The following discussions define the various heating and
cooling notions __for entry stagnation heating__, in terms of very simple models that are known
to be well inside the ballpark. How to
achieve the energy conservation balance among them is also addressed. *This is more of an “understand how it
works” article than it is a “how to actually go and do” article.*

**Convective Stagnation Heating**

The stagnation point heating model is proportional to
density/nose radius to the 0.5 power,
and proportional to velocity to the 3.0 power. The equation used here is H. Julian Allen’s
simplest empirical model from the early 1950’s,
converted to metric units. It is:

q_{conv}, W/sq.cm = 1.75 E-08 (rho/r_{n})^0.5
(1000*V)^3.0, where rho is kg/cu.m, r_{n} is m, and V is km/s

The 1000 factor converts velocity to m/s. This is a very crude model, better correlations are available for various
shapes and situations. However, this is very simple and easy to use, and it has been "well inside the
ballpark" since about 1953. This is
where you start. See Figure 1.

Figure 1 – Old,
Simple Model for Entry Stagnation Convection Heating

**Plasma Radiation Stagnation Heating**

There are all sorts of correlations for various shapes and
situations. However, to get started, you just need a ballpark number. That comes from the widely-published notions
that (1) *radiational heating varies with the 6th power of velocity,* and (2) *radiation dominates over
convection heating at entry speeds above 10 km/s. *What that means is you can use a very
simple radiational heating model, and
"calibrate" it with your convection model:

q_{rad},
W/sq.cm = C (1000*V)^6, where V
is input as km/s

The 1000 factor converts speed to m/s. The resulting units of the constant C are
W-s^6/sq.cm-m^6. You have to
"calibrate" this by evaluating C with your convective heating result
at 10 km/s, and a "typical"
entry altitude density value, for a
given nose radius for your shape:

find q_{conv} per above at V = 10 km/s with
"typical" rho and r_{n},
then

C, W-s^6/sq.cm-m^6 = (q_{conv} at 10 km/s)(10^-24)

This should get you into the ballpark with both convective
and radiation heating. Figure both and then
sum them for the total stagnation heating.
Below 10 km/s speeds, the
radiation term will be essentially zero.
Above 10 km/s it should very quickly overwhelm the convective heating
term.

As an example, I had
data for an Apollo capsule returning from low Earth orbit. I chose to evaluate the peak stagnation
heating point, which occurred about 56
km altitude, and about 6.637 km/s
velocity. See Figure 2.
Dividing that convective heating value of 55.72 W/sq.cm by the velocity
cubed, and then multiplying by 10 km/s
cubed, I was able to estimate stagnation
convective heating at 10 km/s and 56 km altitude as 190.59 W/sq.cm.

Dividing that value by 10 km/s to the 6^{th} power
gave me a C value of 1.90588 x 10^{-4},
__to use directly with velocities measured in km/s__, for estimating radiation heating from the
plasma layer adjacent to the surface.
The resulting trends of convective,
radiation, and total stagnation
heating versus velocity (at 56 km) are shown in Figure 3.

Figure 2 – Relevant Peak Convective Stagnation Heating Data
for Apollo From LEO

Figure 3 – Estimated Apollo Stagnation Heating Trends Versus
Velocity at 56 km Altitude

**Radiational Cooling**

This is a form of Boltzmann's Law. The power you can radiate away varies as the
4th power of the surface temperature,
but gets modified for an effective temperature of the surroundings
receiving that radiation (because that gets radiated back, and emissivity is equal to absorptivity):

q_{rerad},
BTU/hr-ft^2 = e sig (T^4 – T_{E}^4) for T’s in deg R and sig =
0.1714 x 10-8 BTU/hr-ft^2-R^4

For this equation, T
is the material temperature, T_{E}
is the Earthly environment temperature (near 540 R = 300 K), e is the spectrally-averaged emissivity (a
number between 0 and 1), and sig is
Boltzmann’s constant for these customary US units.

*This radiation model *__presumes transparency__ of
the medium between the radiating object and the surroundings. That assumption *fails rapidly above 10
km/s speeds*, as the radiating
plasma in the boundary layer about the vehicle *becomes more and more
opaque to those wavelengths*.

Therefore, *do
not attempt radiationally-cooled refractory heat protection designs for entry
speeds exceeding about 10 km/s.*
They won't work well (or at all) in practice. __Ablative protection becomes pretty much your
only feasible and practical choice__.

**Heat Conduction Into The Interior**

This is a cooling mechanism for the exposed surface, and a heating mechanism for the interior
structure. In effect, you are conducting heat from the high surface
temperature through multiple layers of varying thermal conductivity and
thickness, to the interior at some
suitable "sink" temperature.

