Author Topic: So, who wants to win 1 million Euro?  (Read 581232 times)

Offline Glom

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Re: So, who wants to win 1 million Euro?
« Reply #315 on: December 30, 2012, 03:48:02 PM »
Ok, I tried the calculation his way, but doing it relatively properly.  In other words, if I'm going to compare kinetic energy before and kinetic energy afterwards, it needs to work like this.

Kinetic energy of spacecraft after LOI + Kinetic energy of exhaust from LOI = Kinetic energy of spacecraft before LOI + Enthalpy change of combustion

Incidentally, I tried it both in the selenocentric frame and in a frame centred on the spacecraft before LOI (ie kinetic energy of spacecraft before LOI is 0) and glorious conservation of energy and momentum are observed.

Anyway, the enthalpy change due to combustion comes out at around 50GJ, so that means that the specific enthalpy change of the reactants is around 5 MJ/kg of reactants.  Does anyone have the specific enthalpy change of combustion of aerozine 50 and nitrogen tetroxide?  It looks a pretty rubbish combination.  Methane is more than twice that.

Offline Echnaton

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Re: So, who wants to win 1 million Euro?
« Reply #316 on: December 30, 2012, 04:01:25 PM »
Hi Everyone

Welcome to the forum, Mag40.
The sun shone, having no alternative, on the nothing new. —Samuel Beckett

Offline nomuse

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Re: So, who wants to win 1 million Euro?
« Reply #317 on: December 30, 2012, 04:07:40 PM »
I get all my numbers from Braunig.  I think Bob has the chemistry of the more common propellants in some detail there, too.  Or at least that's what I remember from the last time I tried to get really detailed with the numbers.

Offline Glom

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Re: So, who wants to win 1 million Euro?
« Reply #318 on: December 30, 2012, 04:23:35 PM »
Well I found a random Google return that allowed me to calculate 7 MJ/kg for hydrazine, which is certainly in the ballpark.  Do we know where Heiwa got his from?

Offline nomuse

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Re: So, who wants to win 1 million Euro?
« Reply #319 on: December 30, 2012, 04:36:07 PM »
I have a suspicion.    ;)

Offline Glom

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Re: So, who wants to win 1 million Euro?
« Reply #320 on: December 30, 2012, 04:37:40 PM »
I have a suspicion.    ;)

A suspicion that we could have all saved ourselves a lot of trouble by simply pointing out his specific enthalpy change figure was wrong?

Offline dwight

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Re: So, who wants to win 1 million Euro?
« Reply #321 on: December 30, 2012, 04:56:06 PM »
Oh darn imagine if we had known about these 56 hours ago. we could have all had them on.

http://www.collectspace.com/ubb/Forum24/HTML/011049.html
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Offline Chew

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Re: So, who wants to win 1 million Euro?
« Reply #322 on: December 30, 2012, 05:32:03 PM »
Well I found a random Google return that allowed me to calculate 7 MJ/kg for hydrazine, which is certainly in the ballpark.  Do we know where Heiwa got his from?

He used the value of hydrazine used as a monopropellant, which is about the lowest Isp you could hope to find. I told him several times he should be using the value of the actual fuel used in the SPS but he ignored me, of course.

Offline Glom

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Re: So, who wants to win 1 million Euro?
« Reply #323 on: December 30, 2012, 05:40:49 PM »
Ah.

Hey, Heiwa! We've solved your problem. Your using the wrong figure for enthalpy change. Use the correct figure and you'll get a propellant consumption that matches.

You still there?

Offline Noldi400

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Re: So, who wants to win 1 million Euro?
« Reply #324 on: December 30, 2012, 07:56:43 PM »
Bear in mind that you're talking to someone who thinks that (among other gems) "A sonic boom only occurs when a jet plane, close to ground, accelerates and pushes air waves ahead of it that cannot escape and then the air produces a sonic boom, when the plane accelerates beyond the local speed of sound. Sonic booms never occur when you decelerate in the other direction", so aircraft travelling faster than Mach 1 don't produce sonic booms.

