Calculating Heat from Drag force on a Rocket Airframe

[LT72884]

Ok, so i have a question, and an approach to my question. Am i going in the correct direction?

Here si the question:

I have a high powered rocket i made. It sometimes hits mach 1.9. I would like to calculate the heat generated from going that fast so i can test other materials.

Here is what i have done

V= 683m/s rho = 1.25 Af= 0.0093m^3 Cd = 0.47 let t= 7 seconds at mach 2 mass = 6.8kg Cp = 1.4j/kg*k for PLA+

Fd = 1274.4 Newtons
work = Fd*t ----> 8920j
Q = mCp(dT)
dT = Q/mCp

let Q be all the drag force that is converted to heat and absorbed by the rocket

dT = w/mCp ----> 0.93K or -457 degrees F

meaning little heat is generated???

Thanks for any help on this

 

[russ_watters]

It's not clear to me what you are doing - what equation are you using for drag? Supersonic drag and associated heating are pretty advanced concepts that don't have a simple calculation like what you seem to be doing. It's not drag, per se, that heats rockets, it's the compression of the air in the shock wave.

 

[LT72884]

It's not clear to me what you are doing - what equation are you using for drag? Supersonic drag and associated heating are pretty advanced concepts that don't have a simple calculation like what you seem to be doing. It's not drag, per se, that heats rockets, it's the compression of the air in the shock wave.

Fd = (0.5)(rho)(A)(Cd)(v^2)

Im just using basic drag, and assuming all the drag force is converted to work, the allowing all the work to be converted to heat

 

[berkeman]

dT = w/mCp ----> 0.93K or -457 degrees F

Presumably dT is the change in temperature, so it's not appropriate to convert the Kelvin dT to a Fahrenheit dT in an absolute fashion. You should do that in a relative fashion instead if you want it in Fahrenheit.

 

[russ_watters]

Fd = (0.5)(rho)(A)(Cd)(v^2)

Im just using basic drag, and assuming all the drag force is converted to work, the allowing all the work to be converted to heat

That equation only works for low speed flight; below about 100m/s.

 

[LT72884]

Presumably dT is the change in temperature, so it's not appropriate to convert the Kelvin dT to a Fahrenheit dT in an absolute fashion. You should do that in a relative fashion instead if you want it in Fahrenheit.

So what would be the best way to convert to F so i know what the temp increase is?

Thanks

 

[berkeman]

What is the ratio of a degree F to a degree K (or degree C)?

 

[LT72884]

That equation only works for low speed flight; below about 100m/s.

Well, openfoam uses it for 210m\s and i know HISA uses for 420m\s when i run cfd

What formula would you recommend for high mach numbers?

 

[LT72884]

What is the ratio of a degree F to a degree K (or degree C)?

32 f is 273k

It seems that for every 2 degrees f, k increases by 1. So 2:1 from F to k

 

[berkeman]

Every F degree is $\frac{9}{5}$ multiplied by a C degree (or K degree).

https://www.cuemath.com/c-to-f-formula/

 

[LT72884]

correct, i do know this equation. and i THOUGHT that is what i used.... or maybe i didnt haha.

Yes, 1 degree increase of C is 1 unit increase of kelvin.

So if my dT is only changing by 1 k, that means the temp change was 1 C as well, meaning the increase was from 32 C to 33C based upon the original problem, regardless of the high mach number. IE if my launch day was 32C (90F) then during the flight, the change in temp would be 1 C so final temp would be 33C or 91.4 degrees F

 

[berkeman]

So if my dT is only changing by 1 k, that means the temp change was 1 C as well

Correct.

meaning the increase was from 32 C to 33C based upon the original problem,

No. If dT = 1[K] = 1[C], then in Fahrenheit it is dT = $\frac{9}{5}[F/K]~1[K]$ = $\frac{9}{5}[F]$

EDIT -- Fixed LaTeX.