The amount of heat flow conducted inward in steady state
depends upon the temperature difference and the effective thermal resistance of
the conduction path. The electrical
analog is quite close, with current
analogous to heat flow rate per unit area,
voltage drop analogous to temperature difference, and resistance analogous to thermal
resistance.

In the electrical analogy to 2-D heat transfer, conductance which is the inverse of
resistance is analogous to a *thermal conductance which is a thermal
conductivity divided by a thickness*.

Resistances in series sum to an overall effective
resistance, so the effective thermal
resistance is the sum of several __inverted__ thermal __conductances__, one for each layer. Each resistance sees the
same current, analogous to each thermal
resistance layer seeing the same thermal flux,
at least in the 2-D planar geometry.

Like voltage/effective resistance = current, heat flow per unit area (heat flux) is
temperature drop divided by effective overall thermal resistance. (The geometry effect gets a bit more
complicated than just thickness in cylindrical geometries.) In 2-D:

q_{cond} = (T_{surf} - T_{sink})/effective
overall thermal resistance

For this the effective overall thermal resistance is the sum
of the individual layer resistances,
each in turn inverted from its thermal conductance form k/t:

eff. th. resistance (2-D planar) = sum by layers of layer
thickness/layer thermal conductivity

*Using the electrical analogy, the current (heat flux)
is the voltage pressure (temperature difference) divided by the net effective
resistance (thermal resistance). The
voltage drop (temperature drop) across any one resistive element (layer) is
that element's resistance (layer thermal resistance) multiplied by the current
(heat flux). *See Figure 4.

Figure 4 – Modeling the Thermal Resistances of Multiple
Layers for Conduction

What that says is that for a given layering with different
thicknesses and thermal conductivities,
there will be a calculable heat flux for a given overall temperature
difference. Each layer has its own temperature drop once the heat flux is
known, and the sum of these temperature
drops for all layers is the overall temperature drop.

*Any layer with a high thermal resistance will have a
high temperature drop, and vice versa*.
High thermal resistance correlates with high thickness, and with low thermal conductivity. A high temperature drop over a short
thickness (a high thermal gradient) requires a very low thermal conductivity
indeed, essentially about like air
itself.

On the other hand, any
high-density material (like the monolithic ceramics) will have high thermal
conductivity, and thus the thermal
gradients it can support are inherently very modest. __Such parts trend toward isothermal
behavior__. Their high meltpoint does
you little practical good, if there is
no way to hang onto the "cool" end of the part. In point of fact, there may not be much of a “cool” end.

**Active Liquid Cooling**

In effect, this is
little different than the all-solid heat conduction into a fixed-temperature
heat sink, as described just above. The heat sink temperature becomes the
allowable coolant fluid temperature, and
the last “layer” is the thermal boundary layer between the solid wall and the
bulk coolant fluid. The thermal
conductance of this thermal boundary layer is just its “film coefficient” (or
“heat transfer coefficient”). The simple
inverse of this film coefficient is the thermal resistance of that boundary
layer. See Figure 5.

Figure 5 – Modifying the Thermal Conduction Model for Active
Liquid Cooling

The main thing to worry about here is the total mass of
coolant m_{coolant} recirculated,
versus the time integral (for the complete entry event) of the heating
load conducted into it. That heat is
going to raise the temperature of the coolant mass and indirectly the pressure at which it must
operate. That last is to prevent boiling
of the coolant.

∫ q_{cond} dt = m_{coolant} Cv (T_{final}
– T_{initial}) where T_{final}
is the max allowable T_{sink}

**Balancing the Heat Flows: Energy Conservation**

A patch of heat shielding area sees convective heating from
air friction, and may see significant
radiation heating if the entry speed is high enough. That same patch can conduct into the
interior, and it can radiate to the
environment, if the adjacent stream
isn't opaque to that radiation. If the
heat shield is an ablative, some of the
heating rate can go into the latent heat of ablation. See Figure 6.

Figure 6 – The Energy Conservation Balance

The correlations for convective and radiation stagnation heating
given above depend upon vehicle speed,
not plasma temperature. The
equation for conduction into the interior depends upon the surface and interior
temperatures. Re-radiation to the
environment depends upon the surface and environmental temperatures. Of these,
both the environmental and heat sink temperatures are known fixed
quantities.

If the heat shield is ablative, then the surface temperature is fixed at the
temperature at which the material ablates;
otherwise, surface temperature is
free to "float" for refractory materials that cool by radiation.

