Atmospheric Burn-up During Re-Entry

[jbriggs444]

Consider this: when you spray dust off your keyboard from a can of compressed air, The can gets cold. A person could wrongly conclude that coldness was “produced“ because of “ kinetic energy” being accumulated back into the cosmos.

You would do well to review the forum rules on personal speculation. Your rigid adherence to false personal positions in the face of careful correction is not appreciated.

The situation when releasing compressed air through a nozzle complicated. See this link for some of the complications. There is a difference between a reversible expansion and a free expansion and between a real gas and an ideal gas.

However, the conclusion in this specific case is correct. The can gets cold because the gas remaining in the can has done work, pushing exhausted gas toward the nozzle opening. The expansion of the portion of the gas that remains in the can counts as "reversible".

 

[256bits]

Does the concept of a boundary layer apply to the very thin atmosphere part, where individual air molecules hit the body?

That's a good and interesting question.
How rarified? and temperature?

See Fig 1, Page 2. of
https://lib.dr.iastate.edu/cgi/view...%3DPERE1#search="rarified gas boundary layer"
Might be from 1972 but it does show that the continuum fluid dynamics becomes applicable further downstream for a long thin plate, with the shock wave being "obviously" produced several steps back from the leading ' point'. Actually the shock is at the tip, or just a bit in front, but due to the kinetic molecular flow predominance near the leading edge, continuum mechanics would not explicitly apply. Perhaps not the best of descriptions.

Table on page 52 is also interesting, showing the rise in pressure and temperature of the gas as we travel along the plate from the leading edge.

 

[Mister T]

I challenge the idea that compressing a gas can somehow “produce heat“.

When work is done to compress a gas, the temperature of the gas increases. Technically, this is indeed not heat.

Consider this: if you had a container that was 10‘ x 10‘ x 10‘ cubed; Full of air at sea level pressure and at room temperature. That container contains a certain amount of heat that can be measured.

It has a certain amount of internal energy.

If you compress one of those thousand cubes into 1/1000 it’s volume, The amount of heat with in your thousand foot container is consistent. No heat was “magically made“.

The amount of work you do is equal to the increase in the internal energy of the air.

I’m just not buying the idea that compressing air “produces” heat.

The compression of the air increases its temperature. Such a thing can be measured with a thermometer. The temperature of the shield increases. This can also be measured with a thermometer. The way we explain the increase in temperature of the air is that work is done on it to increase its internal energy. The way we explain the increase in temperature of the shield is that heat is transferred from the air to the shield, increasing its internal energy.

 

[cjl]

Does the concept of a boundary layer apply to the very thin atmosphere part, where individual air molecules hit the body?

Not really, or at least we have to be a lot more careful about how we treat the problem. This is where the concept of the so-called "knudsen number" becomes important. The Knudsen number is a nondimensional parameter defined as the ratio between the mean free path of the air molecules and some representative scale on the object (often the object's length or diameter). For very large knudsen numbers (Kn >> 1), the flow is considered to be a "free-molecular" flow. At this flow regime, the air molecules have almost no interaction with each other, so the surface just sees the freestream conditions directly. For very small knudsen numbers (Kn<<1), you basically have continuum flow. There is an intermediate region though where people often model the flow as continuum but with some modifications, and one of those modifications is a so-called "slip" boundary layer condition. You still have a boundary layer, but the flow "slips" at the surface, and never reaches zero relative velocity.

If you want to play around with the concept of rarefied flows, there's a fun little 2-d real time simulation program you can download that uses the so-called DSMC (Direct Simulation Monte Carlo) method, and you can see how various flow speeds and densities impact the overall flowfield. The program can be found here:

http://www.gab.com.au/page2.html

 

[cjl]

When work is done to compress a gas, the temperature of the gas increases. Technically, this is indeed not heat.

I agree with the rest of your post, but I disagree with this one. When you do work on the gas, the heat content of the gas does indeed increase, since you have the same mass of gas and it is at a higher temperature.

 

[jbriggs444]

I agree with the rest of your post, but I disagree with this one. When you do work on the gas, the heat content of the gas does indeed increase, since you have the same mass of gas and it is at a higher temperature.

