(compiled from sci.space.tech)

Simple combustion chamber |

Excel Simulation of a combustion chamber |

Formulae Simulation of a one stage rocket going to LEO |

Newton's first law is that force is equal to mass times
acceleration.

F = ma.

In SI units F is measured in Newtons, m
in kilograms, and acceleration in meters per second per second.So, 1
N = 1 kg m/s/s

Kinetic energy of a moving mass is equal to one half the mass of a moving body times the velocity of the body squared.

KE = 1/2 * m * v^2

Again, in SI units KE is measured in Joules, m in kilograms, v in
meters per second.

Note that acceleration is the rate of change
in velocity. This is called the DIFFERENTIAL of velocity with respect
to time.

Power is the differential of energy with respect to time. It is measured in watts. One watt is equal to one joule per second. It measures the rate of change of energy.

One ordinary energy unit is the Kilowatt-Hour. It's useful for comparison between chemical and electrical energy..

1 KWH = 1 Joule /(3,6 * 10exp6) = 1J/(3600000)

The force exerted by a rocket engine is equal to the mass flow rate of the rocket engine times the velocity of the exhaust gases (the stuff coming out of the rocket).

F = mdot * Ve

Where F is measured in Newtons, mdot is the mass flow rate in
kg/sec, and Ve is the exhaust velocity in meters per second.

The
mass flow rate is how much propellant is pumped through the engine
each second.

The power needed to sustain this thrust (or produced when this thrust is achieved) is merely the rate of energy needed to accelerate the mass of propellant to the exhaust speed.

Or applying the formula for kinetic energy; Power is equal to one half the mass flow rate of the engine times the exhaust velocity squared.

P = 1/2 * mdot * Ve ^2

In SI units Power is measured in watts, mdot is measured in kg per sec, and Ve is measured in meters per second.

Since mdot*Ve = F, we can write a relation between power and thrust:

P = 1/2 * mdot * Ve * Ve = 1/2 (mdot * Ve) Ve = 1/2 (F) Ve = 1/2 * F * Ve

Now, there is a relation, which I won't derive, between Ve and Isp, between exhaust velocity and specific impulse:

P = F * Isp * 4.91

In a chemical rocket to generate power all you have to do is burn the propellants in the engine at the desired mass flow rate.

**I'll like to know what is the maximum speed a space vehicle
(rocket) can attained?**

What you're looking for is called the "rocket equation"
.

Basically, it's this : (M+P)/M = Exp(dv/C)

where M is the
dry mass of the rocket, P is the mass of the propellant,

dv is
the delta-V or velocity change, and C is the exhaust velocity of the
rocket's engine. Just for reference, C = ISp * g (g = 10m/s^2), since
rocket specs are often given with specific impulse (ISp) instead of
exhaust velocity.

The reason bigger exhaust velocities make for more efficient
rockets is

because of the rocket equation;

Vf = Ve * LN (1/(1-u))

where Vf = final velocity of the rocket.

Ve = exhaust velocity
of rocket engine.

LN() = natural log function. LN(2.72) = 1.

u
= propellant fraction (fraction of rocket by mass that's propellant).

*Orbital velocity is 7,000 m/sec.*

*Escape velocity is 11,000 m/sec*

*Moon mission is 70,000 m/sec*

*Mars mission is 90,000 m/sec*

If we solve the rocket equation for these Vf knowing the Ve of the
rocket

type given above, its easy to see that propellant fraction
is very very close

to 1 (100% propellant 0% anything else).

**This means the rockets are impractical for other things that
orbiting around Earth, and even for that the propellant fraction is
so out of reach of today economical technology that staging must be
used.**