The way to achieve energy conservation for refractories is
to adjust the surface temperature until q_{conv }+ q_{rad} - q_{cond}
- q_{rerad} = 0. For
ablatives, the surface temperature is
set, and you just solve for the rate of
material ablation (and the recession rate):
q_{abl} = q_{conv} + q_{rad} - q_{cond}
- q_{rerad}.

The heating flux rate q_{abl} that goes into
ablation, divided by the latent heat of
ablation L_{abl} times virgin density ρ, can give you an estimate of
the ablation surface recession rate r (you
will want to convert to more convenient units):

(q_{abl} BTU/ft^2-s)/(L_{abl} BTU/lbm)(ρ lbm/ft^3)
= ft^3/ft^2-s = r, ft/s

(q_{abl} W/m^2)/(L_{abl} W-s/kg)(ρ kg/m^3)
= m^3/m^2-s = r, m/s

**Other Locations**

Those require the use of empirical correlations or actual
test data to get accurate answers.
However, to just get in the
ballpark, any guess is better than no
guess at all! For lateral windward-side
heating, try about half the heat flux as
exists at the stagnation point. For
lee-side heating in the separated wake,
try about 10-20% of the stagnation heating.

**Clarifying Remarks**

*Bear in mind that these equations are for steady-state
(thermal equilibrium) exposure.*
The conduction into the interior is the slowest to respond to
changes. *Transient behavior takes
a finite-difference solution to analyze*. There is __no__ way around that
situation.

But if that conduction effect is small compared to the
applied heating terms because there is lots of re-radiation or there is lots of
ablation to balance them, you can then approximate
things by deleting the conduction-inward term.
__You simply cannot do that if there is no ablation or re-radiation__. And your conduction effect will __not__ be
small compared to the heating, if you
are doing active liquid cooling.

If you are entering from Earth orbit at speeds no more than
8 km/s, you can reasonably ignore the
plasma radiation stagnation heating term.
On the other hand, for entry
speeds above 8 km/s, your re-radiation
cooling term will rapidly zero as the plasma layer goes opaque to thermal
radiation. Its transmissibility must
necessarily zero, in order for its
effective emissivity to become large.
The zero transmissibility is what zeroes the re-radiation term.

If you choose to do a transient finite-difference thermal
analysis, __what you will find in the
way of temperature distributions within the material layers has little to do
with steady-state linear temperature gradients__. Instead,
there will be a “humped” temperature distribution that moves slowly like
a wave through the material layers. This
is called a “thermal wave”. It forms
because heat is being dumped into the material faster than it can percolate
through by conduction. See Figure 7.

Figure 7 – The Thermal Wave Is a Transient Effect

This humped wave of temperature will decrease in height and
spread-out through the thickness, as it
moves through the material. But, __usually its peak (even at the backside of
the heat shield) is in excess of the steady-state backside temperature
estimates__. It may take longer to
reach the backside than the entire entry event duration, but it will certainly tend to overheat any
bondlines or substrate materials.

Material properties such as thermal conductivity are
actually functions of material temperature.
These are usually input as tables of property versus temperature, into the finite-difference thermal
analyses. Every material will have its
own characteristics and property behavior.
I did not include much material data in this article, in the way of typical data, for you to use. As I said at the beginning, this article is more about “understanding”
than it is “how-to”.

The local heating away from the stagnation point is
lower. There are many correlations for
the various shapes to define this variation,
but for purposes of finding out what “ballpark you are playing in”, you can simply guess that windward lateral
surfaces that see slipstream scrubbing action will be subject to crudely half
the stagnation heating rate. Lateral
leeward surfaces that face a separated wake see no slipstream scrubbing
action. Again, there are lots of different correlations for
the various situations, but something
between about a tenth to a fifth of stagnation heating would be “in the
ballpark”.

Low density ceramics (like shuttle tile and the
fabric-reinforced stuff I made long ago) are made of mineral flakes and fibers
separated by considerable void space around and between them. The void space is how minerals with a high
specific gravity can be made into bulk parts with a low specific gravity:

sg_{mineral} * solid volume fraction = sg_{mineral}
*(1 – void volume fraction) = bulk effective sg

The low effective density confers a low thermal conductivity
because the conduction paths are torturous,
and of limited cross-section from particle to particle, through the material. Such thermal conductivity will be a lot
closer to a mineral wool or even just sea level air, than to a firebrick material.

This low effective density also reduces the material
strength, which must resist the wind
pressures and shearing forces during entry.
A material of porosity sufficient to insulate like “mineral wool-to-air”
will thus be no stronger than a styrofoam.