I suspect that @Mister T is using a definition of "heat" as energy transferred due to a temperature difference. One can Google up any number of references to such a definition.

Careful and consistent use of this definition would mean that "heat" (exactly like "work") is not associated with an object but is, instead, associated with an interface. "Heat content" of an object then becomes something of an oxymoron and one might resort to circumlocutions such as "thermal energy" instead.

The distinction is probably wasted on someone who is disputing whether the act of compressing a volume of gas succeeds in increasing the thermal energy of the gas.

 

[boneh3ad]

Second of all I’m doubling down on my statement that and object hitting the earth’s atmosphere is like a bellyflop.

You've yet to make clear exactly why you think it's like a belly flop, but it's not. Entry into the atmosphere doesn't involved a spacecraft hitting a sudden change in density and fluid properties. Instead, it's a spacecraft gradually encountering an increasingly dense atmosphere.

Does the concept of a boundary layer apply to the very thin atmosphere part, where individual air molecules hit the body?

In a sense, yes... depending on just how rarefied the flow is. How well it applies comes down to a parameter called the Knudsen number,
$$Kn = \dfrac{\lambda}{L},$$
where $\lambda$ is the mean free path in the gas and $L$ is a physical length scale representing generally the smallest scale of interest in a given problem. For $Kn\gg 1$, the flow is absolutely not going to obey continuum laws and instead has to be treated molecularly, usually statistically. This is often called free-molecule flow. Usually, anything $Kn\geq 1$ is considered rarefied. For $Kn\ll1$, the flow is a continuum since the mean free path is very small compared to flow scales of interest. Usually, $Kn\leq 0.1$ is a good criterion to use here. In between is sort of a transition region where the behavior is more poorly defined.

Generally, if you are outside of the continuum region, but not by a lot, you still see a boundary layer effect but you get what we call "slip." In a slip flow regime, the no-slip condition that gives rise to the boundary layer is violated and you have to account for that, but you can still reasonably apply many of the continuum laws. There will still be a boundary layer but the flow at the wall will have nonzero velocity (though still smaller than the free-stream velocity). I am not intimately familiar with rarefied flows and don't want to lead you astray, so I won't dig any deeper than that.

Another note is that any flow that is rarefied is necessarily going to be both compressible and low Reynolds number ($Re$). This is because
$$Kn \propto \dfrac{M}{Re},$$
where $M$ is the Mach number. This means that any rarefied flow (with large $Kn$) has a reasonably high $M$ and low $Re$, so incompressible approximations never apply, nor do traditional boundary-layer approximations. Data calculated or acquired in experiments in continuum flows are not good analogues for rarefied flows.

As you might imagine, this is a deep rabbit hole. Some introductory reading on the concepts can be found in:
[1] Liepmann HW, Roshko A. 1957. "Elements of gasdynamics."
[2] Hayes WD, Probstein RF. 1959. "Hypersonic flow theory."
You would need a book dedicated to the topic to get any deeper.

EDIT: @cjl Technically I typed all this up before you posted but a student walked into my office and I never finished, so you beat me to it. Now people are going to have to read nearly the same thing twice.

That's a good and interesting question.
How rarified? and temperature?

See Fig 1, Page 2. of
https://lib.dr.iastate.edu/cgi/view...%3DPERE1#search="rarified gas boundary layer"

This is the link from hell.

Might be from 1972 but it does show that the continuum fluid dynamics becomes applicable further downstream for a long thin plate

This goes back to the idea of choosing a length scale. Near the leading edge, all applicable length scales are quite small (shock thickness, boundary-layer thickness, leading edge radius, etc.) so it takes a lot less rarefaction to result in a free-molecule flow than it would downstream where the boundary layer and other length scales are larger.

with the shock wave being "obviously" produced several steps back from the leading ' point'. Actually the shock is at the tip, or just a bit in front, but due to the kinetic molecular flow predominance near the leading edge, continuum mechanics would not explicitly apply. Perhaps not the best of descriptions.

