I know that chamber pressure is

$$P_0=p \left[ 1+ \frac{1}{2}(k-1)M^2 \right]^\frac{k}{k-1}$$

where $M$ is exit Mach number and according this equation

$$M= \sqrt{ \frac{2}{k-1} \left[ \left( \frac{P_0}{P_e} \right)^\frac{k-1}{k} -1 \right]}$$

But exit Mach number equation uses $P_0$!

So I found another equation of $P_0$

$$P_0= \left[ \frac{A_b}{A^*} \frac{a \rho_p}{\sqrt{ \frac{k}{R T_0} \left( \frac{2}{k+1} \right)^\frac{k+1}{k-1} }} \right]^\frac{1}{1-n} $$

and to calculate it, I need $A^*$, burn rate coefficient $a$, and exponent $n$. I can't find other equations of chamber pressure except this one


So my question is, how to get $a$ and $n$, to get burn rate $r$ and then calculate $A^*$ and finally chamber pressure $P_0$?

And the main question, what $P_0$ value do I have to choose for the equation $r=aP_0^n$ to get proper $a$ and $n$?


1 Answer 1


How to get a and n

The burn rate coefficient and exponent are propellant specific. There are catalogues of different solid propellant mixtures on Richard Nakka's Website (which I gather you're familiar with) under the "Propellants" section. For each propellant he links to a "performance characteristics" page where he tabulates $a$ and $n$ along with other information you'll need like the flame/total temperature $T_o$ and the specific heat ratio $k$ or $\gamma$. (See, for example, these values listed for KN-Sucrose)

These values are typically determined experimentally by measuring burned web and chamber pressure as they vary with time. There is a great deal of research into the development of analytical and computational models of erosive burning (an older paper detailing experimental and numerical methods here), these are not something you're going to be able to calculate without a great deal of pain, which is why tabulated values are the way to go.

Once you pick a propellant and find the corresponding burn rate and exponent, you have all the information you need to close your problem. The third equation you list for steady state chamber pressure can now be used to relate design choices (such as throat and burn area) to chamber pressure.

Between Chamber Pressure, Throat Area, and Burn Area, you get to choose two; that equation will tell you what the third must be.

Note: Be very careful of dimensional consistency for the burn rate coefficient and exponent, they are cited in a variety of units. Also, the burn area is typically variable (a function of web) so for safety margins etc. you will want to use the largest burn area to see what your maximum chamber pressure will be.

  • $\begingroup$ for example if I want to calculate my own combustion chamber's pressure, let's say with APCP (HTPB) propellant, then I have to make propellant, actual solid rocket motor and burn it? as I see from a lot of examples, they are taking small fraction of the propellant and burning it under 1000psi pressure to get a and n values, but why they are choosing 1000psi? I mean how they know that 1000psi will be the real chamber's pressure? $\endgroup$ Commented Jun 21, 2022 at 21:08
  • $\begingroup$ For experimental determination: researchers can place a small amount in a larger pressurized container that wont have a significant pressure change during combustion because the partial pressure of the comustion products will be relatively small. But like I say, you should just look up values. $\endgroup$
    – A McKelvy
    Commented Jun 21, 2022 at 21:22
  • $\begingroup$ You said "larger pressurized container" and my question is, how they know what pressure should be in the container , before they burn that small amount of propellant? $\endgroup$ Commented Jun 21, 2022 at 21:39
  • $\begingroup$ The burn rate coefficient and exponent are generally independent of chamber pressure, so as long as they measure the burn rate at a variety of different pressures, they will be able to fit a curve to the data points. $\endgroup$
    – A McKelvy
    Commented Jun 21, 2022 at 21:55
  • $\begingroup$ are you saying that the burn rate coefficient and exponent will be same with different pressure? really? $\endgroup$ Commented Jun 21, 2022 at 22:12

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