Power available question???

[borgy]

Hi Guys,
I have searched but I'm unable to find out why on the the power required/available graph, power available increases with airspeed. I understand that with turbine engines there is a ram air effect is it the same with pistons???

Thanks in advance

[Pugilistic Animus]

I'm a little confused about what exactly you are asking what flight conditions do you mean best climb best range, best endurance???as a basic begining the best climb rates occur at the thust/velocity point where there's the greatest difference between power required and power available,..

but best speed for climb 'Vy' occurs at the point where the greatest difference between 'Thrust horse power' required and that available, remembering that power is thrust times velocity. I hope that helps but if not keep posting your question some one will get to it and sort it out if I can't---like Old Smokey did:ok:


PA
sorry for my formatting this computer is stupid!

[borgy]

sorry if the question was not that clear. What I was trying to say was if your looking at the graph that would normally be used to illustrate Vy to a student the power available line increases with increasing airspeed. I'm trying to understand what causes this increase in available power.

[eckhard]

PA has the main point, which is that:

Power = Thrust x TAS

Why is that?

Well, working in SI units, we have:

Force = Newtons (N)
Distance = metres (m)
Work = Force x Distance
= Nm
Power = rate of doing work
= Nm/s
But Speed = m/s
Therefore
Power = N x m/s

or in other words, Force x Speed

In our case, the force is the thrust, so

POWER = THRUST x SPEED

The thrust available from a typical piston engine/propeller combination is high at low speed and then tends to reduce slowly with increasing airspeed. The rate of reduction increases as the speed increases, due to propeller inefficiencies. At the high speed end, the reduction of thrust with increasing speed is quite steep.

Countering this effect is the physical fact that Power increases with speed.

The resultant 'Power Available' curve will therefore exhibit an increase with speed at the low speed end of the graph, a peak, and then a reduction at the high speed end. The exact shape depends on the shape of the thrust available curve, which is rarely published in any textbooks.

Here is one graph that I found:

Thrust Available (http://www.tpub.com/content/aviation2/P-202/P-2020097.htm)

Hope this helps!

:8

[borgy]

Thanks that has shed some light on it for me. I think the confusion began with the interpretation of the word power. I naturally think of power as the rated hp of an engine so assumed that according to the graph an engine that produces 240 hp WOT and stationary will produce more than 240hp WOT at 110 kias.

From your explanation I'm getting that the word power is used in a physics sense and the increase has nothing to do with ram air effect (at least in a piston) and is just due to the fact that more work can be done at a higher airspeed, is that correct???

[Tail-take-off]

Surely on aircraft with fixd pitch props there is an optimum speed for the pitch of the prop that will give the most thrust at full throttle. Above or below that speed thrust will reduce. Or have I missed something?

[Old Smokey]

Power = Force X Velocity (Thrust X TAS)

Jet engines DIRECTLY produce Thrust, thus, at a fixed Thrust Setting, Power increases linearly with Speed (TAS).

Piston and Turbo-Prop engines DIRECTLY produce Power, which is translated into Thrust by the propeller. Thus, at a fixed Power Setting, Thrust decreases with increasing speed. (Thrust = Power / TAS). All of this is why we had to invent the jet engine if we wanted to go faster and faster.

Aircraft need Thrust, they always have, and they always will. This applies from the Wright Flyer through to the Concorde). They have never needed Power, except for propeller aircraft which needed power to produce the all-important Thrust.

For the Jet aircraft, all performance considerations come back to Thrust, except in the ABSTRACT sense in evaluating Rate of Climb by comparing Thrust X TAS at varying speeds. Because (for the jet) Power increases with increasing speed, best Rate of Climb speed is very high, often in EXCESS of the cruise speed to follow the climb. The opposite is true for the propeller aircraft.

The original poster alluded to Ram Recovery, it is an important factor, but directly affects Thrust. At a fixed engine setting, due to the effects of the 'Thrust Equation', Thrust declines with increasing speed up to about M0.5, where the decline bottoms out, and then, due to ram recovery, begins to rise again reaching parity with static thrust at about M0.75, and may exceed static thrust thereafter. (If there were no ram recovery, jet aircraft would do no better than propeller aircraft).

As a foot-note, Jet, Turbo-Prop, and piston engined aircraft all produce considerable internal power to drive the engine. This is an engine factor, and not an aircraft performance factor. It's what comes out the back of a Jet engine, or what is delivered to the propeller of a Piston / Turbo-Prop that matters to aircraft performance.

