Forum: General F-22A Raptor forum

I'm having a bit trouble converting this myself, I don't even know if it's possible so forgive me if it's a dumb question.

Could somebody please, if possible, convert the following for me to Watt/Joule?

Purely theoretical: if 2 turbojet engines, each producing 150.000N (150kN) of thrust, propel an aircraft to get-to, and maintain a velocity of Mach 2.5 at altitude, how much Watts/Joules of energy is each engine producing?

IT--

F-16.net Sponsor

150,000,000 Joule's?

http://www.convertunits.com/from/kN/to/joule/metre

You need the Mass of aircraft to calculate Energy, kinetic energy, a basic one, without considering of loss from drag.

Or the other way, you have a constant force, you'll need to know how long does the aircraft move by this force to reach M2.5, the displacement, multiple these 2 you'll get energy. The acceleration is (Thrust force - Drag force) / Mass.

Energy is force times distance, and Power is force times velocity.
The velocity in question would be exit velocity of gas from the engine, in m/s, rather than the velocity of the jet itself.

Energy is conservative so it is neither created nor destroyed but converted to some other form. That converted amount is called as work (done). (Force x distance)

Power is the conversion rate of energy. (Work per time)

Thrust (in your case) is a reaction force.

You can find the power (energy converted per second) if you know the exit velocity of the gas minus the velocity of the aircraft (Net velocity).

I think you all have helped me by a great deal. Thank you very much for educating a newbie such as myself, It is very much appreciated!

IT--

What is it that you were working on?

I was having a discussion with a mate of mine who is studying maritime engineering.
He was talking about how much energy those huge oil-tanker engines each produce and it got me wondering...
I wanted to figure out around how much energy a pair of P&W F119's produce in full afterburner if converted to Joule/Watt/Newton.
My mind finds thrust (weight) hard to comprehend so thats why I wanted to convert it to J/W/N.

IT--

Alrighty:

Power = Force * Velocity

Speed of sound at altitude (>FL300) ~ 300 m/s

= 150,000 N * 2 engines * 300 m/s * 2.5 (Mach)

= 225,000,000 Watts = 225,000,000 Joules/Second = 225 Megawatts = 300,000 HP

Bear in mind that 150,000 N (~35,000 lbs) is likely the ballpark sea level static thrust of the F119, and the actual thrust at altitude, though classified, is almost certainly substantially less. A simple back-of-the-envelope calculation would show it to be roughly 40% thrust at FL300, given the density ratio. Installation losses and ram pressure gains would be competing forces as well.

So 225MW at sea level theoretically?... Are you absolutely sure that your calculations are correct? That sounds absolutely insane!
Let's say that the effective thrust will be 30% of that at altitude, that still gives a result of 67.5MW which is crazy.

Is there no way to use some of this energy to power up a laser to for example disable incoming missiles?

Thank you so much for the effort!

IT--

Velocity must be the net velocity of the exhaust gases.

Here, ''net'' means speed of the exhaust gases minus the speed of the aircraft (That is in scalar form, in vector form it is addition where one sign is negative).

e-dog wrote:

I was having a discussion with a mate of mine who is studying maritime engineering.
He was talking about how much energy those huge oil-tanker engines each produce and it got me wondering...
I wanted to figure out around how much energy a pair of P&W F119's produce in full afterburner if converted to Joule/Watt/Newton.
My mind finds thrust (weight) hard to comprehend so thats why I wanted to convert it to J/W/N.

IT--

Here's a tidbit for your buddy. The Pluto reactor for SLAM produced 513MW which they equated to 35,000lbs thrust. Now before some old codger whips out his slide rule Wink

they didn't show how they got one from the other. SLAM was an 88,000lb missile designed to fly at Mach 3.2 on the deck so maybe you could work the problem that way. Regardless, 513 MW equals about 688,000hp or more than 2 Nimitz class carriers. Shocked

Quote:

Velocity must be the net velocity of the exhaust gases.

Here, ''net'' means speed of the exhaust gases minus the speed of the aircraft (That is in scalar form, in vector form it is addition where one sign is negative).

True, if you want to calculate the work done on the air which will give you the rate of that work (even for a static rig). However, according to conservation of energy and Newton's laws, net power generated must equal the aircraft's velocity multiplied by its thrust. It's not as insightful in terms of the operation of the engine, but it's the easy way to calculate it.

That power is not all useful power, however, since the compressor draws energy from the turbine to keep the engine cycle going. If you extracted too much power for your laser, the engine would wind back and stall. In terms of numbers on that, I'm not sure. Would have to break out the books again. Haven't seen thatengineguy yet, he's probably got the skinny on this.

Then you are saying that:

Exhaust gas velocity is 2x the aircraft's velocity

... which is clearly not the case.

There's a reason jet engine output is measured in units of force, and the output of engines that drive an output shaft is measured in units of power. There are way too many variables in how a jet's output will be translated into work for its output to have any real meaning in units of power when it's being used to push a jet. Really, you don't measure the engine. You measure the drag on the plane, and integrate that over the distance it's moving. The engine can dump energy into stirring the air into a frenzy, and that gives no useful measure of the engine's power. It's strictly how much it's able to move the plane.

If you put a power take-off turbine on an output shaft in the exhaust of a jet engine, you get a gas turbine, where you can integrate the torque on the shaft over cycles of rotation, with output measured in units of power. For an example of why a straight conversion of thrust to power is so relatively meaningless, the CF6 jet engine in different configurations produces about 50-60,000 lbs. thrust. When used as the core of an LM2500 turbine geared to a ship's propeller shaft, it produces a bit over 30,000 horsepower. But hooked to an optimal electrical generator, it produces over 40,000 horsepower. I'm not involved with GE but I'd assume the reason is that the take-off turbine is better able to match the exhaust flow when it can be optimized for turning a generator at one speed, vs. a turbine that needs more attention to low-end torque to spin a prop up. Converting the jet's thrust to power depends on the hardware you use to convert it; there's no one right unit conversion. All you can get is a ballpark number unless you specify an exact configuration.

Forum: General F-22A Raptor forum

Could someone please convert this for me?