The Merlin Rocket Engine: Newsletter Challenge (October 2014)

Rarefied air causes less drag. So in effect the engine does not increase thrust but realizes less drag.

At least that's the way my mind see's it.

I never thought about it that way! Cool!

(things that make you go hmmmmm)

GA; problem solved.

As I see it, the difference here is not the thrust, but the change in thrust when the engine is running (compared to that whenit is not). It would be disappointing if this was all that the question meant.

In this case the effective thrust is the same as change in the net thrust - because there is pressure acting on an equal area that retards the rocket.

But I don't believe that this is the whole story - but I'll try to place other comments somewhere more appropriate.

GA.

Yet another way to gauge the effect of atmospheric pressure on the thrust of a rocket engine is to imagine a situation where you increased the atmospheric pressure outside the engine until it is equal to the pressure inside. What happens the thrust meanwhile? It drops to zero, yes? Increase it further and the atmosphere pushes its way into the engine (you wouldn't have to go very deep into Jupiter's atmosphere to find such conditions, for example).

Atmospheric pressure is the bane of rocket thrust. The less atmosphere the better.

"... imagine a situation where you increased the atmospheric pressure outside the engine until it is equal to the pressure inside. What happens the thrust meanwhile? It drops to zero, yes?..."

.

Are you certain about that 'yes'? Assuming the engine is operating during this pressure increase and propellant is accelerated to several times the speed of sound for the propellant in the nozzle at the beginning of this scenario, how is the increase in pressure communicated upstream?

.

Don't get me wrong, I agree with the final assessment you draw, but I think the velocity of the propellant and therefor the thrust may not drop that quickly once moving at many times its sonic speed.

.

If we increase the pressure before the engine is fired, the reaction that rapidly increases the volume of the propellant when the engine is fired would require a matching increase in outside pressure to insure no thrust.

Does the exhaust remain supersonic indefinitely? No, it does not, even under ordinary conditions. Eventually it slows to transsonic, then subsonic, speeds where external pressures can be 'felt' upstream.

Two things happen to supersonic flows when you increase external pressure. Firstly, the point at which the flow becomes transsonic moves closer to the nozzle, eventually reaching it when the external pressure becomes great enough. Secondly, external pressure *perpendicular* to the flow compresses it laterally, eventually pinching it off - the opposite effect of what you see of rocket exhaust during, say, a Shuttle launch, as the atmospheric pressure decreases with increasing altitude.

As external pressure increases, both effects work together to slow the exhaust and, eventually, if the pressure becomes great enough, halt it completely.

In short, 'yes'.

Is this rocket in motion or on a test stand?

I think the air pressure in back is the same as in front for the sea level case.

Yup, GA. The faster the rocket can throw out reaction mass at the back, the higher the thrust.

That's a bit like saying a car can accelerate faster on ice than a rubber track because the wheels can spin faster. I know it's not quite the same because with the car you're not ejecting anything out the back, but, having some thing to push against is always going to make a rocket more efficient.

My guess is that because there is no resistance to ejecting the exhaust out of the rear of the engine, it is able to burn more fuel in a given unit of time.

The chemical reaction and energy conversion rate should remain closely the same, but the efficiency in terms of thrust generation goes down in air. In space the released energy virtually all goes into two components: the kinetic energy of exhaust gas backwards plus the kinetic energy of the rocket moving forward.

In air, there is a third component that has to come out of the fixed energy conversion rate: kinetic energy of some surrounding air moving backwards, which subtracts from both the other two. Hence, less forward rocket kinetic energy.

I think the kinetic energy of the surrounding air would be more than compensated for by the lower kinetic energy of the exhaust gasses.

As altitude increases, air pressure decreases. Air pressure resists the outflow from the rocket engine. The greater the air pressure, the less thrust.

Thrust is a measure of force not the effect of that force. When testing a rocket or jet engine they will hold the unit under test still to measure the force: the fact that it is not accelerating does not mean that it is producing less thrust just that there are equal and opposite forces in the other direction.

Randall, from your answers it seems that you are challenging the stated question: "Why does the thrust of the Merlin 1D increase with increasing altitude?" Or do I understand what you are saying incorrectly?

Hmmmm not sure.

My assertion is that the efficiency is better at sea level, but, that the engine can produce more thrust in a vacuum because it is able to burn more fuel in a given unit of time.

