Author Topic: Electric/chemical Energy to kinetic energy conversion  (Read 135888 times)

Offline Bob B.

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #30 on: June 19, 2006, 08:21:37 AM »
Bob you are right about chemical rockets. The reason that higher exhaust velocities are desirable in chemical rockets is because chemical rocket exhaust velocities are somewhere between a half and a quarter of the optimised velocity for typical space applications.

I don’t think this is the reason; I believe it is because chemical propellants are their own energy source.  The energy to accelerate the exhaust gases is stored chemically within the propellant.  Burning the propellant releases that energy where it is converted to kinetic energy.   The more energy we can get out of a propellant the faster the exhaust gases can be accelerated.  Higher exhaust gas velocity means greater thrust for a given mass of propellant and higher efficiency.

Increasing exhaust gas velocity always results in less propellant being required to produce the same change in velocity.  Since the only mass in a chemical propulsion system is the engine and the propellant, decreasing the mass of propellant always decreases the overall mass of the system.  If we were to plot the total system mass versus the exhaust gas velocity required to produce a given delta-v, the mass would always decrease as we move toward higher gas velocity.  And since the goal in rocketry is to be as mass efficient as possible, we want to drive the exhaust gas velocity up as high as possible.

The problem with chemical propulsion it that there is a limit to how much energy can be obtained from a given mass of propellant.  The limit seems to have been reached with LOX/LH2 where an exhaust gas velocity of 4,500 m/s is about the maximum.  (There are some other propellants that can do a little better but they are very toxic and problematic.)  However, if there was some super-duper exotic propellant (which there is not) that could produce gas velocities of 20,000 m/s, it would be the ultimate chemical propellant.  We could dramatically reduce the mass of propellant required to produce a certain delta-v, yet that smaller mass contains all the energy required to produce the very high gas velocity.  Even though the system may not be optimized for energy, we don’t care because we’ve gained high mass efficiency by using an extremely energetic propellant.

Electric and nuclear propulsion are different because the energy does not come from the propellant.  In ion propulsion, for example, the propellant does not react chemically; it is ionized and accelerated by an electric field.  There must, therefore, be another power source to generate the electricity.  The total mass of the propulsion system is the engine, the propellant, and the power source.

Higher exhaust gas velocities require more power.  As gas velocity increases, the mass of propellant required to produce a given delta-v decreases but the mass of the power sub-system increases.  We therefore have two system components – propellant and power source – trending in opposite directions.  If we were to plot the total system mass versus the exhaust gas velocity required to produce a given delta-v, there would be an optimum point where the curve bottoms out.  At this point the system is most mass efficient and is the point for which we want to design the propulsion system.

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #31 on: June 19, 2006, 05:32:14 PM »
The energy requirements and not just the power requirements are greater for excessive exhaust velocity.  Chemical propellants only have between a quarter and a sixteenth of the optimum energy dencity.
If it were possible for the optimum energy to be less than the chemical potential energy dencity of the propellants. then the propellant could be diluted with some thing cheep and relatively inert like liquid nitrogen or water. 
In reality as you know chemical rocket propellants are limited to exhaust velocities of about 4,000m/s or less, mission delta-v is usually in excess of 9,000m/s and optimum fixed exhaust velocities as calculated are approximately two thirds the total delta-v.  Optimum varying exhaust velocities max out at the total delta-v. However,true energy optimized exhaust velocities will be higher than what was calculated, do to, among other things, the following three factors:

1 The tank mass was included in the dry mass of the rocket, and included no allowance for the smaller payload fraction due to larger tanks for more propellant. (outher factors that could complicate this calculation are the dencity and temperature controle requirements of the propellant ect.)

2 Gravity was ignored, gravity loss increases with greater propellant mass (gravity loss also increases inversely with acceleration, but the advantages of high acceleration is countered by greater structural and engine mass, by the tolerances of the payload and by greater air resistance since greater velocities are attained in the lower atmosphere)

3 Air resistance was ignored, this could be reduced with smaller propellant tanks and hence a smaller rocket since less air must displaced.  (this is dependent on so many outher factors that I wont even start naming them)

Thus the optimum exhaust velocity from an energy conserving point of view is certainly greater than 6,000m/s. Assuming 6,000m/s is the lower bound for the optimum exhaust velocity we can calculate the lower bound of the optimum energy dencity.

