Author Topic: Medusa concept using chemical explosives  (Read 6793 times)

Offline Twark_Main

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Re: Medusa concept using chemical explosives
« Reply #20 on: 12/05/2022 01:44 am »
[snip]

No matter how efficient your heat transfer, your exhaust velocity can never exceed the limiting value determined by the total amount of energy available.
I will cite as an example the Spitfire fighter, in which the amount of heat entering the radiator was equal to the amount of energy driving the propeller. However, the thrust from the radiator was negligible and barely covered its air resistance, and all because of the temperature of the antifreeze in the region of 100 degrees Celsius.
There is not only energy density, but also energy quality.

Yes, we all know how waste heat works (or more accurately, doesn't).

I can guarantee you the Spitfire didn't violate the Conservation of Energy either.  :D
« Last Edit: 12/05/2022 01:48 am by Twark_Main »
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Offline Beratnyi

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Re: Medusa concept using chemical explosives
« Reply #21 on: 12/05/2022 01:49 am »
[snip]

No matter how efficient your heat transfer, your exhaust velocity can never exceed the limiting value determined by the total amount of energy available.
I will cite as an example the Spitfire fighter, in which the amount of heat entering the radiator was equal to the amount of energy driving the propeller. However, the thrust from the radiator was negligible and barely covered its air resistance, and all because of the temperature of the antifreeze in the region of 100 degrees Celsius.
There is not only energy density, but also energy quality.

Yes, we all know how waste heat (vs useful energy) works.

I can guarantee you the Spitfire didn't violate the Conservation of Energy either.  :D
The quantity and quality of energy are two different things! At the Spitfire, a megawatt of heat passed through the radiator, but even a cutlet on it would be impossible to fry. At the same time, a 1 kilowatt burner can easily fry a kebab.
Energy density refers only to the amount of energy available to you, not its quality.
« Last Edit: 12/05/2022 01:52 am by Beratnyi »

Offline Twark_Main

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Re: Medusa concept using chemical explosives
« Reply #22 on: 12/05/2022 01:55 am »
[snip]

No matter how efficient your heat transfer, your exhaust velocity can never exceed the limiting value determined by the total amount of energy available.
I will cite as an example the Spitfire fighter, in which the amount of heat entering the radiator was equal to the amount of energy driving the propeller. However, the thrust from the radiator was negligible and barely covered its air resistance, and all because of the temperature of the antifreeze in the region of 100 degrees Celsius.
There is not only energy density, but also energy quality.

Yes, we all know how waste heat (vs useful energy) works.

I can guarantee you the Spitfire didn't violate the Conservation of Energy either.  :D
The quantity and quality of energy are two different things!

I know.

At the Spitfire, a megawatt of heat passed through the radiator, but even a cutlet on it would be impossible to fry. At the same time, a 1 kilowatt burner can easily fry a kebab.

I fail to see how this relates to your original proposal, or to my criticism.

My criticism was that you can't violate Conservation of Energy, so total energy still matters. Nothing about the Spitfire changes that.
"The search for a universal design which suits all sites, people, and situations is obviously impossible. What is possible is well designed examples of the application of universal principles." ~~ David Holmgren

Offline Twark_Main

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Re: Medusa concept using chemical explosives
« Reply #23 on: 12/05/2022 01:59 am »
Spitfire

...

kebab

"Hey I heard u like Spitfires, so I made a Spitfire spit fire at ur spit."
"The search for a universal design which suits all sites, people, and situations is obviously impossible. What is possible is well designed examples of the application of universal principles." ~~ David Holmgren

Offline Beratnyi

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Re: Medusa concept using chemical explosives
« Reply #24 on: 12/05/2022 02:03 am »
[snip]

No matter how efficient your heat transfer, your exhaust velocity can never exceed the limiting value determined by the total amount of energy available.
I will cite as an example the Spitfire fighter, in which the amount of heat entering the radiator was equal to the amount of energy driving the propeller. However, the thrust from the radiator was negligible and barely covered its air resistance, and all because of the temperature of the antifreeze in the region of 100 degrees Celsius.
There is not only energy density, but also energy quality.

Yes, we all know how waste heat (vs useful energy) works.

I can guarantee you the Spitfire didn't violate the Conservation of Energy either.  :D
The quantity and quality of energy are two different things!

I know.

At the Spitfire, a megawatt of heat passed through the radiator, but even a cutlet on it would be impossible to fry. At the same time, a 1 kilowatt burner can easily fry a kebab.

I fail to see how this relates to your original proposal, or to my criticism.

My criticism was that you can't violate Conservation of Energy, so total energy still matters. Nothing about the Spitfire changes that.
For example, we have a substance with an energy density of 10 MJ per kg. We can release this energy in a second and get a temperature of 3000 degrees, or we can release this energy in an hour and not even get warm. And hypothetically, we can release this energy in a nanosecond and get millions of degrees. In all three cases, the energy density is the same, but the rate of its release is different. Speed is what matters, which is why the temperature in the center of a nuclear explosion is so high, because the detonation velocity is thousands of times higher than that of any chemical explosive, and not at all because of the energy density.

Offline InterestedEngineer

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Re: Medusa concept using chemical explosives
« Reply #25 on: 12/05/2022 06:48 am »
What are the kinetic and thermal energy components of high grade (e.g. c4 or better) explosives or shaped charges?

Without a nozzle mind you.  A de Laval nozzle converts thermal into kinetic energy, but in doing so gets very hot and and has pressure limits as noted above.

I note Raptor 2 has about a 56/44 ratio of kinetic/thermal energy after expansion on the Vacuum nozzle.  Quite the waste really, but amazing for a thermal engine.