 

[LT72884]

That equation only works for low speed flight; below about 100m/s.

i do know that it also uses pressure drag(Cdp) at higher machnumbers. but doesnt pressure drag end up being simplified down to Fd once you know the Cdp?

 

[LT72884]

Correct.


No. If dT = 1[K] = 1[C], then in Fahrenheit it is dT = $\frac{9}{5}~1[F/K]$ = $\frac{9}{5}[F]$

ok, so if my launch day was 32C (90F) and the change in temp was 1 C, meaning the new temp was now 33C, that would mean the temp was 91.4 degrees F?

Or is the change in temp only referring to the material itself since thats the Cp we are using? so whatever tmep the material was at the beginning of the launch, would experience a 1C change in temp or a 9/5ths F

 

[berkeman]

so if my launch day was 32C (90F) and the change in temp was 1 C, meaning the new temp was now 33C, that would mean the temp was 91.4 degrees F?

9/5=1.8. So if the ambient temp was 90F, the increased temperature would be 91.8F.

 

[LT72884]

google must round differently is all. There 32 was 90.. But really it shold be 89.6

oh well

 

[Mister T]

google must round differently is all. There 32 was 90.. But really it shold be 89.6

It's an issue of significant figures.

 

[TeethWhitener]

work = Fd*t ----> 8920j

Work is force times distance, not time. At 683 m/s, your rocket travels nearly 5 km over 7 seconds.

 

[berkeman]

Thread prefix changed I-->B.

 

[russ_watters]

Well, openfoam uses it for 210m\s and i know HISA uses for 420m\s when i run cfd

What formula would you recommend for high mach numbers?

There isn't a specific formula, there's a procedure involving thermodynamics.

i do know that it also uses pressure drag(Cdp) at higher machnumbers. but doesnt pressure drag end up being simplified down to Fd once you know the Cdp?

View attachment 347685

Supersonic drag is called "wave drag":
https://en.wikipedia.org/wiki/Wave_drag

You don't have a wave drag coefficient, nor is this the only thing impacting heating due to drag.

This problem is not solvable at the level needed here. It's a master's-degree level investigation that doesn't have a simple equation to provide the answer.

That being said....how do you even know your rocket goes supersonic? I'd be surprised if it could.

 

[erobz]

If you are looking to solve the problem you think it is, as opposed to what it actually is (as was pointed out), then apply a power balance:

$$ m_r c_p \frac{dT}{dt} = F v $$

This assumes low speed drag ( no shockwave ), but also no thermal losses from the rocket to the surroundings. (these two errors should offset) MAYBE it's a lower bound? I don't really know, but it is certainly back of the envelope.

At constant velocity, $T$ will just be a linear function in time.

 

[berkeman]

I have a high powered rocket i made. It sometimes hits mach 1.9. I would like to calculate the heat generated from going that fast so i can test other materials.

BTW, do you get any telemetry or other data from the rocket? If you already have a link that you use, maybe you could add a small temperature sensor to the rocket to measure the heating of the nosecone. I guess that may be offset by the cooler air at altitude, though.

 

[Sira distributor]

It's not clear to me what you are doing - what equation are you using for drag? Supersonic drag and associated heating are pretty advanced concepts that don't have a simple calculation like what you seem to be doing. It's not drag, per se, that heats rockets, it's the compression of the air in the shock wave.

Calculating the heat generated from drag force on a rocket airframe involves several factors, primarily the velocity of the rocket and the drag coefficient of its shape. The formula typically used is:

\[ Q = \frac{1}{2} \rho v^3 A C_d \]

where:
- \( Q \) is the heat generated due to drag force,
- \( \rho \) is the air density,
- \( v \) is the velocity of the rocket relative to the air,
- \( A \) is the reference area (usually the cross-sectional area of the rocket),
- \( C_d \) is the drag coefficient.

