# B Atmospheric Burn-up During Re-Entry

#### russ_watters

Mentor
Despite what the Cambridge dictionary may say, boneh3ad is right - aerodynamic drag differs from friction in a number of important ways, and generally should not be considered to fall under the umbrella of "friction". In addition, even if you insist on calling drag "friction", that's still not what causes reentry heating, which is almost entirely a result of compression in the shock ahead of the object.
Let me say this a slightly different way: typically drag is divided into at least two types:
• Friction drag: caused by particles of air essentially rubbing on the surface of the object.
• Pressure or form drag: caused by the pressure changes in the airflow around the object.
• Wave drag: sometimes listed as its own category, sometimes a subset or pressure drag. It's the supersonic version

How much of each of these matters depends on the speed of the object and for this particular thread, pressure or more specifically wave drag is pretty close to the entire issue.

So calling all drag "friction" is at best an overly broad description that for this particular thread misses the key part of the drag. I sometimes lean toward not being pedantic, but on this particular issue I think the difference matters.

#### cjl

It's worth noting that even if you make that distinction, friction drag is not caused by air "rubbing on the surface of the object". It's caused by viscous dissipation in the boundary layer due to shear. At the surface of the object, the fluid has no relative velocity, and therefore cannot "rub".

Gold Member
My argument was not with your physics but your tone. You doubtless know more aerodynamics than I do.....probably I know more about atomic interactions with surfaces than you. So what? I was trying to provide a answer to the OP that would be prove useful.
And I take issue with the fact that apparently correcting someone who is saying something incorrect or misleading is somehow automatically considered rude now.

I even specifically mentioned that your misconception is common and was shared by Newton. You shouldn't take things so personally.

Last edited:

Gold Member
Let me say this a slightly different way: typically drag is divided into at least two types:
• Friction drag: caused by particles of air essentially rubbing on the surface of the object.
• Pressure or form drag: caused by the pressure changes in the airflow around the object.
• Wave drag: sometimes listed as its own category, sometimes a subset or pressure drag. It's the supersonic version
You've mischaracterized "friction" drag, which is often called viscous drag for just that reason. It has nothing to do with air molecules rubbing on the surface. It's based on the fact that real fluids are viscous, meaning intermolecular forces cause them to resist shearing. At the same time, fluids "stick to" surfaces, meaning there must be a velocity gradient and therefore a shear stress. That shear stress at the wall can be integrated over a vehicle to give viscous drag. If you are searching for a nice visual, a fluid "pulls" on objects moving through it.

Wave drag is also entirely separate from what you call pressure drag. It only arises in supersonic (or transonic flows with locally supersonic regions) and does not replace pressure drag. It is in addition to pressure drag.

#### russ_watters

Mentor
You've mischaracterized....
There are clearly different preferences with respect to terminology, and these lecture notes state the problem pretty well:
...basic drag nomenclature is frequently more confused than it needs to be, and sometimes the nomenclature gets in the way of technical discussions.

They include a very complex chart that isn't worth putting a thousand words into dissecting for the purpose of a narrow question in a forum thread:

Gold Member
I do high speed aerodynamics for a living and that chart makes no sense to me at all.

cjl

#### Rensslin

First of all I’d like to thank everyone for helping me with these concepts.
Second of all I’m doubling down on my statement that and object hitting the earth’s atmosphere is like a bellyflop.
Second I may be wrong about not believing that kinetic energy might be somehow changed into heat. Thank you for helping me with this. However, would someone grasp what I am saying about when a volume of gas with a specific heat becomes reduced, that heat (albeit may be increased by the addition of kinetic energy) now occupies a smaller space. Therefore the heat energy becomes greater than it’s adjacent area and becoming a gradient. This is strictly a physical reaction. I’m not considering any kinetic energy that may or may not be introduced. I’m only taking about the amount of heat that was in the previous volume is now in a smaller volume becomes a gradient.
This is how a refrigerator works.
Am I crazy?
Am I the only one who grasps this?
At our lab we compression test concrete.
We put 60,000 psi into a sample when it breaks. There is no increase in temperature because the volume never decreased.

#### Rensslin

I hope I’m not being disrespectful. 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”.
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.
I tell you this is not the case. Just as it is not the case when the gas is compressed I am not convinced that heat is “produced”.

#### A.T.

In fact, right at the surface of the vehicle, the air has zero velocity with respect to the surface (a concept called a boundary layer)...
Does the concept of a boundary layer apply to the very thin atmosphere part, where individual air molecules hit the body?

#### A.T.

There is no increase in temperature because the volume never decreased.
There is no noticeable increase in temperature because you do no work, if the applied force acts over zero distance.

#### jbriggs444

Homework Helper
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

Gold Member
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

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

Gold Member
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.

Last edited:

#### 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:

Last edited:

#### 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

Homework Helper
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.

Last edited:

Gold Member
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.

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.

Last edited:

#### 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

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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)

"Atmospheric Burn-up During Re-Entry"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving