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I've just read Wernher Von Braun's Project Mars: A Technical Tale, and I'm trying to understand the math in the back. I can follow some of what he's doing, but other parts I just can't tell where his numbers come from. In particular, I can't figure out his equation for theoretical Isp.

I've searched all over the internet, and found several equations relating Isp to the temperature, pressure, and geometry of the chamber, along with gas constants, specific heat ratios and molar masses of the propellant, as well as ambient pressure. However, Von Braun doesn't use any of these numbers (or he doesn't mention that he does); his equation is simply:

$I_{sp,th} = 9.323 * \sqrt{H_0 - H_e}$

Does anyone have any idea how that was derived? Or at the very least, what the significance of 9.323 is?

enter image description here

From Goodreads.

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I guess that it goes approximately like this: assume that the enthalpy change (I'll denote it $\Delta H$) is fully converted to the kinetic energy of the exhaust and the exhaust moves with velocity $v$ relative to the engine. Then we have: $$ mv^2/2 = m \Delta H;$$ $$ v = \sqrt{2\Delta H};$$ $$I_{sp} = \sqrt{2\Delta H}/g.$$ But that is assuming that all values are expressed in, for example, SI units ($I_{sp}$ in seconds, $\Delta H$ in J/kg, $g$ in m/s2). Von Braun uses the value of $\Delta H$ in cal/g instead. Using the value of calorie as 4.184 J, we have 1 cal/g = 4.184 J/g = 4184 J/kg, so $$I_{sp}[s] = \sqrt{2\cdot 4184\Delta H[cal/g]}/g[m/s^2] = 91.48\sqrt{\Delta H[cal/g]}/g[m/s^2].$$ Using the value $g = 981~ \text{cm/s}^2 = 9.81~ \text{m/s}^2$, which is used earlier in the page, we get $$I_{sp}[s] = 91.48\sqrt{\Delta H[cal/g]}/9.81 = 9.325\sqrt{\Delta H[cal/g]}.$$ Well, close enough. Von Braun probably used slightly different values for calorie or $g$.

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  • $\begingroup$ You're the best! So this works for any fuel, yes? Though it does seem to rely on the fuel/oxidizer being in a stoichiometric ratio, right? $\endgroup$ – DiceMaster1018 Aug 19 '17 at 2:27
  • $\begingroup$ @DiceMaster1018 Yes, it doesn't depend on the fuel. I would say that the formula is applicable even if the fuel and the oxidizer are not in the stoichiometric ratio, it's just that if the exhaust contains fuel or oxidizer that has not reacted, then the actual enthalpy difference is smaller than the one given by the stoichiometric equation; but I'm not sure. $\endgroup$ – Litho Aug 19 '17 at 16:21

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