calculating thrust at altitude
Thread Starter
Joined: May 2000
Posts: 326
Likes: 1
From: UK mainly
calculating thrust at altitude
Returning to Manchester today above the headwinds we were at FL490 (!) Lear 40; whilst with glazed eyes was wondering ... this question for a moment and it just popped into my head ... how much thrust on a 3500Ib rated engine is actually being produced at this sort of level?
Anyway whatsover to answer this or is it rocket science?
Anyway whatsover to answer this or is it rocket science?
Joined: Mar 2006
Posts: 13
Likes: 0
From: Ormond Beach
Well, keep in mind that anything that uses oxygen for combustion will lose power with increase of altitude. The figure given to us for the CF34 was that at 35,000 feet it's down to roughly 40% rated thrust.
Joined: Jun 2005
Posts: 1,420
Likes: 1
From: AEP
Thrust and TSFC etc.
Hola Dean...
xxx
I could help you a little if you were able to give me the TSFC of your type of engine, and its rated thrust at sea level... How I explain this in classroom for the 747 engines is as follows...
xxx
747-287 - Engine P&W JT9D-7Q
xxx
Maximum rated thrust sea level = 52,700 lbs
TSFC thrust specific fuel consumption = .378
Fuel flow at takeoff = 19,915 lbs (to get 52,700 lbs of thrust)
xxx
So if in cruise, FF is 4,000 lbs, thrust is about 10,500 lbs
xxx
Happy contrails
xxx
I could help you a little if you were able to give me the TSFC of your type of engine, and its rated thrust at sea level... How I explain this in classroom for the 747 engines is as follows...
xxx
747-287 - Engine P&W JT9D-7Q
xxx
Maximum rated thrust sea level = 52,700 lbs
TSFC thrust specific fuel consumption = .378
Fuel flow at takeoff = 19,915 lbs (to get 52,700 lbs of thrust)
xxx
So if in cruise, FF is 4,000 lbs, thrust is about 10,500 lbs
xxx
Happy contrails
Joined: Jan 2000
Posts: 347
Likes: 0
From: sussex
Surely, in level flight at ANY altitude, lift will equal weight ? Since lift and drag are affected by the same parameters, (speed, density etc) the drag is directly related to lift and drag is opposed by thrust.
Therefore, for any given weight and constant mach number, the lift, drag and thrust will unaffected by altitude.
On the 747-400 (GE engines) at a weight of 310.0 kg and mach .86, fuel burn per engine at 39.0 is 2830 kg/hr and at 26.0 is 3175, the 10% improvement at altitude largely due the higher - and more efficient - rpm required to produce the same thrust.
Wouldn't this suggest that the engine thrust was more-or-less the same at both altitudes.................or am I missing something ?
Therefore, for any given weight and constant mach number, the lift, drag and thrust will unaffected by altitude.
On the 747-400 (GE engines) at a weight of 310.0 kg and mach .86, fuel burn per engine at 39.0 is 2830 kg/hr and at 26.0 is 3175, the 10% improvement at altitude largely due the higher - and more efficient - rpm required to produce the same thrust.
Wouldn't this suggest that the engine thrust was more-or-less the same at both altitudes.................or am I missing something ?
Joined: Jun 2004
Posts: 1,843
Likes: 0
From: Australia
dynamite dean,
I would exercise great caution in relating any of your performance numbers to other "Full Fan" aircraft (e.g. B747), at RAS 197 at F/L 490 and Mach Number of M0.799, you would have a very substantial ram recovery factor for your aircraft, which has a centrifugal compressor.
The ram recovery factot at M0.799 for an engine with a centrifugal compressor stage would be very significant indeed.
Regards,
Old Smokey
I would exercise great caution in relating any of your performance numbers to other "Full Fan" aircraft (e.g. B747), at RAS 197 at F/L 490 and Mach Number of M0.799, you would have a very substantial ram recovery factor for your aircraft, which has a centrifugal compressor.
The ram recovery factot at M0.799 for an engine with a centrifugal compressor stage would be very significant indeed.
Regards,
Old Smokey
Joined: May 2003
Posts: 412
Likes: 0
Just some thoughts....
Dynamite, keep in mind the 3500 lb rating for the 731-20 engine on the Lear 40, at SL ISA conditions, is for an N1 of 85% or so. Maximum cruise N1 is normally 100% N1. So to compare thrust at altitude with SL thrust involves more than just pressure and mach differences, but N1 also.
Flyboyike, if the CF34 produces 40% of it’s SL rated thrust at 35,000 feet, that’s quite an engine. Somehow I think that figure should be closer to 25% as a guess, perhaps less.
Rigpiggy, not sure what the “b” stands for in “bsfc”. The term I’m used to is TSFC as in “Thrust specific fuel consumption”. That is, lbs of fuel burned, per lb of thrust produced, per hour.