Table on page 52 is also interesting, showing the rise in pressure and temperature of the gas as we travel along the plate from the leading edge.

What this figure does not show is a shock forming downstream of the leading edge, as you seem to suggest. That line is intended to show the boundary of the free-molecule region, not a shock wave. The line showing the shock is the more upstream one.

Ordinarily, a shock would not form on a perfectly flat plate with a perfectly sharp leading edge. However, due to the presence of the boundary layer, there is a nonzero displacement thickness growing away from the leading edge, meaning the incoming inviscid flow does not see a true flat plate, instead encountering what appears to it as a slowly thickening plate that grows with $\sqrt{x}$ downstream. This causes a shock to form, which will be highly curved near the leading edge where that displacement thickness changes most rapidly. This effect can cause rather dramatic departures from inviscid theory and is one of several major differences between supersonic and hypersonic flows. It is often called the viscous-inviscid interaction.

 

[cjl]

I suspect that @Mister T is using a definition of "heat" as energy transferred due to a temperature difference. One can Google up any number of references to such a definition.

Careful and consistent use of this definition would mean that "heat" (exactly like "work") is not associated with an object but is, instead, associated with an interface. "Heat content" of an object then becomes something of an oxymoron and one might resort to circumlocutions such as "thermal energy" instead.

The distinction is probably wasted on someone who is disputing whether the act of compressing a volume of gas succeeds in increasing the thermal energy of the gas.

That's fair. As you say, I think the misunderstandings here are a bit more fundamental, so the exact nuance of how words are used is probably irrelevant.

 

[cjl]

EDIT: @cjl Technically I typed all this up before you posted but a student walked into my office and I never finished, so you beat me to it. Now people are going to have to read nearly the same thing twice.

In your defense, your writeup is prettier, with much nicer formulas (I usually can't be bothered to do all the TeX formatting, since I don't use it that often day to day).

 

[Mister T]

I agree with the rest of your post, but I disagree with this one. When you do work on the gas, the heat content of the gas does indeed increase, since you have the same mass of gas and it is at a higher temperature.

Why do they call certain types of compressions adiabatic? Adiabatic means "no heat", yet there can be a dramatic increase in temperature. Things do not contain heat. That idea went away with the caloric theory. You can increase the internal energy of a gas with the same result regardless of whether you performed mechanical work on the gas or transferred heat to it. Realizing this is heralded as one of the greatest discoveries, it has led us to our modern understanding of the conservation of energy.

 

[cjl]

Adiabatic always implied no heat transfer across the system boundaries to me, but I think this is just a subtle difference in how we're using the terms. Regardless, internal energy is probably the more accurate phrase, so I agree with your correction there.

 

[davenn]

I hope I’m not being disrespectful.

you are being very disrespectful to the many people here that are trying to help you get over a lot
of false thoughts you have

ul. It would be easy for someone to see a gas being compressed, note that the temperature at the point of compression increased, and wrongly think that heat was “produced”.

Heat is being produced ... have you not pumped up a bicycle tyre with a hand pump and felt how hot
it gets because of the air compression inside the pump ?

Consider this: when you spray dust off your keyboard from a can of compressed air, The can gets cold.

that's the opposite effect to the hand pump, the "air" when released from the can cools
due to expansion

 

[Mister T]

Exhale through your mouth onto your hand. If you do it with the lips wide open the air feels warm. Do it with the lips pursed (so that the air comes out faster) and the air feels cool.

 

[Mister T]

Adiabatic always implied no heat transfer across the system boundaries to me

Right. So one simple model is to consider an imaginary boundary surrounding the air immediately in front of the heat shield, and assume that air is compressed so quickly that negligible heat crosses the boundary. That's an adiabatic compression. Heat is then conducted from that air to the heat shield.

 

[cjl]

Right. So one simple model is to consider an imaginary boundary surrounding the air immediately in front of the heat shield, and assume that air is compressed so quickly that negligible heat crosses the boundary. That's an adiabatic compression. Heat is then conducted from that air to the heat shield.

