.70 × 6,000 = 315,000 kWh/yr → ₹25.2 lakh/yr.
  2. An ultrasonic survey finds leakage equal to 25% of generation. Halving it removes 12.5% of the flow, and with load/unload or VSD control the power follows roughly in proportion: 0.125 × 315,000 = 39,375 kWh ≈ ₹3.15 lakh/yr saved.
  3. Reducing the header setting from 7.0 to 6.0 bar (if every user can work on 6) saves about 7%: 22,050 kWh ≈ ₹1.76 lakh/yr.
  4. Scale of one defect: a single 3 mm hole at 7 bar gauge passes about 0.4 m³/min FAD; at 6 kW per m³/min that is 2.4 kW, or 19,200 kWh and about ₹1.54 lakh a year at 8,000 h. One unrepaired drain valve pays for a leak-survey contract.

Example 5 — centrifugal stage pressure ratio. Impeller tip diameter 0.50 m, 18,000 rpm, radial vanes, slip factor 0.90, power input factor 1.04, inlet 293 K, isentropic efficiency 0.80.

  1. u₂ = π × 0.50 × 18,000/60 = 471.2 m/s.
  2. W = 1.04 × 0.90 × 471.2² = 0.936 × 222,070 = 207.9 kJ/kg → ΔT₀ = 207,900/1,005 = 206.9 K.
  3. Pressure ratio = (1 + 0.80 × 206.9/293)3.5 = (1.565)3.5 = 4.79 — the kind of single-stage ratio an aero-derivative or high-speed process centrifugal reaches; industrial oil-free centrifugals use several lower-ratio stages with intercooling.
ReciprocatingScrew (rotary)Centrifugal
PrinciplePositive displacement, pistonPositive displacement, meshing rotorsDynamic, impeller + diffuser
Flow rangeSmall to mediumSmall to largeLarge only
PressureLow to very high (300 bar+)Up to about 13 bar per package2–8 bar per stage, multistage higher
Flow steadinessPulsating, needs receiverNearly steadySteady
Efficiency at part loadGood with unloadersGood only with VSDPoor below surge margin
MaintenanceValves, rings, more wear partsLow, but oil and element changesLow, but high-speed bearings
Oil-free optionYes (dry, labyrinth)Yes (dry screw, Class 0)Inherently oil-free air path
Typical Indian useWorkshops, PET blowing, breathing air, loco brakesGeneral plant air, textiles, pharmaSteel, cement, large refineries, air separation

Where it is used

Interview questions

Common mistakes

Related topics

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