Revisting Impulse and Energy.

[PB666]

Where does impulse come from and what is specific about it.

Before this is addressed is it useful to begin defining problems in a abstract manner. This is useful in physics because if can frequently make complex problems simpler once you crash through the learning curve. The abstract problem in space can be defined as such, you generally start from a point which is iso-energetic with respect to some central body that is generally defined as an orbit with parameters of semi-major axis and a position vector with respect to some reference point given by Θ. This was discussed in the specific orbital energy thread.

What is desired is to go from that orbit to an orbit of different energy or go from the ground to orbit. We have several ways to describe energy, one way to describe the change of energy is

δE = F * δd.  While this is easily said, its not so easily done accurately without some understanding of physics.

Issues on or near the ground regarding the application of force are muddled by the fact that we are always pushing off something exterior to the energized object. For example a car here on Earth is using its tires to push against the earth, an airplane uses its propeller to push against the air, the earth then pushes back against the tires, the air pushes back against the propeller or the aircraft and eventually everything comes back to rest on the surface of he Earth. On the earth and in the air above the earth these low velocity things tend to have immeasurable and fluid specific impulses. IOW are perception of momentum on the ground is distorted by its complexity because on the ground because we are always in a non-inertial reference frame. There is no object within 120 km of the Earths surface that is in an inertial reference frame, excepting the entire mass of the Earth itself. So that it stands to reason we might think about the physics of motivation when in space differently than we thing about the comparable physics on Earth. For an object at surface rest F imparted by the non-inertial frame = m * ([µ / (altitude + 6371000)²] - [(cos latitude * 4E7/86400)^2/(altitude + 6371000)]) general defined as m*g. To give an example of the thrust problem, imagine a jet during take off, its performance is dependent on many factors. First the inlet air flow (related to the density of air), the percentage of oxygen and its temperature, the fuel. If we set the mass flow as constant we have δF/δV which can be rewritten as δF/δP * δP/δV, δF/δTemp, δF/δOx * δOx/δP * δP/δV etc. Consequently if you look as ISP for a jet, its generally given as an equivalent, since a pure ISP is ephemeral. 

In space this is not the case, in space we have simpler applications for the law of momentum. What is momentum is the first order property of matter that is conserved in an inertial reference frame. For an velocity to change acceleration needs to be changed, for an object momentum to change force needs to be applied.

  δp is the impulse. If we specify impulse with respect to mass then δV = Fδt/m    or   δV = aδt

But what about propulsion two objects are changing velocity at once. Imagine two particles (craft and exhaust) lying on each other in an inertial reference frame 0 = m_cv_c + m_ev_e and both are zero with respect to each other. and momentum is conserved, nor is there earth or air for the two particles to push off of, m_cv_{c = -}m_ev_e. In an impulse we set the two particles in relative motion along a linear vector. We then switch the reference frame (from frame 1 to 2) to c with the direction of the exhaust as positive, therefore the momentum of the craft is 0, but the momentum of the exhaust is now equal to m_eV_{e2 =} m_e(V_c1-V_e1) <-a useful psuedophysics (more useful versions applied below), we can recall that the craft is moving in + direction, the exhaust in "-" therefore we have defined δp of e relative to c. As a consequence of the impulse we now have an m_eV_e.

Some of you will automatically see where this is going, but we have to clear up one dilemma and do some magic. First the impulse is defined with respect to time, where as we simply declared we created an impulse. On a rocket impulses don't magically appear from vacuum energy. But here's the problem with a rocket, what I mean is we could go about examining force on fuel-->exhaust as it moves through a rocket to its final destination. There  multiple energy transitions, starting from thermal and potential energy fuel and ending up as combination of thermal energy, axial velocity, velocities orthogonal to the axis of motion. Consequently in a chemical rocket, what is going on with respect to time is not generally a recordable quantity (see thread on specific impulse help). So the δT in the equation above is basically at some point in time before a rockets 'impulse' starts to some point in time after the 'impulse' ends.

Thus     wikipedia here is using J as impulse, not joules.

But this is also the change of momentum for e relative to c.

and in the last equation, this is initial momentum before the impulse started and final momentum after it was all over.