A surprising amount of it is transferred through radiation from the thermal luminescence of the gas/plasma in and behind the bow shock actually, but yes, that's the basic idea.

(I'm already very familiar with the process)

 

[Rensslin]

You would do well to review the forum rules on personal speculation. Your rigid adherence to false personal positions in the face of careful correction is not appreciated.

The situation when releasing compressed air through a nozzle complicated. See this link for some of the complications. There is a difference between a reversible expansion and a free expansion and between a real gas and an ideal gas.

However, the conclusion in this specific case is correct. The can gets cold because the gas remaining in the can has done work, pushing exhausted gas toward the nozzle opening. The expansion of the portion of the gas that remains in the can counts as "reversible".

Thank you for your careful consideration; thank you also for your thoughtful responses. And thank you for directing me towards the forum rules. I will carefully consider them.
I don’t believe I’m being rigid. I suppose no pigheaded person does. I once argued with my high school Physics teacher about why I couldn’t make a bicycle generator that would push the bicycle until he explained the problem in a way I could understand. I appreciated my teachers patience with me, just as I appreciate you and the others now.
But I don’t see how my original idea has been addressed in a way that I can understand. If you have nine cats in a box and you decrease the size of the box you still have nine cats in that box. If you want to argue that decreasing the size of the box will require adding more cats, I will concede that. You could even argue that the reduction of the box is exactly proportional to the amount of cats being added. My point is that changing the size of the box has increased the gradient of the original amount of cats with respect to those outside the box. Again, irrespective of the cats that were added to reduce the size of the box.
I always try to be patient with people who are not as smart as me. Believe it or not there really are people not as smart as me.

 

[Rensslin]

@boneh3ad.
Well I think entering Earth's atmosphere could be kind of like a bellyflop.
Here you are cruising through space with no aerodynamic drag; then all of a sudden, you have a condition that dramatically restricts your speed. Failure to consider angle could have injurious results.

 

[Rensslin]

They are most certainly not "stationary with respect to the planet". An object in low Earth orbit is moving at about 18,000 miles per hour relative to the ground, and basically the same speed relative to the atmosphere. this is about 12,000 26,000 feet per second, or about 20 40 times as fast as a bullet.

Also, the power dissipated by air friction is proportional to the cube of the speed through the air, so going 20 40 times faster dissipates 8000 64,000 times more power. This is more than enough to melt or even vaporize the components of the object.

Edit, reading marcusl's post I realized that I incorrectly converted from miles/hour to feet/second, so I corrected the numbers.

@phyzguy
I was told that there are communication satellites that are constantly stationary. That is how they can communicate with the ground; if they move around the planet then they quickly lose communication with their station. Is that wrong? If a stationary satellite “falls“ it must start off very slowly.
Also, I saw a video of guy back in the 60s ride a balloon all the way up to Earth's orbit. Yeah it was cool. He videoed himself jumping out. I wonder why he didn’t burn up. I think the guy still alive. Also is there something called maximum velocity? Or something like that?

 

[phyzguy]

I was told that there are communication satellites that are constantly stationary. That is how they can communicate with the ground; if they move around the planet then they quickly lose communication with their station. Is that wrong? If a stationary satellite “falls“ it must start off very slowly.

Geostationary satellites are 26,000 miles up. The atmosphere starts at about 100 miles up. If a satellite falls from 26,000 miles up to 100 miles up, it gains a huge velocity. You should be able to calculate how fast it would be going when it hit the atmosphere by calculating the change in potential energy and equating it to the gain in kinetic energy.

Also, I saw a video of guy back in the 60s ride a balloon all the way up to Earth's orbit. Yeah it was cool. He videoed himself jumping out. I wonder why he didn’t burn up. I think the guy still alive. Also is there something called maximum velocity? Or something like that?

He rode a balloon up to the edge of space, but he didn't ride it into orbit. In order to orbit, he would need a lateral velocity of about 18,000 miles an hour. There is no way to do this with a balloon. Most of the energy expended by a rocket putting a payload into orbit is in accelerating it laterally. This energy requirement is quite a bit larger than what is required to lift it into orbit.

