Specific Impulse Help

[sevenperforce]

1 hour ago, Cheif Operations Director said:

Why even account for seconds eeing as its 1 also where does 9.8 for gravity come into play. Fuel flow right. And thanks I think I got it just did A few can you verify? 

Well, it's a unit. You can't just get rid of it.

1 hour ago, Cheif Operations Director said:

124 Ibs of thrust

1.31 as fuel flow.

I got 95 seconds rounded.

 

another

50 lb of thrust

0.32 as fuel flow 

Answer: 156 seconds (rounded)

Well, your math is correct, but you need to make sure that you are using units. "1.31" and "0.32" are meaningless unless you actually attach units. You should say "1.31 lb/sec" or "0.32 lb/sec". Otherwise, what if you end up getting fuel flow in gallons per second, or kilograms per minute?

[Cheif Operations Director]

1 hour ago, sevenperforce said:

Well, it's a unit. You can't just get rid of it.

Well, your math is correct, but you need to make sure that you are using units. "1.31" and "0.32" are meaningless unless you actually attach units. You should say "1.31 lb/sec" or "0.32 lb/sec". Otherwise, what if you end up getting fuel flow in gallons per second, or kilograms per minute?

Yes Im using units but why d that whole thing with with division you did for the seconds if it is representative of one.

[FleshJeb]

if one unit is divided by a time unit, it represents a rate at which something happens. If you walk 5 kilometers in one hour, you walked 5 km per hour (written as 5km/hr = 5 km ÷ 1 hr). If you know you're walking at that speed, how far do you walk in two hours?

(5km / hr) x 2hr = 10 km because the hours cancel.

It's called https://en.wikipedia.org/wiki/Dimensional_analysis

You can sometimes figure out how to solve a problem by looking at the units you expect to get, and looking at the pieces. If you want to know how fast 5km per hour is in meters per second:

Convert the distance: One-thousand meters happen per one kilometer, so:

5km / hr * 1000m / km = 5000m/hr, because the kilometers cancel.

Convert the time: One hour happens per sixty minutes, and one minute happens per sixty seconds, so:

(5000 m /  1 hr) * (1 hr / 60 min) * (1 min / 60 sec) =

(5000m * 1 hr * 1 min) / (1 hr * 60 min * 60 sec).

The hours cancel, the minutes cancel, and you're left with (5000 m / 3600 sec) = 1.39 m/s

[GoSlash27]

On 3/16/2018 at 6:15 PM, Cheif Operations Director said:

Yes Im using units but why d that whole thing with with division you did for the seconds if it is representative of one.

COD,

 The division sign represents "per". 1.31 lb/sec is 1.31 pounds per second. Every second the engine is running, it consumes 1.31 pounds of fuel.

HTHs,
-Slashy

[sevenperforce]

On 3/16/2018 at 8:15 PM, Cheif Operations Director said:

Yes Im using units but why d that whole thing with with division you did for the seconds if it is representative of one.

One thing to keep in mind is that "seconds" in a specific impulse is a physically meaningful measure.

If you had a rocket with exactly enough thrust to lift the weight of its own propellant on Earth (though not necessarily dry mass or payload), and you fired it at full throttle, then it would burn until all its propellant is expended. The amount of time it would burn is the specific impulse of the engine.

For example, if you had an engine which produces 1,000 lbs of thrust while burning 5 pounds of propellant every second, its specific impulse would be 200 seconds (1000 lbs/5 lbs/sec = 200 sec). If that engine was placed on a rocketplane carrying 1,000 lbs of propellant and the engine was fired at full throttle, the engine burn would last for 200 seconds before running dry.

This applies to any sort of engine, actually, and you can employ various transformations to make it more useful. Suppose you have a jet airliner powered by turbofan engines with a specific impulse of 3500 seconds. If this jet reaches cruising altitude with 2,000,000 pounds of fuel, then we know that it would take 3500 seconds to burn all that fuel if the engines produced 2,000,000 pounds of thrust. Of course, they do not produce anywhere near that amount of thrust; they likely produce around a tenth of that, or 200,000 lbs of trust. So this tells us that the engines can burn at cruise for 3500 seconds * 10, or around 9.7 hours.

