Jet Engine Air Intake

[whitestivo]

No it's not a new intake for the sportivo. I'm always surprised how much ppl know on here, so my question

On a jet engine (say 747), how much air in metres cubed per second is consumed at typical cruising speed?

I'm after an approximate level to see if something I'm working on could work.

If it's a possible, I'd also like to know how much of that air is actually combusted, and the amount of waste product. I understand from current research that up to 44:1 compression can be achieved in a jet engine, which is amazing, so I'm assuming there's a lot of air going in there.

whitestivo

[DJKOR]

Talk about a complex question. So we'll need to be specific here. When you talk about a jet engine, we'll need to stick to a specific usage. You mentioned a 747, so we shall stick to that one then. As well, because the engines on a 747 (and other commercial airliners for that matter) are of the high-bypass turbofan design, did you want the total air that is consumed by just the engine, or the the total air that passes through (the air that is consumed by the engine and the air that bypasses the engine).

I'm curious as well now. Either way, it is bound to be a sht load of air going though the engine. Even on ground level:

[whitestivo]

both

i'd like to know how much air is consumed in the combustion process and how much passes through.

the combustion process is of particular interest to me

I've learned that these engines are more efficient in the sky, as they need the air ramming into the engine, the intakes designed certain fashion, which i think is also interesting for car applications.

i figure if they are travelling at 950km/h, and their surface area of intake is 3m, by 3m, thát's got to be a truckload of air

offline i'll tell you about how much air i'm trying to deal with

whitestivo

[the_random_hero]

both

i'd like to know how much air is consumed in the combustion process and how much passes through.

the combustion process is of particular interest to me

I've learned that these engines are more efficient in the sky, as they need the air ramming into the engine, the intakes designed certain fashion, which i think is also interesting for car applications.

i figure if they are travelling at 950km/h, and their surface area of intake is 3m, by 3m, thát's got to be a truckload of air

offline i'll tell you about how much air i'm trying to deal with

whitestivo

If you're talking about doing something for a car, the bigger you make the opening the more drag you will cause. Best is to stick with whatever size intake piping you're running with a bellmouth facing the oncoming wind.

[Guni]

Whatever you are doing, it better be able to do this when you are done else it ain't worth it!

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[Suprathai]

If its 3m wide then, you just multiply 3.14*1.5*1.5*950000*1000=6711750000L/hour. Thats assuming that the air travelling through the 3m circular opening is at 950km/hr. Then we have to take into account other factors.

Anyone here is an aerospace engineer??

[DJKOR]

If its 3m wide then, you just multiply 3.14*1.5*1.5*950000*1000=6711750000L/hour. Thats assuming that the air travelling through the 3m circular opening is at 950km/hr. Then we have to take into account other factors.

Anyone here is an aerospace engineer??

And then add to the fact that the blades are spinning at such and such RPM and then you have the blade pitch etc. This is definitely a question for the engineers out there because my maths stopped at 950km/h.

[zr nate]

you know my girlfriend calls me a geek for being on a forum. I will not tell her that we are now talking about aeronautical engineering and complex math equations!!

[whitestivo]

surpathai, you sure that's right? As converted, that's 1864 m3 per second in each engine, and that is a huge amount of air

even though originally an accountant, now working in commercial, it's amazing the breadth of work i get involved in.

i still remember being astounded with a previous similar post that the average cai took in 300l plus per second/minute, it was enormous.

with this post, i'm only looking for ballpark as to whether a project i'm thinking of has legs.

whitestivo

[Hiro]

surpathai, you sure that's right? As converted, that's 1864 m3 per second in each engine, and that is a huge amount of air

even though originally an accountant, now working in commercial, it's amazing the breadth of work i get involved in.

i still remember being astounded with a previous similar post that the average cai took in 300l plus per second/minute, it was enormous.

with this post, i'm only looking for ballpark as to whether a project i'm thinking of has legs.

whitestivo

Engineers wouldn't scoff at those numbers at all. We're used to seeing things x10^9 (giga) all the time.

[Suprathai]

bah, engineers crunch numbers all day. I'm one of them, but I'm a chemical engineer and work in the water/wastewater side of things with regards to membrane development.

Think of desalination and I have my part in it.

Don't let big numbers get to you, there not that important unless its to do with your personal savings :D

[Hiro]

I'm a drafty/design engineer in the mining vehicle industry, so I see millimetres combined with tonnes all the time, and that gives big big numbers...