If you get the Thrust of my argument, it will add a lot of Power to your argument!:ok:

Regards,

Old Smokey

[eckhard]

An outstanding explanation from Old Smokey!

The original poster alluded to Ram Recovery, it is an important factor, but directly affects Thrust. At a fixed engine setting, due to the effects of the 'Thrust Equation', Thrust declines with increasing speed up to about M0.5, where the decline bottoms out, and then, due to ram recovery, begins to rise again reaching parity with static thrust at about M0.75, and may exceed static thrust thereafter. (If there were no ram recovery, jet aircraft would do no better than propeller aircraft).

Could you elaborate a bit please? I assume this applies only to jets? Do you have any diagrams or references to illustrate this?

Thanks!

Eckhard

[Old Smokey]

eckhard,

I do have an excellent diagram (which I plaigiarised from Lufthansa), but unfortunately do not have the means of posting diagrams (My son does, so I'll have him post for me within a few days).

In the interim, in words, rather than mathematical formulae or diagrams........

Jet thrust is achieved by Newton's law, "For every action there is an equal and opposite reaction". So, by expelling jet efflux rearwards at high velocity, we achieve an equal and opposite reaction, i.e. forwards thrust. As Force is the product of Mass and Velocity, we arrive at -

Force (Thrust) = Mass (of the jet efflux) X Velocity (of the jet efflux RELATIVE to aircraft speed)

The word RELATIVE is very important. If, for example, the jet efflux speed was 600 Kt, but the aircraft stationary on the brakes, Thrust would equal the efflux mass X 600. If aircraft speed increased to 200 Kt, the relative speed would only be 400 Kt, and Thrust would only equal the efflux mass X 400. As speed further increased, the aircraft would "run out of chuff" at about 300 Kt or so (when the thrust was half of the static thrust). That's why I stated that if it were not for ram recovery, Jets would do no better than props.

RAM RECOVERY - About two thirds of the power (hate that word for jets) extracted from the fuel is used to drive the compresser/s via the turbine/s. The remaining third is used to expel the exhaust gases producing thrust. As Mach Number increases, air impacting the aircraft, and entering the engines, is COMPRESSED. This compression relieves the turbine of some of the work in driving the compressor, leaving more energy to expel the exhaust gases. Thus, at all speeds there is Ram Recovery. The degree of compression is exponential (related to Mach Number), whereas the decline in Thrust due to the thrust equation is linear. At low mach numbers this compression is not very significant, but does partially compensate for 'Thrust Equation' losses. They are about equal at approximately M0.5 (300 Kt in ball-park figures). With further Mach Number increase, the exponentially increasing Ram Effect exceeds the linearly decreasing 'Thrust Equation' effect, and at about M0.75, Thrust is back to static thrust. (Much more of the engine's internal power is now used to propel the exhaust gases, much less is used to drive the compressor). Above this Mach Number, Net Thrust can and does exceed static thrust.

That's it in a nut-shell, excepting that later generation High Bypass Turbofans are much more 'proppy', and the Mach number where parity with static thrust will be somewhat higher than the M0.75 quoted.

Ram recovery certainly does assist Piston and Turbo-Prop engines (Higher Manifold pressure for an unsupercharged aircraft, or less Supercharger work for a Supercharged aircraft), but as Mach Numbers are relatively low, it could only be considered as a small bonus. On RR Dart powered aircraft which I once flew, it was common to observe increasing Torque as the aircraft accelerated from 140 KIAS or so up towards 200 KIAS at a fixed Power setting. (14,200 RPM at 730 degrees TGT, how's that for 30 year memory?)

Flak jacket on, the purists are about to wade in with their high fallutin' mathematical formulae:eek: Consider this as the 'wordy' layman's explanation.

Regards,

Old Smokey

[411A]

(14,200 RPM at 730 degrees TGT, how's that for 30 year memory?)


Excellent...and, you are spot on!:ok:

[eckhard]

Thanks Old Smokey.

A very illuminating reply! It's a bit late where I am now (Hangzhou) to digest in detail, but I will peruse it further tomorrow.

Looking forward to your son's posting of the Lufty diagram.

I too find the word 'power', when applied to a jet, grates more than a little.

P1: 'Setting Power' (presses TOGA)
P2: (Observing correct EPR obtained): 'Power Set' (thinks: No it isn't, the THRUST is set, the POWER is increasing as we roll down the runway!):ugh:

I thought I was the only one with such (over)sensitivities!