And that's the magic. We transformed the equation from units that are roughly useless to units that are useful. Time is important in other steps, but not as important as change of momentum per unit mass. We can imagine it like this, suppose the the craft had infinite mass, then its velocity changes microscopically, then a nicer psuedo-physics is that only the velocity of the exhaust changes. We can change the units of mass we use such that we can achieve this.

In our case m_ev_ef - m_ev_ei since we are imagining our exhaust at rest relative to the craft for some huge spacecraft δp = m_ev_ef. Where do we measure v2? V2 has to be measured at the point where work done by the exhaust on the craft is no longer continuing. For a chemical rocket in the vacuum of space we can argue that no more work is being done as the exhaust gas passes from within the nozzle to not-within the nozzle. This means more specifically that a nozzle has an axis of symmetry, stated as Z axis. At a rocket-proximal point a along Z axis that is a greater distance than any surface point on the nozzle (i.e the bell opening) in the z direction from the rocket end of the nozzle. At this point there is   an orthogonal plane, all the exhaust gas should pass through this plane (for turbochargers and alternators it may not). I have laboriously defined this plane for a reason. If we use chemical energy, for example we can define the chemical energy hypothesized from the burning then the burnt molecules must pass through the plane, the energy of the reaction, if it does not transform into some type of thermal heating of the rocket must also cross the plane.  As I have pointed to in many threads  . . . . . .never, in space, forget where our energy is coming from and going tp because energy is difficult to obtain and heat energy is difficult to get rid of.  

[handwaving] 2H₂ + O2 ------> 2H20  [handwaving - 2]H₂ as  SPE 10 MJ/Kg. The how will be discussed later. From the perspective of the rocket the overwhelming majority of the energy does not go into pushing the rocket forward but pushing the fuel backwards.  The point were most of the energy goes into rocket is, relative to a common reference (a central body) the exhaust gas is moving slower relative to the rocket than the rocket is moving forward with respect to the reference. In the perfect situation our rocket is traveling forward at 4560 m/s and it is ejecting gas relative to itself at 4560 thus the gas has no KE, but it does have thermal energy that it gives into zero vector momentum. So as stated the energy conversion moves the rocket toward a momentum were it can more easily gain energy per unit of mass expelled.

The ISP of a hydrolox rocket with the [handwaving] fuel ratio is 450 sec * 9.8 m/sec. This quantity is converted to 4413 m/s of exhaust velocity. We can determine the KE in the fuel for an infinitely heavy rocket, relative to the infinitely heavy rocket. Thus KE = 1/2 * M * V² deprecated by mass SKE = 1/2 V² = 9,740,000 MJ/Kg. The difference is 262KJ/kg   Lets investigate where might the energy went. Since our space craft is infinitely heavy and is the frame of reference KE = F * d, and while the force on the craft is equal to the force on the gas, it change in distance over the period is negligible. It gets almost none of the energy directly,  but can capture the energy in other forms (such as energy accumulates). So if we look at a SpaceX rocket video we note that just after MaxQ is past we see the exhaust flare away from the Z axis. Aerodynamically there are two things that coincide there is a shockwave at the front of the rocket, the atmosphere curls back toward the rocket, but the rocket is traveling so fast that after maxQ the recombination is well below the tail this creates negative pressure, plus the rocket is higher, pressure is lower. The pressure no longer confines the gas. But still that doesn't explain the motion of the gas. The gas also has thermal motion PV = nRT. Such that in order for a gas traveling in direction Z (down the side of the nozzle) to change its direction there must be a motivational pressure on that gas. In this case the pressure is caused by T (kelvin) which creates thermal motion. IOW there are gas particles already in motion along XY that can expand in the orthogonal plane.  These are cosine losses of the form due to V_e,z = SQRT(Speed² - (V_e,X² + V_e,Y²)). As these gases expand they cool and commit work to the gas where the forces of the gas are countering other forces of the gas, IOW no work done along the Z axis that impart a force on the rocket. Thus we can see from this simplified illustration that we can deduce the efficiency of the energy conversion from chemical energy to work.  There is actually more energy in the fuel than here reported. Energy also exists in the differentials of fluids. For example the temperature differential of a cryogenic liquid is a source of energy. The more compact a fuel is, the more work that is done as the fuel expands, and less energy goes into heat and more energy goes into work. Liquid hydrogen is very cold, it boils at a few degrees above K, if you heat liquid hydrogen to 800K when it was at 23K at SP then it expands 34.78 times at SP. During the flight the gas wants to expand, it is passed through the nozzles shell, stealing heat from the exhaust and accelerating hydrogen within the engine and further heating it. So (handwaving -2) hydrogen is an amount of extra hydrogen that is added to convert an optimal amount of heat into work.

