Specific Impulse Help

[Cheif Operations Director]

        Ve

Isp=——

          G

 

Isp specific_impulse (s)

ve average exhaust speed along the axis of the engine (m/s)

g standard acceleration due to gravity(for an object in a vacuum near the surface of the Earth : 9.80665 m/s^2) (m/s2)

 

Is this correct for calculating specific impulse  and what is "the average exhaust speed along the axis of the engine".

how do you calculate/find it, is that just pound of force or newtons?

[tater]

The velocity of the exhaust is V sub e in your equation. The units of I_sp are seconds.

You have to look up the exhaust velocity, or calculate one for some notional engine.

[sevenperforce]

Specific impulse is the number of pounds of thrust per pound of propellant you burn each second. Or the number of tonnes of thrust per tonne of propellant. 

If you have an engine which produces 100 tonnes of thrust (9800 N) while burning 1 tonne of propellant per second, then you divide 100 tonnes by 1 tonne / second; tonnes cancel, and you get 100 seconds. If you have an engine which produces ten pounds of thrust while burning one ounce of propellant per second, you recognize that one ounce is 1/16th of a pound, so you fix the units and you get 160 seconds. If you have an engine which produces 60,000 pounds of thrust while burning 3 gallons of fuel per second, and you know your fuel weighs 6.68 pounds per gallon, then you can do the math and find out that your specific impulse is 3000 seconds.

Of course, those numbers are only going to make sense on Earth, really, because you're using weight instead of mass. If you want to be a little more scientific about it, you can divide the thrust (in units of Newtons, preferably) by the mass flow (kg/sec) and you end up with a value measured in meters/second. For example, consider an engine which produces 800,000 N of force while consuming 262 kg of propellant every second. 800 kN / 252 kg/s works out to 3,050 m/s. 

As it turns out, 3,050 m/s happens to be the average exhaust velocity of the propellant coming out of the back end of a Merlin engine. That makes sense; the faster your exhaust is moving, the harder it is pushing against your nozzle as it leaves, and the more thrust you get out of it.

Of course measuring specific impulse in seconds looks a little cleaner. So you can convert m/s to s by dividing by 9.8 m/s², since that's how you convert between weight and mass. 3,050 m/s divided by 9.8 m/s² is 311 seconds, which is the specific impulse of a Merlin engine.

[PB666]

6 minutes ago, tater said:

The velocity of the exhaust is V sub e in your equation. The units of I_sp are seconds.

You have to look up the exhaust velocity, or calculate one for some notional engine.

 

Ve = Thrust/(δM/δt) where δM/δt = equal to the change of mass per unit of time in which the the average thrust is measured.

They can be obtained on Earth for MSL values in which the rocket is placed horizontally mounted to a vertical plate with stress sensors between the plate and the its ground mounts. When the rocket it throttled up the flow of liquids can be measured with a precalibrated flow meter most probably, after the mixture ratio is optimized only the flow of one (the most stable) component will be measured, e.g. kerosene. They can be extrapolated to vacuum based on the chamber pressure, the expansion ratio and atmospheric pressure (its a very rough estimate) and refined after monitoring performance in space. For example the Falcon 9 second stage engine (Merlin 1 D vacuum) only flies in vacuum, so that once it reaches its goal orbit, it might have fuel left that can be burn't for x-amount of seconds after payload detachment (such a deorbit burn need not be precise) so that the rocket equation can be used as a two step process to determine how much dV was obtained on two legs.

dV =  Ve * ln (starting mass / final mass)

 

[tater]

I know how to derive I_sp.

If he doesn't have the data for the exhaust velocity, he also likely doesn't have mass flow data, either. If he can look up one, he can look up the other.

[PB666]

2 minutes ago, tater said:

I know how to derive I_sp.

If he doesn't have the data for the exhaust velocity, he also likely doesn't have mass flow data, either. If he can look up one, he can look up the other.

Sorry quoted the wrong post.

 

[tater]

That it simplifies to c/g (usually that's how we wrote it, where c is exhaust velocity) is nice, since we were often talking about notional engines (what if you could throw atoms out the back, really fast? (ion) ) where we did not have specific mass-flow vs thrust data.

[sevenperforce]

25 minutes ago, tater said:

That it simplifies to c/g (usually that's how we wrote it, where c is exhaust velocity) is nice, since we were often talking about notional engines (what if you could throw atoms out the back, really fast? (ion) ) where we did not have specific mass-flow vs thrust data.

