Turbo Machinery

Why it matters

Turbo machinery is crucial in various industries, including power generation, aviation, and manufacturing. Understanding how these machines work helps in designing efficient systems for energy conversion and propulsion.

Key ideas

• Turbo Machinery: Machines that transfer energy between a rotor and a fluid. They include turbines, compressors, and pumps.
• Turbines: Convert fluid energy into mechanical energy. Common types include steam turbines and gas turbines.
• Compressors: Transfer shaft work to a gas to raise its stagnation enthalpy and pressure. Dynamic compressors use rotating blades and diffusion; they are not simply shrinking containers. Types include axial and centrifugal compressors.
• Pumps: Move fluids by mechanical action, typically converting electrical energy into hydraulic energy.
• Performance Parameters: Efficiency, pressure ratio, and flow rate are key performance indicators.

Formulas

• η = (Output Power / Input Power) × 100

  • η: Efficiency (percentage)
  • Output Power: Power delivered by the machine (W)
  • Input Power: Power supplied to the machine (W)

• P = ρ × g × H × Q

  • P: Power (W)
  • ρ: Density of fluid (kg/m³)
  • g: Acceleration due to gravity (9.81 m/s²)
  • H: Head (m)
  • Q: Flow rate (m³/s)

For an ideal rotor, Euler specific work added to the fluid is w = U₂Vθ₂ - U₁Vθ₁, where U is blade speed and Vθ is the absolute tangential velocity component, using a consistent rotation sign convention. Turbine work out has the opposite sign. Velocity triangles relate absolute, relative and blade velocities.

For an incompressible pump, ρgHQ is hydraulic output power. Shaft input is ρgHQ/η_p with η_p as a fraction, not a percentage; motor efficiency is separate. For a turbine, shaft output is η_tρgHQ. H is total head, which may include pressure, elevation, velocity and system losses.

Worked example

Given: A pump supplies a total head of 20 m at a flow rate of 0.05 m³/s. The density of water is 1000 kg/m³.

1. Calculate the hydraulic power delivered to the water.

   Formula: P = ρ × g × H × Q

   Substitute the values:

   P = 1000 kg/m³ × 9.81 m/s² × 20 m × 0.05 m³/s

   P = 9810 W

   Answer: 9810 W hydraulic power. If pump efficiency is specified as 80%, required shaft input is 9810/0.8 = 12,262.5 W.

Common mistakes

• Confusing the types of turbo machinery and their applications.
• Incorrectly calculating efficiency by not converting units properly.
• Neglecting losses in the system, leading to overestimated performance.

For GATE ME

Questions often involve calculating efficiency, power, and flow rates in turbines and compressors. Practice problems on energy conversion and performance analysis.

Quick check

1. What is the primary function of a turbine?
2. Name two types of compressors.
3. What is the unit of flow rate?

Answers: 1. Convert fluid energy into mechanical energy. 2. Axial and centrifugal. 3. m³/s.