Let's take a very simplistic example over a very short period of time, so that, the change in mass of the rocket and several other factors can be ignored. Also lets assume that the rocket starts from zero velocity with respect to our coordinate system; this means that as long as the end velocity is very low wind resistance can be ignored for the one at sea level.

Let's assume that the rocket's mass is M and that after the "burn" the rocket in both cases is travelling at V, and that the ejected gas in the vacuum has mass m and is travelling at v.

You can choose any combination of air/exhaust gas mixtures and velocities but to keep the sums easy let's assume that the air and gas which are moving backwards have a mass of 2m and must therefore have a velocity of ½v

because for the rocket in a vacuum MV = mv

For the other rocket MV is the same, and, MV = 2m x ½v

Now calculate the total energy of the two systems

Total energy of system with rocket in vacuum = ½MV² + ½mv²

Total energy of system with rocket at sea level = ½MV² + ¼mv²

That's a lot more energy for the "total system" for the rocket in the vacuum. But clearly the thrust in both cases was the same because both rockets had the same acceleration.

The extra energy must have come from somewhere, and can only have come from burning more fuel.

Whatever way you divi up the mass of the air and exhaust gas and relative velocities, you end up with a result which is in the same "logical direction", because momentum is proportional to velocity whereas energy is proportional to the square of velocity.

There is some form of a convergent-divergent nozzle through which the velocity of the propellant transitions from subsonic before the throat to supersonic after the throat into the divergent portion.

.

Since the propellant is traveling too fast for any pressure waves to make progress against its flow, how would you suggest that the ambient pressure outside the rocket has any effect on the burn rate of the propellant.

.

Atmospheric pressure at sea level limits the expansion and velocity of the exiting propellant. The same mass is coming out per unit time, it is just more dense, hotter, and not quite as fast as when it is fired in a vacuum.

Good answer.

So the critical thing there is that at sea level more energy is "tied up" in the heat of the exhaust gasses than in a vacuum?

Yeah, at sea level more energy remains in the exhaust as heat and pressure, meaning less is available as bulk kinetic energy.

This question is fun and it is very simple. In short. There is 14.7 PSI less pressure outside of the Rocket Engine. Thus increasing the engine output by 14.7 PSI. 14.7 PSI does not sound like much. But in a nozzle the size of these engines it is a great deal of extra power due to Newtons simple law of each action has an = and opposite reaction. In this case 10% based on the info from the question. Or if you want to look at it from another direction. In the Vacuum of Space the rocket reaches full output. But at sea level in the earths atmospear where there is 14.7 PSI of pressure compared to zero pressure in space. The extra 14.7 PSI outside of the engine lowers the pressure difference from inside the engine to outside the engine. Less pressure difference = less propellant shot from the engine per second ( Or what ever time measure you want to use ) = less thrust. jim Davison

"...Less pressure difference = less propellant shot from the engine per second..."

.

Things look pretty good up until that second to the last sentence. The burn rate need not differ for the difference in thrust. In fact it would be very difficult for a difference in ambient pressure to affect the pressure upstream of the convergent-divergent nozzle.

The formula for rocket thrust = {(rate at which mass of propellant is decreasing with time) X (velocity at which propellant is leaving the nozzle exit)} + {[(pressure of propellant at nozzle exit) - (ambient pressure at nozzle exit)] X (Area of nozzle exit)}.

The first part of the formula is just the reaction force due to the mass X speed of propellant being thrown out the back of the rocket. The second part is due to the fact that the ambient pressure is pushing equally on all parts of the rocket except for the nozzle exit where the pressure is the pressure of the propellant. In the atmosphere this can be greater than, equal to, or (surprisingly) even lower than the atmospheric pressure. In any case since the ambient pressure in space is zero this means that the thrust is greater in space than at sea level by an amount equal to the pressure at sea level X the area of the nozzle exit.

In rocket design there is a trade-off between pressure at the nozzle and speed of the propellant. The highest thrust is obtained when the exit speed of the propellant is maximised which is at the lowest practical exit pressure. Most rocket engines are designed so that the propellant pressure is close to the ambient pressure as if it is higher the rocket is less efficient and if it is too low the ambient pressure causes instability in the exhaust flow.

"...In any case since the ambient pressure in space is zero this means that the thrust is greater in space than at sea level by an amount equal to the pressure at sea level X the area of the nozzle exit...."

.

This sentence ignores an important part of the formula that you posted initially.

.

