Centrifugal and axial compressors; turbochargers

Why it matters

Almost every modern diesel and downsized petrol engine is turbocharged. A small centrifugal compressor driven by an exhaust-gas turbine packs more air into the cylinders, so a smaller engine gives the same power with better fuel economy. Axial compressors do the same job at much larger flows in gas turbines and jet engines. Understanding compressor work, efficiency, surge and choke — and the turbine–compressor power balance — is essential for engine performance work and for GATE thermal-fluids questions.

Key ideas

• Compressor types.
  • Centrifugal (radial): air enters axially at the eye, is whirled outward by the impeller, and is diffused in a vaneless or vaned diffuser and volute. A high pressure ratio per stage (about 3–5 in turbochargers), compact, robust and wide-ranging — ideal for turbochargers and small gas turbines.
  • Axial: many stages of rotor blades followed by stator blades, each giving a modest pressure ratio (about 1.15–1.3 per stage). A high mass flow per frontal area and high efficiency — used in large gas turbines and aero-engines.
• Stagnation (total) properties. Gas velocities are large, so use stagnation temperature T₀ = T + V²/(2c_p) and stagnation pressure p₀. The steady-flow energy equation for an adiabatic compressor gives work per unit mass w = c_p(T₀₂ − T₀₁).
• Isentropic (total-to-total) efficiency. η_c = ideal temperature rise / actual temperature rise, where the ideal rise reaches the same pressure ratio isentropically. A real compressor heats the air more than ideal, which is one reason for intercooling.
• Centrifugal impeller work. With axial (whirl-free) entry, the Euler work is u₂·V_w2. With radial vanes at the tip, the ideal V_w2 equals u₂, but the finite number of vanes gives slip, so V_w2 = σ·u₂ (slip factor σ ≈ 0.85–0.92). A power input factor ψ ≈ 1.03–1.06 accounts for disc friction and windage. So w = ψ·σ·u₂². Tip speed is limited by impeller stress (roughly 450–550 m/s for aluminium turbocharger wheels; check material data).
• Axial stage. A rotor adds whirl and so does work; the following stator removes the whirl and diffuses the flow. With constant axial velocity V_a and blade speed u, the Euler work is u·V_a·(tan β₁ − tan β₂), with relative angles β measured from the axial direction. A work-done factor λ (about 0.85–0.98, falling with stage number) allows for the non-uniform axial-velocity profile.
• Degree of reaction R = enthalpy rise in the rotor / enthalpy rise in the stage. A 50 % reaction stage has symmetrical velocity triangles (α₁ = β₂, α₂ = β₁) and shares diffusion equally between rotor and stator, which is good for efficiency.
• Characteristics and limits. Plotted as pressure ratio against corrected mass flow m·√T₀₁/p₀₁ for lines of corrected speed N/√T₀₁.
  • Surge (left limit): at low flow and high pressure ratio, the flow separates in the diffuser or blades and the system oscillates violently with flow reversal. It is dangerous to the machine and must be avoided.
  • Rotating stall in axial compressors: stalled cells rotate around the annulus — often a precursor of surge.
  • Choke (right limit): somewhere in the passage the flow reaches sonic speed, so mass flow cannot increase further.
• Turbocharging.
  • Exhaust gas drives a radial-inflow turbine on a common shaft with the compressor. In steady running, turbine power × mechanical efficiency = compressor power.
  • Boost raises intake density, so more fuel can be burnt per cycle. An intercooler (charge-air cooler) removes the compression heat, raising density further and reducing knock (petrol) and NOx (diesel).
  • A wastegate bypasses exhaust around the turbine to limit boost at high speed. A variable-geometry turbine (VGT) changes nozzle-vane angle to give good boost at low engine speed and avoid over-boost at high speed.
  • Turbo lag: delay in boost because the rotor must accelerate and exhaust energy must build up. It is reduced with small low-inertia rotors, VGT, twin-scroll housings, two-stage or electrically assisted turbochargers.
  • Matching: the engine's air-flow line must lie in the compressor map's high-efficiency region with adequate surge margin at low speed and full load, and away from choke at high speed.

Formulas

T₀ = T + V²/(2·c_p) Stagnation temperature (K); c_p ≈ 1005 J/(kg·K) for air.

w_c = c_p·(T₀₂ − T₀₁) P_c = ṁ·w_c Compressor work (J/kg) and power (W); ṁ mass flow (kg/s).

T₀₂s/T₀₁ = (p₀₂/p₀₁)^((γ−1)/γ) η_c = (T₀₂s − T₀₁)/(T₀₂ − T₀₁) Isentropic temperature and efficiency; γ = 1.4 for air.

p₀₂/p₀₁ = [1 + η_c·(T₀₂ − T₀₁)/T₀₁]^(γ/(γ−1)) Pressure ratio from the actual temperature rise.

w = ψ·σ·u₂² u₂ = π·D₂·N/60 Centrifugal compressor work with axial entry; ψ power input factor, σ slip factor, D₂ tip diameter (m), N (rpm).

w = λ·u·V_a·(tan β₁ − tan β₂) Axial stage work (J/kg); λ work-done factor; β₁, β₂ rotor relative inlet and outlet angles from axial.