 

[boneh3ad]

@boneh3ad.
Well I think entering Earth's atmosphere could be kind of like a bellyflop.
Here you are cruising through space with no aerodynamic drag; then all of a sudden, you have a condition that dramatically restricts your speed. Failure to consider angle could have injurious results.

As I previously stated, that isn't what happens, though. You don't suddenly have a condition that dramatically restricts speed. It's more like driving into the edge of a rainstorm where you get near the edge and there are a few stray raindrops hitting your windshield here and there. As you drive closer to the center, the frequency of raindrops hitting the wind increases until it's a constant pitter patter of rain on the window. That's a better illustration of molecules during reentry. The craft first encounters atmosphere that is so thin that it makes little difference to how it is flying, and as it continues on lower into the atmosphere, the frequency of molecular collisions with the vehicle increases as the density increases. It's much more gradual than you describe.

 

[cjl]

Thank you for your careful consideration; thank you also for your thoughtful responses. And thank you for directing me towards the forum rules. I will carefully consider them.
I don’t believe I’m being rigid. I suppose no pigheaded person does. I once argued with my high school Physics teacher about why I couldn’t make a bicycle generator that would push the bicycle until he explained the problem in a way I could understand. I appreciated my teachers patience with me, just as I appreciate you and the others now.
But I don’t see how my original idea has been addressed in a way that I can understand. If you have nine cats in a box and you decrease the size of the box you still have nine cats in that box. If you want to argue that decreasing the size of the box will require adding more cats, I will concede that. You could even argue that the reduction of the box is exactly proportional to the amount of cats being added. My point is that changing the size of the box has increased the gradient of the original amount of cats with respect to those outside the box. Again, irrespective of the cats that were added to reduce the size of the box.
I always try to be patient with people who are not as smart as me. Believe it or not there really are people not as smart as me.

Temperature isn't like "cats" in this analogy though. You have a box with 1 kg of air in it. You shrink the box. It still has 1 kg of air in it, but in the process of shrinking it, you had to apply force, so now the air that is in the box has more energy, and thus is at a higher temperature. The equivalent to the cats in your box is the quantity of air.

 

[Mister T]

If the cats are instead supposed to indicate the quantity of energy, then the only way to increase the number of cats is to increase the energy. There are two ways of doing this: work (which is a mechanical transfer of energy) and heat (which is a thermal transfer of energy).

But you can get the same result, either way. Transfer in a cat by doing work and now you have one extra cat. If instead you had transferred in a cat by transferring heat, you still end up with one extra cat. It makes no difference how you do it. Energy is energy.

 

[Mister T]

I was told that there are communication satellites that are constantly stationary.

They complete one Earth orbit every day, therefore they are not stationary. They are, however, always in the same place in the sky so that once you point your antenna at one it will stay pointed at it.

 

[DaveC426913]

I challenge the idea that compressing a gas can somehow “produce heat“. My mind simply rejects it.
Consider this: if you had a container that was 10‘ x 10‘ x 10‘ cubed; Full of air at sea level pressure and at room temperature. That container contains a certain amount of heat that can be measured. If you compress one of those thousand cubes into 1/1000 it’s volume, The amount of heat with in your thousand foot container is consistent. No heat was “magically made“.

"Amount of heat" is an absolute value, yes. But if you reduce the volume, the temperature will increase proportionally (because the same "amount of heat" is now in a smaller space).The gas laws are specific and well understood. They show the direct and inverse correlations between volume, pressure and temperature of an ideal gas.

https://en.wikipedia.org/wiki/Gas_laws

 

[jbriggs444]

"Amount of heat" is an absolute value, yes. But if reduce the volume, the temperature will increase proportionally (because the same amount of heat is now in a smaller space).

This is not correct. If you reduce the volume but carefully avoid adding energy to the contents (draining as much in thermal energy as you are injecting by performing mechanical work) the result is an isothermal compression.

The same amount of thermal energy is in a smaller space, but the temperature is unchanged.

