View Full Version : question about reaction engines

raaaid

11-09-2005, 05:39 AM

i heard that in reaction engines they use nozzles to concentrate the spurt of gases

this is similar to having a fluid in a tube and half the section of the tube doubling the speed of the fluid to keep constant the volume

but you can half the section tending to 0 so youll be doubling the speed to the infinite

arent resistance of materials then the limit?

i have water in a squared tube with a section of 4 m*m and suddenly make the section 2 m*m reducing the section and the mass 4 times and increasing the speed 4 times

so if i had a mass of 1 kg at a speed of 1 m/s and decrease the nozzle 4 times i will have a mass of 0.25 kg at a speed of 4 m/s

from an initial impulse of 0.5*1kg*1*1m/s=0.5 i get an impulse of 0.5*0.25*4*4m/s=2

by decreasing the nozzle 4 times i get 4 times more kinetic energy or impulse

im just using what im taught

WWSensei

11-09-2005, 07:08 AM

Originally posted by raaaid:

i heard that in reaction engines they use nozzles to concentrate the spurt of gases

this is similar to having a fluid in a tube and half the section of the tube doubling the speed of the fluid to keep constant the volume

but you can half the section tending to 0 so youll be doubling the speed to the infinite

arent resistance of materials then the limit?

i have water in a squared tube with a section of 4 m*m and suddenly make the section 2 m*m reducing the section and the mass 4 times and increasing the speed 4 times

so if i had a mass of 1 kg at a speed of 1 m/s and decrease the nozzle 4 times i will have a mass of 0.25 kg at a speed of 4 m/s

from an initial impulse of 0.5*1kg*1*1m/s=0.5 i get an impulse of 0.5*0.25*4*4m/s=2

by decreasing the nozzle 4 times i get 4 times more kinetic energy or impulse

im just using what im taught

Well, you are using parts of what you are taught.

You left out pressure. Kinetic energy formula is Energy = Pressure + 1/2Mass*Velocity^2

Also, nozzle efficiency isn't a linear progression. Simply reducing it's size doesn't equate to it maintaining the same efficiency. As you approach supersonic speeds air starts acting more like a solid than a gas and it's compression efficiency is very low. It's why jet engines aren't equipped with infinitely small nozzles. Not just due to material construction of the nozzles but because at some point the efficiency just drops off.

The force of generated by a rocket engine works on the formula of:

F = m(Ve)+((Pe) - (Pa))(Ae)

where m = mass flow rate, Ve is velocity, Pe is static pressure, Pa is ambient pressure and Ae is the area at nozzle exit. Best performance is achieved when ambient and ext pressures are equal. If exit pressure is less than ambient then there is a loss in thrust. If exit pressure exceeds ambient the full potential of thrust isn't reached.

For those that just went Huh?

Another to think of it is that the thrust of a rocket engine can be expressed as the imbalance of pressure forces. You have to have more pressure in the engine wanting to come than you have external pressure wanting to go in. The area of the throat of a nozzle has a direct effect on that. The thrust could be expressed as F = P*A*C where P is the internal pressure, A is the area of the nozzle and C is a thrust coeefficient. C isn't really important for this discussion but suffice to say it varies in value from 1.6 to 2.

raaaid. A hint. If your arguments start off with "halfing infinitly toward zero" you can rest assured it is because the formula you are using is leaving something--usually a key variable--out of the equation.

BSS_Goat

11-09-2005, 07:14 AM

AAAAAAAAAHHHHHHHHHHHHHHHHH MY BRAIN EXPLODED!!!!

Ritter_Cuda

11-09-2005, 08:37 AM

this is how hydo mineing was done as well as water motors large intake that narrows down as the line falls. starting with 4' intake ending with a 2" out and a 10'-1000' drop and you have a hell of a blast.

Cuda

fordfan25

11-09-2005, 03:32 PM

"this is how hydo mineing was done as well as water motors large intake that narrows down as the line falls. starting with 4' intake ending with a 2" out and a 10'-1000' drop and you have a hell of a blast.

Cuda"

have you been peeking in on me while i take a leak cuda http://forums.ubi.com/images/smilies/88.gif

Jumoschwanz

11-10-2005, 08:24 PM

A smaller nozzle will take longer to pass the same weight of particles that a larger nozzle will. So vs. time, you do not get as much power some might guess intuitively.

A body moving through a gas has to have not twice the energy to double it's speed, but four times the energy, friction increases at a square. So an airplane that needs a thousand horsepower to go 300mph, would not go 600mph with 2000hp, it would need a thousand times a thousand horsepower to do the job! This is why it is much easier to get speed increases with improvements in aerodynamics than with adding hp.