BelArgUSA, I’m not sure fuel flow is linear with thrust as altitude changes, so your calculation may give a ballpark figure only. TSFC of .378 sounds too good to be true for cruise altitudes, perhaps it’s SL only. In high altitude cruise, .58 is about as low as you go.
XP Morten, For this graph to be of any help, one need to know what “efficiency” means. I’m quite sure it does not mean TSFC.
Hawk
Flyboyike, if the CF34 produces 40% of it’s SL rated thrust at 35,000 feet, that’s quite an engine. Somehow I think that figure should be closer to 25% as a guess, perhaps less.
Rigpiggy, not sure what the “b” stands for in “bsfc”. The term I’m used to is TSFC as in “Thrust specific fuel consumption”. That is, lbs of fuel burned, per lb of thrust produced, per hour.
BelArgUSA, I’m not sure fuel flow is linear with thrust as altitude changes, so your calculation may give a ballpark figure only. TSFC of .378 sounds too good to be true for cruise altitudes, perhaps it’s SL only. In high altitude cruise, .58 is about as low as you go.
XP Morten, For this graph to be of any help, one need to know what “efficiency” means. I’m quite sure it does not mean TSFC.
Hawk
Joined: Jan 2005
Posts: 319
Likes: 0
From: north
Hawk,
Thats the overall total efficency;
n = (speed)/(TSFC * Specific energy fuel)
As for the TFE731-2 engine (not -20) I have some data on it.
SLS SFC = 0,493
Cruise SFC = 0,87
With alatriste's FF of 817, that should give the TFE731
about 940 lb thrust in cruise - or about 25% of rated like you suggested
Small typo there;
TSFC = lb/hr/lbf or kg/hr/kN
Cheers,
M
Thats the overall total efficency;
n = (speed)/(TSFC * Specific energy fuel)
As for the TFE731-2 engine (not -20) I have some data on it.
SLS SFC = 0,493
Cruise SFC = 0,87
With alatriste's FF of 817, that should give the TFE731
about 940 lb thrust in cruise - or about 25% of rated like you suggested
I’m used to is TSFC as in “Thrust specific fuel consumption”. That is, lbs of fuel burned, per lb of thrust produced, per hour.
TSFC = lb/hr/lbf or kg/hr/kN
Cheers,
M
Last edited by XPMorten; 9th July 2007 at 19:14.
Joined: Jun 2005
Posts: 1,420
Likes: 1
From: AEP
TSFC of some jet engines
If of any interest to you, here is the TSFC of some engines...
Sea level, max takeoff power...
xxx
CF6-50E (747) = .36
JT3D-3B (707 - DC8) = .53
JT8D-7A (727 - 737 - DC9) = .58
JT3C-6 (old 707, old DC8) = .74
CJ610-4 (Learjet 20 Series) = .99
xxx
Happy contrails
Sea level, max takeoff power...
xxx
CF6-50E (747) = .36
JT3D-3B (707 - DC8) = .53
JT8D-7A (727 - 737 - DC9) = .58
JT3C-6 (old 707, old DC8) = .74
CJ610-4 (Learjet 20 Series) = .99
xxx
Happy contrails
Joined: May 2003
Posts: 412
Likes: 0
Quote:
I’m used to is TSFC as in “Thrust specific fuel consumption”. That is, lbs of fuel burned, per lb of thrust produced, per hour.
Small typo there;
TSFC = lb/hr/lbf or kg/hr/kN UNQUOTE
XP, I don't see any typo in what I had written in the first line. It appears to me to be the same as what you wrote, although for clarity I might write your TSFC equation as:
TSFC = (lb/hr)/lbf
Here's a link that explains it a bit more if anyone is interested:
http://www.grc.nasa.gov/WWW/K-12/airplane/sfc.html
Hawk
I’m used to is TSFC as in “Thrust specific fuel consumption”. That is, lbs of fuel burned, per lb of thrust produced, per hour.
Small typo there;
TSFC = lb/hr/lbf or kg/hr/kN UNQUOTE
XP, I don't see any typo in what I had written in the first line. It appears to me to be the same as what you wrote, although for clarity I might write your TSFC equation as:
TSFC = (lb/hr)/lbf
Here's a link that explains it a bit more if anyone is interested:
http://www.grc.nasa.gov/WWW/K-12/airplane/sfc.html
Hawk
Joined: Jul 2007
Posts: 21
Likes: 0
From: Germany
I hope you consider the change the change of TSFC over altitude and Mach number as well as thrust setting. Thrust and TSFC are quite ugly functions of both. For a JT9D I would set the TSFC at common cruise altitudes and speeds to about 0.65 to .7
Thrust decays with altitude, linearly with density. On the other hand it changes with Mach number, strongly depending on the bypass ratio (high bypass > decay; no bypass > increase). For a low (<3.5) bypass engine I would say 20 to 30% is the right ballpark, but exact values are hard to obtain, normally only from manufacturer charts (which are well hidden).