Allegedly, a rotating detonation system doesn't need such a nozzle but you only get about 20-30% more Isp.

Can an explosive system bypass those limits by creating more initial kinetic energy and less thermal energy?

If I did read this chart and do the math correctly, the ratio of kinetic energy to thermal energy for most high grade explosives is 5:1:

https://www.hindawi.com/journals/ijce/2019/4017068/tab3/

IOTW, 83% kinetic 17% thermal.  Much more efficient.
« Last Edit: 12/05/2022 06:56 am by InterestedEngineer »

Offline ppnl

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Re: Medusa concept using chemical explosives
« Reply #26 on: 12/05/2022 07:45 am »

For example, we have a substance with an energy density of 10 MJ per kg. We can release this energy in a second and get a temperature of 3000 degrees, or we can release this energy in an hour and not even get warm. And hypothetically, we can release this energy in a nanosecond and get millions of degrees. In all three cases, the energy density is the same, but the rate of its release is different. Speed is what matters, which is why the temperature in the center of a nuclear explosion is so high, because the detonation velocity is thousands of times higher than that of any chemical explosive, and not at all because of the energy density.

This is wrong on so many levels.

First, say you have a kg of fuel with 10 MJ of energy. Burn it in any device you wish to produce a jet of gas. The kinetic energy of that gas cannot exceed 10 MJ otherwise you have violated conservation of energy and can make a perpetual motion device. One kg of gas traveling at 4472.13 m/s has just about exactly 10 MJ of energy. It is simple math. Nothing you do can get a faster exhaust velocity.

Second, lets look at the temperature. A hot gas contains more energy than a cold gas. If you burn a kg of fuel producing 10 MJ of energy then the hot gas has exactly 10 MJ of heat more than the fuel you started with. That exactly defines the maximum temperature that gas can reach.  If you exceed that temperature then you have violated conservation of energy and can make a perpetual motion device.

Now lets look at chemicals. Say you have a kilogram of hydrogen/oxygen mix. You say that if we burn that fast enough then we could have temperatures in the millions of degrees. But a kg of water heated to millions of degrees has far more energy than you can ever get from the kg of hydrogen/oxygen. Again a perpetual motion device exactly as stated above. But worse than that at any temperature above 2182 C the water will start to disassociate back into hydrogen and oxygen absorbing heat from the gas and so cooling it. That gives us the maximum temperature we can ever get from burning hydrogen. Millions of degrees would dissociate all of the water and reduce the gas to a plasma.



Offline Beratnyi

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Re: Medusa concept using chemical explosives
« Reply #27 on: 12/05/2022 08:11 am »

For example, we have a substance with an energy density of 10 MJ per kg. We can release this energy in a second and get a temperature of 3000 degrees, or we can release this energy in an hour and not even get warm. And hypothetically, we can release this energy in a nanosecond and get millions of degrees. In all three cases, the energy density is the same, but the rate of its release is different. Speed is what matters, which is why the temperature in the center of a nuclear explosion is so high, because the detonation velocity is thousands of times higher than that of any chemical explosive, and not at all because of the energy density.

This is wrong on so many levels.

First, say you have a kg of fuel with 10 MJ of energy. Burn it in any device you wish to produce a jet of gas. The kinetic energy of that gas cannot exceed 10 MJ otherwise you have violated conservation of energy and can make a perpetual motion device. One kg of gas traveling at 4472.13 m/s has just about exactly 10 MJ of energy. It is simple math. Nothing you do can get a faster exhaust velocity.

Second, lets look at the temperature. A hot gas contains more energy than a cold gas. If you burn a kg of fuel producing 10 MJ of energy then the hot gas has exactly 10 MJ of heat more than the fuel you started with. That exactly defines the maximum temperature that gas can reach.  If you exceed that temperature then you have violated conservation of energy and can make a perpetual motion device.

Now lets look at chemicals. Say you have a kilogram of hydrogen/oxygen mix. You say that if we burn that fast enough then we could have temperatures in the millions of degrees. But a kg of water heated to millions of degrees has far more energy than you can ever get from the kg of hydrogen/oxygen. Again a perpetual motion device exactly as stated above. But worse than that at any temperature above 2182 C the water will start to disassociate back into hydrogen and oxygen absorbing heat from the gas and so cooling it. That gives us the maximum temperature we can ever get from burning hydrogen. Millions of degrees would dissociate all of the water and reduce the gas to a plasma.
Why did you decide that 1 kg of water is subject to heating to millions of degrees? The greater the specific impulse, the lower the mass of exhaust gases for a given amount of energy spent on heating.
In other words, using a limited energy source of 10 MJ/kg for fast heat transfer to the reaction mass, we will increase the specific impulse, but reduce the ejected mass. There is no violation of the laws of physics here.
« Last Edit: 12/05/2022 08:18 am by Beratnyi »

Offline edzieba

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Re: Medusa concept using chemical explosives
« Reply #28 on: 12/05/2022 10:43 am »

For example, we have a substance with an energy density of 10 MJ per kg. We can release this energy in a second and get a temperature of 3000 degrees, or we can release this energy in an hour and not even get warm. And hypothetically, we can release this energy in a nanosecond and get millions of degrees. In all three cases, the energy density is the same, but the rate of its release is different. Speed is what matters, which is why the temperature in the center of a nuclear explosion is so high, because the detonation velocity is thousands of times higher than that of any chemical explosive, and not at all because of the energy density.

This is wrong on so many levels.