As ka9q commented above, this guy makes Hunchbacked look positively sane.
"The sane understand that human beings are incapable of sustaining conspiracies on a grand scale, because some of our most defining qualities as a species are... a tendency to panic, and an inability to keep our mouths shut." - Dean Koontz

Offline ka9q

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Re: So, who wants to win 1 million Euro?
« Reply #325 on: December 30, 2012, 08:19:48 PM »
According to your own methods, and assuming you have the average mass of 71 kg for a European human, there is a difference of 68.5 J when walking from a standing start and about 22 KJ while on the plane. Are your legs suddenly really 320 times more powerful during flight?
The interesting thing here, and I'm sure it'll go way, way over Haiwa's head, is that, relative to the earth (and to the air if it's stationary) you really do have 22 kJ more kinetic energy in your body when you walk forward on the plane.

It's just that almost all of this 22 kJ comes from the plane's engines, not from your legs. When you begin to walk forward, your feet momentarily push rearward on the floor of the plane. That increases the force that the engines must overcome to maintain a constant forward velocity. Because the plane is moving so fast, this relatively small rearward impulse requires the engines to produce a fairly large amount of energy (nearly 22 kJ) until you have finished accelerating relative to the plane and are moving forward in the aisle at a constant velocity.

When you walk into the forward bulkhead (or simply stop walking) you produce a momentary forward force that momentarily reduces the required engine power and the plane gets back most of that 22 kJ.

Offline cjameshuff

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Re: So, who wants to win 1 million Euro?
« Reply #326 on: December 30, 2012, 08:52:27 PM »
The interesting thing here, and I'm sure it'll go way, way over Haiwa's head, is that, relative to the earth (and to the air if it's stationary) you really do have 22 kJ more kinetic energy in your body when you walk forward on the plane.

A related (and similarly confusing) phenomenon is the Oberth effect. A given spacecraft with a given amount of propellant will always be able to give itself the same immediate delta-v, but when doing something like departing Earth, doing the burn deeper in the gravity well is far more effective. This is because because the specific orbital energy of the spacecraft is increased more when the burn is done at higher orbital velocity, leaving more kinetic energy once the craft has climbed out of the gravity well. Including the exhaust in the calculations (released deeper in the well, at lower relative velocity) shows that energy is still conserved, but it's a rather unintuitive result.

This is also a clear example of how spaceflight is not like operating a boat...spacecraft can travel between regions of wildly different gravitational potentials, trading kinetic and potential energy back and forth and transferring momentum between themselves and other objects.

Offline ka9q

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Re: So, who wants to win 1 million Euro?
« Reply #327 on: December 30, 2012, 08:56:39 PM »
You claim that to calculate the energy requirement to change speed on Apollo 11 you need to work out the difference between the kinetic energy before and after the burn using KE = 1/2 mv^2. So you need the starting speed and the final speed, from which you calculate the difference in kinetic energy between those two speeds for a spacecraft of given mass. That determines the energy change you need to affect with the engine to achieve the end result. Yes?
Haiwa keeps overlooking the kinetic energy stored in the rocket propellant before the burn. At the high speeds involved in space flight, the kinetic energy, per kilogram of propellant, is often considerably greater than the stored chemical energy! And when the rocket is fired, depending on the direction the kinetic energy in the exhaust can be greater or less than the kinetic energy in the stored propellant, with much of the difference exchanged with the spacecraft.

For example, after TLI Apollo 11 was moving at 35,546 ft/s (10,834 m/s), just under earth escape velocity. Its specific kinetic energy was therefore 1/2 * 10834^2 = 58.7 MJ/kg. The chemical energy stored in Aerozine-50/N2O4 is only about  6 MJ/kg! (When you burn them in an ideal rocket engine, the kinetic energy in the exhaust relative to the rocket is about 5.6 MJ/kg, with some extra energy from the propellants lost heating the exhaust, rocket nozzle, etc).