[Cheif Operations Director]

8 hours ago, sevenperforce said:

One thing to keep in mind is that "seconds" in a specific impulse is a physically meaningful measure.

If you had a rocket with exactly enough thrust to lift the weight of its own propellant on Earth (though not necessarily dry mass or payload), and you fired it at full throttle, then it would burn until all its propellant is expended. The amount of time it would burn is the specific impulse of the engine.

For example, if you had an engine which produces 1,000 lbs of thrust while burning 5 pounds of propellant every second, its specific impulse would be 200 seconds (1000 lbs/5 lbs/sec = 200 sec). If that engine was placed on a rocketplane carrying 1,000 lbs of propellant and the engine was fired at full throttle, the engine burn would last for 200 seconds before running dry.

This applies to any sort of engine, actually, and you can employ various transformations to make it more useful. Suppose you have a jet airliner powered by turbofan engines with a specific impulse of 3500 seconds. If this jet reaches cruising altitude with 2,000,000 pounds of fuel, then we know that it would take 3500 seconds to burn all that fuel if the engines produced 2,000,000 pounds of thrust. Of course, they do not produce anywhere near that amount of thrust; they likely produce around a tenth of that, or 200,000 lbs of trust. So this tells us that the engines can burn at cruise for 3500 seconds * 10, or around 9.7 hours.

1000/5 = 200 so it 200 seconds

1000/5/sec (S) = 200 

so why even take seconds into account if you just use the unit at the end.

what I mean is, is will the third variable second (S) ever be used in the forumula or It it just thief is you have an excuse to call it 200 seconds instead of 200.

Specific Impulse to to measure efficency per second not burn time right? 

[GoSlash27]

1 hour ago, Cheif Operations Director said:

what I mean is, is will the third variable second (S) ever be used in the forumula or It it just thief is you have an excuse to call it 200 seconds instead of 200.

Everything you do in physics has a unit, unless it's a conversion factor or is dimensionless. It is important to keep track of the units and state what they are. Specific impulse is a measure of efficiency and it's unit is "seconds". If you convert the thrust of a rocket engine to the equivalent mass in fuel, then the engine will fire for "I_sp" seconds. This handy fact allows you to directly compare one engine's efficiency to another's and calculate all sorts of other helpful things.

1 hour ago, Cheif Operations Director said:

Specific Impulse to to measure efficency per second not burn time right? 

Not 'efficiency per second', just efficiency. But it can be used to calculate burn times, changes in thrust with atmospheric density changes, the efficiency of ganged engines of varying types, exhaust velocity for the Tsiolkovsky rocket equation... Most of your math in KSP will involve this figure.

Best,
-Slashy
 

[Cheif Operations Director]

3 minutes ago, GoSlash27 said:

Everything you do in physics has a unit, unless it's a conversion factor or is dimensionless. It is important to keep track of the units and state what they are. Specific impulse is a measure of efficiency and it's unit is "seconds". If you convert the thrust of a rocket engine to the equivalent mass in fuel, then the engine will fire for "I_sp" seconds. This handy fact allows you to directly compare one engine's efficiency to another's and calculate all sorts of other helpful things.

Not 'efficiency per second', just efficiency. But it can be used to calculate burn times, changes in thrust with atmospheric density changes, the efficiency of ganged engines of varying types, exhaust velocity for the Tsiolkovsky rocket equation... Most of your math in KSP will involve this figure.

Best,
-Slashy
 

How can it be used to figure out a burn time hen it doesn't take into account how much mass of propellant their is. 

