The chevrons on a 787 are there exactly to decrease a type of drag that makes the engine louder, but the chevrons form vortices causing more drag. How much thrust is lost from the vortices on the chevrons?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ Can somebody please edit the question and add a picture of a 787 chevrons on the engine. Sorry I'm using my laptop this morning and it can't upload images. $\endgroup$– EthanAug 28, 2015 at 11:58
-
$\begingroup$ When you are unable to upload image, you can put links to images taken from well known websites (airliners.net, wikimedia,...). $\endgroup$– Manu HAug 28, 2015 at 13:07
-
1$\begingroup$ also the imgur interface to upload images in not machine-dependant, afaik. $\endgroup$– FedericoAug 28, 2015 at 13:32
-
$\begingroup$ SImilar Questions: aviation.stackexchange.com/q/16500/1289. aviation.stackexchange.com/q/12832/1289 This Q is different mainly in that it asks (implicitly) for numbers. $\endgroup$– RedGrittyBrickAug 28, 2015 at 14:18
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
The main purpose of the Chevrons is the reduction of engine noise. During research into this, NASA tested a number of configurations under simulated flying conditions and measured the nozzle thrust coefficient given by,
$C_{T_{r}} = \frac{F^{d}_{g}}{F^{i}_{g}}$
where,
$F^{d}_{g}$ is the measure gross thrust and
$F^{i}_{g}$ is the ideal gross thrust
We have,
$F^{i}_{g}$ = .{m}$_{c} v_{i,c}$ + .{m}$_{f} v_{i,f}$
where .{m}$_{c}$ is mass flow rate of core, .{m}$_{f}$ is mass flow rate of fan and $v_{i,c}$ and $v_{i,f}$ are core and fan ideally expanded jet velocities respectively.
The reduction in nozzle thrust coefficient was compared with baseline (no modifications) configuration.
Source: NASA/TM—2000-209948
The configuration used in engine (Chevrons in the fan and nothing in the nozzle produced the lowest loss of thrust coefficient, 0.18.
However, the noise reduction was lesser in this configuration compared to the others. Further modifications have been carried out in the chevrons used in production engines (they are rounded, for one). So, the noise reduction has been obviously improved, though the minimal loss of thrust would've been maintained.
Note: All data from Acoustics and Thrust of Separate-Flow Exhaust Nozzles With Mixing Devices for High-Bypass-Ratio Engines, Naseem H. Saiyed, Kevin L. Mikkelsen and James E. Bridges, Prepared for Prepared for the Sixth Aeroacoustics Conference and Exhibit, Lahaina, Hawaii, June 12–14, 2000
answered Aug 28, 2015 at 16:41
-
1$\begingroup$ With $C_T$ defined as actual thrust over ideal thrust, wouldn't $C_T = 0.18$ mean you're only getting 18% of the ideal thrust? Or do you mean that $C_T$ decreased by 0.18? That's still huge, isn't it? If the starting value was perfect efficiency ($C_T = 1.0$), losing 0.18 would take you down to 82% efficiency. i.e. you've lost nearly 1/5th of the engine's thrust! That seems like a lot, and not something you'd call minimal. (I didn't try to find a copy of the paper. If you have it, maybe you can clarify what the actual thrust ratio is between original and noise-reduced.) $\endgroup$ Apr 11, 2018 at 22:26