This formula gives the rate at which heat is generated due to drag. To determine the total heat generated over a specific time or during a flight, this rate can be integrated over the flight duration or the period of interest.

It's important to note that this calculation assumes steady-state conditions and neglects other factors like heating due to combustion gases or atmospheric re-entry effects, which would require more complex modeling.

 

[bigfooted]

\[ Q = \frac{1}{2} \rho v^3 A C_d \]

It is actually $$q_w = \frac{1}{2}\rho_\infty V^3_\infty C_H$$,
where $C_H$ is the Stanton number. The book of Anderson, "Hypersonic and high-temperature gas dynamics" is a nice resource. But it requires some background knowledge, I'd say 2 years of Uni engineering including linear algebra, mechanics, calculus, fluid dynamics and partial differential equations.
Anyway, the main message is that for supersonic and hypersonic flows, the aerodynamic heating increases with the cube of the velocity.
The Stanton number can be derived for some simple cases. For incompressible flow over a flat plate, it is simply
$$C_H = C_f \frac{1}{2}Pr^{-2/3}$$, with $C_f$ the skin friction. For incompressible flows, $$C_f = \frac{0.664}{\sqrt{Re_x}}$$.
According to Anderson, these approximations for $C_H$ and $C_f$ are still OK to use for hypersonic flows as well.

Another thing to point out is that in general, you do not obtain the temperature as a result of the heat transfer analysis, but the heat flux. The temperature depends on the conjugate heat transfer towards other connecting parts, thermal conductivity of the material etc... And possibly radiation if the temperature is very high. A good resource for incompressible heat transfer is Lienhard & Lienhard, A Heat Transfer Textbook (for free on his website).

If you want to know the skin temperature, I believe that you are better off inserting a couple of small thermocouples for temperature measurements, they are relatively cheap and light-weight. You could connect it to a small Arduino or Raspberry Pi board to log the data. The stagnation point in the nose of your rocket will be the most interesting location.

 

[boneh3ad]

That equation only works for low speed flight; below about 100m/s.

That's not true. The drag equation can be used at any speed provided it's a continuum flow. $C_D$ is not necessarily constant, however.

At for the rest of this discussion, assuming that 100% of the work done to overcome drag gets converted into heat that is absorbed by the rocket is not going to be accurate. Not all heat generated is going to be absorbed (some gets "left behind" in the wake), plus there is plenty of kinetic energy generated, for example, when the wake behind the object is still moving after it passes and/or the flow becomes turbulent.

Accurate prediction of aerodynamic heating is actually remarkably complex and is typically a graduate level topic. You can get a decent estimate of stagnation point heating rate from methods like Fay & Riddell (1958), but then you need to work out the heating over other portions of the body. One of the best (and still most frequently cited) sources on aerodynamic heating estimates was by Van Driest (1956). Still, this requires a lot of knowledge to implement.

Ultimately, aerodynamic heating has to do with the velocity (Mach number), the shape of the vehicle (which dictates things like the pressure distribution over the surface, and therefore the boundary layer), and flight attitude.

As @bigfooted mentioned, there are relationships between the skin friction coefficient and the Stanton number (heat transfer coefficient). This is referred to as the Reynolds analogy. The relationship can be justified theoretically on a laminar flat plate but becomes semi-empirical for more complicated geometries or turbulent flows.

I agree that Anderson (2019) is a good source

[1] Fay, J. A., & Riddell, F. R. (1958). Theory of Stagnation Point Heat Transfer in Dissociated Air. J. Aeronaut. Sci., 25(2), 73–85. https://doi.org/10.2514/8.7517
[2] Van Driest, E. R. (1956). The Problem of Aerodynamic Heating. Aeronautical Engineering Review, 15(10), 26–41.
[3] Anderson, Jr., J. D. (2019). Hypersonic and High-Temperature Gas Dynamics, Third Edition. American Institute of Aeronautics and Astronautics, Inc. https://doi.org/10.2514/4.105142