Thrust and Power (Redux)

[StrongEagle]

Hello Folks,

While doing a Google search of ‘thrust versus horsepower’ I came upon this PPRuNe article dating back to 2005: http://www.pprune.org/tech-log/194018-pounds-thrust-horsepower.html. This article, along with others I read provided valuable information into the entire subject but missed the point of why this question is asked in the first place (although the original poster did ask for information for the ‘ice cream lickers’).

Ice cream lickers (I am one) want to know how much POWER it will take to push that big machine up to the point it will take off because, when expressed in this manner, the ‘power’ of an airplane can be compared to the power of a top fuel dragster in the quarter mile or the quarter mile/top speed times of the latest Ferrari, or maybe the power required to drive the largest container ships in the world. I don't think many are concerned about the power required to keep the plane at 33,000 ft.

So, I think this question is not answered by trying to relate thrust directly (which as the article above and several others imply, is essentially a fruitless endeavor) but rather to approach it from a simple perspective. If it were the wheels that drove the airplane down the runway, how much power would be required to accelerate to the point of Vr?

To answer this question, I went to the Boeing website and pulled up some info on the 747-400. I also did a bunch of web searches to develop this back of the envelope approach to answering the power question.

I came up with the following.

MTOW – 875,000 lbs (396,890 kg) for the non ER version and for the purposes of this calculation, I’ll take the max weight.

Time from V0 to Vr – Varies from reference to reference but around 45 seconds, as little as 20 seconds for an empty airplane into a good headwind.

Estimated acceleration – Not much out there but I found 7.3 ft/sec^2 (2.23 m/sec^2).

Take off roll distance (not F.A.R. runway length requirements) – Estimates of 6000 ft (1829 m) to 7000 ft (2134m), admittedly all SWAG’s, and hence part of the reason for this post.

Speed at Vr – 178 mph to 184 mph = 261 ft/sec to 270 ft/sec.

Engine thrust – 60,000 lbs to 63,000 lbs, irrelevant for this computation.

Assumption – Acceleration will be uniform from V0 to Vr – Probably not exactly correct since parasitic drag increases with speed, somewhat offset by more thrust once the engine is moving faster than V0, but probably OK for this back of the envelope exercise. Also, there should not be any induced drag until Vr when the weight of the airplane transfers to the wings.

Computation

The first step is to see how much variation there is in the estimates above since they are all related and to see if they properly relate to one thing known with fair certainty: Vr.

Acceleration and time, solving for velocity: V = a * t
Vr = 7.3 * 45 = 328.5 ft/sec = 224 mph (hmmm – What! It aint right on Google???)

But, if I use the 7.3 number and compute for time I get
t = Vr / a = 270 / 7.3 = 37 secs

I’ve got a range of Vr speeds and 2 take off roll distances so I see what this means in terms of acceleration.

Velocity and distance and solving for acceleration: V^2 = 2a * X
a = V^2 / 2X = (4 possible combinations as below)
a = 261^ 2/ 2 * 6000 = 5.7 ft/sec^2
a = 270^ 2/ 2 * 6000 = 6.1 ft/sec^2
a = 261^ 2/ 2 * 7000 = 4.9 ft/sec^2
a = 270^ 2/ 2 * 7000 = 5.2 ft/sec^2

Let’s see what Vr looks like with the highpoint of these acceleration numbers and the 45 seconds estimate found on the web

Vr = a * t = 6.1 * 45 = 274.5 ft/sec = 187 mph – a bit too fast but after playing around I find that if I make the take off distance I get

a = 270^ 2/ 2 * 6100 = 6.0 ft/sec^2

and

Vr = a * t = 6.0 * 45 = 270 ft/sec = 184 mph – on the money!

For fun, I took the 37 seconds and 7.3 ft/sec^2 computed above and used it to compute what the rollout distance must be to achieve a Vr of 184 mph (270 ft/sec).

X = Vr^2 / 2*a = 270^2 / 2*7.3 = 4993 ft take off length

There are lots of possible combinations of acceleration, time, and distance that provide the correct Vr – for simplicity’s sake, I’ll use just these two.

Computing Horsepower.

F = M * a
But an 875,000 lb airplane has an English units mass of
875,000 lbs / (32 ft / sec^2) = 27,344 slugs (why am I doing this in English units?)

For both examples:
F = 6.0 * 27,344 = 164064 lbs
F = 7.3 * 27,344 = 199 611 lbs

Power = (F * X) / t = (164, 064 * 6100) / 45 = 22, 239,787 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
40, 436 HP or

Power = (F * X) / t = (199, 611 * 4993) / 37 = 26,936,695 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
48,976 HP – a pretty large variation!