I like explaining it to newbies in terms of mass flow because it's more intuitive than momentum impulse and allows an immediate comparison of jets and rockets without having to worry about nonsense like "effective exhaust velocity". He seemed confused by "the axis of the engine" so I thought it best to stick to thrust and mass flow.

[PB666]

22 minutes ago, sevenperforce said:

I like explaining it to newbies in terms of mass flow because it's more intuitive than momentum impulse and allows an immediate comparison of jets and rockets without having to worry about nonsense like "effective exhaust velocity". He seemed confused by "the axis of the engine" so I thought it best to stick to thrust and mass flow.

The problem with ISP is weight is relative to earth's surface and there in no cogent metric equivalent (the slug, not an international unit). When you are in orbit you have no concern of surface weight at all.

 

[Cheif Operations Director]

1 hour ago, sevenperforce said:

Specific impulse is the number of pounds of thrust per pound of propellant you burn each second. Or the number of tonnes of thrust per tonne of propellant. 

If you have an engine which produces 100 tonnes of thrust (9800 N) while burning 1 tonne of propellant per second, then you divide 100 tonnes by 1 tonne / second; tonnes cancel, and you get 100 seconds. If you have an engine which produces ten pounds of thrust while burning one ounce of propellant per second, you recognize that one ounce is 1/16th of a pound, so you fix the units and you get 160 seconds. If you have an engine which produces 60,000 pounds of thrust while burning 3 gallons of fuel per second, and you know your fuel weighs 6.68 pounds per gallon, then you can do the math and find out that your specific impulse is 3000 seconds.

Of course, those numbers are only going to make sense on Earth, really, because you're using weight instead of mass. If you want to be a little more scientific about it, you can divide the thrust (in units of Newtons, preferably) by the mass flow (kg/sec) and you end up with a value measured in meters/second. For example, consider an engine which produces 800,000 N of force while consuming 262 kg of propellant every second. 800 kN / 252 kg/s works out to 3,050 m/s. 

As it turns out, 3,050 m/s happens to be the average exhaust velocity of the propellant coming out of the back end of a Merlin engine. That makes sense; the faster your exhaust is moving, the harder it is pushing against your nozzle as it leaves, and the more thrust you get out of it.

Of course measuring specific impulse in seconds looks a little cleaner. So you can convert m/s to s by dividing by 9.8 m/s², since that's how you convert between weight and mass. 3,050 m/s divided by 9.8 m/s² is 311 seconds, which is the specific impulse of a Merlin engine.

For your one tonne example:

Do 100 tonnes divided by one tonne of propellant divided by burn time? 

So if I had 10 tonnes of propellant in the rocket it would be

100÷1÷10 = 10 that doesn't seem right 

[Cheif Operations Director]

Is it This

 

          Ve

isp= ------

           G

       100

X= --------

            1

 

X=100

Again where does gravity play into this?

[sevenperforce]

31 minutes ago, Cheif Operations Director said:

For your one tonne example:

Do 100 tonnes divided by one tonne of propellant divided by burn time? 

So if I had 10 tonnes of propellant in the rocket it would be

100÷1÷10 = 10 that doesn't seem right 

No, it's 100 tonnes divided by your fuel flow rate. The amount of propellant in your rocket is unrelated; that's the great thing about specific impulse. Specific impulse tells you about the performance of your engine, independent of how big a stage it is attached to.

Specific impulse is thrust divided by propellant flow rate.

Your propellant flow rate is usually measured in "pounds per second" or "tonnes per second". Suppose your engine burns 1 tonne of propellant every second at full throttle. In that case, fuel flow rate is 1 tonne per second, or 1 tonne/sec.

Divide tonnes by tonnes/sec and you get seconds, because the units cancel and the reciprocal flips.

You just have to make sure your units cancel properly.

If your thrust is measured in kN rather than tonnes, then you need to work in the force of gravity to figure out how many tonnes of thrust you're producing.

Are you very good at algebra? Rocket science (at least, at this level) doesn't require calculus, but ya gotta be pretty darn good at algebra.

[Starman4308]

18 minutes ago, Cheif Operations Director said:

Again where does gravity play into this?

It goes back to the early 20'th century. People in the US were using feet/second for their exhaust velocities, and people using reasonable units were using meters/second.

To simplify things, the decided to divide exhaust velocity by the standard gravitational acceleration at Earth's surface, about 32 ft/sec^2 or 9.81 m/sec^2, and instead discuss specific impulse, which is in seconds.

So, if you have an engine with a specific impulse of 315 seconds, exhaust velocity is either 10080 ft/sec or 3090 m/sec.