KE/m = (.5mv^2)/m = .5v^2 = .5*(6,000m/s)^2 = 18MJ/Kg

Compair to chemical propellants at 3,000m/s to 4,000m/s
KE/m = .5*(3,000m/s)^2 = 4.5MJ/Kg
KE/m = .5*(4,000m/s)^2 = 8.0MJ/Kg

So we can conclude that for maximum energy efficiency a perfect propellant should have over twice the energy dencity of the best chemical propellants.

On a side note, neather minimum propellant nor minimum energy are good goals of a rocket design engineer since outher factors must also be considered.
Joshua D. Benner Associate in Arts and Sciences in General Science

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #32 on: June 19, 2006, 05:58:56 PM »
If I have a given fixed amount of fuel, then there is a fixed amount of chemical energy stored in the rocket. Why would I not attempt to maximize the specific impulse of my engine to get the most delta-V for a given amount of stored energy?


Don, you bring up a good point.
For the sake of argument let us say that our propellent has twice the optimum energy dencity for a given mission. 

In this case it is less wastefull to convert all the energy to kinetic energy,obtaining more momentum then if half of the chemical(or nuclear, anti-matter ect.) potential energy remained with the exhaust.

However it is even less wastfull to mix your propellent with an equal quantity of something cheep like water. Now your energy dencity is optimum and half of your propellant is allmost free.
Joshua D. Benner Associate in Arts and Sciences in General Science

Offline DonPMitchell

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #33 on: June 19, 2006, 09:42:03 PM »
You always want to impart the maximum momentum to a vehicle, that is to say the maximum impulse.  But I think different propulsion systems create different optimization problems, and it is not just the case that there is always an optimum exhaust velocity.

For a chemical rocket, you want to optimize the momentium per fuel mass, and that will always mean carrying the most energetic fuel you can find (not watering it down), and driving an engine with the highest chamber pressure and exhaust velocity you can achieve.

Now for a situation like an ion engine, you might have a finite energy supply E.  For example, let's say you have 100 kg of plutonium to burn.  Now you could talk about the maximum momentum per energy used.  That would lead to a different strategy.
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Offline Bob B.

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #34 on: June 19, 2006, 10:47:41 PM »
For the sake of argument let us say that our propellent has twice the optimum energy dencity for a given mission. 

In this case it is less wastefull to convert all the energy to kinetic energy,obtaining more momentum then if half of the chemical(or nuclear, anti-matter ect.) potential energy remained with the exhaust.

However it is even less wastfull to mix your propellent with an equal quantity of something cheep like water. Now your energy dencity is optimum and half of your propellant is allmost free.

I don’t think this works, JD.  Anytime you dilute the propellant you lower the specific impulse and that results in less delta-v.

Let’s say we’re burning N2O4 and UDMH at a mixture ratio of 2.10 and a chamber pressure of 50 atm expanded to sea level pressure.  According to my calculations we have a theoretical performance of:

C = 2,637 m/s
Isp = 268.9 s

Let’s say we need to produce a delta-v of 3,600 m/s and our final mass is 1,000 kg.  We therefore have,

Mo = Mf*e^(dV/C)
Mo = 1,000*e^(3,600/2,637)
Mo = 3,916

So the propellant mass is,

Mp = Mo–Mf
Mp = 3,916–1,000
Mp = 2,916 kg

And the kinetic energy of the expelled gas is,

KE = 0.5mv^2
KE = 0.5*2,916*2,637^2
KE = 1.0139E+10 J

You said the energy of the exhaust gas is minimum when C = 2/3*dV.  Suppose I reformulate my propellant by diluting with water to satisfy this condition.  Let’s say we use 50% UDMH and 50% water for the fuel and add N2O4 at a mixture ratio of 1.40.  We now have,

C = 2,401 m/s
Isp = 244.8 s

Mo = 1,000*e^(3,600/2,401)
Mo = 4,479

Mp = 4,479–1,000
Mp = 3,479 kg

KE = 0.5*3,479*2,401^2
KE = 1.0028E+10 J

As you demonstrated earlier, we’ve lowered the kinetic energy of the exhaust gas but we’ve done so at a significant reduction in performance.  And even though some of our propellant is cheap water, we have more propellant so there is only slightly less N2O4+UDMH in the second example than there was in the first (2,754 kg vs. 2,916 kg).  Furthermore, since we have more propellant in the second example we’ll need larger tanks, therefore the final mass will be greater than 1,000 kg.  Let’s say Mf is 1,050 kg, we then have,

Mp = 1,050*e^(3,600/2,401) –1,050
Mp = 3,652 kg

This means we now have 2,892 kg of N2O2+UDMH, essential equal to the first example, plus 760 kg of water.  We end up with a bigger, heavier, less efficient rocket with significantly reduced payload capacity.