Conclusion

I realize this computation is completely unrelated to thrust mechanics, engine shaft speed and torque, etc, but it does provide a ‘feel’, relative to other ‘understandable’ engine sizes, of what it takes to get a large airplane into the air.

I’d appreciate any insight into the variables used to make the computations because there was always something missing. For example, the 45 second roll out time as for a 747-400 but made no reference to weight, altitude, headwinds, temperature, etc. Similarly, I have no idea if a 6100 (or 4993) foot rollout distance is reasonable for a standard day at max weight. For that matter would 184 mph (160 kts) really be the Vr for a fully loaded aircraft?

Many thanks.

[Microburst2002]

:}Is this some kind of homework?
I feel like :8 very often, but that is too much for me!
It is a shame I don't have the book here. The thrust-power conversion was clearly explained.

I think you are calculating the power developed by the airplane in a take off. It is OK that you are trying to compare with a Ferrari to have an idea of the magnitude. However I think that it makes no sense.

Any body being pushed by a force and moving is developing power, or "has" power. That power is directly proportional to the force applied and the speed, exclusively.
If at take off (about 150 kt ground speed with 60,000 pounds of thrust) an airplane develops a given power, how much power has the same airplane in cruise, flying at 500 kt groundspeed? Well, we need to know cruise thrust, I don't know, maybe 40,000 pounds?. Much more HP!
A Ferrari has to burn fuel to produce the power within the engine and then transfer it to its propulsive system (the wheels). The Ferrari cannot obtain more power than that (no wind, no slope). Twice as much fuel flow gives it twice as much power. And it has to "share" it between speed and thrust.

The 747 burns the fuel to get thrust in its propulsive system directly. Twice as much fuel flow, twice as much thrust. There is no limit to the power the airplane can reach, other than the limits to its speed and thrust. If it keeps accelerating for a constant thrust, its power will keep increasing.

Imagine a car with a 1000 hp engine. At full throttle, brakes on, it "has" no HP yet (zero speed) but it has, say, 1000 pounds of push force. After releasing brakes and having accelerated to its maximum speed (when the thrust is equal to drag plus friction) it has the 1000 HP the engine can provide it, but no longer 1000 pounds of thust. It has less, because now it has speed. The propulsive system has less and less ability to produce thrust with the horsepower from the engine as it speeds up until the thrust cannot overcome the opposing force and speed becomes constant. It could never get more than 1000 Hp even in case there were no opposing forces at all.
The same car with a 1000 pounds jet engine (trying to have a page in the Guinnes Book) also reduces somewhat its thrust with speed, but due to other factors. Thrust becomes less an less but power will increase with speed as long as speed increases. However if there are compressibility effects (high mach number) thrust will start increasing with speed! Then we would have high airspeeds and high thrust as well. That means the car would "have" more and more power.

Simply put, jet engines are thrust producing machines, they don't give us power for fuel. You cannot compare them.

Maybe you can compare fuel burnt during the take off roll of a jet airplane with that burnt by a car running the same distance. If they use different fuels, you need to find the energy released by burning a kilo of each. That would be comparing the energy required to cover that distance.
If you then find their kinetic energies you can calculate and compare the efficiency of the engines, too. You could also find the same for a whole flight/trip from A ot B, and find what is more fuel efficient per each 1,000 kg transported ¿By air or by car? ¿What contaminates more, 200 people going to the beach 400 miles away in their cars or in a 737?

haha That is homework! And of course I am no physics teacher so I cab be radically wrong!
I just felt very very :8

[barit1]

I'll take a different tack, based only on these numbers:
TOGW = 875K#
Acceleration = 7.3 ft/sec^2
Gravitational accel = 32.2 ft/sec^2
Vr = 270 ft/sec
Speed at which thrust numerically equals HP = 550 ft/sec

Then:
Acceleration in G's = 7.3/32.2 = 0.227 G
Thrust to achieve .227 G = 875000*0.227 = 198400 lb
Vr as a fraction of 550 = 270/550 = 0.491
HP at Vr = 198400*0.491 = 97381

IMPORTANT: The horsepower in this example is NOT constant. At the instant after brake release, horsepower is ZERO. It increases linearly with speed; the value cited is the number at Vr.

[StrongEagle]

haha That is homework! And of course I am no physics teacher so I cab be radically wrong!
I just felt very very http://images.ibsrv.net/ibsrv/res/src:www.pprune.org/get/images/smilies/nerd.gifYes, you are very, very radically wrong. I was on the ferry from Tanah Merah (near Changi Airport) on my way to play golf in Batam. There were 4 Koreans with me on the way to play golf as well. We were riding on the top deck, under open skies.

