About the 737 engine power
Thread Starter
Join Date: Mar 2009
Location: Shanghai
Posts: 9
Likes: 0
Received 0 Likes
on
0 Posts
About the 737 engine power
hello all
I have a question that seem a little bit trick.
what's the power for the 737-800? I looked up in the operation manual, it doesn't mentioned. It just tell that the thrust of the engine is 26K lbs.
I'm thinking why the turbofan engine need to express in thrust.
But in daily life, the car's engine is usually express in horsepower or KW, or torque.
thank you for the help.
I have a question that seem a little bit trick.
what's the power for the 737-800? I looked up in the operation manual, it doesn't mentioned. It just tell that the thrust of the engine is 26K lbs.
I'm thinking why the turbofan engine need to express in thrust.
But in daily life, the car's engine is usually express in horsepower or KW, or torque.
thank you for the help.
Well you car engine has to turn a gearbox to get to your tires to move you, so torque against this rotating resistance is important. Once you get up to a healthy speed you need energy expressed in horsepower to keep you moving against windage and powertrain friction.
In an aircraft you basically need force (thrust) to overcome gravity thus providing both forward speed and lift to the wings. Tis true that you can use formale to convert any instantaneous energy balance to horsepower and torque, but since its changing all the time there is no meaningful number.
Watch for any updates from the engineers on this forum
In an aircraft you basically need force (thrust) to overcome gravity thus providing both forward speed and lift to the wings. Tis true that you can use formale to convert any instantaneous energy balance to horsepower and torque, but since its changing all the time there is no meaningful number.
Watch for any updates from the engineers on this forum
Thread Starter
Join Date: Mar 2009
Location: Shanghai
Posts: 9
Likes: 0
Received 0 Likes
on
0 Posts
Thank you for the help.
I have find a formula that I can not totally understand which is:
Fnet = m(vjfe - va )
m = intake mass flow
vjfe = fully expanded jet velocity (in the exhaust plume)
va = aircraft flight velocity
accroading to my memory， the formula for the force in basic physics is, F*t=m*v
But here, it show's that F=m*v.
Plz help me slove my comfuse, thanks!
I have find a formula that I can not totally understand which is:
Fnet = m(vjfe - va )
m = intake mass flow
vjfe = fully expanded jet velocity (in the exhaust plume)
va = aircraft flight velocity
accroading to my memory， the formula for the force in basic physics is, F*t=m*v
But here, it show's that F=m*v.
Plz help me slove my comfuse, thanks!
Join Date: Nov 2008
Location: The New Forest, UK
Posts: 228
Likes: 0
Received 0 Likes
on
0 Posts
Form a physical stand point engine thrust is just a force - the max thrust rating is the max force the engine can develop static thrust is when engine is still. This is not power.
The power developed by the engine(s) depends on airplane speed
Engine Power= Thrust x Speed = (total drag) x speed
For low drag designs and cruising in thin air a lower thrust is required for a given flight speed than when flying at sea level in a high drag aeroplane.
Drag is composed of profile and aerodynamic induced drag. Added togther gives (total drag) which equals the thrust required from the engine.
So you can see that thrust on its own is meaningless.
Watch out for units, in MKS system , Thrust is in Newtons, speed in meters/s gives power in watts ( divide by 1000 for kilowatts)
Where does this power ( energy) go? It goes into moving air ( air kinetic energy) and heating the airplane skin thru friction- in outer space where there is no drag no engine power is required.
Even with zero profile drag keeping an airplane in the air requires power becuase lift is a result of momentum exchange from the downwash created by the wings. So lift is no free lunch.
The power developed by the engine(s) depends on airplane speed
Engine Power= Thrust x Speed = (total drag) x speed
For low drag designs and cruising in thin air a lower thrust is required for a given flight speed than when flying at sea level in a high drag aeroplane.
Drag is composed of profile and aerodynamic induced drag. Added togther gives (total drag) which equals the thrust required from the engine.
So you can see that thrust on its own is meaningless.
Watch out for units, in MKS system , Thrust is in Newtons, speed in meters/s gives power in watts ( divide by 1000 for kilowatts)
Where does this power ( energy) go? It goes into moving air ( air kinetic energy) and heating the airplane skin thru friction- in outer space where there is no drag no engine power is required.
Even with zero profile drag keeping an airplane in the air requires power becuase lift is a result of momentum exchange from the downwash created by the wings. So lift is no free lunch.
Last edited by b377; 11th Mar 2009 at 10:25.
Join Date: Jan 2008
Location: London, England
Age: 56
Posts: 300
Likes: 0
Received 0 Likes
on
0 Posts
Quite a good explanation here:
Aerospaceweb.org | Ask Us - Convert Thrust to Horsepower
Aerospaceweb.org | Ask Us - Convert Thrust to Horsepower
Thread Starter
Join Date: Mar 2009
Location: Shanghai
Posts: 9
Likes: 0
Received 0 Likes
on
0 Posts
Thank you all guys.