Ok am I doing this right

1. Rocket Thrust:150 lbs.         Fuel Flow Rate: 0.5

 

150÷0.5 = 300 Seconds

2. Rocket Thrust:1,400,500 Lbs.                Fuel Flow Rate 5.0 

 

1,400,500 ÷ 5.0= 280, 100 seconds

To confirm for 1. My Isp is 300 seconds

and for 2. It is 280100 seconds

[GoSlash27]

Because the value assumes equality between the thrust and the fuel supply. You can express a ratio between how much fuel you give it and the thrust, and your resultant burn time will hold the same ratio to Isp. Likewise, you can change the throttle setting and have the inverse apply to the burn time.

Example: The LV-T45 produces 200 kN of thrust and an Isp of 320s. 200 kN is equivalent to 20.4 tonnes. If I were to feed this engine 20.4 tonnes of fuel, it would run for 320 seconds. But if I instead gave it 6 tonnes of fuel, it would run for (6 tonnes/ 20.4 tonnes) x 320 seconds = 94.1 seconds.

 Likewise, if I were to set the throttle to half and keep the same 6 tonne fuel supply, it would run twice as long; 188.2 seconds.

 

 

[Cheif Operations Director]

 

Me raging about Specific Impulse  

6 minutes ago, GoSlash27 said:

Because the value assumes equality between the thrust and the fuel supply. You can express a ratio between how much fuel you give it and the thrust, and your resultant burn time will hold the same ratio to Isp. Likewise, you can change the throttle setting and have the inverse apply to the burn time.

Example: The LV-T45 produces 200 kN of thrust and an Isp of 320s. 200 kN is equivalent to 20.4 tonnes. If I were to feed this engine 20.4 tonnes of fuel, it would run for 320 seconds. But if I instead gave it 6 tonnes of fuel, it would run for (6 tonnes/ 20.4 tonnes) x 320 seconds = 94.1 seconds.

 Likewise, if I were to set the throttle to half and keep the same 6 tonne fuel supply, it would run twice as long; 188.2 seconds.

 

 

How can that account for a better fuel type I mean kerosene is different from hydrogen plus 20.4 tones of thrust doesn't mean I'm putting 20.4 pounds of propellant. Especially though a De Laval Nozzle.

I mean Take model rockets for example 20 lb of thrust. Those engine do not even weight 5 pounds 

[GoSlash27]

1 hour ago, Cheif Operations Director said:

How can that account for a better fuel type I mean kerosene is different from hydrogen plus 20.4 tones of thrust doesn't mean I'm putting 20.4 pounds of propellant. Especially though a De Laval Nozzle.

 Not according to Newtonian physics. You're just taking a mass and chucking it out the back. It doesn't care if you're throwing hydrolox, or solid propellants, or powdered doughnuts.
 The changes in efficiency from using different fuel types are reflected in the (wait for it) specific impulse itself

 

[Cheif Operations Director]

6 minutes ago, GoSlash27 said:

 Not according to Newtonian physics. You're just taking a mass and chucking it out the back. It doesn't care if you're throwing hydrolox, or solid propellants, or powdered doughnuts.
 The changes in efficiency from using different fuel types are reflected in the (wait for it) specific impulse itself

 

If they are on fire and going from liquid to gas and expanding the gases expand in increase and speed and tempeture. That's why rockets have nozzles. Or solid to gas. They have a de laval nozzle. I'm not talking about RCS 

[GoSlash27]

1 hour ago, Cheif Operations Director said:

plus 20.4 tones of thrust doesn't mean I'm putting 20.4 pounds tonnes of propellant. Especially though a De Laval Nozzle.

As I said before, what's important is that you know how the fuel mass you *are* using relates to the fixed value of 20.4 tonnes. If it's half, you'll run for half the time. If it's double, you'll run twice as long.

 And I think you're still confuzzled. You can run an infinite mass of propellant through the nozzle. All you need is an infinite stretch of time.
 You see... Propellant mass flow rate (stated in tonnes per second) is a fixed, finite value. At least until you go monkeying with the throttle.

