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Some months ago I was doing research on Tsiolkovsky's rocket equation, and I created a PDF "cheat sheet" (using $\LaTeX$) with the various parameters re-arranged ($m_{\text{i}}$, $m_{\text{f}}$, $I_{\text{sp}}$, $\Delta v$, etc.) to help me better understand the basics of rocket propulsion. Some of the equations I compiled for my document are shown below:

Delta-V: $~~~~\Delta v = v_{\text{exh}} \cdot \ln\left(\dfrac{m_{\text{i}}}{m_{\text{f}}}\right) $

Final mass: $~~~~m_{\text{f}} = m_{\text{i}} \cdot \exp\left(\dfrac{-\Delta v}{v_{\text{exh}}}\right) $

Initial mass: $~~~~m_{\text{i}} = m_{\text{f}} \cdot \exp\left(\dfrac{\Delta v}{v_{\text{exh}}}\right) $

Propellant mass: $~~~~ m_{\text{p}} = m_{\text{i}} - m_{\text{f}}$

Mass fraction of propellant: $~~ \dfrac{m_{\text{p}}}{m_{\text{i}}} = \dfrac{m_{\text{i}} - m_{\text{f}}}{m_{\text{i}}} = 1 - \exp\left(\dfrac{-\Delta v}{v_{\text{exh}}}\right)$

Specific impulse (mass): $~~~~I_{\text{sp}} = 4.55368\sqrt{\dfrac{\text{heat released (kJ/mol)}} {\text{mass of products (kg/mol)}}}$

Specific impulse (thrust): $~~~~I_{\text{sp}} = \dfrac{v_{\text{exh}}}{g_{\oplus}}$

Exhaust velocity: $ ~~~~v_{\text{exh}} = I_{\text{sp}} \cdot g_{\oplus} $

Granted, these equations are hideously redundant, but the stoichiometric equation of mass-based $I_{\text{sp}}$ (units of velocity) had me curious:

What explains the 4.55368 constant?

I realize now that I either (1) found that number somewhere online several months ago when doing my research, or (2) I calculated it somehow. But now (embarrassingly) I cannot remember where, and Google searching earlier today has come up empty.

UPDATE

The below equations are in response to the comments: I claim that the constant value has no units, and I'm defending this claim below. Please note that the radical term reduces to $\dfrac{m}{s}$, which is how the mass-based $I_{sp}$ is measured.

1 Joule is

$$ 1 \text{kg} \cdot \dfrac{\text{m}^{2}}{\text{s}^{2}}$$

From my equation above,

$$I_{sp} \left(\text{as}~\dfrac{\text{m}}{\text{s}}\right) = const \cdot \sqrt{\dfrac{~~\frac{J}{mol}~~}{~~\frac{kg}{mol}~~}}$$

The moles in each denominator (inside the radical) cancel, giving

$$\dfrac{m}{s} = const \cdot \sqrt{\dfrac{J}{kg}} = const \cdot \sqrt{\dfrac{kg \cdot \frac{m^{2}}{s^{2}}}{kg}}$$

Kilograms cancel out, giving

$$ \dfrac{m}{s} = \text{const} \cdot \sqrt{\dfrac{~\dfrac{m^{2}}{s^{2}}~}{1}} = \text{const} \cdot \sqrt{\dfrac{m^{2}}{s^{2}}} = \text{const} \cdot \sqrt{\left( \dfrac{m}{s}\right)^{2}}$$

$$ \dfrac{m}{s} = \text{const} \cdot \dfrac{m}{s} $$

pr1268
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  • Are you looking for why the constant is specifically 4.44368 or are you asking why there is a constant there? – Lio Elbammalf Jan 14 '17 at 21:54
  • Actually, both. – pr1268 Jan 15 '17 at 09:45
  • The constant is there as a conversion factor and takes in some selection of constants (or things you assume as constant). Just doing rearranging your equation with the constant and the v/g0 one dimensional analysis gives your constant dimensions of $\frac{1}{[L][T]^{-2}}$ which looks like one over an acceleration (perhaps gravity). This doesn't answer your question but I thought it might jog your memory of what constants may have gone in there. – Lio Elbammalf Jan 15 '17 at 14:02
  • The constant has no dimensions, and the radical term reduces to m/s (velocity). This is a well-documented, but uncommonly-used way of expressing $I_{sp}$ according to Wikipedia. – pr1268 Jan 15 '17 at 14:06
  • Also, I'm not sure of the accuracy of the value, but when I use that many significant digits, it means I either calculated it carefully, or researched/found it (likely online). The value and the meaning of this constant is what I'm curious about. – pr1268 Jan 15 '17 at 14:09
  • $const = \frac{v_{exh}}{g_{0}}\sqrt{\frac{Mass (kg/mol)}{heat(kJ/mol)}}$ is, dimensionally $\frac{[L][T]^{-1}}{[L][T]^{-2}}\sqrt{\frac{[M]}{[M][L]^{2}[T]^{-2}}}$ and a bit of canceling leaves $\frac{1}{[L][T]^{-2}}$ You are probably better acquainted with your equations but it does seem to me that the constant has dimensions unless I'm missing something. – Lio Elbammalf Jan 15 '17 at 14:20
  • @LioElbammalf: Perhaps my edit to the original question will elucidate things. – pr1268 Jan 15 '17 at 23:08
  • You have not defined $g_\oplus$? Hard to do dimensional analysis without all definitions. The few times I had to struggle with magical numbers, the equation was usually full of dimensions commonly used in the field (none of them written out, of course, leaving days of guesswork before things worked...) – Emil Feb 04 '21 at 07:14
  • @Emil: $g_{\oplus}$ refers to the gravity of Earth, i.e. $9.80665 \frac{m}{s^2}$. – pr1268 Feb 07 '21 at 09:11

2 Answers2

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Specific impulse is defined through $$F_{thrust}={\dot m}\,g\,I_{sp}$$ with $\dot m$ the mass flow rate in $\mbox{kg/s}$, $g$ gravitational acceleration, $g\approx9.81\mbox{m/s}^2$, and $F_{thrust}$ the thrust of the rocket engine in $\mbox{kg/m s}^2$. Clearly specific impulse thus has units of $\mbox{s}$.

I'm not sure exactly where your factor of $4.55368$ is coming from. Basically the maximum theoretical exhaust speed $v_{exh}$ is equal to $\sqrt{R\,T}$, with $R$ the specific gas constant and $T$ the temperature in the combustion chamber. Obviously its value will depend on the composition of the exhaust gas.

Pirx
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  • Well, that makes both of us not sure where the 4.55368 comes from... While I do realize $I_{sp}$ is measured in seconds, the mass-based variant is m/s to eliminate the need for Earth-specific gravity. Thank you for the gas constant- and temperature-based clarification. – pr1268 Jan 17 '17 at 17:48
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If all the heat released goes into the kinetic energy of the combustion products $$ \frac 12 mv^2 = m\times \hbox{heat per unit mass} $$ you have $$ v=\sqrt 2 \sqrt{\dfrac{\text{heat released (kJ/mol)}} {\text{mass of products (kg/mol)}}} $$ so $I_{sp} = v/ g= \sqrt 2/g\approx 0.144 \sqrt{\ldots} $. Thus the conversion number is neither dimensionless nor is it 4.5.

mike stone
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