[AR] Re: Thermal/Cooling Questions in Liquid Rocket Engines
- From: "Graham Sortino" <dmarc-noreply@xxxxxxxxxxxxx> (Redacted sender "gnsortino" for DMARC)
- To: "arocket@xxxxxxxxxxxxx" <arocket@xxxxxxxxxxxxx>
- Date: Mon, 28 Nov 2016 15:09:07 +0000 (UTC)
Thanks David –
You can't separate the thickness of the layer from the thermal conductivity
without knowing one or the other This is somewhat what I had expected. I was
kind of hoping that I could synthetically define the depth by extracting the
component of in2*sec*oR/Btu that referred to depth but I guess it doesn’t
seem like it contains a depth component so it’s a bit futile. Also, RPA seems
to use a different calculation (see p19)
http://propulsion-analysis.com/downloads/2/docs/RPA_ThermalAnalysis.pdf to ;
apply the thermal barrier coating, which makes this even harder.
I did calculate this manually using 4-18 but generally I like to use RPA to
double-check my calculations with RPA to make sure I haven’t made any mistake.
However, I guess in this case it’s not possible.
Appreciate the help as always.
Best,
Graham
On Sunday, November 27, 2016, 12:48:15 PM EST, David Gregory
<david.c.gregory@xxxxxxxxx> wrote:The resistance in H&H is in units of
Energy/time/area/temperature, or alternately power/area/temperature (actually
the inverse of that, on the graph). So in the MKS system, its W/m^2/K. You
can't separate the thickness of the layer from the thermal conductivity without
knowing one or the other. In the absence of further data or another reference,
you simply have to use the value of the resistance as shown with the equation
4-18.
As far as the conversion:
https://www.wolframalpha.com/input/?i=Btu%2Fs%2Fin%5E2%2FF+to+W%2Fm%5E2%2FC
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