Physical basis for high-bypass turbofans

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SUMMARY

High-bypass turbofans achieve lower specific fuel consumption (SFC) by utilizing the relationship between momentum and velocity. By adding a bypass stage, the turbofan reduces the jet's velocity, which decreases momentum but significantly reduces kinetic energy, allowing the engine to power a larger volume of slower-moving air. This results in a higher net impulse for a given amount of fuel. Understanding the thrust equation and specific thrust is crucial for grasping the efficiency benefits of turbofans over turbojets.

PREREQUISITES
  • Understanding of the thrust equation in aerodynamics
  • Knowledge of specific thrust and its implications
  • Familiarity with the principles of momentum and kinetic energy
  • Basic concepts of turbofan and turbojet engine design
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  • Research the Thrust equation and its components in detail
  • Study the differences between turbofan and turbojet engine designs
  • Explore the concept of specific fuel consumption (SFC) in aviation
  • Learn about the physics of momentum and kinetic energy in fluid dynamics
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Aerospace engineers, aviation students, and professionals involved in aircraft propulsion systems will benefit from this discussion on the efficiency of high-bypass turbofans.

mikeph
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Hi,

Can someone double check I understand this correctly?

The turbofan has lower specific fuel consumption because a gas's momentum is proportional to its velocity, whereas a gas's kinetic energy is proportional to its squared velocity. Therefore a turbojet can be made more efficient by adding a bypass stage, which effectively removes some of the jet's velocity (via a turbine powering the fan), removing some of the momentum (and therefore thrust), but removing also a much higher proportion of kinetic energy. If this kinetic energy powers the front fan, then it can be added to the slower moving inlet air, which increases its speed and therefore the bypass air gains disproportionally more backwards momentum than was lose at the jet. The net impulse on the air is therefore higher for a given amount of fuel.

Is this correct? It's a "slow the fast thing down and use the energy to speed the slow thing up" idea?

Do pressure ratios come into it as well?

Cheers.
 
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mikeph said:
...because momentum is proportional to velocity, whereas kinetic energy is proportional to squared velocity.

Agreed. Accelerate much air a bit consumes less than accelerate little air a lot.

Hence by-pass. If the aeroplane speed allows it, prefer even wider turboprops. And for a helicopter, take a huge rotor.
 
Your statement describing that the kinetic energy is added to the incoming air thereby speeding it up is wrong...
 
In order to understand the need of bypass turbofan engine you have to first of all understand the Thrust equation and the term Specific thrust..For a same amount of thrust at same inlet velocity condition if u design a turbojet and turbofan engine...
F=mair(Vj-Vi)
Turbojet concentrates more on the (Vj-Vi) on the thrust equation term.But to produce same amount of the thrust turbofan increases the mair entering the engine..turbofan accomplishes this by having a fan infront of it,more work is extracted from the gas with the help of the turbine to drive the fan.By increasing the mass flow rate rather concentrating on the (Vj-Vi),specific thrust decreases which decreases SFC(specific fuel consumption)...
 
Anish Aravinth said:
Your statement describing that the kinetic energy is added to the incoming air thereby speeding it up is wrong...

Not at all - in fact, it's exactly correct. A high-bypass turbofan (and a turboprop for that matter) remove kinetic and thermal energy from the core flow, and then use the extracted energy to increase the kinetic energy of a much larger volume of air. As stated in the OP, this allows for a much larger momentum flux for the same power output, and thus, more thrust for the same fuel consumption.
 

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