1 hour ago, Cheif Operations Director said:

If they are on fire and going from liquid to gas and expanding the gases expand in increase and speed and tempeture. That's why rockets have nozzles. Or solid to gas. They have a de laval nozzle.

Yeah. And? Trying to figure out the hangup here, so please bear with me

[Cheif Operations Director]

25 minutes ago, GoSlash27 said:

Because the value assumes equality between the thrust and the fuel supply. You can express a ratio between how much fuel you give it and the thrust, and your resultant burn time will hold the same ratio to Isp. Likewise, you can change the throttle setting and have the inverse apply to the burn time.

Example: The LV-T45 produces 200 kN of thrust and an Isp of 320s. 200 kN is equivalent to 20.4 tonnes. If I were to feed this engine 20.4 tonnes of fuel, it would run for 320 seconds. But if I instead gave it 6 tonnes of fuel, it would run for (6 tonnes/ 20.4 tonnes) x 320 seconds = 94.1 seconds.

 Likewise, if I were to set the throttle to half and keep the same 6 tonne fuel supply, it would run twice as long; 188.2 seconds.

 

 

Which numbers are the fuel flow rate and thrust, furthur what does how much fuel being used have to do with the efficency. If I light a car and run it for 10 miles or 20 miles on a race track the efficency should stay about the same.

5 minutes ago, GoSlash27 said:

As I said before, what's important is that you know how the fuel mass you *are* using relates to the fixed value of 20.4 tonnes. If it's half, you'll run for half the time. If it's double, you'll run twice as long.

 And I think you're still confuzzled. You can run an infinite mass of propellant through the nozzle. All you need is an infinite stretch of time.
 You see... Propellant mass flow rate (stated in tonnes per second) is a fixed, finite value. At least until you go monkeying with the throttle.

Yeah. And? Trying to figure out the hangup here, so please bear with me

I think I mis understood you 

[GoSlash27]

Okay, let's try this: Let's back up and pretend the term "specific impulse" doesn't exist.
 

 A chemical rocket works by combusting it's propellant and throwing it out the back as fast as possible. Some propellants will be able to accelerate to higher velocities than others due to "chemistry" or "technology". Ideally you want an extremely light molecule that undergoes a very violent chemical reaction, so you can throw it out the back at a higher speed, because moar speed = moar thrust, right?

 Well, this means we can define the efficiency of a rocket engine another way; thrust (kN)/ propellant mass flow rate (tonnes/ second). If we put them in like terms, 200 kN of thrust becomes 20.4 tonnes thrust/ 0.06375 tonnes per second mass flow rate = 320 seconds. We've just invented the "specific impulse" figure.

 Or we could define the efficiency another way; the velocity at which it kicks the exhaust out the back. An LV-T45 accelerates the exhaust to 3,139 m/sec velocity. If you divide that by g0, you get... 320 seconds. Isp is really pretty unavoidable.

 If you were to use a more efficient fuel or method, the resultant specific impulse would increase as a result. See wikipedia for examples of how changes in fuel affect an otherwise identical engine's stats.

HTHs,
-Slashy

 So at the very basic level, let's "invent" a rocket engine together. We hook it up to a test stand that records the thrust as tonnes of force. We supply it with a fixed mass of propellant. We fire it up, and time how long it runs.

 You pick the values. How much thrust, how much fuel, and how long did it run?

 

Edited March 20, 2018 by GoSlash27

[Cheif Operations Director]

12 minutes ago, GoSlash27 said:

Okay, let's try this: Let's back up and pretend the term "specific impulse" doesn't exist.
 

 A chemical rocket works by combusting it's propellant and throwing it out the back as fast as possible. Some propellants will be able to accelerate to higher velocities than others due to "chemistry" or "technology". Ideally you want an extremely light molecule that undergoes a very violent chemical reaction, so you can throw it out the back at a higher speed, because moar speed = moar thrust, right?