First, say you have a kg of fuel with 10 MJ of energy. Burn it in any device you wish to produce a jet of gas. The kinetic energy of that gas cannot exceed 10 MJ otherwise you have violated conservation of energy and can make a perpetual motion device. One kg of gas traveling at 4472.13 m/s has just about exactly 10 MJ of energy. It is simple math. Nothing you do can get a faster exhaust velocity.

Second, lets look at the temperature. A hot gas contains more energy than a cold gas. If you burn a kg of fuel producing 10 MJ of energy then the hot gas has exactly 10 MJ of heat more than the fuel you started with. That exactly defines the maximum temperature that gas can reach.  If you exceed that temperature then you have violated conservation of energy and can make a perpetual motion device.

Now lets look at chemicals. Say you have a kilogram of hydrogen/oxygen mix. You say that if we burn that fast enough then we could have temperatures in the millions of degrees. But a kg of water heated to millions of degrees has far more energy than you can ever get from the kg of hydrogen/oxygen. Again a perpetual motion device exactly as stated above. But worse than that at any temperature above 2182 C the water will start to disassociate back into hydrogen and oxygen absorbing heat from the gas and so cooling it. That gives us the maximum temperature we can ever get from burning hydrogen. Millions of degrees would dissociate all of the water and reduce the gas to a plasma.
Why did you decide that 1 kg of water is subject to heating to millions of degrees?
Water is the combustion product of Hydrogen and Oxygen.
Quote
The greater the specific impulse, the lower the mass of exhaust gases for a given amount of energy spent on heating.
No. Exhaust mass will not vary (if you throw 1kg of propellant out the back 1kg of propellant goes out the back as exhaust, no matter how its combusted.
Quote
In other words, using a limited energy source of 10 MJ/kg for fast heat transfer to the reaction mass, we will increase the specific impulse, but reduce the ejected mass. There is no violation of the laws of physics here.
No, magically losing mass is indeed an egregious violation of conservation of mass. You burn 1kg of propellant, you get 1kg of reaction products.

Offline Beratnyi

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Re: Medusa concept using chemical explosives
« Reply #29 on: 12/05/2022 05:09 pm »

For example, we have a substance with an energy density of 10 MJ per kg. We can release this energy in a second and get a temperature of 3000 degrees, or we can release this energy in an hour and not even get warm. And hypothetically, we can release this energy in a nanosecond and get millions of degrees. In all three cases, the energy density is the same, but the rate of its release is different. Speed is what matters, which is why the temperature in the center of a nuclear explosion is so high, because the detonation velocity is thousands of times higher than that of any chemical explosive, and not at all because of the energy density.

This is wrong on so many levels.

First, say you have a kg of fuel with 10 MJ of energy. Burn it in any device you wish to produce a jet of gas. The kinetic energy of that gas cannot exceed 10 MJ otherwise you have violated conservation of energy and can make a perpetual motion device. One kg of gas traveling at 4472.13 m/s has just about exactly 10 MJ of energy. It is simple math. Nothing you do can get a faster exhaust velocity.

Second, lets look at the temperature. A hot gas contains more energy than a cold gas. If you burn a kg of fuel producing 10 MJ of energy then the hot gas has exactly 10 MJ of heat more than the fuel you started with. That exactly defines the maximum temperature that gas can reach.  If you exceed that temperature then you have violated conservation of energy and can make a perpetual motion device.

Now lets look at chemicals. Say you have a kilogram of hydrogen/oxygen mix. You say that if we burn that fast enough then we could have temperatures in the millions of degrees. But a kg of water heated to millions of degrees has far more energy than you can ever get from the kg of hydrogen/oxygen. Again a perpetual motion device exactly as stated above. But worse than that at any temperature above 2182 C the water will start to disassociate back into hydrogen and oxygen absorbing heat from the gas and so cooling it. That gives us the maximum temperature we can ever get from burning hydrogen. Millions of degrees would dissociate all of the water and reduce the gas to a plasma.
Why did you decide that 1 kg of water is subject to heating to millions of degrees?
Water is the combustion product of Hydrogen and Oxygen.
Quote
The greater the specific impulse, the lower the mass of exhaust gases for a given amount of energy spent on heating.
No. Exhaust mass will not vary (if you throw 1kg of propellant out the back 1kg of propellant goes out the back as exhaust, no matter how its combusted.
Quote
In other words, using a limited energy source of 10 MJ/kg for fast heat transfer to the reaction mass, we will increase the specific impulse, but reduce the ejected mass. There is no violation of the laws of physics here.
No, magically losing mass is indeed an egregious violation of conservation of mass. You burn 1kg of propellant, you get 1kg of reaction products.
Fuel and reaction mass are two different things! In chemical rocket engines, this is almost the same thing, which is misleading people. But for example in ion and nuclear thermal engine these things are clearly separated.
Ask why engineers are so proud of the amount of pressure in the combustion chamber? This does not have a significant effect on the TWR (the Merlin has a better indicator than the RD-180), and the smaller effect of backpressure in atmosphere harms the rocket as a whole rather than helps (since it increases gravitational losses). The only reason to increase this parameter is to accelerate the processes of heat transfer in the combustion chamber from the fuel (for example, Methalox) to the reaction mass (water plus CO2), which leads to a higher temperature and, accordingly, to a higher exhaust gas velocity (specific impulse improves).
There is no magic here, the faster the transfer of heat from the energy source to the reaction mass, the faster the speed of the outflow of gases from the engines and the lower the mass flow. That is why rockets with a large specific impulse, other things being equal, have more Delta-V.