I’m sorry but I just don’t see how there is any advantage to what you propose.  If you take some of the excess energy in the exhaust gas to accelerate the added water (or whatever you add), you do so at the expense of slowing down the highly energetic gas.  You end up with a larger mass moving more slowly rather than a smaller mass moving very fast.  This is the exact opposite of what we want to achieve.  The faster the gas is moving the higher the specific impulse will be and the less propellant we'll need.
« Last Edit: June 20, 2006, 12:04:27 AM by Bob B. »

Offline ijuin

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #35 on: June 20, 2006, 01:12:39 AM »
Let's say that you are carrying a ton of stabilized metahelium (theoretical exhaust velocity of 30 km/s, but nobody has figured out how to keep it stable for more than a few hours), and a ton of water. Adding the water to the metahelium WILL gain you more delta-v than if you just expended the metahelium by itself, BUT you would get even MORE delta-v by just using two tons of metahelium instead. Thus, it is only if your primary fuel is ultra expensive/rare enough that you can not simply pack in as much as you can carry (e.g. tritium/helium 3, or antimatter) that you are better off diluting it.

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #36 on: June 20, 2006, 07:35:39 PM »
Let's say that you are carrying a ton of stabilized metahelium (theoretical exhaust velocity of 30 km/s, but nobody has figured out how to keep it stable for more than a few hours), and a ton of water. Adding the water to the metahelium WILL gain you more delta-v than if you just expended the metahelium by itself, BUT you would get even MORE delta-v by just using two tons of metahelium instead. Thus, it is only if your primary fuel is ultra expensive/rare enough that you can not simply pack in as much as you can carry (e.g. tritium/helium 3, or antimatter) that you are better off diluting it.

Again, I must say that minimizing energy expenditure is rarely the primary goal in engineering.  But your extreme case of an exhaust velocity in range of the total mission delta-v for an interplanetary mission, I must say that for launching earth satellites efficiently, it may be desirable to slow down the exhaust.

Let us see what we can do with one ton of metahelium (assuming of course that we can produce it in ton batches, and that it really has the stated energy density). We will also subtract 10% of P from M to correct for propellant tank mass.

The rocket equation is:
(M+P)/M = e^ (dv/c)
Solving for M yields:

M= P/ (e^ (dv/c) – 1)


P1 = 1,000Kg
dv = 10,000m/s
c1 = 30,000m/s
10%P1 = 100Kg

M1 = P1/ (e^ (dv/c1) – 1) = 1,000Kg/ (e^ (10,000m/s/30,000m/s) – 1) = 2,528Kg
Subtract the 100Kg mass of the propellant tanks.
Total useful mass to orbit is 2,528Kg – 100Kg = 2,428Kg

Remember 2,428Kg

Now let us water down the methahelium and see what we can get out of it.
First we must find the energy density of methahelium. 
KE = .5mv^2
KE/m = .5v^2
KE/Kg = .5*(30,000m/s) ^2 = 450MJ/Kg

Now let us calculate the energy density of a propellant that produces an exhaust velocity of 7,000m/s.

KE/Kg = .5*(7,000m/s) ^2 = 24.5MJ/Kg
Now we can find the ratio of water to methahelium, by dividing methahelium’s energy density by our desired energy density and subtracting one.

(450MJ/Kg)/ (24.5MJ/Kg)-1 = 17.4

Let us check this calculation
450MJ/(1Kg methahelium + 17.4Kg water)= 450MJ/18.4Kg
450MJ/18.4Kg = 24.

Now we can recalculate payload to orbit:

M= P/ (e^ (dv/c) – 1)

P2 = 18.4*P1 = 18,400Kg
dv = 10,000m/s
c2 = 7,000m/s
10%*P2 = 1,840Kg
M2 = P2/ (e^ (dv/c2) – 1) = 18,400Kg/ (e^ (10,000m/s/7,000m/s) -1) = 5,799Kg
Subtract the 1,840Kg mass of the propellant tanks.
5,799Kg – 1,840Kg = 3,959

So our final payload mass to orbit is 63% greater using the same amount of methahelium.