"...rocket thrust = {(rate at which mass of propellant is decreasing with time) X (velocity at which propellant is leaving the nozzle exit)} + {[(pressure of propellant at nozzle exit) - (ambient pressure at nozzle exit)] X (Area of nozzle exit)}..."

.

The reduction in ambient pressure has an effect on the velocity of the propellant, right?

.

The other thing that will generally keep the change in thrust from equaling the change in ambient pressure X the nozzle area, is that the nozzle will never be big enough to prevent under expansion in a vacuum. As a result, the difference in thrust will only be close to the difference in ambient pressure X the nozzle area if the loss of efficiency due to under expansion exactly off sets the increase in thrust to reduced ambient pressure.

In space the velocity at which the propellant is leaving the nozzle exit will indeed be higher however the pressure of the propellant at the nozzle exit will be correspondingly lower (but not zero hence there will be some under expansion thereby reducing efficiency). You are correct that I have ignored the inneficiency due to under expansion but I believe that this is not a significant factor. If someone knows the surface area of the end of the nozzle for a merlin 1D rocket engine we can check whether air pressure at sea level (approx 100 kN/m2) X surface area of nozzle exit is close to the 70kN difference in thrust observed. Equates to about 0.7m diamater nozzle I think.

0.7 m diameter would be an area of about 0.4 m^2.
Sea level pressure is close to 100 kN / m^2 .

.

....the thing is, I'm not sure if knowing the area helps us determine if the nozzle over/under expansion plays a significant role, or which part of the equation plays a dominant role. My gut tells me the increase in propellant velocity plays the biggest role, but without some idea of the magnittude of the effect of the nozzel expansion , i'm just guessing.

Sorry that should have been 0.94m diameter to get 70kN at sea level. If that is close to the actual nozzle size then we know that the pressure X area accounts for most of the difference in thrust.

I agree that the propellant velocity plays the biggest role, however, it still leads to the same result.

The equation for the propellant velocity is quite complicated but has a factor related to the difference in pressure between the combustion chamber and the exhaust that looks like (1 - (pressure of exhaust)/(pressure of combustion chamber)). The factor approaches 1 (maximum velocity) when the pressure of the exhaust is zero and 0 (no velocity) when the pressure of the exhaust equals the combustion pressure. At sea level the pressure of the exhaust is close to atmospheric pressure whereas in space it is close to zero so there is a direct correlation between this presure difference and the exhaust velocity which in turn has a direct correlation to thrust given a constant burn rate.

You and I are on the same page for pretty much everything, with the one exception being:

.

"....If that is close to the actual nozzle size then we know that the pressure X area accounts for most of the difference in thrust...."

.

There are other factors, the specifics of which we know too little, to be certain this simplification is reasonable. The design of the nozzle is one of those factors. The degree of underexpansion in vacuum and the amount of any overexpansion at sea level affects the pressure of the propellant as it exits the nozzle. At a minimum, this leads to the realization that the pressure of the propellant is not a constant and so cannot be reasonably cancelled out in the comparison between thrust at sea level and vacuum.

I think that the complexity arises from trying to look at the reaction to the force that is driving the rocket rather than the force itself. The propellant being shot out of the back is the equal and opposite reaction to the force which is pushing the rocket which is the internal pressure acting on the combustion chamber and nozzle.

You indicated that the one of the main reasons that you didn't feel the pressure X area difference was a complete picture was..

this leads to the realization that the pressure of the propellant is not a constant and so cannot be reasonably cancelled out in the comparison between thrust at sea level and vacuum.

Although it is not constant across the area of the Nozzle I believe that the pressure of the propellant does have the same pressure profile within both the combustion chamber and the Nozzle whether it is in the atmosphere or in space. The convergent/divergent nozzle is designed so that the speed of the propellant remains above Mach 1 for some distance outside of the nozzle at sea level pressure. As long as the area where the speed of the propellant drops below Mach 1 remains outside the nozzle (not significantly over expanded) the pressure profile inside the nozzle doesn't change. This means that the net force on the combustions chamber and Nozzle area from the internal pressure is the same in space as in the atmosphere. The only effect on the thrust therefore is due to the air pressure pressing inwards everywhere except the nozzle exit (where the supersonic propellant flow prevents it from exerting any force there). The net result is a force equal to area X air pressure acting in the opposite direction to the net force from the internal pressure.