Offline edzieba

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Re: Medusa concept using chemical explosives
« Reply #30 on: 12/05/2022 05:40 pm »

For example, we have a substance with an energy density of 10 MJ per kg. We can release this energy in a second and get a temperature of 3000 degrees, or we can release this energy in an hour and not even get warm. And hypothetically, we can release this energy in a nanosecond and get millions of degrees. In all three cases, the energy density is the same, but the rate of its release is different. Speed is what matters, which is why the temperature in the center of a nuclear explosion is so high, because the detonation velocity is thousands of times higher than that of any chemical explosive, and not at all because of the energy density.

This is wrong on so many levels.

First, say you have a kg of fuel with 10 MJ of energy. Burn it in any device you wish to produce a jet of gas. The kinetic energy of that gas cannot exceed 10 MJ otherwise you have violated conservation of energy and can make a perpetual motion device. One kg of gas traveling at 4472.13 m/s has just about exactly 10 MJ of energy. It is simple math. Nothing you do can get a faster exhaust velocity.

Second, lets look at the temperature. A hot gas contains more energy than a cold gas. If you burn a kg of fuel producing 10 MJ of energy then the hot gas has exactly 10 MJ of heat more than the fuel you started with. That exactly defines the maximum temperature that gas can reach.  If you exceed that temperature then you have violated conservation of energy and can make a perpetual motion device.

Now lets look at chemicals. Say you have a kilogram of hydrogen/oxygen mix. You say that if we burn that fast enough then we could have temperatures in the millions of degrees. But a kg of water heated to millions of degrees has far more energy than you can ever get from the kg of hydrogen/oxygen. Again a perpetual motion device exactly as stated above. But worse than that at any temperature above 2182 C the water will start to disassociate back into hydrogen and oxygen absorbing heat from the gas and so cooling it. That gives us the maximum temperature we can ever get from burning hydrogen. Millions of degrees would dissociate all of the water and reduce the gas to a plasma.
Why did you decide that 1 kg of water is subject to heating to millions of degrees?
Water is the combustion product of Hydrogen and Oxygen.
Quote
The greater the specific impulse, the lower the mass of exhaust gases for a given amount of energy spent on heating.
No. Exhaust mass will not vary (if you throw 1kg of propellant out the back 1kg of propellant goes out the back as exhaust, no matter how its combusted.
Quote
In other words, using a limited energy source of 10 MJ/kg for fast heat transfer to the reaction mass, we will increase the specific impulse, but reduce the ejected mass. There is no violation of the laws of physics here.
No, magically losing mass is indeed an egregious violation of conservation of mass. You burn 1kg of propellant, you get 1kg of reaction products.
Fuel and reaction mass are two different things! In chemical rocket engines, this is almost the same thing, which is misleading people. But for example in ion and nuclear thermal engine these things are clearly separated.
Irrelevant for an 'explosive driven' engine: either the explosive reaction products are the remass (in which case the situation is identical to monoprop or biprop chemical rockets) or you're trying to heat some other remass (which which case you have a catastrophically inefficient thermal rocket).
Quote
Ask why engineers are so proud of the amount of pressure in the combustion chamber? This does not have a significant effect on the TWR (the Merlin has a better indicator than the RD-180), and the smaller effect of backpressure in atmosphere harms the rocket as a whole rather than helps (since it increases gravitational losses).
False. Chamber pressure is proportional to TWR, for a given engine.
Quote
The only reason to increase this parameter is to accelerate the processes of heat transfer in the combustion chamber from the fuel (for example, Methalox) to the reaction mass (water plus CO2), which leads to a higher temperature and, accordingly, to a higher exhaust gas velocity (specific impulse improves).
False. Chamber temperature is what is most strongly coupled to exhaust temperature (and this ISP), which is why the Raptor with it's stupendous 300 Bar combustion chamber pressure has a vacuum ISP of ~363s, whereas the RL-10's puny 24 bar chamber pressure still beats it up and down the street with an ISP of 465s.
Quote
There is no magic here, the faster the transfer of heat from the energy source to the reaction mass, the faster the speed of the outflow of gases from the engines and the lower the mass flow.
False. 'Speed of heating' has nothing whatsoever to do with either thrust or ISP. About the only thing it affects is physical length of the combustion chamber.
« Last Edit: 12/05/2022 05:42 pm by edzieba »

Offline Beratnyi

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Re: Medusa concept using chemical explosives
« Reply #31 on: 12/05/2022 09:27 pm »

For example, we have a substance with an energy density of 10 MJ per kg. We can release this energy in a second and get a temperature of 3000 degrees, or we can release this energy in an hour and not even get warm. And hypothetically, we can release this energy in a nanosecond and get millions of degrees. In all three cases, the energy density is the same, but the rate of its release is different. Speed is what matters, which is why the temperature in the center of a nuclear explosion is so high, because the detonation velocity is thousands of times higher than that of any chemical explosive, and not at all because of the energy density.

This is wrong on so many levels.

First, say you have a kg of fuel with 10 MJ of energy. Burn it in any device you wish to produce a jet of gas. The kinetic energy of that gas cannot exceed 10 MJ otherwise you have violated conservation of energy and can make a perpetual motion device. One kg of gas traveling at 4472.13 m/s has just about exactly 10 MJ of energy. It is simple math. Nothing you do can get a faster exhaust velocity.

Second, lets look at the temperature. A hot gas contains more energy than a cold gas. If you burn a kg of fuel producing 10 MJ of energy then the hot gas has exactly 10 MJ of heat more than the fuel you started with. That exactly defines the maximum temperature that gas can reach.  If you exceed that temperature then you have violated conservation of energy and can make a perpetual motion device.

