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As a matter of interest, how are you arriving at these figures.....?



Power or thrust?

Thanks for starting a new thread Riggy.


My prior assumption of 100k pounds was based on 2 engines, so 220k made reasonable sense for a 4 engined craft

I have watched the displayed groundspeed during takeoff, it’s usually around 180mph.

220000 / 2.2 = 100000kg of thrust, * 9.80665 = 980665 Newtons of thrust
180mph = 80.5 meters per second
Power = Force * Velocity = 79MW, / 745 (from memory) = 105k bhp



Power. The useful mechanical power output of a stationary reactive jet engine at full thrust is 0 (assuming the thrust isn’t used to drive other mechanical devices)

Power is power, no matter how it’s generated or transferred.
Power = Force (thrust or torque) * Velocity (with correct unit conversion)

Consider the case with of a rear wheel drive car with the rear brakes disabled; the power transmission/transfer path will be the same as the aircraft system.


Same applies to car brakes, the drive torque won’t necessarily change as you brake (until the gear is changed or the engine under-revs).

Ok, thanks mate, that calculation looks about right to me; may be even a little conservative on the thrust per engine score in fact!
I actually had something else in mind when I asked the question but now I'm not sure it's actually relevant to the point you were making The thrust of a jet (in BHP or SHP) can only be compared to other forms of power generation at precisely 375 mph; at all other speeds its not valid. I'd need to get back to work to dig out my Pratt and Witney textbook to explain why, but as I said, I probably missed your point anyway.



Agreed, as work = force * distance and power is the rate of doing work, then when a jet is stationary it is doing no work therefore developing no power, even at full throttle...try explaining that to a layman

However, I believe theres a merging of ideas in this statement.



We know from Newtons second law that Force (Thrust) = Mass * Acceleration; for a jet engine we can read Thrust = Mass Airflow * How much it is accelerated by through the engine. Acceleration can, in this instance, be simplified to the difference in engine intake entry speed and exhaust nozzle exit speed.
As the aircraft forward speed increases so the thrust produced increases because mass airflow increases due to the ram effect; it will eventually peak when the engine reaches its swallowing capacity.
Once this point is reached, a further increase in forward speed will see power continue to increase, but thrust will then begin to fall off as the difference between the entry velocity of the air at the intake and the exit velocity at the propelling nozzle get closer together. In fact it will in theory at least fall to zero once the aircraft's forward speed equals the engine exhaust speed - not actually possible because some thrust must be used to overcome drag.

I can't get my head around this jet engine case in the wider context. Suppose our 'static' jet engine is fixed at the equator firing west. Surely the power output will accelerate the earth's spin - (albeit by not very much)?

And then there's relative velocity. We might be static relative to the surface, but we're doing 1,000mph relative to the centre of the earth. Does the power output change with position datum???

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Yeah, I would go with that. When stationary, the jet must ‘suck’ in air, the air pressure just prior to the blades would be reduced, hence there will be less airmass to ‘grab’ and throw rearwards. Therefore it follows that the thrust would initially increase with increasing velocity.

I once read that the maximum blade tip speed for a commercial jet is around mach 1.5. From what I remember, the angle of attack at the blade tip is around 45 degrees, hence the jet won’t be able to generate any thrust once the airspeed reaches 1100mph. Therefore it follows that the thrust would eventually decrease with increasing velocity (this is the classic loading effect where full power is achieved roughly midway to maximum of a characteristic)

The concepts of work and power are generally expressed within the context of the immediate environment in which they are observed.
If I move an object with a force of 1N a distance of 1m I am considered to have done 1Nm or 1 Joule of work. The fact that the earth may have rotated whilst I was moving the object and I am now 2 miles (in relative space) from where I started is not factored.
Thus, if a jet engine producing 10,000lbs of thrust is running in an aircraft with its brakes applied and there is no forward speed the work done and hence the power) is considered to be zero. Weird I agree.

[Added following an edit and a thought!]

We might consider that the engine is producing power because, by the laws of thermodynamics, it is doing work on the air passing through it by changing its pressure and volume whilst moving the air a given distance.
However, in the field of propulsion and fluid dynamics, we only consider the effect the engine is having on the aircraft.

The jet is accelerating the planet and the air in opposite directions. These will (eventually) cancel each other - the jetstream will eventually strike other objects thereby decelerating the air, the resulting (opposing) force needed to decelerate the air will be transferred back to the planet, thereby decelerating it again. System momentum remains conserved.


We’re doing 67,000mph relative to our solar system. What about the solar system relative to the rest of our galaxy?

Relative power/energy is not a straightforward area. I’ll return to this - assuming I find a good answer!