Thrust decays with altitude, linearly with density. On the other hand it changes with Mach number, strongly depending on the bypass ratio (high bypass > decay; no bypass > increase). For a low (<3.5) bypass engine I would say 20 to 30% is the right ballpark, but exact values are hard to obtain, normally only from manufacturer charts (which are well hidden).
Joined: Jan 2005
Posts: 319
Likes: 0
From: north
Made some very unofficial calculations based on data that I have.
FF's were based on NOMINAL weights cruise at FL370 for the
various acf types.
So, cruise thrust in percent of rated
B752 18,0%
B773 14,7%
MD80 21,4%
B744 14,9%
B763 15,0%
A319 19,6%
A340 16,8%
Cheers,
M
FF's were based on NOMINAL weights cruise at FL370 for the
various acf types.
So, cruise thrust in percent of rated
B752 18,0%
B773 14,7%
MD80 21,4%
B744 14,9%
B763 15,0%
A319 19,6%
A340 16,8%
Cheers,
M
Joined: Mar 2008
Posts: 6
Likes: 0
From: L
There is an empirical relationship for thrust versus altitude. It is based on the fact that decreasing air density causes a decrease in thrust i.e. there is physically less air to compress and expand hence the thrust drops.
The simple equation for this is taken from:
Martin E Eshelby, "Aircraft Performance - Theory & Practice", Arnold, 2000
F/Fsea level=a^x
F= Thrust at chosen altitude
Fseal level = Thrust value at sea level
a = air density ratio (known as sigma)
x =0.7 in troposphere
x=1.0 in stratosphere (~36000ft)
Page 6 in this link will give the the density ratio for different altitudes:
http://home.anadolu.edu.tr/~mcavcar/common/ISAweb.pdf
So for 3500lb of thrust at 40,000 ft we have:
F = (3500)(0.2462)^1 = 861.7lb
Of course this is just an empirical relationship but can give you a rough idea.
Hope it helps!
The simple equation for this is taken from:
Martin E Eshelby, "Aircraft Performance - Theory & Practice", Arnold, 2000
F/Fsea level=a^x
F= Thrust at chosen altitude
Fseal level = Thrust value at sea level
a = air density ratio (known as sigma)
x =0.7 in troposphere
x=1.0 in stratosphere (~36000ft)
Page 6 in this link will give the the density ratio for different altitudes:
http://home.anadolu.edu.tr/~mcavcar/common/ISAweb.pdf
So for 3500lb of thrust at 40,000 ft we have:
F = (3500)(0.2462)^1 = 861.7lb
Of course this is just an empirical relationship but can give you a rough idea.
Hope it helps!
Joined: Jun 2025
Posts: 1
Likes: 0
From: Mandurah
Returning to Manchester today above the headwinds we were at FL490 (!) Lear 40; whilst with glazed eyes was wondering ... this question for a moment and it just popped into my head ... how much thrust on a 3500Ib rated engine is actually being produced at this sort of level?
Anyway whatsover to answer this or is it rocket science?
Anyway whatsover to answer this or is it rocket science?
You could calculate how much maximum thrust the engines of an aircraft has knowing absolute ceiling at a certain weight and speed, and knowing the lift over drag ratio at that speed.
For example, take a theoretical airliner that at a weight of 200,000lb, having a lift over drag ratio of 20:1 at its absolute ceiling would have 10,000lb of thrust in total. Any less thrust, and it would lose altitude, any more and it could gain either altitude or speed.
Joined: Apr 2004
Posts: 2,166
Likes: 86
From: Planet Earth
Not sure of the actual answer, but most of the answers here are attempts based on cruise thrust, not maximum thrust.
You could calculate how much maximum thrust the engines of an aircraft has knowing absolute ceiling at a certain weight and speed, and knowing the lift over drag ratio at that speed.
For example, take a theoretical airliner that at a weight of 200,000lb, having a lift over drag ratio of 20:1 at its absolute ceiling would have 10,000lb of thrust in total. Any less thrust, and it would lose altitude, any more and it could gain either altitude or speed.
You could calculate how much maximum thrust the engines of an aircraft has knowing absolute ceiling at a certain weight and speed, and knowing the lift over drag ratio at that speed.
For example, take a theoretical airliner that at a weight of 200,000lb, having a lift over drag ratio of 20:1 at its absolute ceiling would have 10,000lb of thrust in total. Any less thrust, and it would lose altitude, any more and it could gain either altitude or speed.
The question was asked 18 years ago, is this a record ?