Now lets look at chemicals. Say you have a kilogram of hydrogen/oxygen mix. You say that if we burn that fast enough then we could have temperatures in the millions of degrees. But a kg of water heated to millions of degrees has far more energy than you can ever get from the kg of hydrogen/oxygen. Again a perpetual motion device exactly as stated above. But worse than that at any temperature above 2182 C the water will start to disassociate back into hydrogen and oxygen absorbing heat from the gas and so cooling it. That gives us the maximum temperature we can ever get from burning hydrogen. Millions of degrees would dissociate all of the water and reduce the gas to a plasma.
Why did you decide that 1 kg of water is subject to heating to millions of degrees?
Water is the combustion product of Hydrogen and Oxygen.
Quote
The greater the specific impulse, the lower the mass of exhaust gases for a given amount of energy spent on heating.
No. Exhaust mass will not vary (if you throw 1kg of propellant out the back 1kg of propellant goes out the back as exhaust, no matter how its combusted.
Quote
In other words, using a limited energy source of 10 MJ/kg for fast heat transfer to the reaction mass, we will increase the specific impulse, but reduce the ejected mass. There is no violation of the laws of physics here.
No, magically losing mass is indeed an egregious violation of conservation of mass. You burn 1kg of propellant, you get 1kg of reaction products.
Fuel and reaction mass are two different things! In chemical rocket engines, this is almost the same thing, which is misleading people. But for example in ion and nuclear thermal engine these things are clearly separated.
Irrelevant for an 'explosive driven' engine: either the explosive reaction products are the remass (in which case the situation is identical to monoprop or biprop chemical rockets) or you're trying to heat some other remass (which which case you have a catastrophically inefficient thermal rocket).
Quote
Ask why engineers are so proud of the amount of pressure in the combustion chamber? This does not have a significant effect on the TWR (the Merlin has a better indicator than the RD-180), and the smaller effect of backpressure in atmosphere harms the rocket as a whole rather than helps (since it increases gravitational losses).
False. Chamber pressure is proportional to TWR, for a given engine.
Quote
The only reason to increase this parameter is to accelerate the processes of heat transfer in the combustion chamber from the fuel (for example, Methalox) to the reaction mass (water plus CO2), which leads to a higher temperature and, accordingly, to a higher exhaust gas velocity (specific impulse improves).
False. Chamber temperature is what is most strongly coupled to exhaust temperature (and this ISP), which is why the Raptor with it's stupendous 300 Bar combustion chamber pressure has a vacuum ISP of ~363s, whereas the RL-10's puny 24 bar chamber pressure still beats it up and down the street with an ISP of 465s.
Quote
There is no magic here, the faster the transfer of heat from the energy source to the reaction mass, the faster the speed of the outflow of gases from the engines and the lower the mass flow.
False. 'Speed of heating' has nothing whatsoever to do with either thrust or ISP. About the only thing it affects is physical length of the combustion chamber.
I explain point by point:
1) The fact that in chemical fuel the energy source and the reaction mass are combined only means that the mass of the energy source is discarded along with the reaction mass, and does not remain a constant load like solar panels for spacecraft with ion engines
2) The pressure in the combustion chamber is not proportional to the TWR, because with a decrease in the volume of the chamber, it is necessary to thicken its walls. That is why the NK-33 has a better TWR than the RD-180 (specially took engines with the same cycle as an example)
3) RL-10 has a higher specific impulse due to a different fuel pair, despite the fact that the temperature in the combustion chamber is lower than that of the Raptor. This is because the reaction mass for hydrolox engines is only water, while for methalox engines it is water plus heavier CO2.
4) The heating rate is of key importance, because the gas is not in a closed volume, it constantly leaves the combustion chamber and the faster we supply heat, the greater the exit gas velocity will be
Always double-check and question your knowledge, and only then can you come up with something new.

Offline ppnl

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Re: Medusa concept using chemical explosives
« Reply #32 on: 12/05/2022 11:39 pm »

For example, we have a substance with an energy density of 10 MJ per kg. We can release this energy in a second and get a temperature of 3000 degrees, or we can release this energy in an hour and not even get warm. And hypothetically, we can release this energy in a nanosecond and get millions of degrees. In all three cases, the energy density is the same, but the rate of its release is different. Speed is what matters, which is why the temperature in the center of a nuclear explosion is so high, because the detonation velocity is thousands of times higher than that of any chemical explosive, and not at all because of the energy density.

This is wrong on so many levels.

First, say you have a kg of fuel with 10 MJ of energy. Burn it in any device you wish to produce a jet of gas. The kinetic energy of that gas cannot exceed 10 MJ otherwise you have violated conservation of energy and can make a perpetual motion device. One kg of gas traveling at 4472.13 m/s has just about exactly 10 MJ of energy. It is simple math. Nothing you do can get a faster exhaust velocity.

Second, lets look at the temperature. A hot gas contains more energy than a cold gas. If you burn a kg of fuel producing 10 MJ of energy then the hot gas has exactly 10 MJ of heat more than the fuel you started with. That exactly defines the maximum temperature that gas can reach.  If you exceed that temperature then you have violated conservation of energy and can make a perpetual motion device.

Now lets look at chemicals. Say you have a kilogram of hydrogen/oxygen mix. You say that if we burn that fast enough then we could have temperatures in the millions of degrees. But a kg of water heated to millions of degrees has far more energy than you can ever get from the kg of hydrogen/oxygen. Again a perpetual motion device exactly as stated above. But worse than that at any temperature above 2182 C the water will start to disassociate back into hydrogen and oxygen absorbing heat from the gas and so cooling it. That gives us the maximum temperature we can ever get from burning hydrogen. Millions of degrees would dissociate all of the water and reduce the gas to a plasma.
Why did you decide that 1 kg of water is subject to heating to millions of degrees? The greater the specific impulse, the lower the mass of exhaust gases for a given amount of energy spent on heating.
In other words, using a limited energy source of 10 MJ/kg for fast heat transfer to the reaction mass, we will increase the specific impulse, but reduce the ejected mass. There is no violation of the laws of physics here.

