A question about $I_{s}=\frac{F}{\dot{w}}=\frac{c}{g}$ with an example

Question

I'm new to rocket propulsion elements and I'm reading Rocket Propulsion Elements by George P. Sutton, 4th Edition.

In the second chapter, it introduced me the following expression,

$$I_{s}=\frac{F}{\dot{w}}=\frac{c}{g}$$

$I_{s}: \text{Specific impulse}$

$F: \text{Thrust}$

$\dot{w}: \text{Weight flow rate}$

$c: \text{Effective exhaust velocity}$

$g: \text{Gravitational constant}$

Now, I created an example myself. I considered that I have an amateur rocket engine that contains $1kg$ of propellant and I ignite this amateur rocket engine on the ground of the Earth (so, $g$ is basically $~9.8m/s^2$). So the initial weight of the propellant is $w_{i}=m_{i}g=1\times 9.8=9.8 kg.m/s^2$. I consider that the burn time of the propellant is $0.3$ seconds. So, the average weight flow rate is,

$$\frac{\Delta{w}}{\Delta{t}}=\frac{w_{f}-w_{i}}{t_{f}-t_{i}}=\frac{0-9.8}{0.3}=32,66kg.m/s^3$$

I give $F=50N$, so I can find the $I_{s}$ of my amateur rocket engine.

$$I_{s}=\frac{F}{\Delta{w}/ \Delta{t}}=\frac{50}{32,66}=0,153s$$

Now, I want to find out the effective velocity of my amateur rocket engine,

$$I_{s}=\frac{c}{g}\Rightarrow c=I_{s}g=0,153\times 9.8=1,5 m/s$$

Are my calculation correct and logic correct?

Answer 1

I think your calculations are correct, but your example is not representative of real rockets — an Estes E9-6 model rocket engine produces 25N of thrust, but burns for 2.8 seconds with only 36 grams of propellant.

Note that g used to convert between Isp in seconds and exhaust velocity is an artifact of the definition of pounds-force and pounds-mass, and doesn’t depend on the local gravity when you fire the engine. It’s not technically correct to use seconds for Isp when working with metric units, but we do it anyway.