We watched an SQ 777 take off from Changi and the Koreans were talking about it. I mentioned that it had about 90,000 lbs of thrust per engine, or 180,000 lbs total. A few minutes an SQ 747-400 took off. We watched it and they asked my 'how much'. About 60,000 lbs and engine or 240,000 total, I said. Then a Quantas A380 took off and naturally the question arose again. I replied that each engine was around 80,000 lbs of thrust for 320,000 total.

Then they wanted to know how much 'power' that was. I've read lots about power/thrust and lack of direct relationship and I could not answer them. But yesterday it occurred to me that a satisfactory answer could be obtained in the form a the laws of motion. What I needed was some numbers to plug into them.

So, -2 for you... not homework, an attempt to answer a question for the next set of Korean golfers that ride over to Batam.

Simply put, jet engines are thrust producing machines, they don't give us power for fuel. You cannot compare them.I am assuming you read my post wherein I stated at the beginning "So, I think this question is not answered by trying to relate thrust directly (which as the article above and several others imply, is essentially a fruitless endeavor) ..."? You see, I am well aware that there is no simple relationship between thrust and power. You might try Aerospaceweb.org | Ask Us - Convert Thrust to Horsepower (http://www.aerospaceweb.org/question/propulsion/q0195.shtml) fopr a decent write up on some of the difficulties.

But the remainder of your quote is simply not correct. Of course, engines give us power for fuel. It is the power that creates the thrust. It's just not a nice neat relationship like converting miles per hour to furlongs per fortnight. This is why I chose to use the simple equations of motion with constant acceleration to model how much power it takes to get a 747-400 to the point that it will fly.

And of course I am no physics teacherI get that. ALL I want to do is compute how much power it takes to roll an 875,000 lb airplane from V0 to Vr (whatever that is). If it were a textbook physics problem it could just as easily be a big block of lead being accelerated down a frictionless plane.

To do this I must know at least 2 or 3 variables (with some reasonable guestimate of temperature (standard day plus 30F?), wind (none?) and other factors that actually go into power settings, rollout, etc.

The point of this whole exercise is not to try to relate power to thrust but to have some 'fun' talking points about how cool it is to fly big iron.

Cheers.

[StrongEagle]

IMPORTANT: The horsepower in this example is NOT constant. At the instant after brake release, horsepower is ZERO. It increases linearly with speed; the value cited is the number at Vr.You are correct. I used a non time dependent equation to compute average HP. Yours is the better answer. While I was typing that it I forgot the old adage that HP is the rate at which torque or force is applied. Thanks.

[Microburst2002]

I am so sorry that I have disturbed you :(

I did not read all the thread, that is why I said it was too much homework. And I missed the "if it was a car"
I'm sorry for having bothered you. -1 for me.
I just think it is fun thinking about these matters. I like it. I used to do it a lot, with paper, pen and a calculator. Sometimes with a friend, writing in the back of a newspaper. I was not trying to teach you anything new, only to share the fun with you, because I like it too.

As for the -2... Why -2?

I still remain convinced that piston engines are power producing machines. Yes, we have to take into account a lot of losses, thermal efficiency, etc... But they deliver power per each pound per hour. That's a fact. After that you can do whatever you want with that power. Like producing pounds of thrust and speed, for instance.

In stead of remarking my own humble "I am no physics teacher" you should study take off physics a little bit more, in the first place, if you are so interested in the subject. You would learn, for instance, that you need integrals to compute thrust, acceleration, distance and Lift as they change all throughout the take of roll (Why did you think there is no lift until VR?).

But you want to know how much power would a 747 engine require to take off if it was a big car engine coupled to the wheels, right?

I suggest you to find the take off roll distance and the average thrust during the take off roll for a typical 747. Multiply them (energy delivered to the 747) and divide the result by the time required. That is the power used (rate at which Work is applied).

:)

[barit1]

I still remain convinced that piston engines are power producing machines. Yes, we have to take into account a lot of losses, thermal efficiency, etc... But they deliver power per each pound per hour. That's a fact. After that you can do whatever you want with that power. Like producing pounds of thrust and speed, for instance.

You are obviously correct - pistons and crankshafts produce RPM and torque, which is of course power.

But it's the propeller that creates thrust to make the plane accelerate. As a very rough rule of thumb, 1.0 hp from the engine creates 2.5 lb. prop thrust at static conditions. And thrust drops off as the plane accelerates.

So if we want to take that 875K# 747 and power it with enough pistons to create 198400# thrust, the answer is 198400/2.5 = 79360 shp.

The difference (79360 vs 97381 hp) is that the typical prop, at static-to-Vr conditions, is pretty efficient compared to a turbofan. At higher speeds, not so much. :ooh:

[BubbaMc]

Piston engines produce TORQUE. Power is simply a way of describing the rate at which work is done, and can be derived directly from the torque vs rpm curve.

[Capt Pit Bull]

The problem here is that the folks that asked the OP what the 'power' was probably were not using the term in its formal sense anyway.