Having defined energy flow out of the rocket the next problem is how does a rocket gain energy. The real question however is how does the payload gain energy, because if we can get the PL to gain energy without having to carry a bunch of fuel mass and heavy fuel tanks and compensatory engines and structures, then we are more successful in our rocketry. That's what we want, our rocket is in simple space, we want it to go to a place (Or set of places, an elliptical) that has a different orbital energy than the elliptical we are currently in. If we were to graph the δE with respect to time we would quickly see that launch is a very bad place with regard to craft kinetic energy. δE = F * δd. This can be rewritten as
δE/δt = F * δd/δt = Power = F * V  and δE = F * V_average * t  . Outside of the first few seconds of flight we gain energy by applying force while moving. Most of the CB-referenced kenetic energy in a rocket comes from this. Suppose we create a rocket with a TWR of 1.001 that never changes. The rocket is producing energy, energy is going into the exhaust and heat, and the remainder is pushing against gravity. This rocket creates alot of energy, but none gets into KE.

    So if we surmise based on experience that 10,000 dV is needed to get to LEO, and we can see that the theoretical amount of dV required for a perfect rocket from a vacuum earth is to say orbit δV of say 7700. The 2300 dV was wasted getting to orbit. If we consider the change of energy.  (10,000dV + 407_sr) - (7700 + 407_sr) = 42MJ/kg of energy for a hypothetical rocket that loses trivial mass. . . about 40%. Mostly, this is not kinetic energy lost (exception = drag), we never had it in the rocket, its potential energy that was somehow wasted trying to get to space from the ground. This is a theoretical finding about spaceflight, that once you have structure in great motion, reuse it. Its a justification for the ISS but also space factories and fuel storage in space. This is to say that there are more energy efficient and less energy efficient ways, with respect to the chemical potential energy in fuel (or electric potential in solar panels) to gain (or lose) energy. An excellent example of this is the Oberth effect, or burning to interplanetary transfer from as low of earth orbit as possible. These are the abstractions we want to get better at.
As we saw above, one way is to get off the launch pad quickly, the idea would be to have a structure so perfect (an light), and a PL fairing so strong and light that we can get to the point where basically F_drag = F_gravity  any faster than this and we are wasting force in drag, any slower and we are wasting gravitational potential energy.  Keeping the argument short, drag is a tempest that is hard to beat, trying to beat drag becomes marginally less useful as the mass of drag accommodating structures increase in mass. And mass is not what we want, we want light PLs into orbit and that can go deep into space with structure based on structural needs in microgravity.

Thus abstraction of the problem. Marking efficient engines we want to use for long periods (like RL10b-2, ka-$hing), particularly in space, versus inefficient thristy engines we want to toss after seconds of use (SFRBs). This defines engines into two groups, thrifty space engines with low thrust, and cheap powerful ground engines with low thrift. For intermediate stages we want mostly thrifty engines but with substantially more thrust ($$ME or RL10c). If life were simple we could use thrifty engines everywhere, but the problem is thrifty engines are often expensive, heavier, complicated, hard to start, the gases boil off. So that application may limit the engine one has. On the other hand we could recycle more expensive engines, such as Merlin-1-D or SSME.

Lets look at efficiency, where does it come from, is there any metric, could we make an engine more efficient? The key abstraction here is that there is physics of motion and physics of chemistry. The second of which we almost never concern ourselves with, but is the basis for specific impulse. 

So for δp/δm (V_e) as mass flows out of the rocket at rate δm/δt * δp/δm = Force imparted on our rocket. But the energy for the impulse is in conventional chemical energy potentials. This is to say, by defining the chemistry and converting the moles to kilograms we generate a parameter specific chemical energy that can be converted to work, specific kinetic energy, and the efficiency of the process can be determined and maybe improved upon by designing better engines.

Lets take simple example. Hydrazine. There are 31.06 moles of hydrazine in a kilogram. It decays in the presence of a metal catalyst into nitrogen and hydrogen.

The formation energies are 945 + 2 * 436 = 1817,000 J per mole

The dissociation energies are 243 + 2*(385 + 351) = 1,715,000 J per mole.
The complete disintegration of hydrazine relaeases = 102,000 J per mole.
Therefore * 102000 J/mole * 31.06 moles/kg = 3223200 J per kg. (rating potential chemical energy per kilogram)
If all of the energy went to the exhaust the we have KE = 1/2 (m=1) v² = yields a δp = 2538 kg * m/s and since the energy was rated per kilogram then V_e also is 2538 m/s . The actual V_e of the products is 220 * 9.8 = 2156 the efficiency of the conversion is 84.5%. There is some value that can be gained by increasing ISP while maintaining thrust and rocket weight.

The lesson here is that potential Ve drops right out of the chemical thermodynamics using on silly equation what Ve is SQRT (2SCE * efficiency). If the units of specific chemical energy are given as kilograms, then change of momentum is the same as change in velocity.

Not only this be we have a potential force that is a function of the absolute rate of chemical reaction (i.e. mass flow) per unit of time. The most efficient reaction of chemical propellant is the conversion of 2.72H₂ + O₂ ---> 2H₂0 + 0.72H₂  with 26.67 moles/kg. The derived specific energy per this reaction is about ~10,500,000 J/kg (don't get hung up in the specifics there are other factors), but this is not all the energy, there is the energy of expansion in the fuels, these can be factored out because they are recycling waste heat from the nozzle into the combustion chamber. Expansion cycle engines can be very efficient. Its theoretical maximum is 4582 meters per second. Its actual is 4530 ergo its basically operating at maximum. The RL10-b2 is operating very close to the theoretical maximum for any chemical rocket. How it does this is by having a bell whose area is very large given the size of the combustion chamber, this allows the gas to expand and cool. It further steals thermal energy from the sides of the rocket and pipes it to the combustion chamber. A third reason is that the H2 + 02 conversion is basically flameless, flames coming from a rocket are basically indicators that the fuel is incompletely burnt. What this means is some of the fuel in the rocket is still in a higher energy state, effectively its a cold plasma. Chemically speaking this means that that energy was added to dissociate the fuel in to smaller plasma molecules or atomic plasma, but instead of reforming and giving energy back that can perform work, the energy escaped the orthogonal plane of the nozzle. This is a problem with fuels like Kerosene, the molecule does not fall apart all at once, and intermediates form other chemicals that are then oxidized. In addition most rockets run fuel rich, intentionally incompletely oxidized. Kerosene in not cryogenic, it does not gain as much as hydrogen does from the expansion of a liquid at 24'K to a gas at 800'K or so.

Thrust = SQRT (SCE * 2 * efficiency) * δm/δt

Lets compare this with other engines.

Electric propulsion ion thruster:

Thrust = 2 * δE/δt * efficiency / Ve.  So what is Ve? its the change of momentum/mass.

Thrust = 2 * δE/dt * efficiency * δm / δp

Photonic thruster:

Thrust = δE/δt / 300MW/N

Comparing a chemical energy rocket with say a great ion drive, the ion drive has about 20 times the Ve and a photon drive has hundred thousand more. But by accelerating more mass per unit of thrust, the chemical rocket in more energy efficient versus mass efficient. The photon rocket is completely mass efficient and the most energy inefficient. For ION drives power requirements = mass. For every N of thrust requires a kg of solar panel (ideally)/ km/s of dV.
Just that in itself limits ion drives to acceleration of less than 1 since it makes no sense to use an Ion drive at ISPs within the chemical range due to the density of energy in fuel versus that of solar panels.

So that now, thinking about efficiency, the design process is how to improve the efficiency without adding detrimental characteristics to the rocket (lower thrust, increased engine mass, engine complexity and cost for single use rockets).

I noted from the other thread that changes in ISP widely during a burn is suggestive of some sort of design issues. Most of the rockets in RL will have chemical energy potentials relatively close to kinetic energy, suggestive that ISP is approaching but never achieving perfection for the chemical conversion. Ve is an intermediate in what we are doing. Energy goes into the black box, we hope energy comes out and the best we can do is to create KE with a velocity vector overwhelmingly along one axis.

 

 

[tonates]

Thanks for a short article that helps to understand some moments with the emergence of impulse.