Dude, you are so tangled up it's hard to see how to untangle you. If you don't expel all of your reaction mass then what do you do with it? Surely you understand that mass is conserved?

Lets try to simplify. You have one kg chemical gas with an energy of 10 MJ of chemical energy. You burn it producing a hot gas that contains 10 MJ of thermal energy. You expel the gas and get a kg of comparatively cool gas with 10 MJ of kinetic energy. You attach this to a rocket and the rocket is pushed absorbing some of the kinetic energy. The kinetic energy of the rocket plus the kinetic energy of the gas will be 10 MJ. Thats all you can do. Ever.

Are you suggesting that you could heat half your fuel to a higher temperature and get a higher specific impulse? Well you can't really do that with burning a chemical fuel but lets say you can by some other mechanism. Will this increases your specific impulse? Yes. The specific impulse is just the impulse per unit mass so half the mass exhausted at a higher temperature will give you a higher specific impulse. But so what you only have half the reaction mass? Worse, the rocket is actually going to get a smaller share of the energy because more energy is being carried away by the higher velocity exhaust. Worse still, what did you do with the other half of the fuel that you didn't expel? If you keep it on board that increases the mass of your rocket and so reduces your velocity even more.

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The greater the specific impulse, the lower the mass of exhaust gases for a given amount of energy spent on heating.       

Lets untangle this hopeless knot because it really makes my brain itch. This is true but you have tortured it to the point of being a crime against thought. Lets restate it:

The hotter you heat a given amount of gas the higher the specific impulse you can get from it.

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In other words, using a limited energy source of 10 MJ/kg for fast heat transfer to the reaction mass, we will increase the specific impulse, but reduce the ejected mass.

For the love of mike what happened to the mass that you didn't eject? And what does the rate of heat transfer have to do with it?

Offline Beratnyi

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Re: Medusa concept using chemical explosives
« Reply #33 on: 12/06/2022 02:54 am »

For example, we have a substance with an energy density of 10 MJ per kg. We can release this energy in a second and get a temperature of 3000 degrees, or we can release this energy in an hour and not even get warm. And hypothetically, we can release this energy in a nanosecond and get millions of degrees. In all three cases, the energy density is the same, but the rate of its release is different. Speed is what matters, which is why the temperature in the center of a nuclear explosion is so high, because the detonation velocity is thousands of times higher than that of any chemical explosive, and not at all because of the energy density.

This is wrong on so many levels.

First, say you have a kg of fuel with 10 MJ of energy. Burn it in any device you wish to produce a jet of gas. The kinetic energy of that gas cannot exceed 10 MJ otherwise you have violated conservation of energy and can make a perpetual motion device. One kg of gas traveling at 4472.13 m/s has just about exactly 10 MJ of energy. It is simple math. Nothing you do can get a faster exhaust velocity.

Second, lets look at the temperature. A hot gas contains more energy than a cold gas. If you burn a kg of fuel producing 10 MJ of energy then the hot gas has exactly 10 MJ of heat more than the fuel you started with. That exactly defines the maximum temperature that gas can reach.  If you exceed that temperature then you have violated conservation of energy and can make a perpetual motion device.

Now lets look at chemicals. Say you have a kilogram of hydrogen/oxygen mix. You say that if we burn that fast enough then we could have temperatures in the millions of degrees. But a kg of water heated to millions of degrees has far more energy than you can ever get from the kg of hydrogen/oxygen. Again a perpetual motion device exactly as stated above. But worse than that at any temperature above 2182 C the water will start to disassociate back into hydrogen and oxygen absorbing heat from the gas and so cooling it. That gives us the maximum temperature we can ever get from burning hydrogen. Millions of degrees would dissociate all of the water and reduce the gas to a plasma.
Why did you decide that 1 kg of water is subject to heating to millions of degrees? The greater the specific impulse, the lower the mass of exhaust gases for a given amount of energy spent on heating.
In other words, using a limited energy source of 10 MJ/kg for fast heat transfer to the reaction mass, we will increase the specific impulse, but reduce the ejected mass. There is no violation of the laws of physics here.

Dude, you are so tangled up it's hard to see how to untangle you. If you don't expel all of your reaction mass then what do you do with it? Surely you understand that mass is conserved?

Lets try to simplify. You have one kg chemical gas with an energy of 10 MJ of chemical energy. You burn it producing a hot gas that contains 10 MJ of thermal energy. You expel the gas and get a kg of comparatively cool gas with 10 MJ of kinetic energy. You attach this to a rocket and the rocket is pushed absorbing some of the kinetic energy. The kinetic energy of the rocket plus the kinetic energy of the gas will be 10 MJ. Thats all you can do. Ever.

Are you suggesting that you could heat half your fuel to a higher temperature and get a higher specific impulse? Well you can't really do that with burning a chemical fuel but lets say you can by some other mechanism. Will this increases your specific impulse? Yes. The specific impulse is just the impulse per unit mass so half the mass exhausted at a higher temperature will give you a higher specific impulse. But so what you only have half the reaction mass? Worse, the rocket is actually going to get a smaller share of the energy because more energy is being carried away by the higher velocity exhaust. Worse still, what did you do with the other half of the fuel that you didn't expel? If you keep it on board that increases the mass of your rocket and so reduces your velocity even more.