This results in a reduction in thrust equal to area of nozzle X air pressure. In order for this to make sense from a reaction point of view we also need the propellant to be shot out at a higher velocity (since the mass flow is constant) as external pressure is reduced. This is achieved because the propellant continues to accelerate for a longer distance past the nozzle exit due to the pressure difference between the nozzle exit and ambient pressure.

I'm no expert here - so
Are you certain that the velocity is still supersonic when the engines are operating into a back-pressure of one atmosphere?

(Subsonic would presumably allow the ambient pressure to cause a difference in flow/burn rate - and so a difference in actual thrust rather than in net thrust)

The gasses are typically well above Mach 1 at the nozzle exit.

"....Although it is not constant across the area of the Nozzle I believe that the pressure of the propellant does have the same pressure profile within both the combustion chamber and the Nozzle whether it is in the atmosphere or in space...."

.

I agree the pressure profile within the combustion chamber and throat of the nozzle won't differ appreciably comparing sea level to vacuum. At the nozzle exit however, ambient pressure, does have a noticeable and important effect on the speed of the propellant.

.

The propellant is traveling at greater than Mach 1, but the effects of ambient pressure need not propagate upstream to affect the exit velocity. Different ambient pressure conditions result in different effective cross sections at the nozzle exit: a convergence over expansion or a divergence for under expansion. A constriction results in a decrease in velocity in this compressible flow regime.

Although I am not a rocket expert the little bit of research I have done on this (always a dangerous thing) indicates that if you were to connect the rocket combustion chamber, throat and nozzle between two air pressure reservoirs P+ at the chamber side and P- at the Nozzle end (initially the same as P+) and then look at the air flow as P- was reduced that you would observe the following:

Initialy as P- was lowered the air would flow from P+ to P- at some average rate and the flow would be highest at the throat of the engine (smallest cross sectional area means faster flow to get same average flow rate) and the pressure would be lowest there (highest flow speed = lowest pressure). After the throat the speed of the flow would decrease and the pressure would increase until it reached P-

As P- was lowered further the average flow rate would continue to increase until the speed of the air at the throat reached Mach 1. A fact that I didn't know before was that the zone at which the air reaches Mach 1 cannot move past the smallest cross-sectional area (ie, the throat of the engine) which means that the flow rate is maximised (choked) at this point regardless of how much lower P- is reduced. This is why the rate at which the propellant mass is reduced doesn't change between operation at sea level and operation in space. The only diference in thrust therefore is obtained from the difference in propellant flow rate through the nozzle which is in turn determined by the pressure profile in the nozzle.

From this point onwards as P- is reduced the velocity of the air increases past the throat with a subsequent drop in pressure until it reaches a shock zone where the velocity almost instantaneously drops below Mach 1 with a corresponding sharp increase in pressure. This shock zone moves further down the Nozzle the lower P1 gets. When it is just outside the nozzle exit the shock wave is no longer perpendicular to the flow and the higher external pressure deflets it inwards initially causing waves in the flow which can cause separation of the flow from the nozzle near the edges. This is known as overexpansion. It should be noted, however, that unless there is separation the internal pressure gradient along the walls of the combustion chamber, throat and nozzle no longer change no matter how much lower P- gets. This means that the flow accelerates all the way to the nozzle exit. Not only that but the accelertion within the nozzle remains the same regardless of how much lower P- gets.

When P- is lowered further it gets to the design pressure of the engine (normally the pressure at the altitude that the engine is designed to operate in) where the pressure at the end of the nozzle is the same as the external pressure at which point the waves in the jet disapear and the flow is uniformly supersonic (but not accelerating) past the nozzle exit.

Further reducing P- leads to a new inbalance between the exit pressure of the jet and the external pressure resulting in expansion waves initially turning the flow at the jet edges outward and setting up a different type of complex wave pattern. This however does not change the acceleration of the propellant inside the engine or the pressure of the propellant at the nozzle exit but does cause an additional force on the propellant accelerating it beyond the nozzle exit. The magnitude of this additional force is area of nozzle X (pressure of propellant at nozzle exit - external pressure). This additional force also adds to the thrust of the engine. At sea level it is 0 because exit pressure of propellant equals external pressure and the propellant is not acceleated past the nozzle exit.

In the case of the Merlin engine which (I believe) is designed for operation at ground level external air pressures its pressure profile is such that it achieves optimal flow at this altitude. When it gets into space it is still exerting the same internal pressure and there will also be underexpansion in the flow (because the engine was designed for operation in atmosphere) but the internal pressure profile is the same as when it operated at ground level. The velocity of the propellant will be higher than at sea level because it will be accelerated past the nozzle exit by the difference between the pressure at the nozzle exit (approximately sea level pressure) and external pressure (none in space). This results in a corresponding increase in thrust equal to area of nozzle X pressure of propellant at exit (should be sea level pressure).