As far as answering the question.... well, since the velocity of the aircraft is variable, the power being developed at different points during the take off run is also variable.

If you want an answer for 'the' (singular) power I guess you could give an average value reasonably easily:

Kinetic Energy = 1/2 M V^2 (all in S.I. approximations)
= 1/2 x 400,000 x 93^2 = 1,730,000,000 (1.73 Gigajoules)

Power = Energy / Time = 1,730,000,000 / 45 = approx 38.5 Megawatts = approx 51,600 horsepower.

Meerkats says "Simples!"

Power = (F * X) / t = (164, 064 * 6100) / 45 = 22, 239,787 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
40, 436 HP or

Power = (F * X) / t = (199, 611 * 4993) / 37 = 26,936,695 ft-lbs / sec
1 HP = 550 ft-lbs / sec so this is
48,976 HP – a pretty large variation!


I wouldn't sweat it ;)

If you were an astrophysicist anything within 3 orders of magnituade would be bang on :)

pb

[Microburst2002]

Barit1
The thrust produced by a propeller (or by wheels) is limited by the power delivered to it in the form of torque and rpm. It follows that if the instantaneous power delivered to a propeller (or wheels) remains constant, the thrust has to linearly decrease as the vehicle speeds up, as you said.

The thrust produced by a jet engine, however, is not limited in that way. Speed affects the net thrust output indeed, but in a different way. It can even increase it by compressibility effects.

The power delivered by a piston engine to a vehicle propulsive system is constant with speed. The power acquired by a jet engine vehicle is not. That is why we cannot relate them easily. We cannot compare them at brake release, because the jet engine has no speed ergo no power. You can compare a Ferrari with a Super Constellation, but not with a 747.

Using the simple formulas relating Energy, Work, Power, Force, Distance and Time we can calculate how much power has been required to carry out any given take off roll, as Pit Bull and I said. And let's also apply the propulsive efficiency and the figures will increase to the Koreans golfers amazement.

By the way. When skipping in the gym the machine shows the calories burnt during the excercise. Now, these calories are not taking into account the human body thermal efficiency, are they? I hope not, otherwise it means that, if I eat that one of those delicious 200 Kcal cookies that call me from the fridge I'll have to run 10 km to burn it! What is the human body "efficiency" more or less?

[Capt Pit Bull]

My take on this whole power / thrust thing

Any powerplant can do work, and the rate that it does so is power. Any power plant can deliver a driving force. Theres nothing special about torque, its just the rotational equivalent of a force.

When it comes to an aircraft power plant, the issue is: How does its behaviour vary with speed.

With the engines throttle/ thrust/ rpm lever/s etc left alone it is a simple approximation to view Propellor engined aircraft as having roughly constant power output with regards to speed. Similarly, Jet engines can be viewed as having roughly constant thrust output with regards to speed.

Both produce power, both produce thrust. Its simply a case of what is a simple way to visualise the engine in question.

Unfortunately the 'sound bite' for this is "Jet engines produce thrust. Propeller engines produces power"

As you can see, a few words too many have been taken out

"When you vary speed, jet engines produce approximately constant Thrust, propeller engines produce approximately constant power."

The other factor with powerplants is producing instrumentation. Measuring thrust for a prop isn't that simple. Measuring torque is a bit easier. Measuring RPM easier still. For a jet, measuring thrust is practical (via EPR).

pb

[Microburst2002]

The explanations about this matter have never satisfied me completely.
Thanks for your post, Pb.

We all seem to know the basic principles, work, power. Of course anything that produces force or work produces power. But I have difficulties in grasping why the difference between them is better visualised labelling them as "power producing" and "thrust producing" engines, respectively.
I always thought it had to do with the way energy was put into the engine, either in strokes or continuously, which made FF to be proportional to power output in pistons and to thrust in jets.
But in the end the fuel flow is pretty continuous in both engines.
Besides if fuel is energy then fuel flow is power. Since fuel is the source of energy of powerplants, the power achieved by any vehicle powered by an engine is limited by its fuel flow and its efficiency (thermal, mechanical and propulsive). All engines are "power producing". So why cannot we properly compare jet and piston airplanes?

Answered:
On the one hand, the behaviour of thrust with speed with a fixed throttle/thrust lever: in the pistons it decreases noticeably with speed whereas in the jets it is approximately constant.
On the other hand, the difficulties in measuring thrust in piston engines (or power in jet engines).

I guess we can always compare fuel flows, although taking into account efficiency and fuel heat power.

By the way, I have read somewhere that the A380 has Thrust gauges, along with the usual N1 or EPR ones. Does anyone know how they calculate thrust? Does its thrust lever angle demands N1, EPR or thrust? I would love to have them in my airplane.