# How do aircraft engine manufacturers achieve a higher bypass ratio while still meeting the thrust requirements for a given aircraft?

I understand that a higher bypass ratio leads to greater efficiency as less air moves through the core, and therefore less fuel is burned. However, how is the thrust requirement for the same aircraft met with a newer engine that has a higher bypass ratio? Since most of the mass flow will be provided by the front fan, I want to understand what changes are made to turbofan engines that allow the same amount of thrust while minimizing airflow through the core.

• Does your question boils down to "For the same engine thrust, how can BPR be increased without increasing fuel consumption in the core"?
– mins
Jun 18 '17 at 8:01
• The core is mostly a fixed size but you grow the fan. Tough part is to harvest energy from the gas to drive the fan. Jun 18 '17 at 21:14

Early turbojets were so inefficient that adding a fan was considered but not implemented because that would had made the engines even more sluggish and narrowed the operating limits even more. If you compare the Jumo 004 with the EJ200, you will find that both are of similar size, have 8 compressor stages, but vary widely in their compression ratio and thrust (3.2 vs. 26 compression ratio and 9 vs. 60 kN dry thrust).

When compressor flow was better understood, turbine temperatures and pressure ratios rose, the excess power delivered by the engine core rose and allowed to drive a second turbine and shaft; initially with bypass ratios as low as 0.25. If you compare the overall pressure ratio with the bypass ratio, you will certainly spot a trend:

Engine             compressor       pressure ratio     bypass ratio
RR Conway        7LP, 9HP stages       14  :1             0.25:1
P&W JT3D      2fan, 6LP, 7HP stages    12.5:1             1.42:1
GE CF6-6      1 fan, 1LP, 16HP stages  25  :1             5.8 :1
RR RB211-535  1 fan, 6IP, 6HP stages   25.8:1             4.3 :1
GE 90         1 fan, 4LP, 9HP stages   42  :1             9   :1
RR Trent XWB  1 fan, 8IP, 6HP stages   52  :1             9.3 :1
RR Ultra Fan                          >70  :1           >15   :1 (projected)


A more efficient core allows to leave more energy in the exhaust past the high pressure turbine, so the low pressure turbine can extract more power to drive a larger fan.

• Good answer! How is more energy extracted with a smaller core region and less fuel? Jun 19 '17 at 2:44
• By using fuel and air more efficiently. Higher compression ratio, better burners, more efficient turbines. Jun 19 '17 at 4:11
• Understood, so the goal for turbofan engines is always to have the highest bypass ratio possible while maximizing core efficiency and decreasing core size? Jun 19 '17 at 20:52
• @flextempers: If you stay firmly in the subsonic realm, and want to minimise fuel consumption on long range flights, then yes. Note that increasing bypass ratio will give you diminishing returns and a bigger, heavier nacelle with higher drag. At some point you might want to artificially keep the nacelle flow laminar for longer to still achieve positive gains from that higher bypass ratio. Jun 20 '17 at 9:49
• This is fascinating. What do you mean by 'artificially' keep nacelle flow laminar? Can't the thickness of the nacelle remain relatively constant even with a large increase in fan diameter? Jun 21 '17 at 7:53

The gas generator is the core of the jet engine: the combination of compressor - combustor - turbine. The difference is in extraction of power by the turbine. If the turbine is larger, it can extract more power from the gas stream to drive the compressor + fan. If there is no bypass fan, less power is extracted from the gas flow by the turbine, and the exhaust speed is higher.

The power of a gas generator is converted into thrust by accelerating a gas flow. Efficiency increases if more gas is accelerated to a lower velocity:

• Thrust of a jet engine is given by $$F = \dot m * (v_e - v_0)$$ with $\dot m$ = gas stream in kg/s, $v_e$ = exit velocity of the gas stream in m/s, and $v_0$ = aircraft air speed.
• The propulsion efficiency of a jet engine (with complete expansion) is defined as $$\eta_p = \frac {2}{1 + v_e/v_0}$$

So for a given thrust, if $V_e$ is reduced and $\dot m$ proportionally increased, the propulsion efficiency increases.

Notes:

1. Complete expansion of the exhaust stream takes place if it does not reach the local speed of sound. If it does, it means there is a choked exhaust and part of the thrust is produced by higher pressure behind the exhaust. This can be accounted for by computing an effective $v_{e_{off}}$ that would produce the same thrust.
2. Propulsive efficiency is 1 = 100% if $v_e$ = $v_0$, but unfortunately there is no thrust anymore.