13
$\begingroup$

In modern VTOL aircraft energy is usually directed through nozzles to achieve vertical lift. How much energy is lost when changing the direction of flow in static thrust conditions?

X-35B Rolls-Royce Pegasus 30

$\endgroup$
3
  • 3
    $\begingroup$ Reminds me of that Mythbusters episode where they bent the barrel of a rifle 90 degrees and it was still lethal. One of my favorites! 🙂 $\endgroup$ – Kirk Woll Aug 28 '20 at 0:24
  • 2
    $\begingroup$ Note that the Rolls-Royce Pegasus always forces the exhaust around a bend, no matter which direction the nozzle points... $\endgroup$ – DevSolar Aug 28 '20 at 11:08
  • 1
    $\begingroup$ My naive thought would be that the direction of the vectoring force (the "bending" force) is in this case upwards (remember those rotating lawn sprinklers using some of the water pressure to propel their rotation) which means that not too much power would be lost to the vectoring, only rather small amount to increased friction/compression in the bend. $\endgroup$ – Pavel Aug 28 '20 at 12:23
16
$\begingroup$

For the F135 engine in this photo, thrust in hover is only about one per cent less than maximum thrust, if Pratt & Whitney's data sheet is to be believed.

Maximum Thrust Class 41,000 lbs
...
Hover Thrust 40,650 lbs

The Hawker Harrier's maximum thrust was about 20,280 pounds. An approximation of its maximum vertical thrust at low airspeed is given by its maximum vertical landing weight, 19,918 pounds, again just about one per cent less.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks for the info and the links! On the F135 it lists the lift fan as part of the hover thrust so it's hard to know for sure as the engine splits it's maximum 27000 lb thrust (military power). 18,680 lbs of thrust comes directly from the rear nozzle. Something like 8320 lbs worth of thrust is sacrificed to drive the more efficient lift fan which then provides once again 18,680 lbs. $\endgroup$ – usernamechecksout Aug 27 '20 at 23:11
  • 1
    $\begingroup$ The trick seems to be to constrict the flow at the nozzle enough that the effects of the 90 degree bend on the flow are minimized. In effect it is like blowing up a crooked balloon and comparing its "thrust" to a perfectly round one. The flow at the bend is much slower than at the nozzle. $\endgroup$ – Robert DiGiovanni Aug 28 '20 at 1:32
  • 1
    $\begingroup$ "a few percent"? Based on those numbers, the loss is 0.86% for the F135 , so "a fraction of a percent" would be more accurate. (It's a 1.79% loss for the Harrier, so I'll cut you slack on that one.) :D $\endgroup$ – FreeMan Aug 28 '20 at 18:16
6
$\begingroup$

A 90 degree bend in a pipe where the radius of the bend is of order ~one pipe diameter creates the same pressure drop as a length of that same pipe of order ~ten to fifteen times the pipe diameter.

$\endgroup$
2
  • 6
    $\begingroup$ Which is how much, compared to an unmodified pipe? Five per cent? Fifty per cent? $\endgroup$ – Camille Goudeseune Aug 27 '20 at 18:03
  • $\begingroup$ do not know in terms of percent. $\endgroup$ – niels nielsen Aug 27 '20 at 21:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.