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
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
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
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.