What is the fan tip speed at maximum thrust? Is it any different for a Rolls-Royce Trent 900 (powers the Airbus A380) and Pratt & Whitney PW6000 (powers the Airbus A318)?
Are they efficient at maximum thrust?
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asked Dec 25, 2017 at 20:16
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$\begingroup$ Compute it, approximately. It's just v = fan radius * maximum rotational speed $\endgroup$– user7241Commented Dec 25, 2017 at 20:25
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$\begingroup$ You'd have to pick the engines for those two planes that you're interested in. $\endgroup$– user7241Commented Dec 25, 2017 at 23:14
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$\begingroup$ I wrote "approximately". If you want the other speed, you have to take into account the air intake and what it does to the air going into the fan. $\endgroup$– user7241Commented Dec 26, 2017 at 10:16
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$\begingroup$ It is difficult to impossible to figure out the latter. Stick with the former. It already requires a bit of thinking. $\endgroup$– user7241Commented Dec 26, 2017 at 10:34
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1 Answer
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Second question first, the higher the jet engine thrust, the more efficient it is (in terms of thrust for fuel rate, bang for buck).
(Source: Boeing)
On to the Trent 900 and PW6000, the big one and small one. The bigger fan will have slower RPM. From the type certificates the max permissible RPM's are 2818 (takeoff) and 6350 (undefined), respectively.
The fan diameters are 2.95 m and 1.435 m, which translate to 435 and 477 m/s. Speed of sound in dry air at 20°C is 343 m/s.
Here's the caveat, the fan RPM (%N1) at takeoff does depend on the ambient air pressure and temperature (among other operational considerations). But as you can see, the smaller fan can rotate faster—read: max permissible—because it has less mass and the centrifugal forces are more manageable.
And since the 1960's or so, the fan RPM is rarely a limiting factor for the jet engine operation, so the engine manufacturer will have designed for the optimum blade tip speeds.
- N1 RPM limit
It might be necessary to limit the engine RPM in order to avoid overloading the fan blades in centrifugal force, and to control the fan blade tip speeds. This is purely an N1 limit and is not a function of altitude.
The N1 limit is rarely more restrictive than the pressure and temperature limits.
(Emphasis mine; Jet Transport Performance Methods)
So much so—as you can see below from a 777 manual—the hotter and higher the airport up to about 40-45°C, the faster the takeoff N1 is to compensate.
(Click to view)
That answers the efficiency part, and the blade tip comparison as requested.
Related: Why does lower atmospheric pressure produce higher EPR (thrust)?
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$\begingroup$ This answer is too simple - you forgot to add the flow speed through the fan. Even in the static case it will be at least Mach 0.3 (approx. 100 m/s) at full thrust. $\endgroup$ Commented Apr 20, 2018 at 5:54
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$\begingroup$ @PeterKämpf - So for the Trent 900, a 435 m/s (M1.27) tip speed in a static test with a M0.3 axial flow should result in 447 m/s tip speed (a 12 m/s -- 2.8% increase), is that correct? As I don't have the tip angle of attack I just used the velocity vectors of M1.27 and M0.3. I checked the reference I linked in the answer and the web for info but they overlooked it. If the above is correct let me know and I'll add it. And would the axial vector be the cruise Mach number for the subsonic inlet? Many thanks. $\endgroup$– user14897Commented Apr 20, 2018 at 23:07
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$\begingroup$ Yes, vector addition is the correct way to add the two speeds, and knowledge of the angle of attack is not needed. Maybe you will find the mass flow for the static case; then it should be easy to get a precise number. It should not surprise you that other pages on the Interwebs contain wrong results, should it? $\endgroup$ Commented Apr 21, 2018 at 19:07