And this make me enlightened. Also I think over the formula I mentioned about the relationship between the force(thrust) and the velosity.
Fnet = m(vjfe - va )
The m is the mass of air in a certain time goes into the turbo.That means in a length of time t，the total mass of the air should be m*t, which I called it M. So accroading the Momentum law：
F*t=Mv
－－ F*t=m*t*v
－－ F＝mv
So the formula Fnet = m(vjfe - va ) make sense.
And this make me enlightened. Also I think over the formula I mentioned about the relationship between the force(thrust) and the velosity.
Fnet = m(vjfe - va )
The m is the mass of air in a certain time goes into the turbo.That means in a length of time t，the total mass of the air should be m*t, which I called it M. So accroading the Momentum law：
F*t=Mv
－－ F*t=m*t*v
－－ F＝mv
So the formula Fnet = m(vjfe - va ) make sense.
Last edited by jjleo; 11th Mar 2009 at 15:51.
Join Date: Jun 2004
Location: Australia
Posts: 1,843
Likes: 0
Received 0 Likes
on
0 Posts
Aircraft don't need Power, they never have, and probably never will.
Aircraft need Thrust, they always have, and probably always will.
This applies from the Wright Flyer through to the Airbus A380.
Take a look at Page 1 or 2 of any Private Pilot's Theory of Flight text-book, the 4 basic forces are shown, Thrust, Lift, Drag, and Weight. Power does not rate a mention.
Piston engines produce Power DIRECTLY, applying torque to the propeller enabling the propeller to produce Thrust, what the aircraft needs.
Jet engines produce Thrust DIRECTLY, and without any added complication of conversion, is what the aircraft needs.
In both types of engines, there's a great deal of Power needed internally to drive the engine, but this power is not available for aircraft performance.
Power = Force (Thrust) X Velicity (TAS).
For the piston engine then, Thrust = Power divided by TAS. As Speed increases, Thrust decreases.
For the Jet engine, Thrust is produced directly, but if interested, Power = Thrust X TAS. As speed increases, Power increases (but who cares, aircraft don't need Power).
For the jet engine then, and with ONE exception, no-one gives a damn about the Power produced, it's academic.
The exception is Rate of Climb. All aircraft's Climb Rate depends upon excess Power available. For the jet, Power, and excess Power increase as the aircraft fly at higher speeds, with excess Power reducing at very high speed as Drag builds up, and the Power Required increases. That's why jet aircraft Climb at very high speed, typically at, and sometimes above, normal cruising speed. For the same reason in REVERSE, best Climb speed for a piston aircraft is somewhat below cruising speed.
All of this also explains why 'prop' aircraft generally have superior Takeoff performance (High low speed Thrust) with modest cruise performance, whilst their jet cousins experience have inferior Takeoff performance, but superior cruise performance.
To drive the point home, a jet at full thrust held on the brakes at zero speed is producing Zero (useable) Power. A piston aircraft in the same circumstances is producing Maximum Power on the brakes.
If you get the Thrust of my argument, it will add much Power to your understanding.
Regards,
Old Smokey
Aircraft need Thrust, they always have, and probably always will.
This applies from the Wright Flyer through to the Airbus A380.
Take a look at Page 1 or 2 of any Private Pilot's Theory of Flight text-book, the 4 basic forces are shown, Thrust, Lift, Drag, and Weight. Power does not rate a mention.
Piston engines produce Power DIRECTLY, applying torque to the propeller enabling the propeller to produce Thrust, what the aircraft needs.
Jet engines produce Thrust DIRECTLY, and without any added complication of conversion, is what the aircraft needs.
In both types of engines, there's a great deal of Power needed internally to drive the engine, but this power is not available for aircraft performance.
Power = Force (Thrust) X Velicity (TAS).
For the piston engine then, Thrust = Power divided by TAS. As Speed increases, Thrust decreases.
For the Jet engine, Thrust is produced directly, but if interested, Power = Thrust X TAS. As speed increases, Power increases (but who cares, aircraft don't need Power).
For the jet engine then, and with ONE exception, no-one gives a damn about the Power produced, it's academic.
The exception is Rate of Climb. All aircraft's Climb Rate depends upon excess Power available. For the jet, Power, and excess Power increase as the aircraft fly at higher speeds, with excess Power reducing at very high speed as Drag builds up, and the Power Required increases. That's why jet aircraft Climb at very high speed, typically at, and sometimes above, normal cruising speed. For the same reason in REVERSE, best Climb speed for a piston aircraft is somewhat below cruising speed.
All of this also explains why 'prop' aircraft generally have superior Takeoff performance (High low speed Thrust) with modest cruise performance, whilst their jet cousins experience have inferior Takeoff performance, but superior cruise performance.
To drive the point home, a jet at full thrust held on the brakes at zero speed is producing Zero (useable) Power. A piston aircraft in the same circumstances is producing Maximum Power on the brakes.
If you get the Thrust of my argument, it will add much Power to your understanding.
Regards,
Old Smokey
Thank you very much!