 Well, this means we can define the efficiency of a rocket engine another way; thrust (kN)/ propellant mass flow rate (tonnes/ second). If we put them in like terms, 200 kN of thrust becomes 20.4 tonnes thrust/ 0.06375 tonnes per second mass flow rate = 320 seconds. We've just invented the "specific impulse" figure.

 Or we could define the efficiency another way; the velocity at which it kicks the exhaust out the back. An LV-T45 accelerates the exhaust to 3,139 m/sec velocity. If you divide that by g0, you get... 320 seconds. Isp is really pretty unavoidable.

 If you were to use a more efficient fuel or method, the resultant specific impulse would increase as a result. See wikipedia for examples of how changes in fuel affect an otherwise identical engine's stats.

HTHs,
-Slashy

 So at the very basic level, let's "invent" a rocket engine together. We hook it up to a test stand that records the thrust as tonnes of force. We supply it with a fixed mass of propellant. We fire it up, and time how long it runs.

 You pick the values. How much thrust, how much fuel, and how long did it run?

 

1500 lbs of thrust

1200 pounds of fuel

burn time 50 seconds.

[GoSlash27]

3 hours ago, Cheif Operations Director said:

1500 lbs of thrust

1200 pounds of fuel

burn time 50 seconds.

Okay, we're dealing in pounds, but no matter.

1500 pounds of thrust is equivalent to 1500 pounds of fuel. But we only fed it 1200 pounds of fuel, so it would've burned 15/12 times longer if we had used 1500 pounds of fuel instead of 1200. That's 5/4 or 1.25 times more. We got 50 seconds of burn time out of 1200 pounds of fuel, so if we had used 1500 pounds of fuel it would've burned for 50 seconds x (1500 pounds/1200 pounds) = 62.5 seconds of specific impulse.

Our propellant's mass flow rate is 1500 pounds/ 62.5 seconds =  24 pounds per second. Conveniently, 1200 pounds in 50 seconds is also 24 pounds per second. Because the mass flow rate doesn't change.

Our exhaust velocity is 62.5 seconds x 32 ft/ sec²= 2,000 ft/ sec.

 Not a very efficient motor, even worse than a "Dawn" at sea level, but we have been able to derive all of it's characteristics by a simple test.

 See how it all ties together?

 

Edited March 20, 2018 by GoSlash27

[Cheif Operations Director]

2 hours ago, GoSlash27 said:

Okay, we're dealing in pounds, but no matter.

1500 pounds of thrust is equivalent to 1500 pounds of fuel. But we only fed it 1200 pounds of fuel, so it would've burned 15/12 times longer if we had used 1500 pounds of fuel instead of 1200. That's 5/4 or 1.25 times more. We got 50 seconds of burn time out of 1200 pounds of fuel, so if we had used 1500 pounds of fuel it would've burned for 50 seconds x (1500 pounds/1200 pounds) = 62.5 seconds of specific impulse.

Our propellant's mass flow rate is 1500 pounds/ 62.5 seconds =  24 pounds per second. Conveniently, 1200 pounds in 50 seconds is also 24 pounds per second. Because the mass flow rate doesn't change.

Our exhaust velocity is 62.5 seconds x 32 ft/ sec²= 2,000 ft/ sec.

 Not a very efficient motor, even worse than a "Dawn" at sea level, but we have been able to derive all of it's characteristics by a simple test.

 See how it all ties together?

 

 

 

give me a second to sort through all of it tommorow when I have more time and am better rested

Edited March 20, 2018 by Cheif Operations Director

[GoSlash27]

Is that a "good" facepalm (is it really that simple) or a "bad" facepalm (just shoot me now)? If it's a "bad" facepalm, we'll keep working at it.

Best,
-Slashy

[Cheif Operations Director]

4 minutes ago, GoSlash27 said:

Is that a "good" facepalm (is it really that simple) or a "bad" facepalm (just shoot me now)? If it's a "bad" facepalm, we'll keep working at it.

