# This concerns the GE90X fanjet engine which generates 130,000 LBS of thrust at takeoff [duplicate]

This concerns the GE90X fanjet engine which generates 130,000 LBS of thrust at takeoff.

But for this example the engine is secured to a test mount, and under "full throttle", the thrust generated is still 130,000 lbs. I believe that 80% of the thrust is from the fan itself.. but let's go ahead and assume 100% of the thrust is generated by the fan only.

If the large 12-foot ducted fan were instead connected by shaft to an external electric motor or gas engine, how much horsepower would be needed, to sufficiently spin the large ducted fan, to generate 130,000 pounds of thrust ?

• FYI, there is no "GE90X" engine. There is a GE90 and a GE9X. The former is rated to 115klbf at takeoff, the later is just slightly less at 110k. The GE9X does have the world record for thrust at 134,300 pounds, but that was on a test stand during a very unique test condition. Normal rated takeoff is less. – Daniel K Jan 31 at 3:05

So we have this equation relating power to thrust: $$T_0 = \sqrt[\LARGE{3\:}]{P^2\cdot\eta_\text{fan}^2\cdot\pi\cdot \frac{d_\text{fan}^2}{2}\cdot\rho}$$ which can be written as: $$P^2 = \frac{T_0^3}{\eta_\text{fan}^2\cdot\pi\cdot \frac{d_\text{fan}^2}{2}\cdot\rho}$$ Using proper SI units, we have $$\rho$$ = 1.225 kg/m³, $$T_0$$ = 578.27 kN and $$d_\text{fan}$$ = 3.4 m. The efficiency $$\eta_\text{fan}$$ must be guessed at this point - propellers have less than 50% in the static case. With those 50% the power required is indeed 66 MW. However, it is fair to assume that the GE90 fan has a lower efficiency than a lossless propeller, so the required power of this hypothetical electric motor should be closer to 80 MW which confirms @Therac's answer.