A gas through a nozzle the same thing, it is all relative, you would get much more thrust by optimizing the shape of the nozzle, than by doubleing the pressure of the combustion chamber.

Probably the best answer to the original question though, is that it takes longer for the gas to escape through the tiny nozzle. So in the same period of time, although you have gas with higher speed, it is less gas, thus less wieght being thrown in one direction, to propel the engine in the opposite direction. The rocket would accelerate slower, but have a higher top speed. Ideally they have the variable nozzles on the backs of some jet and rocket engines that alter the speed of the gas jet, and this has the same effect in a sense as altering the pitch of a prop plane!

If you are going to be in engineering and physics in this life, I hope at least you enjoy it, because you may not make much money at it......

Jumoschwanz

Treetop64

11-10-2005, 09:09 PM

Originally posted by WWSensei:

Well, you are using parts of what you are taught.

You left out pressure. Kinetic energy formula is Energy = Pressure + 1/2Mass*Velocity^2

Also, nozzle efficiency isn't a linear progression. Simply reducing it's size doesn't equate to it maintaining the same efficiency. As you approach supersonic speeds air starts acting more like a solid than a gas and it's compression efficiency is very low. It's why jet engines aren't equipped with infinitely small nozzles. Not just due to material construction of the nozzles but because at some point the efficiency just drops off.

The force of generated by a rocket engine works on the formula of:

F = m(Ve)+((Pe) - (Pa))(Ae)

where m = mass flow rate, Ve is velocity, Pe is static pressure, Pa is ambient pressure and Ae is the area at nozzle exit. Best performance is achieved when ambient and ext pressures are equal. If exit pressure is less than ambient then there is a loss in thrust. If exit pressure exceeds ambient the full potential of thrust isn't reached.

For those that just went Huh?

Another to think of it is that the thrust of a rocket engine can be expressed as the imbalance of pressure forces. You have to have more pressure in the engine wanting to come than you have external pressure wanting to go in. The area of the throat of a nozzle has a direct effect on that. The thrust could be expressed as F = P*A*C where P is the internal pressure, A is the area of the nozzle and C is a thrust coeefficient. C isn't really important for this discussion but suffice to say it varies in value from 1.6 to 2.

raaaid. A hint. If your arguments start off with "halfing infinitly toward zero" you can rest assured it is because the formula you are using is leaving something--usually a key variable--out of the equation.

Thanks, @sshole, for making me feel like a complete dumb@ss now...

http://forums.ubi.com/groupee_common/emoticons/icon_biggrin.gif

Stackhouse25th

11-10-2005, 09:33 PM

*runs away* us pilots cant handle this much information!

raaaid

11-11-2005, 01:20 PM

but i believe that what determines the force of reaction is momentum not kinetic energy

because a bullet at 1200 km/h has the same kinetic energy than a 10 ton train at 1 km/h

reaction will push you back slightly when you shoot a bullet but if you are strong enough to send away a 10 tons train at 1 km/h reaction would push you back hard as hell

LEBillfish

11-11-2005, 02:15 PM

raaaid, if I can make a suggestion that I think would really help out your work and I think you two would work well together......

Contact Tagert on these forums and have him work up some graphs http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

Professor_06

11-11-2005, 02:32 PM

Heeha hee hee; raaid studied physics from Timothy Leary illustrated by Salvador Dali. Gold Star for humor.

Zeus-cat

11-11-2005, 03:48 PM

raaaid,

Do your math again. I come up with a 16.67kg bullet using your 10 ton train moving at 1km/h example.

Zeus-cat

raaaid

11-11-2005, 04:30 PM

a 0.01 kg bullet at 300 m/s has a kinetic energy of 0.5*0.01*300*300 =450

10000kg at 0.3 m/s gives 5000*0.3*0.3=450

for me is undiscussible that if you send away 10 grams at 300 m/s you dont get the same thrust that if you send away a 10 tons train at 0.3 m/s

Zeus-cat

11-11-2005, 05:59 PM

Oops, I calculated momentum, not kinetic energy. I missed where you said kinetic energy. Sorry.

Zeus-cat

GAU-8

11-12-2005, 01:22 AM

"teacher!"

im gonna go home now, ..

my brain is FULL.

polak5

11-12-2005, 08:31 AM

Originally posted by BSS_Goat:

AAAAAAAAAHHHHHHHHHHHHHHHHH MY BRAIN EXPLODED!!!!

ahhah Dito http://forums.ubi.com/images/smilies/35.gif