From a cost point of view however, the benefit may not be so apparent.  For example, if our propellant tanks are disposable they must cost less than 38% the cost per Kg of methahelium if we are going to brake even(for this specific propellant ratio).  And, if the designers are building a missile, storage and transport issues may rule in favor of the pure methahelium since the weight of the propellant and its tanks will be over an order of magnitude lower than our chosen ratio.

Which only goes to show that energy expenditure is not the only factor to conceder.
Joshua D. Benner Associate in Arts and Sciences in General Science

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #37 on: June 20, 2006, 07:53:40 PM »
For the sake of argument let us say that our propellent has twice the optimum energy dencity for a given mission. 

In this case it is less wastefull to convert all the energy to kinetic energy,obtaining more momentum then if half of the chemical(or nuclear, anti-matter ect.) potential energy remained with the exhaust.

However it is even less wastfull to mix your propellent with an equal quantity of something cheep like water. Now your energy dencity is optimum and half of your propellant is allmost free.

I don’t think this works, JD.  Anytime you dilute the propellant you lower the specific impulse and that results in less delta-v.

Let’s say we’re burning N2O4 and UDMH at a mixture ratio of 2.10 and a chamber pressure of 50 atm expanded to sea level pressure.  According to my calculations we have a theoretical performance of:

C = 2,637 m/s
Isp = 268.9 s

Let’s say we need to produce a delta-v of 3,600 m/s and our final mass is 1,000 kg.  We therefore have,

Mo = Mf*e^(dV/C)
Mo = 1,000*e^(3,600/2,637)
Mo = 3,916

So the propellant mass is,

Mp = Mo–Mf
Mp = 3,916–1,000
Mp = 2,916 kg

And the kinetic energy of the expelled gas is,

KE = 0.5mv^2
KE = 0.5*2,916*2,637^2
KE = 1.0139E+10 J

You said the energy of the exhaust gas is minimum when C = 2/3*dV.  Suppose I reformulate my propellant by diluting with water to satisfy this condition.  Let’s say we use 50% UDMH and 50% water for the fuel and add N2O4 at a mixture ratio of 1.40.  We now have,

C = 2,401 m/s
Isp = 244.8 s

Mo = 1,000*e^(3,600/2,401)
Mo = 4,479

Mp = 4,479–1,000
Mp = 3,479 kg

KE = 0.5*3,479*2,401^2
KE = 1.0028E+10 J

As you demonstrated earlier, we’ve lowered the kinetic energy of the exhaust gas but we’ve done so at a significant reduction in performance.  And even though some of our propellant is cheap water, we have more propellant so there is only slightly less N2O4+UDMH in the second example than there was in the first (2,754 kg vs. 2,916 kg).  Furthermore, since we have more propellant in the second example we’ll need larger tanks, therefore the final mass will be greater than 1,000 kg.  Let’s say Mf is 1,050 kg, we then have,

Mp = 1,050*e^(3,600/2,401) –1,050
Mp = 3,652 kg

This means we now have 2,892 kg of N2O2+UDMH, essential equal to the first example, plus 760 kg of water.  We end up with a bigger, heavier, less efficient rocket with significantly reduced payload capacity.

I’m sorry but I just don’t see how there is any advantage to what you propose.  If you take some of the excess energy in the exhaust gas to accelerate the added water (or whatever you add), you do so at the expense of slowing down the highly energetic gas.  You end up with a larger mass moving more slowly rather than a smaller mass moving very fast.  This is the exact opposite of what we want to achieve.  The faster the gas is moving the higher the specific impulse will be and the less propellant we'll need.


Bob, I believe we have reached the point where it only appears that we are debating.  While it is true that you know more about rockets than I do, in this post you have failed to teach me any thing new.  And I have already addressed your last post.

However,true energy optimized exhaust velocities will be higher than what was calculated, do to, among other things, the following three factors:

1 The tank mass was included in the dry mass of the rocket, and included no allowance for the smaller payload fraction due to larger tanks for more propellant. (outher factors that could complicate this calculation are the dencity and temperature controle requirements of the propellant ect.)

2 Gravity was ignored, gravity loss increases with greater propellant mass (gravity loss also increases inversely with acceleration, but the advantages of high acceleration is countered by greater structural and engine mass, by the tolerances of the payload and by greater air resistance since greater velocities are attained in the lower atmosphere)

3 Air resistance was ignored, this could be reduced with smaller propellant tanks and hence a smaller rocket since less air must displaced.  (this is dependent on so many outher factors that I wont even start naming them)



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Offline Bob B.

