Question on heat/ablator, help check physics

[FancyMouse]

So I found I'm almost running out of ablator when aerocapturing/braking at Eve. Curious if I can spare a little bit more, I started playing calculus around various physics quantities (mainly aero drag F ~ rho*v^2 and heat power P ~ rho*v^3), until I figured out all the calculation can be summarized as one phrase - energy conservation. Another typical day that I realize I'm more stupid than I think I am.

So now here are my questions:

1. Is it true that for the same ship, same initial speed (usually from a interplanetary transfer) and same target speed (usually low-orbit speed), basically same everything except trajectory, I'll always end up with roughly the same heat produced into the ship, no matter how I aerobrake (different altitude/multiple pass/etc.)? Note I'm talking about the heat in Joules, not ablation consumption for this question. In other words, does almost all the kinetic energy of the ship gets converted to heat into the ship?

2. Does KSP model the heat in a similar way, so that the above estimate is close enough to what actually happens in the game?

3. Is there any way to reduce ablator consumption, other than making too many aerobrake passes to prevent ablator kicking in (I'm fine with two or three but probably impatient for more), and reducing mass (usually payload will be fixed thus hard to optimize)? I can for sure add ablators but I want to know if can reduce its consumption first.

Thanks.

Edited October 1, 2015 by FancyMouse

[NathanKell]

That's very close, the difference is that:

1. Heat transfer coefficient is rho^.5 * v^3 but

2. If you go too fast at too low, you move from laminar to turbulent flow, and the heat transfer spikes up (over 30-100x IIRC)

That means that you *mostly* have equal heat loads no matter your entry angle, but that (assuming you can stand the peak heat *flux* without the shield over-temping) it's better to go steeper than shallower.

(That's ignoring the fact that that's merely the transfer coefficient, the temperature difference is (shock temp - part skin temp) and the shock temp is...fairly proportional to v)

Only real way to lower ablator loss is to lower the lossconst in the MODULE, or try to keep the shield at a lower skin temp, but that's not terribly controllable because you need high temps from low Pes in order to capture.

[FancyMouse]

Wow didn't expect NathanKell you come to the thread. Time to start learning

>Heat transfer coefficient is rho^.5 * v^3

From dimension analysis point of view, this ^.5 sounds weird - rho is of unit kg/m^3 and this will introduce m^.5 which you'll need to cancel somewhere. Is it just an empirical formula like other complicated thermo formulas or there's a theory behind it? If latter, just toss me some terms and I'll happy to search myself.

Of course, this power of rho doesn't seem affect my original problem since v is still at third power.

>you move from laminar to turbulent flow, and the heat transfer spikes up (over 30-100x IIRC)

Do you mean the part of the equation independent of rho and v would spikes 30-100x? That sounds horrifying. But I guess since I'm dealing with things like Eve entry, I'll have to deal with turbulent flow for the most part anyway?

Thanks again. The above post is by itself a wealth of information, and let me think slowly what consequence I can expect.

[NathanKell]

Happy to help.

The sqrt of rho tl;dr is 'ferram said so'; it actually comes from dealing with Reynolds numbers, because heating is approximated as 0.5 * rho * v^3 * Stanton number, where the latter is proportional (in laminar flow) to rho / sqrt(Re).

Since Re itself is proportional to density, you end up with rho^.5 * v^3 as the final proportionality for heating in laminar flow.

KSP implements a "pseudo Reynolds number" which is just velocity (in m/s) times density (in kg/m^3). In Physics.cfg, you'll see a Turbulent Convection Start and an End value; that determines the pseudoRe when the convection coefficient multiplier starts shifting from 1.0 to (turbulent convection mult--it's stated as 100 in the cfg, but I think the value may not be read in 1.0.4--I think same for all the values, though start and end are the correct values, but mult may be 30 no matter what you put), and it reaches the full multiplier when pseudoRe reaches the End value (it's a linear interpolation).

You'd be surprised at how fast / how low you need to go before you get over 100 pseudoRe.

[Kobymaru]

That means that you *mostly* have equal heat loads no matter your entry angle, but that (assuming you can stand the peak heat *flux* without the shield over-temping) it's better to go steeper than shallower.

I think I read the explanation from you in another thread, but I didn't get it unfortunately.

Can you walk me through the logic of why it is better to go steeper than shallower?

Because so far, my strategy was to stay in upper atmosphere as long is possible out of the reasoning that if I reduce heat spikes as much as possible, I have a longer time to radiate off all heat.

[nekogod]

I think I read the explanation from you in another thread, but I didn't get it unfortunately.

Can you walk me through the logic of why it is better to go steeper than shallower?

Because so far, my strategy was to stay in upper atmosphere as long is possible out of the reasoning that if I reduce heat spikes as much as possible, I have a longer time to radiate off all heat.

There is a sweet spot where heating is higher but so is drag so you decellerate faster leading to less heating overall. Deeper than this and aero/gforces/heating destroy you and above this you'll experience lower heating but it will last much longer as you're decellerating much slower and overall leads to more heating.

In simple terms 1200K for 5 seconds is less damaging than 800K for 45 seconds (finger in the air numbers but show the point I'm making). Think of it like your oven, put in a cake at 220C for 5 minutes and it'll barely be warm, put it in at 180C for an hour and it'll be burnt to a cinder.

[NathanKell]

Drag is proportional to density. Heat is proportional to sqrt(density).

Exampe: Density = 0.01

Every second, you slow down 0.01 units, but you take .1 units of heat.

Example 2: Density = 0.25

Every second, you slow down .25 units, but you take 0.5 units of heat.

So in case 1, you take 10x as much heat as slowing, in case 2 only double.