How much power is drawn by a turbofan to power its own compressor?

In a modern commercial airliner how much percentage of its engine power is used to power its own compressor, the ducted fan and other ancillaries like the airconditioning etc? Maybe it can't be calculated with accuracy but I'm looking for at least an approximate figure.

To answer this question, it is best to start with the compression work which has to be performed on the air flowing through the engine. If you just want to know how much power is needed to run the fan on a mid-sized passenger jet, you can look at industrial installations of the Rolls-Royce RB-211, two of which power the Boeing 757 and four the Boeing 747-200. The electrical output comes directly from the low pressure turbine which would drive the fan in an airliner engine, and is up to 32 MW for the RB-211 (after generation losses).

Now to the core compressor: If we assume that no heat is exchanged between the gas and the surroundings we can use the formulas for the adiabatic process as presented in Wikipedia. The example given is actually a nice fit for the RB-211-535 or an EJ-200 engine:

Initial state (Index 0): 27°C/300 K and 1 bar/100 kPa

Final state (Index 1): 477°C/751 K and 25 bar/2500 kPa, 10% of the initial volume.

When we use this Wikipedia page, we get: $$W = \frac{\kappa\cdot R\cdot(T_1 - T_0)}{\kappa - 1} = \frac{\kappa\cdot R\cdot T_0}{\kappa - 1}\cdot\left(\left(\frac{P_1}{P_0}\right)^{\frac{\kappa - 1}{\kappa}} - 1\right)$$

where $R$ = 287 $\frac{J}{kgK}$ is the gas constant and $\kappa$ = 1.4 for diatomic gasses like nitrogen and oxygen.. $V_0$ of 1 kg of air is 0.84 m³ and $V_1$ is 0.084 m³. The result is $W$ = 454 kJ per kg of air, and that has now to be multiplied with the mass flow. For the RB-211 I find mass flow data only for industrial applications where the engine runs at a reduced pressure ratio, but I will use the 203 - 212 kg/s regardless because it fits with the initial pressure and temperature data of my example. The result is in the order of 95 MW in a 200 kN thrust class engine.

Considering that the above equations are for the ideal, friction-less case, the true power need is probably 10% - 20% higher. However, at altitude the air density is only a quarter of that at sea level, so the compressor power needs are also only a quarter.

To put it into proportion, we need flight speed as well: At 240 m/s the 50 kN thrust at altitude translate into 12 MW of propulsive power, so the compressor power under the same conditions is twice as high.

• I'm a bit baffled. So what percentage of it's total power is used in powering it? – user12782 Dec 29 '15 at 2:51
• If I understand the reasoning, the 95 MW figure is what's needed for a compressor at 200 kN trust. At 240 m/s, 50 kN of trust translates to 12 MW propulsive, so 200 kN is 48 MW propulsive. This means that the percentage at 240 m/s is 95/(95+48) which is 2/3, 66.4% – MSalters Dec 29 '15 at 23:06
• @MSalters: No, the 200kN are not possible at Mach 0.85. Power comes from speed (thrust x speed). The compressor power is approx. 24 kW at Mach 0.85 (neglected pre-compression here!), which explains the lower fuel flow at altitude and full throttle. But the percentage is still similar. – Peter Kämpf Dec 30 '15 at 8:18
• But actually these 66% are reconverted for the most part into accelerated air which bypasses the core by the fan. It would be interesting to know which part of the energy from combustion is not used for accelerating the cold flow, but for moving accessories, pressurizing bleed air, or countering internal friction. I think that's the spirit of the question. But maybe this is already contained in the answer, and I read it wrongly, not sure. – mins Dec 30 '15 at 14:26
• @mins: Actually, the mass flow is only for the core (I took it from the industrial RB-211), because I understood the question this way. The fan would consume another 10 - 12 kW which is the power output of the industrial RB-211. Its lower pressure ratio is precisely because it does not have the fan of the aviation version. Accessories consume rarely more than 10%, and internal friction is very hard to guess without real numbers. With what I found on the web I cannot divide it up into more detailed fractions. – Peter Kämpf Dec 30 '15 at 19:43