Even though not all of the propellant is being accelerated in a direction perpendicular to the nozzle exit in space the internal pressure profile within the rocket is still the same as at ground level it is just that the external pressure prevented the propellant from shooting off in sideways directions there.

Yes

You are correct. The Time Part of my statment is not needed. It has no bearing on the Question at hand. I only added it due to the overall power of a rocket is NOT just How Much Mass and at what Velocity ( Giving the Amount of thrust ) But also the time of the burn or how long it can burn ( If it is even Burning but most rockets (Not all) Burn the fuel for the extra pressure it creates)).

For example. The Space shuttle has several Oxy/Hydrogen Rockets and two solid Fuel Rockets. I forgot what the total output of all of these rockets are but it is close to 5 Million Lbs of thrust. Dam that is so cool. Too Bad we will never get to see the space shuttle take off again with all that thrust and fuel. But my point in adding time was as you stated NOT needed. But I put it there as it does relate Not to the question given. Just to the overall ability of the Rocket. For example Lets say the Shutle is in fact in total exactly 5 Million Lbs of thrust. IF it can only burn for 60 seconds it is not going to make it to orbit.

So that was what was on my mind. Overall a Rocket is rated at NOT only its thrust but also how long it can keep up that thrust. Even if it has a throtle.

When we were kids we used to use compressed Air and water to power our rockets.

The Water gave extra Mass AND extra time for the rocket to run.

With Just Air alone it would just go a few feet. The Extra Weight of the water made the rocket heavyer as far as fuel but it also made for extra Mass and it made for extra time to run as the general notion is that Water is not compresable it took longer to get out of the nozel and thus the rocket burned longer and due to extra mass even though slower

Velocity at the nozzel the overall power of Velocity and mass = More Power and more time = more altitude. But you are correct. Time is not needed for the question given.

PS No one I can find makes those toy rockets that used Compressed Water and air or if they do I can not find them. As we got older we switched to solid fuel rockets. And they cam in not only a rating of thrust but also average burn time. So I am very used to adding time to the equation. A habit I guess.

Thanks.

My uneducated guess is that at sea level you have air pressure and gravity against you. The further you travel away from sea level these two elements lessen, reducing friction, increasing the opportunity for increased thrust. I believe gravity plays its part in this equation, something most have left out.

gravity will affect the acceleration of the rocket, but not the thrust.

In this case F= ma force on the rocket(thrust) = Mass flow of exhaust gas X acceleration the exhaust gas.

the mass glow does not change significantly but the acceleration of the exhaust gas through the nozzle does. In an atmosphere it is resisted but the atmospheric pressure as several people have been at pains to point out, but in a vacuum or near vacuum this resistance is removed.

Also a rocket engine is a reaction engine, the exhaust gas does not 'push' against anything as someone else suggested!

Good Answer.

But.

If the exhaust gas does not push against anything, then how does the anything affect the exhaust gas?

Thrust is proportional to the acceleration imparted to the mass flow of exhaust gas, thus anything which reduces the final exhaust velocity has reduced the acceleration imparted and thus reduced the thrust achieved, therefore placing anything behind the rocket engine reduces the thrust!

Consider the launch pads at Cape Canaveral, behind the exhaust nozzles there are huge ducts to re-direct the exhaust gasses sideways while minimising any restriction to their flow. Obviously there is some restriction in the ducts which cannot be avoided, but they are made as large as possible to minimise that restriction.

with increasing altitude the density of the environmential air decreases.

the resistance of the environmental air decreasas with the density to zero in absolut vacuum.

the difference ist the thrust increase from sea level to vacuum thrust.

need mathematics?

contact me

I think there are quite a few factors at work here.

As the rocket ascends,it becomes lighter,due to fuel being consumed.

At the same time, the density of the air is decreasing,reducing drag and turbulence.

The decrease in external resistance will result in a higher velocity discharge rate,

even though the fuel consumption(mass) rate remains the same.

Also a slight change in the pull of gravity has an effect.

(By the way,the greatest G force felt by astronauts is not at take off,but after the

ejection of the 2nd stage booster.)

Of course,that is just my opinion,I could be wrong.