Many people know about the ridiculously poor "fuel mileage" of the Saturn V as it lifts slowly off the pad, burning many tons of propellants each second just to move a few meters. But those propellants aren't being used as inefficiently as you might think; much of their energy was actually being "invested" in the kinetic energy of the propellants not yet burned. In fact, by the time the S-IVB stage fired, the power being applied to the Apollo spacecraft (i.e., its increase in kinetic energy per unit time) was considerably greater than the power being released by the combustion of H2 and O2 in the J-2 engine. I.e., the efficiency appeared well over 100%! The difference came from the release of stored kinetic energy in the propellants as they were burned and ejected in the opposite direction of flight. This extra energy came from the propellants of the lower stages as they accelerated the S-IVB along with the spacecraft.

I once computed the overall efficiency of the Saturn V in terms of the mechanical energy applied to the Apollo spacecraft vs the stored chemical energy in the propellants of all three stages. I expected a truly tiny number but I got about 6%, which I thought was amazingly high. Maybe rockets aren't quite so bad after all.
« Last Edit: December 30, 2012, 09:02:22 PM by ka9q »

Offline ka9q

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Re: So, who wants to win 1 million Euro?
« Reply #328 on: December 30, 2012, 09:09:17 PM »
A related (and similarly confusing) phenomenon is the Oberth effect. A given spacecraft with a given amount of propellant will always be able to give itself the same immediate delta-v, but when doing something like departing Earth, doing the burn deeper in the gravity well is far more effective. This is because because the specific orbital energy of the spacecraft is increased more when the burn is done at higher orbital velocity, leaving more kinetic energy once the craft has climbed out of the gravity well. Including the exhaust in the calculations (released deeper in the well, at lower relative velocity) shows that energy is still conserved, but it's a rather unintuitive result.

Yes, this is an excellent example. The way I think of it is that when you carry the propellants deep into the gravity well, you release the very large amount of gravitational potential energy stored in them, turning it into kinetic energy. By doing the burn deep in the planet's gravity well and leaving the propellants behind, you get to keep most of that energy as you climb back upstairs.


Offline ka9q

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Re: So, who wants to win 1 million Euro?
« Reply #329 on: December 30, 2012, 09:24:22 PM »
Anyway, the enthalpy change due to combustion comes out at around 50GJ, so that means that the specific enthalpy change of the reactants is around 5 MJ/kg of reactants.  Does anyone have the specific enthalpy change of combustion of aerozine 50 and nitrogen tetroxide?  It looks a pretty rubbish combination.  Methane is more than twice that.
The kinetic energy in the exhaust of an ideal rocket burning these propellants in vacuum is about 5.6 MJ/kg. The specific enthalpy of combustion has to be greater than this because even an ideal rocket is not 100% efficient at turning the stored chemical energy into the kinetic energy of the exhaust. There is additional energy in exhaust heat, i.e., random as opposed to linear motion of the exhaust molecules and energy in their useless internal degrees of freedom (rotation, etc.)

If I had time right now I'd try to work this out by first writing the chemical formula for the combustion of these propellants and then tallying the enthalpies of formation for the propellants and their exhaust products. You have to remember, however, that real rocket engines invariably run rich mixtures so you have quite a bit of incompletely burned combustion products like H2, CO, and the like.

A rocket is simply a heat engine that converts the heat of combustion into the kinetic energy of the combustion products. Like any heat engine, it cannot be 100% efficient. Besides the usual Carnot limits that depend on the source and sink temperatures, the efficiency depends heavily on the composition of those exhaust products, particularly their molecular weight. Nearly every engine has a peak efficiency that occurs when the fuel/oxidizer mixture is run rich, not at stoichoimetric, because the loss in released chemical energy is more than made up for by the increased conversion efficiency from heat to kinetic energy that results from a lower average molecular weight in the exhaust.

This is why hydrogen is the ideal propellant for a nuclear thermal rocket (or any rocket using an external source of heat).


« Last Edit: December 30, 2012, 09:31:12 PM by ka9q »