So, typically, these low density ceramic materials are weak. And they are very brittle. The brittleness does not respond well to stress-
or thermal expansion-induced deflections in the substrate, precisely because brittle materials have
little strain capability. *That
fatal mismatch has to be made up in how the material is attached to the
substrate.* If bonded, considerable flexibility is required of the
adhesive.

**Related Thermal/Structural Articles On This Site**

Not all of these relate directly to entry heat
transfer. The most relevant items on the
list are probably the high __speed aerodynamics and heat transfer__
article, and the article taking a look
at__ nosetips and leading edges__.

The “trick” with Earth orbit entry using refractories
instead of ablatives is to maximize bluntness in order to be able to use low
density ceramics at the stagnation zone without overheating them. That leaves you dead-broadside to the
slipstream, and thus inevitably ripping
off your wings, unless you do something way
“outside the box”.

The __pivot wing spaceplane concept__ article is a
typical “outside the box” study restricted to 8 km/s or less. The older __folding wing spaceplane__ article
is similar. In both studies, the wings are relocated out of the slipstream
during entry, which is conducted dead
broadside to the oncoming flow.

The reinforced low density ceramic material that I made long
ago is described to some extent in the article near the bottom of the list
(about __low density non-ablative ceramic heat shields__). It is like the original shuttle tile material
that inspired it, but is instead a
heavily reinforced composite analogous to fiberglass. This information was also presented as a paper
at the 2013 Mars Society convention.

The fastest way to access any of these is to use the search
tool left side of this page. Click on the
year, then on the month, then on the title. It is really easy to copy this list to a txt
or docx file, and print it.

1-2-20…On High Speed Aerodynamics and Heat Transfer

4-3-19…Pivot Wing Spaceplane Concept Feasibility

1-9-19…Subsonic Inlet Duct Investigation

1-6-19…A Look At Nosetips (Or Leading Edges)

1-2-19…Thermal Protection Trends for High Speed Atmospheric
Flight

7-4-17…Heat Protection Is the Key to Hypersonic Flight

6-12-17…Shock Impingement Heating Is Very Dangerous

11-17-15…Why Air Is Hot When You Fly Fast

6-13-15…Commentary on Composite-Metal Joints

10-6-13…Building Conformal Propellant Tanks, Etc.

8-4-13…Entry Issues

3-18-13…Low-Density Non-Ablative Ceramic Heat Shields

3-2-13…A Unique Folding-Wing Spaceplane Concept

1-21-13…BOE Entry Analysis of Apollo Returning From the
Moon

1-21-13…BOE Entry Model User’s Guide

7-14-12…Back of the Envelope Entry Model

There are also several studies for reusable Mars landers
that I did not put in the list. They are
similar to the two spaceplane studies,
but entry from Mars orbit is much easier than entry from Earth
orbit. Simple capsule shapes work fine
without stagnation zone overheat for low density ceramics, even for the large ballistic coefficients
inherent with large vehicles.

**Final Notes**

I’ve been retired for some years now, and I have been retired out of aerospace work
much longer than that. I’ve recently
been helping a friend with his auto repair business, but that won’t last forever.

Not surprisingly, I
am not so familiar with all the latest and greatest heat shield materials, or any of the fancy computer codes, and I have little beyond these
paper-and-pencil-type estimating techniques to offer (which is exactly “how we
did it” when I first entered the workforce long ago). Yet these simple methods are precisely what
is needed to decide __upon what__ to expend the effort of running computer
codes! Today’s fresh-from-school
graduates do not know these older methods.
But I do.

Sustained high speed atmospheric flight is quite distinct
from atmospheric entry from orbit (or faster),
but if you looked at that high speed aerodynamics and heat transfer article
cited in the list, then you already know
that I can help in that area as well.

Regardless, it should
be clear that I do know __what to worry about as regards entry heat protection__, and __how to get into the ballpark__ (or
not) with a given design concept or approach in either area (transient entry or
sustained hypersonic atmospheric flight)!
*I can help you more quickly screen out the ideas that won’t work, from those that might. *

And if you look around on this site, you will find out that I also know enough to
consult in ramjet and solid rocket propulsion,
among many other things. I’m
pretty knowledgeable at alternative fuels in piston and turbine engines, too.

If I can help you,
please do contact me. I do consult in these things, and more.

More? __More__: I also build and sell cactus eradication farm
implements that really work easier,
better, and cheaper than anything
else known. I turned an accidental
discovery into a very practical family of implements. It’s all “school of hard knocks” stuff.