Best,
-Slashy

It's a Im exhausted and feel like I can't add 1+1 moth equation and need to review it tomorrow.  seriously I little sleep yesterday and have been busy today. Let me review it tommorow when I can do basic addition  

Or type math correctly 

[GoSlash27]

1 hour ago, Cheif Operations Director said:

It's a Im exhausted and feel like I can't add 1+1 moth equation and need to review it tomorrow.  seriously I little sleep yesterday and have been busy today. Let me review it tommorow when I can do basic addition  

Or type math correctly 

Okay. We've got all the time we need.

Best,
-Slashy

[sevenperforce]

Suppose Alice has an engine which produces 200 pounds of thrust while consuming 2 pounds of propellant every second, while Bob has an engine which produces 100 pounds of thrust while consuming 2 pounds of propellant every second. Which engine is more efficient?

Obviously, Alice's engine is more efficient, because they both consume the same amount of propellant every second but Alice's engine gets twice as much thrust. 

But what if you add Carol? Carol's engine produces 100 pounds of thrust while consuming 1 pound of propellant every second. Now her engine doesn't produce as much thrust as Alice's...but it does use less fuel. So which one is more efficient?

To answer this question, you ask a new question. "How much thrust (or 'impulse') is produced for each pound of fuel burned every second?" In other words, what is the impulse at a specific fuel consumption rate?

That's a question you can answer. Carol's engine produces 100 pounds of thrust while burning 1 pound of fuel per second; Alice's engine produces 200 pounds of thrust while burning 2 pounds of fuel per second. 100 pounds / 1 pound per second comes to 100 seconds, and 200 pounds / 2 pounds per second comes to 100 seconds. So the specific impulse of both engines is 100 seconds, so they have the exact same efficiency.

Note that you can't just drop "seconds" here. What if Dave comes along and his engine produces 50 pounds of thrust and consumes 2 pounds of propellant per minute? If you divide 50 by 2, you get 25...but if you follow the units, that's a specific impulse of 25 minutes, which converts to 1,500 seconds, far more efficient than the engines of Alice, Bob, and Carol. So the units are important.

And like I said before, the units are meaningful. If Dave's engine produces 50 pounds of thrust and has a specific impulse of 25 minutes, then you know it would take him 25 minutes to burn 50 pounds of propellant, or 50 minutes to burn 100 pounds of propellant, or 5 minutes to burn 10 pounds of propellant. Alice's engine would take 100 seconds to burn 200 pounds of propellant. Carol's engine would take 100 seconds to burn 100 pounds of propellant.

[Cheif Operations Director]

20 hours ago, GoSlash27 said:

Okay, we're dealing in pounds, but no matter.

1500 pounds of thrust is equivalent to 1500 pounds of fuel. But we only fed it 1200 pounds of fuel, so it would've burned 15/12 times longer if we had used 1500 pounds of fuel instead of 1200. That's 5/4 or 1.25 times more. We got 50 seconds of burn time out of 1200 pounds of fuel, so if we had used 1500 pounds of fuel it would've burned for 50 seconds x (1500 pounds/1200 pounds) = 62.5 seconds of specific impulse.

Our propellant's mass flow rate is 1500 pounds/ 62.5 seconds =  24 pounds per second. Conveniently, 1200 pounds in 50 seconds is also 24 pounds per second. Because the mass flow rate doesn't change.

Our exhaust velocity is 62.5 seconds x 32 ft/ sec²= 2,000 ft/ sec.

 Not a very efficient motor, even worse than a "Dawn" at sea level, but we have been able to derive all of it's characteristics by a simple test.

 See how it all ties together?

 

Ok I think I got it so you are using another formula other than 

axis of engine exhaust divided by fuel flow correct 

 

You did thrust (lbs) divided amount of propellant (Ibs) x burn time correct? 

 

 

8 hours ago, sevenperforce said:

Suppose Alice has an engine which produces 200 pounds of thrust while consuming 2 pounds of propellant every second, while Bob has an engine which produces 100 pounds of thrust while consuming 2 pounds of propellant every second. Which engine is more efficient?

