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BA 77
4th Jan 2010, 18:41
I wasn't sure where to ask this so I'll just place it here.

I've bought "The Jet Engine" book by Rolls Royce and it states that 75% of engine trust is generated by the air going through the bypass duct. I was wondering if that figure falls as you climb (as the air gets thinner) or whether that 75% figure remains roughly constant regardless of altitude.

Thanks, BA 77.

vikrant soni
4th Jan 2010, 19:52
BA 77,


At high altitudes the total thrust produced by the engines will be less. Altitude will not affect the 75% ratio of thrust produced by bypass.

However the airspeed will affect it. At high airspeeds the high bypass turbo fans experience more MOMENTUM DRAG than the rise in thrust due to RAM EFFECT.

whereas low bypass engines at high speeds still can impart much more additional velocity to air, as compared to high bypass whose fan can't accelerate air to velocities,the inner jet engine can.

So for a low bypass engine the thrust remains almost constant at all speeds whereas a high bypass engine produces less thrust at high airspeeds.

So i guess the Percentage for high bypass turbofans decreases with airspeed.

Vicky :)

BA 77
6th Jan 2010, 06:44
Thanks for the quick reply.

Basically, does that mean that the core of the engine produces a higher proportion of thrust, the faster you fly and the only reason the bypass doesn't produce as high a proportion is because it can't accelerate the air as much as the core?

Thanks again, BA 77.

blackmail
6th Jan 2010, 08:16
hello all,
the bypass ratio, because it is a ratio or a fraction, by definition has no units.
eg. a bypass ratio(called the greek letter of lambda) of 8, means that for 1 unit of air going through the core engine for combustion, you have 8 units of free(no money required)air going through the fan. altitude & density affects both terms in the same way, so the bypass ratio remains the same. there is a simple formula(don't ask me to prove it) that says : delta thrust(F) = square root of(1 + bypass ratio). taking same example from above : delta F = sq.root of (1+8) = 3. so the high by pass engine produces 3 times more thrust than a straight(noisy) jet engine.
ram drag(= massflow x True aispeed)indeed as said in another post, reduces thrust output, but this is compensated a bit by the saturated thrust that comes into play only at higher altitudes for high bypass engines. so it is more or less a matter of give & take, resulting in thrust output remaining +/- constant with speed at high altitude. you will find definition of saturated thrust in the roll's book.
bm.

BA 77
6th Jan 2010, 17:05
Thanks a lot that really helped, especially the formula. So, just to clarify, the amount of thrust produced by the bypass air proportionally is not affected by the altitude/air density but is affected marginally by changes in airspeed.

Thanks again, BA 77.

vikrant soni
6th Jan 2010, 18:04
BM

please correct me if i am wrong.

The bypass ratio basically remains same along all conditions, what varies is the thrust produced by the engine, and that depends on speed of the high bypass turbofans. In case of low bypass its thrust is constant along all speeds approx.

The faster the high bypass engine travels the lower the thrust due to increase in the momentum drag due to the Fan. (jeppesen ATPL Vol. 9 chapter 3)

So the bypass ratio remains same but the thrust ratio will change with speed (as what i understand).
:)

Vicky

Ocampo
8th Jan 2010, 01:47
The faster the high bypass engine travels the lower the thrust due to increase in the momentum drag due to the Fan.

And you could climb a bit higher so that air is a bit less dense which in turn would create less drag, correct?

vikrant soni
8th Jan 2010, 08:55
Ocampo,

Well not exactly, the momentum Drag is not due to the air molecules, but due to the forward velocity of the airplane.

The jet engine has to impart additional velocity to air to produce thrust.

When an aircraft is at rest all the velocity that is imparted to the molecules is turned into thrust, but when the aircraft is moving it has to impart additional velocity which has to be in excess of the aircraft velocity to produce thrust.

IF THE VELOCITY IMPARTED TO EXHAUST IS LESS THAN THE AIRCRAFT VELOCITY IT WILL PRODUCE DRAG NOT THRUST AS IN ENGINE FAILURE.

So just the fan is unable to accelerate air at that high velocities because of lack of jet mechanism as in the inner core, hence less thrust at high speeds.

Vicky

Ocampo
8th Jan 2010, 14:43
Thank you very much for the clarification :)

BA 77
8th Jan 2010, 15:47
Is the drop in thrust from the bypass duct a significant amount or is it a relatively small drop in thrust output?

Thanks, BA 77.

singpilot
8th Jan 2010, 18:14
Seems to me, I remember a GE CF34 number like 8500ish lbs thrust at sea level, 1000ish at altitude.

I did find a chart for the cruise thrust of several engines.

CFM56-7B18 SL= 18,500 CRUISE= 5905

TFE731-60 SL= 5,000 CRUISE= 1120

I do know the overall bypass ratio will come into play, as well as the dynamic efficiency of the fan itself.

Seems that the rule of thumb was that much above FL350 (area of max-Q), most of the work was being done by the core.

vikrant soni
9th Jan 2010, 13:23
@ Ocampo

You are most welcome.

@ BA77

The thrust drop is compensated by RAM RISE, however it is not enough in high bypass engines.

For initial high altitudes like FL300 there is a drop but not very significant. However for high altitudes like above FL350 this drop might be significant.

At low altitudes most of the work is done by the FAN, at high altitudes most of the work is done by the jet engine core.

Vicky :)

BA 77
9th Jan 2010, 17:12
Thanks, that's been really helpful.

BA 77.

Lightning Mate
9th Jan 2010, 17:19
"you have 8 units of free(no money required)air going through the fan"

Really!!!!!!

Please explain the source of energy to drive the fan.......

Ocampo
9th Jan 2010, 18:02
Please explain the source of energy to drive the fan.......

Errr...the N2 turbine that is driven by the hot gases exiting the combustion chamber?

barit1
10th Jan 2010, 01:38
Please explain the source of energy to drive the fan.......
Errr...the N2 turbine that is driven by the hot gases exiting the combustion chamber?

Not quite. Consider the nomenclature:

On a two-spool machine (P&W, GE, CFMI, GP7000...) it's the final multi-stage turbine that drives N1 (fan rpm). Also called the LP turbine.

Upstream of that, at the combustor exit, is the HP turbine which is either 1 or 2 stages. It turns the HP compressor, a.k.a. N2.

In the R-R machine there's an intermediate turbine - and the gas flow though the turbine passes through N3(HP), then N2(IP), then N1(LP).