I'm not sure what you're getting at here. A compressor rotor blade 'sees' the arriving at a relative angle of attack which is a vector derived from the axial speed at which the air arrives at the compressor face and the rotational speed of the rotor disc. Tip speed of mach 1.5 are, as you say typical today thanks to some real fancy blade aerodynamics, but the axial speed isn't necessarily a direct function of the aircraft forward speed. The intake serves to reduce the air's speed to a velocity the compressor can accept. If it is necessary, intake design can reduce the speed of the incoming air from mach 2 down to around 0.5 mach, indeed this was absolutely necesary in early engine swhich couldn't handle air arriving at a higher speed.

I was just stating why I agree with your previous post (and I don't disagree with your last).

A blade tip (of the main fan) will be moving (and limited) at a given rotational velocity (orthogonal to the airflow) V.
The tip will be at an angle A (A = 90 for blade parallel to the airflow, A = 0 for blade orthogonal to the airflow).
For a given V, there will be an airspeed S for which this blade will simply cut through and will not exert any mechanical force on the air (produce no thrust). This speed will be:
Smax = V Tan (A)

I realise it’s not quite as straight forward as that (curved blades, friction) but it does support the principle that mechanical thrust will decrease towards 0 as aircraft speed increases.

Oh
Sorry mate, got carried away yakking about my own line of speciality

However, to show what a sad geek I am, I've remembered why horsepower and jet thgrust can only be compared at 375mph but I think I need a cure for my insomnia

Does this make me really geeky?
(I honestly didn't know this but it just felt right so I tried it)

1 pound of thrust at 375mph:

1 / 2.204622 = 453.59g, * 9.80665 = 4.4482 Newtons, * (375/(3600/1609.344)) = 745.700W, / 745.700 = 1.00000bhp

If it does, then this makes me plain mad

I think your calculation works because its derived from standard units of measure the origins of which I outline below. I'm sure you know some of this BTW, but for our other viewers....

To understand the why it is so difficult to make a comparison between the power rating of a reaction jet engine and a reciprocating piston engine (or turbo prop for that matter) we need to look behind the scenes.

A piston engine or turbo prop generates a mechanical output or drive. We know that P = F * D /t, however F * D is also expressed as torque; these are effectively torque producing engines. Thus, the power output of a torque producing engine is, effectively, a measure of its ability to produce a ceratin amount of torque over a given time. This value will be fixed for a given design and measured in bhp for piston engine, s(shaft)hp for turboprops.

Reaction jet engines only produce thrust, which is a force. Ergo we only have one component part of the power formula, any power (torque) generated in the traditional sense as discussed above, is consumed entirely within the engine itself to turn the compressor and its accesories. So how can we make a meaningful comparison, well it is possible but first we've got to dig a little deeper into the concept of horsepower

Scottish engineer James Watt determined that one of his pit ponies raising coal from the mines could do 22,000 ftlbs of work per minute, i.e. it could raise 22 lbs of coal 1,000 ft in a minute. For reasons best known to himself he elected to increase this value by a nominal 50% (1.5 pit ponies !) giving us the unit of measure for 1hp i.e. 33,000 ftlbs per minute or 550 ftlbs per second.
So, 1 hp is required to raise 33 lbs of coal 1,000 ft or 330 lbs 100 ft in one minute. In order to raise 1 lb of coal 33,000 ft our poor pit pony would need to gallop at a phenomenal 375 mph. Thus, one pound of weight being moved at 375 mph is demanding 1hp, or 1 lb of force moving an object at 375 mph is equivalent to 1 hp. We now have a datum against which a comparison can be made, 1 Thrust Horsepower = Lbs of thrust * speed / 375, i.e. an engine developing 20,000 lbs of thrust flying at 375 mph is producing 20,000 hp. paradoxiacally, fit the same engine into another sleeker aircraft capable of 750 mph, and we get 20,000 * 750/375 = 40,000 hp!

OK, thats enough geeky stuff for now

I had guessed that horsepower was something to do with the power output of 1 horse but I didn’t know any specifics. Thanks for the explanation, it makes sense.
I assume the figure of 375mph has no real physical significance (other than the 550ftlb/s) even though it’s almost exactly mach 0.5?

Why the ‘b’ in bhp?

Willcove (I know you can see this )

I refuse to post anything to do with planes in the BMW thread because it isn’t relevant to the topic at hand.


I agree, but the original context was with the jet/car at full forward thrust/power with the brakes trying to counteract the forward mechanical power.


