## Propulsion## How does propulsion work? |

To calculate the specific impulse, we first need to calculate the exhaust velocity. Since the real exhaust velocity is exceeding complex to calculate, we will be using some simplifying assumptions to make a simpler equation. Assume that the exhaust velocity follows the following formula:

V_{e}^{2}= kR_{gas}T_{c}[1 - (p_{e}/p_{c})^{(k-1)/k}] / (k-1)

where

k = ratio of specific heats, c

_{p}/c_{v}

p_{e}= nozzle exit pressure

p_{c}= combustion chamber pressure

T_{c}= combustion chamber temperature

R_{gas}= exhaust flow specific gas constant RR/MM

RR = universal gas constant

MM = exhaust gas molecular weight

Note that k is typically between 1.21-1.26 for a wide range of fuels and oxidizers.
Also note that the subscripts (e, c) stand for exit and chamber. Therefore,
p_{c} stands for temperature of chamber and p_{e} stands for
pressure of exit.

Because we are just approximating, we can simplify the above equation. The second half of the equation:

[1 - (p_{e}/p_{c})^{(k-1)/k}] / (k-1)

typically ends up as a small number which we will approximate with 1. Since anything multplied by one is itself then a rough guess of a specific impulse is:

V_{e}^{2}= kR_{gas}T_{c}

and since k is typically between 1.21-1.26 then we will also remove that term and approximate it as 1 giving us:

V_{e}^{2}= R_{gas}T_{c}

This result gives us some interesting observations:

- to get a high exhaust velocity and therefore a high isp we want both a high
temperature in the chamber (T
_{c}) and a large exhaust flow gas constant (R_{gas}) - for a given temperature (T
_{c}), because R_{gas}is RR/MM, then if we reduce the molar weight of the exhaust gas then we get a higher velocity.

This then tells us that for better performance (high exhaust velocity), we want fuel-oxidizer mixtures that burn very hot and have a low molar mass exhaust.

The specific impulse is:

I

_{sp}= u_{eq}/g_{e}

where

I

_{sp}= specific impulse

u_{eq}= total impulse / mass of expelled propellant

g_{e}= acceleration at Earth's surface (9.8 m/s^{2})

And since we are approximating the speed of a gas with a constant velocity; the momentum of the escaping gas is:

p = mv

where

p = momentum (kg m/s)

m = mass (kg)

v = velocity (m/s)

Notice how masses cancel out and therefore we the I_{sp} is just the
velocity of the exhaust gas (V_{e}) divided by the gravitational attraction
of the Earth (g_{e}).

What is thrust?

What are the types of rocket propulsion?

Why is mass important?

What is chemical propulsion?

What is specific impulse?

What are some rocket propellants?

How do you calculate rocket engine performance?