 

[DaveC426913]

This is not correct. If you reduce the volume but carefully avoid adding energy to the contents (draining as much in thermal energy as you are injecting by performing mechanical work) the result is an isothermal compression.

The same amount of thermal energy is in a smaller space, but the temperature is unchanged.

Did you not see the reference to gas laws? They don't involve bleeding off excess.

The OP believes that compressing a volume of gas will not raise its temperature. That is not true.

You can always complicate the experiment to obfuscate the principle being demonstrated if you want.

 

[jbriggs444]

Did you not see the reference to gas laws? They don't involve bleeding off excess.

The OP believes that compressing a volume of gas will not raise its temperature. That is not true.

You can always complicate the experiment to obfuscate the principle being demonstrated if you want.

Both your post and the point you were responding to were incorrect. It does little to help correct a misconception if you do so by promulgating a new one.

 

[DaveC426913]

Both your post and the point you were responding to were incorrect.

How is my drawing attention to the ideal gas laws incorrect? The OP seems not to be aware of them.

(OK, the first half of my post phrased it using the OP's terms, but the point is made that the ideal gas laws show that compressing a gas will raise its temperature - other factors being equal.)

 

[jbriggs444]

the point is made that the ideal gas laws show that compressing a gas will raise its temperature - other factors being equal.)

And that point is incorrect.

In fact, per the ideal gas law, PV=nRT, compressing a gas (reducing its volume) while holding pressure and amount of substance constant can only be achieved by reducing temperature. If you are planning to have all factors equal, you'd better spell out which ones you are holding constant and how.

 

[davenn]

They are, however, always in the same place in the sky so that once you point your antenna at one it will stay pointed at it.

not quite ... they actually move in an oscillation north and south of the equator by a small amount
an example ...

​

 

[Rensslin]

Temperature isn't like "cats" in this analogy though. You have a box with 1 kg of air in it. You shrink the box. It still has 1 kg of air in it, but in the process of shrinking it, you had to apply force, so now the air that is in the box has more energy, and thus is at a higher temperature. The equivalent to the cats in your box is the quantity of air.

Got it: that makes sense. Thanks

 

[DaveC426913]

And that point is incorrect.

In fact, per the ideal gas law, PV=nRT, compressing a gas (reducing its volume) while holding pressure and amount of substance constant can only be achieved by reducing temperature. If you are planning to have all factors equal, you'd better spell out which ones you are holding constant and how.

< sidebar >
OK, I take full responsibility for not grokking this - after all, as a diver, I'm supposed to know the gas laws down pat.

Yes, it is obvious (even to me) that - if you plan to keep pressure constant - you'll have to reduce the temperature to get the volume to reduce. This I know.

But if you were to take a volume of gas, in a closed container - and compress it (reduce its volume) - both temperature and pressure will go up. After all, this is how diesel engines work. (Please tell me I'm right about this, or I'm going to set my laptop on fire, and go live in a shack in the woods)

So what am I missing? Is it simply that I didn't specify that pressure doesn't have to remain constant?

< /sidebar >

 

[Mister T]

But if you were to take a volume of gas, in a closed container - and compress it (reduce its volume) - both temperature and pressure will go up. After all, this is how diesel engines work. (Please tell me I'm right about this, or I'm going to set my laptop on fire, and go live in a shack in the woods)

Well, that is certainly one of the possibilities. In fact the quantity $pT$ must go up, but there are ways of doing that that will make $p$ or $T$ go down.

 

[phyzguy]

@DaveC426913, I think what people are objecting to is the following statement you made:

"Amount of heat" is an absolute value, yes. But if you reduce the volume, the temperature will increase proportionally (because the same "amount of heat" is now in a smaller space).

When you compress a gas, the temperature does not increase because, "the same "amount of heat" is now in a smaller space". It increases because the act of compression does work on the gas and increases the "amount of heat" in the gas. The "amount of heat", also called the internal energy, goes up. If the internal energy is held constant, the temperature of the gas will not go up, even if it occupies a smaller volume

 

[DaveC426913]

Cool. I learned some new things today.

One of them is that I had a naive understanding of the source of the temperature rise when you compress a gas.