Completely agree but is irrelevant in the context of the discussion. The ‘engine’ produces a force at a given velocity (power output); the brakes provide an equal (in this discussion) and opposing force at the same velocity (power transfer, usually in the form of heat). If the craft/vehicle speed remains constant (opposing forces equal, disregarding drag/rolling losses), the brakes will be dissipating the mechanical engine output – end!

Feel free to PM me if you take issue, I’m happy to post your response in this thread.

Couple of thoughts while I have me morning "brew"...

1. Is it fair to say that if the plane isn't moving the engines aren't developing any power? I agree that the definition of "power" requires something to move but I'd suggest that it's the air out of the back of the engine's that's moving. If I sat in my car, floored the throttle and dumped the clutch so that the wheels spun, I don't think it would be valid to say the engine wasn't developing any power because the car wasn't moving!

2. Just had a quick look on a plane-nerd website and it quotes the maximum take-off weight of a 747 200 as being about 375,000kg. It also quotes the maximum take-off distance as being 3100m. Assuming linear acceleration (is this fair?) we get about 1.05 m/s^2 (or about 0.1G) of acceleration to get to 80.5 m/s in 3100m. Using F=Ma it comes out as about 40,000kg of thrust. Would it be reasonable to assume the rest of the thrust was used up in overcoming air and rolling resistance?

You assume correct sir. 375mph is simply the speed at which you would have to travel in order to move Mr Watts 1lb weight over a distance of 33,000 ft in one minute....

33,000 ft = 11,000 yards = 6.25 miles. 6.25 miles in 1 minute = 375 mph. As James Watt appeared to pluck the figure 33,000 out of thin air in the first instance, it has no real significance with respect to other physical parameters.



Brake. Not sure why piston engines are rated in brake horespower, think its something to do with the power available at a brake if applied to the system.



Its an odd concept to get ones head around I agree. By strict adherence to the Power formula (Force * distance / time), a jet engine when stationary is producing no power in relation to the job the exhaust gas stream is expected to do, i.e. move the aircraft forward. Its what the aircraft/engine does thats important, and because the aircraft doesn't go anywhere, distance = 0 therefore the whole formula resolves back to zero! This is what propulsion students are taught and is the generally accepted concept within the discipline.
Power is being produced and consumed within the engien itself because the compressor is being spun by the turbine, but this is not contributing to the thrust being developed as the exhaust gasses exit the nozzle so isn't considered.



Again, it is relative concept. The engine of your car is driving the wheels which demands a certain amount of torque (force * distance), distance in this sense is that covered by the wheels as they rotate against the tarmac, therefore power is being demanded and but it is all being consumed. If your car doesn't go anywhere as a result, all of the power is being wasted, net result car goes nowhere and we're back to distance = zero, power = zero again.



Yes, there is always some air resistance and rolling resistance whilst the aircraft is on the ground. Therefore some engine thrust will alaways be consumed to overcome this 'drag'; aerodynamic drag increases with the square of speed i.e. the faster it goes, the worse the problem becomes.
Airliners fly at 40,000 for a very good reason, there is less air resistance and hence less drag, and the thinner air demands less fuel to effect combustion therefore fuel consumption is considerably reduced. Our 'simulator' realises savings of over 50% when we take it up to 40,00o ft.

BTW, did that nerdy plane website tell you just how little aerodynamic lift (per square inch of wing surface area) is being produced by the 747 at take-off speed You might be shocked to learn by just how fine the margins are as you settle into your seat and head off for sunny Lazarote

Nah! I've never studies aeronautical engineering so I can't argue that this isn't what's taught in the profession but it really doesn't work in automotive engineering. The engines ARE doing work, they're moving something with a mass through a distance in a particular time. Thefact that it's just hot air and is of no benefit to the job in hand is neither here nor there as far as the engine is concerned (and let's face it, we'd have a lot of redundant politicians if moving hot air from one place to another was not regarded as useful work)!

Hmmm. seem to have just scuppererd my ow argument there!

But getting back to the car, I could take the argument a stage further and say that I drove it to the shops to get a loaf of bread but the shop was shut so the car did no useful work and therefore didn't develop any power!

I completely agree with Riggy’s answer, but I can elaborate a little more


Correct. It’s producing a lot of hot air and thrust, the latter exerting a velocity*force on the airmass around it. This will eventually be dissipated as frictional heat and some ruffling of leaves.


The power is being used to shear the rubber compound from the tyre (torque at rotational velocity) as well as a lot of heat, some of the latter used to oxidise the compound.

The total mechanical energy will be used to accelerate the surrounding airmass and warm the brake disks (or as friction within the engine if using engine braking) when you need to decelerate again, as well as warming the tyres/gears (rolling/transmission losses). In general, eventually it all gets converted to heat.