Quote
The greater the specific impulse, the lower the mass of exhaust gases for a given amount of energy spent on heating.       

Lets untangle this hopeless knot because it really makes my brain itch. This is true but you have tortured it to the point of being a crime against thought. Lets restate it:

The hotter you heat a given amount of gas the higher the specific impulse you can get from it.

Quote

In other words, using a limited energy source of 10 MJ/kg for fast heat transfer to the reaction mass, we will increase the specific impulse, but reduce the ejected mass.

For the love of mike what happened to the mass that you didn't eject? And what does the rate of heat transfer have to do with it?
If you follow your logic, then the specific impulse does not matter at all, and it’s not even clear why rocket scientists around the world are trying to improve it with all their might, because, in your opinion, the same amount of kinetic energy will be transferred to the rocket anyway.

I'll try to explain again. Imagine that the fuel in the rocket is not used as a reaction mass and is used solely to heat water vapor, which creates thrust. We can heat the steam up to 2000 С and get a high Isp, or heat it up to 200 C and get a low Isp. No matter how much energy you have stored in the fuel, if you only heat the steam up to 200 degrees you will end up with a bad rocket.

You also do not take into account that with an increase in specific impulse, fuel consumption decreases while maintaining the same level of thrust. In other words, for a given amount of energy, it is always more profitable for a rocket to eject less reaction mass at a higher velocity than to eject more reaction mass at a low velocity.

Everything I say is actually obvious to anyone with even the slightest knowledge of the subject. I am very surprised that I have to explain elementary things here.
« Last Edit: 12/06/2022 02:57 am by Beratnyi »

Offline ppnl

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Re: Medusa concept using chemical explosives
« Reply #34 on: 12/06/2022 08:04 am »

If you follow your logic, then the specific impulse does not matter at all, and it’s not even clear why rocket scientists around the world are trying to improve it with all their might, because, in your opinion, the same amount of kinetic energy will be transferred to the rocket anyway.

No, specific impulse matters but you misunderstand why and how it matters. If I am targeting a deltaV of one meter per second then I would be happy for a low Isp fuel to avoid wasting energy. If I am targeting a hundred kilometers per second deltaV then I need a very high Isp to avoid the crushing penalty of the weight of my fuel. And that is the thing you are missing despite the fact that I have asked about it several times. You have a blind spot here that you need to break through.  Think hard about the weight of your fuel.

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I'll try to explain again. Imagine that the fuel in the rocket is not used as a reaction mass and is used solely to heat water vapor, which creates thrust. We can heat the steam up to 2000 С and get a high Isp, or heat it up to 200 C and get a low Isp. No matter how much energy you have stored in the fuel, if you only heat the steam up to 200 degrees you will end up with a bad rocket.

Fine, if you heat water enough you can achieve as high an Isp as you want. How do you heat your water? Lets say you use a battery and do it electrically. You can get high Isp but how much does your battery weigh? If you do the math it turns out that the weight of your battery will absolutely destroy the performance of your rocket. The weight of your rocket will be more than 99% battery. It turns out that your rocket will perform much better at a much lower Isp due to the smaller battery pack. But that lower Isp does put an upper practical limit on the speed the rocket can reach due to the weight of the water.

Well there are denser ways to store energy like chemicals for example. Say your rocket burns hydrogen and oxygen to heat some water. Well burnt hydrogen and oxygen is water so... Your hydrogen and oxygen is your battery and reaction mass combined in one efficient package. You expel your "battery" as reaction mass as you use up its energy. You can't do better than that.

Quote
You also do not take into account that with an increase in specific impulse, fuel consumption decreases while maintaining the same level of thrust. In other words, for a given amount of energy, it is always more profitable for a rocket to eject less reaction mass at a higher velocity than to eject more reaction mass at a low velocity.

But until you account for the weight of the "battery" or whatever energy storage system you are using you will not understand. That's why I keep asking what you did with the mass you didn't eject? What did you do with your spent battery? When you account for the weight of your spent battery you may find that a lower Isp would be better.

Quote
Everything I say is actually obvious to anyone with even the slightest knowledge of the subject. I am very surprised that I have to explain elementary things here.

We all think the Dunning-Kruger effect only affects other possibly dumber people. The horrifying fact is that it applies to everyone. You are in a forum inhabited by many people who actually work in the field. None of them are defending you. You need to stop and think really really hard about that fact. There are some counter intuitive aspects to this subject but you seem reasonably intelligent. I believe you can do it.

Remember Feynman's first principle:

“The first principle is not to fool yourself – and you are the easiest person to fool.”

Offline pk67

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Re: Medusa concept using chemical explosives
« Reply #35 on: 12/06/2022 09:31 am »

If you follow your logic, then the specific impulse does not matter at all, and it’s not even clear why rocket scientists around the world are trying to improve it with all their might, because, in your opinion, the same amount of kinetic energy will be transferred to the rocket anyway.

No, specific impulse matters but you misunderstand why and how it matters. If I am targeting a deltaV of one meter per second then I would be happy for a low Isp fuel to avoid wasting energy. If I am targeting a hundred kilometers per second deltaV then I need a very high Isp to avoid the crushing penalty of the weight of my fuel. And that is the thing you are missing despite the fact that I have asked about it several times. You have a blind spot here that you need to break through.  Think hard about the weight of your fuel.

...

Quote
Everything I say is actually obvious to anyone with even the slightest knowledge of the subject. I am very surprised that I have to explain elementary things here.