[whitestivo]

Let me make sure i got this

surface area circle is piR2 = 7.065m2 if we assume depth of 1m, then we can assume 7.065 m3

where i get stuck is at 950kmh, how quickly does that void get filled, if i say once a second, then i'm dealing with 7.065m3ps volume. This is the bit I get stuck at as I expect the number to be around 70-100m3 per second.

i think what suprathai is saying, is multiply by 950000 (not sure why?), then to covert m3 to litres by * 1000

whitestivo

Edited May 27, 2009 by whitestivo

[Hiro]

Let me make sure i got this

surface area circle is piR2 = 7.065m2 if we assume depth of 1m, then we can assume 7.065 m3

where i get stuck is at 950kmh, how quickly does that void get filled, if i say once a second, then i'm dealing with 7.065m3ps volume. This is the bit I get stuck at as I expect the number to be around 70-100m3 per second.

i think what suprathai is saying, is multiply by 950000 (not sure why?), then to covert m3 to litres by * 1000

whitestivo

You're going about it the wrong way - think of flow as a cross-sectional area moving at a speed. So you'll have roughly 7m^2 moving at roughly 260m/s...multiply the two and you'll have (7x260)(m^2 x m/s), giving you 1820 cubic metres / second, or 1820000L/s

[Suprathai]

following on from that, all you have to do is divide 7065 by 1820000 and you'll get your anyswer in seconds. which means its really small by the way.

[Evo7]

I GIVE UP <_<

[Hiro]

These numbers dont seem right...

If the engine is travelling at 950km/hr = 950,000m

950,000/3.6 = 263888.8889 m/s

263888.889/1000 = 263.9m^2/s

Where did that 1000 come from? And why did it suddenly change to m^2/s?

So, we assume that the engine diameter is 3m

pi*R^2 = 3.14*1.5^2

=7.065m^2

101.3 KPa = atmospheric air pressure 101.3Kpa =

10.33Kg/m^2

So what i have is this:

263.9m^2/s*7.065m^2*(10.33Kg/m^2/1000)

= 19.26m^2kg/s

A bit out there, but i think there is more to it... i have to get back to work lol

we also have to include the variable of vacuum or compressed air being 'sucked in'

but thats all from me for now,

Evo

Your numbers are screwed - you've changed from m/s to m^2/s for no apparent reason, and then decided to introduce kilograms. Why are you bringing atmospheric pressure into it?

1) That will vary depending on altitude

2) As you said, you will get venturi effect inside the engine

3) Atmoshperic pressure isn't the density of air (they are related though)

You're ending up with meaningless numbers...m^2kg/s? What use is that? You'll either want volumetric (m^3/s) or mass (kg/s) flow rates

Edited May 29, 2009 by Hiro Protagonist

[whitestivo]

my original post

surface area circle is piR2 = 7.065m2 if we assume depth of 1m, then we can assume 7.065 m3

where i get stuck is at 950kmh, how quickly does that void get filled, if i say once a second, then i'm dealing with 7.065m3ps volume. This is the bit I get stuck at as I expect the number to be around 70-100m3 per second.

i think what suprathai is saying, is multiply by 950000 (not sure why?), then to covert m3 to litres by * 1000

---------------------------------------------------------------------------------------------------

OK, i got it now, if we take 950km/h back to metres ps, we come to 263.88 metres ps.

So if you are multiplying 263.88m/s by 7.065 m2, then you get 1864 m3ps.

This is significantly more than i expected, as I know that an Hitachi Jet turbine (net info) will consum 72m3ps at sea level, standing still, through the turbine only.

Evan, where does the atmospheric pressure come into it?

This pressure drop is approximately exponential, so that pressure decreases by approximately half every 5.6 km (18,000 ft) and by 63.2% (1 − 1 / e = 1 − 0.368 = 0.632) every 7.64 km (25,100 ft),

So using that, I can assume at 30,000 ft, when a jet engine is operating, the density of air would be 70% less, so taking that into account, then

1864*30% = 559.2m3 ps. Now I also know that only 20% of air is actually taken through the engine, and that 80% is used as compression to generate heat (not sure exaclty why), so that bring me down to 111.84m3ps in the actual turbine, getting back to closer to the hitachi numbers.

I appreciate everyones assistance as this has helped me immensely, and grown legs to a project I'll start to look at pre feasibility.

I wish I'd taken physics over biology now. Also note I've quick edited this about 8 times now to make sure it's right.

whitestivo

Edited May 29, 2009 by whitestivo

[SuperDave]

These numbers dont seem right...

If the engine is travelling at 950km/hr = 950,000m

950,000/3.6 = 263888.8889 m/s

263888.889/1000 = 263.9m^2/s

Much simpler to just go;

(60*60)/1000=3.6

950/3.6 = 264m/s

Phil, if you're going to factor in pressure changes with altitude you'll also have to factor in temperature changes. There is however no point in including these corrections if the figures you are trying to emulate are based on a different altitude.