R = V_a·(tan β₁ + tan β₂)/(2u) Degree of reaction for an axial stage with constant V_a.

η_m·ṁ_t·c_pg·(T₀₃ − T₀₄) = ṁ_c·c_p·(T₀₂ − T₀₁) Turbocharger power balance; subscript t turbine (exhaust gas, c_pg ≈ 1150 J/(kg·K), γ ≈ 1.33), η_m mechanical efficiency.

Worked examples

Example 1 (standard) — turbocharger compressor. Given: tip diameter D₂ = 80 mm, N = 100 000 rpm, slip factor 0.9, power input factor 1.04, isentropic efficiency 0.75, inlet T₀₁ = 300 K, p₀₁ = 1 bar, ṁ = 0.15 kg/s, axial entry. Find the temperature rise, pressure ratio and power.

1. u₂ = π × 0.08 × 100 000/60 = 418.9 m/s.
2. w = ψσu₂² = 1.04 × 0.9 × 418.9² = 164 230 J/kg.
3. ΔT₀ = w/c_p = 164 230/1005 = 163.4 K → T₀₂ = 463.4 K.
4. p₀₂/p₀₁ = (1 + 0.75 × 163.4/300)^3.5 = (1.4085)^3.5 = 3.32.
5. P_c = ṁ·w = 0.15 × 164 230 = 24 630 W. Answer: ΔT₀ ≈ 163 K, pressure ratio ≈ 3.32, power ≈ 24.6 kW. The air leaves at about 190 °C — hence the intercooler.

Example 2 (GATE level) — axial stage and turbine matching. Given: (a) an axial compressor stage has mean blade speed 220 m/s, axial velocity 160 m/s, rotor relative angles β₁ = 50° and β₂ = 30° from axial, work-done factor 0.88, stage efficiency 0.88, T₀₁ = 300 K. Find the stage work, pressure ratio and degree of reaction. (b) The turbine driving the compressor of Example 1 receives 0.155 kg/s of exhaust at T₀₃ = 900 K (c_pg = 1150 J/(kg·K), γ = 1.33), with η_t = 0.70 and η_m = 0.95. Find the turbine outlet temperature and expansion ratio.

1. (a) w = λ·u·V_a·(tan 50° − tan 30°) = 0.88 × 220 × 160 × (1.1918 − 0.5774) = 0.88 × 21 627 = 19 032 J/kg.
2. ΔT₀ = 19 032/1005 = 18.94 K.
3. Pressure ratio = (1 + 0.88 × 18.94/300)^3.5 = 1.208.
4. R = 160 × (1.1918 + 0.5774)/(2 × 220) = 0.643.
5. (b) Turbine temperature drop: ΔT₀,t = 24 630/(0.95 × 0.155 × 1150) = 145.5 K → T₀₄ = 754.5 K.
6. Isentropic drop = 145.5/0.70 = 207.8 K → T₀₄s = 692.2 K.
7. Expansion ratio = (900/692.2)^(1.33/0.33) = (1.3002)^4.03 = 2.88. Answer: (a) w ≈ 19.0 kJ/kg, stage pressure ratio ≈ 1.21, R ≈ 0.64; (b) T₀₄ ≈ 755 K, turbine expansion ratio ≈ 2.88.

Common mistakes

• Using static instead of stagnation temperatures and pressures in efficiency and work.
• Inverting the isentropic-efficiency definition (compressor: ideal/actual; turbine: actual/ideal).
• Using γ = 1.4 and c_p = 1005 J/(kg·K) for hot exhaust gas.
• Forgetting the slip factor, or forgetting to convert rpm to m/s at the tip.
• Measuring axial-compressor blade angles from the tangential direction while using formulas written for axial angles.
• Calling surge and choke the same thing — surge is the low-flow instability, choke is the high-flow sonic limit.

For GATE ME

Expect: compressor work and power from c_pΔT₀, isentropic efficiency and outlet temperature, pressure ratio from the temperature rise, centrifugal impeller work with slip, axial-stage work and degree of reaction from velocity triangles, and conceptual questions on surge, choke, intercooling and turbocharger power balance. Practise the isentropic relation T₂/T₁ = (p₂/p₁)^((γ−1)/γ) until it is automatic.

Quick check

1. What is the work per kg for an air compressor with T₀₁ = 300 K and T₀₂ = 450 K?
2. Which compressor limit is caused by sonic flow in the passages?
3. What does a slip factor of 0.9 mean?
4. Why is an intercooler used after a turbocharger compressor?
5. For a 50 % reaction axial stage, which angles are equal? Answers: 1. 1005 × 150 = 150.75 kJ/kg. 2. Choke. 3. The actual whirl at the impeller tip is 90 % of the blade speed (radial vanes). 4. To cool the compressed air, raising its density and reducing knock and NOx. 5. α₁ = β₂ and α₂ = β₁ (symmetrical triangles).