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #38 on: June 20, 2006, 09:46:13 PM »
So our final payload mass to orbit is 63% greater using the same amount of methahelium.
That’s what ijuin said...

Adding the water to the metahelium WILL gain you more delta-v than if you just expended the metahelium by itself
ijuin also said...

BUT you would get even MORE delta-v by just using two tons of metahelium instead.

Which is true.  Using JD’s example, 1,631 kg of methahelium will deliver the same payload to orbit as 1,000 kg of methahelium + 17,400 kg of water.

M = P / (e^(dv/c1) – 1) = 1,631 / (e^(10,000/30,000) – 1) = 4,123

Payload = M – 0.1*P = 4,123 – 0.1*1,631 = 3,960 kg

The total liftoff mass of this rocket is,

4,123 + 1,631 = 5,754 kg

versus the one using diluted methahelium,

5,799 + 18,400 = 24,199 kg

So unless building a rocket with over eleven times the tank size and more the four times the thrust costs less than the addition of 631 kg (63%) of methahelium, it is cheaper not to dilute, or to dilute to a lesser degree.

Offline Bob B.

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #39 on: June 20, 2006, 10:26:06 PM »
Bob, I believe we have reached the point where it only appears that we are debating.
I'm just trying to get to the bottom of what the misunderstanding is between us.  I think I’m beginning to see where we differ.

In your previous example you said we could deliver more payload to orbit using the same amount of metahelium by diluting it with water.  I don’t necessarily dispute that, however by my definition the additional water IS propellant.  I consider anything expelled out the nozzle to produce thrust propellant.  You seem to consider propellant only that which liberates energy.

By my definition, reducing exhaust gas velocity by dilution is most definitely not more efficient.  In your example it took 18,400 kg of propellant to deliver the same payload as 1,631 kg of metahelium alone (a mass ratio of 4.2 versus 1.4).  However, if I’m understanding you correctly, you seem to be saying dilution is more efficient because you can deliver more payload with the same amount of metahelium (not counting the water as propellant).

The argument seems to be more over how we define “propellant” and “efficiency”.  My argument from the beginning has been that to get the most delta-v out of a kilogram of propellant (my definition) we need to get the exhaust gas velocity as high as possible, and I still believe that.

Offline DonPMitchell

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #40 on: June 21, 2006, 12:18:24 AM »
You always want to maximize delta-V.  I think there are just several different optimziation problems that can be posed.  In particular, if the energy supply E and the reaction mass M are separate, then I think it may be a different game.  You can decide then to generate impulse by expending energy to get high exhaust speed or to expell more mass.  I wonder if jbrenner's argument might work then.

But I do not think it applies to a chemical rocket, where the energy and reaction mass are bound together.  In that case, I am sure you want to maximize specific impulse (that is, exhaust velocity).  And you never want to carry water in a rocket, when you can replace that payload with more fuel.
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Offline ijuin

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #41 on: June 21, 2006, 02:32:32 AM »
Whether or not to water down your fuel depends on your goal.

Let us suppose for the following examples that you have a water-augmented metahelium rocket that can consume any fuel/water ratio that you desire. In pure metahelium mode you consume the most metahelium per unit of delta-v, but consume the least gross reaction mass. The more you water down the metahelium, the more reaction mass you consume, but the more delta-v you gain from each unit of metahelium--at least until the metahelium is so diluted that the water does not fully evaporate and expand in the rocket nozzle.

Example 1:
You are launching from a spaceport on Earth, where metahelium fuel is readily available for the relatively low price of US$5.00 per liter. You want to go to LEO and back. Your payload is limited solely by your launch weight (i.e. if you can lift off with it, you can reach orbit with it--note that I said weight, not mass, since Earth has the highest surface gravity of any body in the solar system that you could land upon and thus has the most difficult liftoffs).

In this example, you are best off using only metahelium so that you can maximize your payload mass as a fraction of your launch mass, since pure-metahelium mode gives you the highest energy density.


Example 2:
Same as Example 1 above, but your payload is limited chiefly by volume (the size of your payload bay), so extra propulsive capacity will not increase the payload that you can carry.

Since your payload limit is fixed at a value that will not strain your launch weight limit, you are better off saving on the metahelium and topping off your tanks with water--you will have to determine the optimum ratio depending on your exact payload mass and delta-v requirement.