We all think the Dunning-Kruger effect only affects other possibly dumber people. The horrifying fact is that it applies to everyone. You are in a forum inhabited by many people who actually work in the field. None of them are defending you. You need to stop and think really really hard about that fact. There are some counter intuitive aspects to this subject but you seem reasonably intelligent. I believe you can do it.

Remember Feynman's first principle:

“The first principle is not to fool yourself – and you are the easiest person to fool.”

Blind spot is a really conservative diagnose imho. As long as trolling is not allowed here these kind of efforts should be reported to mod i'm sure.  But deleting some trolling efforts is not the best way we can deal with trolls. Keeping these kind of posts in humor department without possibility to delete or edit by author would be much better solution imho. This way we could gain a lot of fun  but it is separate topic.  Some funny jokes can make a day.   

Offline Beratnyi

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Re: Medusa concept using chemical explosives
« Reply #36 on: 12/06/2022 03:19 pm »

If you follow your logic, then the specific impulse does not matter at all, and it’s not even clear why rocket scientists around the world are trying to improve it with all their might, because, in your opinion, the same amount of kinetic energy will be transferred to the rocket anyway.

No, specific impulse matters but you misunderstand why and how it matters. If I am targeting a deltaV of one meter per second then I would be happy for a low Isp fuel to avoid wasting energy. If I am targeting a hundred kilometers per second deltaV then I need a very high Isp to avoid the crushing penalty of the weight of my fuel. And that is the thing you are missing despite the fact that I have asked about it several times. You have a blind spot here that you need to break through.  Think hard about the weight of your fuel.

Quote
I'll try to explain again. Imagine that the fuel in the rocket is not used as a reaction mass and is used solely to heat water vapor, which creates thrust. We can heat the steam up to 2000 С and get a high Isp, or heat it up to 200 C and get a low Isp. No matter how much energy you have stored in the fuel, if you only heat the steam up to 200 degrees you will end up with a bad rocket.

Fine, if you heat water enough you can achieve as high an Isp as you want. How do you heat your water? Lets say you use a battery and do it electrically. You can get high Isp but how much does your battery weigh? If you do the math it turns out that the weight of your battery will absolutely destroy the performance of your rocket. The weight of your rocket will be more than 99% battery. It turns out that your rocket will perform much better at a much lower Isp due to the smaller battery pack. But that lower Isp does put an upper practical limit on the speed the rocket can reach due to the weight of the water.

Well there are denser ways to store energy like chemicals for example. Say your rocket burns hydrogen and oxygen to heat some water. Well burnt hydrogen and oxygen is water so... Your hydrogen and oxygen is your battery and reaction mass combined in one efficient package. You expel your "battery" as reaction mass as you use up its energy. You can't do better than that.

Quote
You also do not take into account that with an increase in specific impulse, fuel consumption decreases while maintaining the same level of thrust. In other words, for a given amount of energy, it is always more profitable for a rocket to eject less reaction mass at a higher velocity than to eject more reaction mass at a low velocity.

But until you account for the weight of the "battery" or whatever energy storage system you are using you will not understand. That's why I keep asking what you did with the mass you didn't eject? What did you do with your spent battery? When you account for the weight of your spent battery you may find that a lower Isp would be better.

Quote
Everything I say is actually obvious to anyone with even the slightest knowledge of the subject. I am very surprised that I have to explain elementary things here.

We all think the Dunning-Kruger effect only affects other possibly dumber people. The horrifying fact is that it applies to everyone. You are in a forum inhabited by many people who actually work in the field. None of them are defending you. You need to stop and think really really hard about that fact. There are some counter intuitive aspects to this subject but you seem reasonably intelligent. I believe you can do it.

Remember Feynman's first principle:

“The first principle is not to fool yourself – and you are the easiest person to fool.”
There seems to be some misunderstanding here. To refute me, you added the rocket mass ratio to the discussion, but I only talked about the specific impulse. For the same mass ratio, a rocket with a high Isp will always have more Delta-V than a rocket with a small Isp. I explain that we are talking about movement in a pure vacuum outside of gravitational fields, because with a vertical launch of a rocket from the Earth, it can really be more profitable to have a lower specific impulse for faster emptying of the tanks and thus reducing gravitational losses. But these losses are taken into account separately and thus only the mass ratio and the magnitude of the specific impulse affect the total number of Delta-V, see the rocket equation.

I will give a simple example to confirm my words. 1 kg of Kerolox mixture contains 9.6 MJ of chemical energy. 1 kg of RDX contains 9.66 MJ of energy - almost complete equality. Both substances, when a chemical reaction is initiated, turn into hot gases. But when 1 kg of Kerolox burns in the combustion chamber, the speed of the outflow of gases is 3.5 km / s, while during the detonation of 1 kg of RDX, the speed of propagation of gases is 8.7 km / s! There is no violation of the laws of physics here, since the release of the same amount of energy occurred at different rates and, accordingly, led to such a significant difference.

Now imagine 2 rockets. The first of them runs on Kerolox and is equipped with a classic rocket engine with a gas exhaust velocity of 3.5 km/s. Its mass ratio is 10. The second rocket runs on RDX shaped charges and its outgassing velocity is 8.7 km/s. This is a conservative assumption, since the velocity of gases becomes higher with the Munroe effect. Its mass ratio is also 10, but let's assume that only half of the fuel substance accelerates to 8.7 km / s, and the rest remains dead weight on the rocket. In reality, this will not be the case, but we will simulate the situation not in our favor. As a result, its mass ratio will be equal to 5. Using the rocket equation, the final speed of the first rocket will be 8 km / s, the second rocket - 14 km / s. Absolutely different result with the same amount of energy in the tanks of both rockets!
« Last Edit: 12/06/2022 03:30 pm by Beratnyi »

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