Quick explaination on specific impulse

[Secuas]

I know for quite a while that a higher specific impulse is good for use in space. But I never knew what it actually was. So just out of curiosity I would like to know what a specific impulse is.

Thanks in advance

Edited December 15, 2014 by Kaname

[capi3101]

There's a good article on the topic on the wiki. Basically it's a mathematical expression of the efficiency of an engine.

[OhioBob]

Mathematically it is simply the thrust of the engine divided by the weight of propellant ejected per second. In other words, an engine with a high specific impulse will produce more thrust per unit of propellant burned.

Edited December 15, 2014 by OhioBob

[Streetwind]

For the simplest possible explanation, without any maths or formulas, you can always fall back on this:

"If this engine is throttled to exactly 1 kN of thrust, then it will take this long to consume 1 t worth of fuel."

A PB-ION, for example, set to 50% thrust will take 4200 seconds to consume 1 t worth of xenon gas. If it was set to full thrust (2 kN), it would run for 2100 seconds because it would consume fuel twice as fast to deliver twice the thrust. You can measure this with a stopwatch ingame. And that is why something describing fuel efficiency is measured in seconds.

Derived from this, you get the mass flow rate = (1 t / Isp) * thrust, which is how much reaction mass is consumed per second while the engine is producing this amount of thrust. Divide by propellant density and the gravity constant g0 = 9.82, and you get the volume flow rate describing how many units of this specific propellant are consumed per second. Which is what KSP shows you in the VAB in the info tabs for engine parts (based on the lowest Isp value for that engine, since it shows the maximum possible flow rate).

Edited December 16, 2014 by Streetwind

[Red Iron Crown]

"If this engine is throttled to exactly 1 kN of thrust, then it will take this long to consume 1 t worth of fuel."

Almost. If it is throttled to exactly 9.82kN of thrust it will take that long to consume 1t of fuel.

[Streetwind]

Pah, details!

(Okay, okay, I apologize )

[Rusty6899]

Almost. If it is throttled to exactly 9.82kN of thrust it will take that long to consume 1t of fuel.

I just saw the comment above and was going mad trying to work out how that was true. This makes sense. Although I'm not sure I understand the need for standard gravity in the equation at all (Tsiolkovsky rocket equation cancels it out after all). I suppose it does fix the units. I'm guessing that's the answer.

[Red Iron Crown]

The inclusion of g₀ is there to facilitate easy conversion between metric and Imperial units. Seconds are valid units for both, and g₀ is a constant most engineers have memorized that allows conversion of I_sp in seconds to appropriate speed units. It's basically a convention for convenience.

[OhioBob]

Derived from this, you get the mass flow rate = (1 t / Isp) * thrust, which is how much reaction mass is consumed per second while the engine is producing this amount of thrust. Divide by propellant density and the gravity constant g0 = 9.82, and you get the volume flow rate describing how many units of this specific propellant are consumed per second. Which is what KSP shows you in the VAB in the info tabs for engine parts (based on the lowest Isp value for that engine, since it shows the maximum possible flow rate).

That's not entirely correct. What you call the mass flow rate is actually the weight of the mass ejected, measured in kN. Dividing by g₀ then gives you the mass flow rate, measured in t. To get the volume flow rate you then have to divide by the propellant density.