Example 3:
You are launching from Earth heading for Ganymede.

Here you want to maximize your available mission delta-v. Since metahelium is readily available and provides the highest energy density, you are best off carrying as much metahelium as your launch weight limit allows--you don't want to be left stranded if you need a major course correction or miss your return launch window!

Example 4:
You are on Ganymede, having traveled from Earth. Metahelium is not available for refueling, but ice is available in unlimited quantities free of charge as long as you can get it into your tanks and melt it.

Here, since there is no chance of getting more metahelium, your best bet for maximizing your available delta-v is to top off your water tanks (if you ever find that you have TOO MUCH water weighing you down, you can always dump some).

Offline Bob B.

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #42 on: June 21, 2006, 09:14:58 AM »
Which only goes to show that energy expenditure is not the only factor to conceder.
I agree.  In fact, I think you are actually proposing cost optimization rather than energy optimization. 

From our earlier examples we saw that 1,000 kg of metahelium + 17,400 kg of water could deliver to orbit the same mass as 1,631 kg of metahelium alone.  If we’re designing for a given payload capacity, using watered down metahelium is cost effective only when the cost of the added water along with the resulting increases in tank and engine cost is less than the cost of the metahelium saved.  It is really an economics problem and not one of energy optimization.  The designer must find the point at which the cost of increased rocket size begins to outweigh the savings in metahelium.

I can think of plenty of reasons for mass optimization.  For instance, an earlier example showed that a payload of 3,959 kg could be delivered to orbit.  But let’s say this mass includes an upper stage that must provide another 4 km/s to inject a science payload on a trajectory to Mars.  Since the mass delivered to orbit is finite, we must maximize the Mars payload by minimizing the propellant mass required to TMI.  This means using 100% metahelium and not some lower-performing watered down version.  When we have a fixed starting mass, the only way to maximize final mass is by maximizing exhaust gas velocity.

I can think of no situation in which optimizing energy alone would be the primary goal of the designer, unless of course it is part of a larger cost optimization problem.

Offline jdbenner

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #43 on: June 21, 2006, 11:38:02 AM »
Bob, I believe we have reached the point where it only appears that we are debating.
I'm just trying to get to the bottom of what the misunderstanding is between us.  I think I’m beginning to see where we differ.

In your previous example you said we could deliver more payload to orbit using the same amount of metahelium by diluting it with water.  I don’t necessarily dispute that, however by my definition the additional water IS propellant.  I consider anything expelled out the nozzle to produce thrust propellant.  You seem to consider propellant only that which liberates energy.

By my definition, reducing exhaust gas velocity by dilution is most definitely not more efficient.  In your example it took 18,400 kg of propellant to deliver the same payload as 1,631 kg of metahelium alone (a mass ratio of 4.2 versus 1.4).  However, if I’m understanding you correctly, you seem to be saying dilution is more efficient because you can deliver more payload with the same amount of metahelium (not counting the water as propellant).

The argument seems to be more over how we define “propellant” and “efficiency”.  My argument from the beginning has been that to get the most delta-v out of a kilogram of propellant (my definition) we need to get the exhaust gas velocity as high as possible, and I still believe that.


Yes the argument is about how to define efficiency. 

Is efficiency measured in mass of propellant required to move a given mass to a given delta-v?  If so we should try to set our exhaust velocity as close to the speed of light as possible.

Is efficiency measured in energy expended to move a given mass to a given delta-v?  If so we should try to set our exhaust velocity to the current cumulative delta-v.

Is efficiency measured in dollars to move a given mass to a given delta-v?  If so we may look at endless factors that effect the cost of design, construction, operating expenses and time to mission accomplishment.

P.S.  I was counting metahelium as both a fuel and a propellant, water was a propellant, but not a fuel.  I was calculating for minimum fuel (energy)expenditure, not for minimum propellant expenditure.  And for the special case of a reusable single stage to orbit rocket, assuming fuel expenses are the primary cost driver, the dilution with water could save significant amounts of money.

« Last Edit: June 21, 2006, 04:29:26 PM by jdbenner »
Joshua D. Benner Associate in Arts and Sciences in General Science

Offline DonPMitchell

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Re: Electric/chemical Energy to kinetic energy conversion
« Reply #44 on: June 21, 2006, 12:53:56 PM »
Is meta-helium something real?
Never send a human to do a machine's job.
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