Obviously, Alice's engine is more efficient, because they both consume the same amount of propellant every second but Alice's engine gets twice as much thrust. 

But what if you add Carol? Carol's engine produces 100 pounds of thrust while consuming 1 pound of propellant every second. Now her engine doesn't produce as much thrust as Alice's...but it does use less fuel. So which one is more efficient?

To answer this question, you ask a new question. "How much thrust (or 'impulse') is produced for each pound of fuel burned every second?" In other words, what is the impulse at a specific fuel consumption rate?

That's a question you can answer. Carol's engine produces 100 pounds of thrust while burning 1 pound of fuel per second; Alice's engine produces 200 pounds of thrust while burning 2 pounds of fuel per second. 100 pounds / 1 pound per second comes to 100 seconds, and 200 pounds / 2 pounds per second comes to 100 seconds. So the specific impulse of both engines is 100 seconds, so they have the exact same efficiency.

Note that you can't just drop "seconds" here. What if Dave comes along and his engine produces 50 pounds of thrust and consumes 2 pounds of propellant per minute? If you divide 50 by 2, you get 25...but if you follow the units, that's a specific impulse of 25 minutes, which converts to 1,500 seconds, far more efficient than the engines of Alice, Bob, and Carol. So the units are important.

And like I said before, the units are meaningful. If Dave's engine produces 50 pounds of thrust and has a specific impulse of 25 minutes, then you know it would take him 25 minutes to burn 50 pounds of propellant, or 50 minutes to burn 100 pounds of propellant, or 5 minutes to burn 10 pounds of propellant. Alice's engine would take 100 seconds to burn 200 pounds of propellant. Carol's engine would take 100 seconds to burn 100 pounds of propellant.

I just meant don't do that drawn out process you did with crossing it out and just divided and tack the unit in the end

[sevenperforce]

3 minutes ago, Cheif Operations Director said:

Ok I think I got it so you are using another formula other than 

axis of engine exhaust divided by fuel flow correct 

You did thrust (lbs) divided amount of propellant (Ibs) x burn time correct? 

Not quite.

"Axis of engine exhaust divided by fuel flow" doesn't tell anyone anything. Forget that.

Now, if you happen to know the average velocity of the exhaust molecules, you can use that to calculate specific impulse. But it's not exactly an easy thing to measure, so let's forget it, too.

"thrust (lbs) divided amount of propellant (lbs) x burn time" isn't quite there, either.

To begin with, you need to know the propellant flow rate. The propellant flow rate tells you how fast your rocket engine (or any other engine) uses fuel. A big engine probably uses fuel quickly; a small engine probably uses fuel more slowly. An efficient engine uses fuel more slowly than an inefficient engine of the same size. 

The propellant flow rate is very easy to calculate. Run your engine for a few minutes, then shut it off. How much fuel did you use? If your engine used 10 gallons of fuel in 5 minutes, then the flow rate is 2 gallons per minute. If it used 20 gallons of fuel in 5 minutes, then the flow rate is 4 gallons per minute. Assuming you know the density of your fuel, it should be trivial to convert gallons into pounds and minutes into seconds. If your fuel has approximately the same density as water, then 4 gallons per minute is the same as 0.53 pounds per second.

Once you know your propellant flow rate, then dividing your total thrust by your flow rate will tell you your specific impulse.

[GoSlash27]

1 hour ago, Cheif Operations Director said:

You did thrust (lbs) divided amount of propellant (Ibs) x burn time correct? 

I did thrust (lbs) divided by the mass flow rate (pounds per second). That yields the specific impulse.
I could've also done exhaust velocity (feet per second) divided by g0 (feet per second per second). That yields specific impulse too.

 Either approach is a valid measure of the efficiency of an engine, and both result in the same answer; specific impulse.

*edit* @sevenperforce beat me to it

Edited March 20, 2018 by GoSlash27