[sevenperforce]

thrust = exhaust velocity * fuel mass flow rate

isp = thrust / fuel weight flow rate

thus 

isp = exhaust velocity / acceleration of gravity

4 minutes ago, Starman4308 said:

It goes back to the early 20'th century. People in the US were using feet/second for their exhaust velocities, and people using reasonable units were using meters/second.

To simplify things, the decided to divide exhaust velocity by the standard gravitational acceleration at Earth's surface, about 32 ft/sec^2 or 9.81 m/sec^2, and instead discuss specific impulse, which is in seconds.

So, if you have an engine with a specific impulse of 315 seconds, exhaust velocity is either 10080 ft/sec or 3090 m/sec.

To be a pedant -- I believe specific impulse came first, as the reciprocal of thrust-specific fuel consumption. "Pounds force divided by pounds per second" is a very straightforward construct.

Edited March 14, 2018 by sevenperforce

[tater]

Well, for someone who can measure stuff, it’s ideal. You measure force in test stand, and you can watch the props drain out over time (and you know that mass). Very empirical.

If you don’t have a test stand and a real rocket, I still think c/g is easier to deal with.

[sevenperforce]

9 minutes ago, tater said:

Well, for someone who can measure stuff, it’s ideal. You measure force in test stand, and you can watch the props drain out over time (and you know that mass). Very empirical.

Indeed.

Thrust-specific fuel consumption curves arose because airbreathing engines do not run at the same efficiency at all speeds, all throttle settings, and all altitudes. Users of Imperial units found it very natural to look at pounds-per-second of fuel flow when the engine is producing a certain amount of thrust. But dividing fuel consumption by thrust ends up giving you seconds^-1, which makes very little sense. However, if you take thrust-specific fuel consumption as constant (invariant with respect to throttle, at least across the test domain), then you can flip your axes and talk about fuel-consumption-specific thrust, which is thrust divided by fuel consumption, which gives you units of seconds. Much easier to work with.

[Cheif Operations Director]

1 hour ago, sevenperforce said:

No, it's 100 tonnes divided by your fuel flow rate. The amount of propellant in your rocket is unrelated; that's the great thing about specific impulse. Specific impulse tells you about the performance of your engine, independent of how big a stage it is attached to.

Specific impulse is thrust divided by propellant flow rate.

Your propellant flow rate is usually measured in "pounds per second" or "tonnes per second". Suppose your engine burns 1 tonne of propellant every second at full throttle. In that case, fuel flow rate is 1 tonne per second, or 1 tonne/sec.

Divide tonnes by tonnes/sec and you get seconds, because the units cancel and the reciprocal flips.

You just have to make sure your units cancel properly.

If your thrust is measured in kN rather than tonnes, then you need to work in the force of gravity to figure out how many tonnes of thrust you're producing.

Are you very good at algebra? Rocket science (at least, at this level) doesn't require calculus, but ya gotta be pretty darn good at algebra.

Ok how do you calculate this is a solid rocket the fuel flow rate. It would simpily the burn time right. So let's say a rocket engine burns for 6 seconds is how much fuel is consumed in 1 second is that number correct? Also how would you figure coring if it was solid fuel. For example if you have an increase in thrust an that is how the core is designed.

[PB666]

Loaded mass = Empty mass + Fuel mass

Flow rate = Fuel mass / total burn time

dV = Ve * ln (Loaded mass/Empty Mass)

Ve = dV / ln(Loaded mass/Empty Mass)

Therefore if you know the loaded mass and the empty mass, and if you measure the change of velocity. ( A good place to do this is from the orbit of a large gas giant where the trajectory would be flat over the burn)

Also you can determine thrust (say at 10 seconds after the burn starts).

[sevenperforce]

12 hours ago, Cheif Operations Director said:

Ok how do you calculate this is a solid rocket the fuel flow rate. It would simpily the burn time right. So let's say a rocket engine burns for 6 seconds is how much fuel is consumed in 1 second is that number correct? Also how would you figure coring if it was solid fuel. For example if you have an increase in thrust an that is how the core is designed.

Specific impulse is easier to calculate for liquid rockets because thrust and fuel flow can be controlled. Solid rockets are not nearly so refined.

Just an aside -- are you talking about a commercial solid rocket booster, a hobby/amateur solid rocket motor, or a homemade rocket?

If you can assume roughly constant thrust on your rocket, then you can simply take the total propellant and divide by burn time. For example, let's say you have a hobby rocket motor which produces 20 pounds of thrust, weighs 1.1 pounds "full", and weighs 0.1 pounds "empty". Let's say that it takes 6 seconds to burn out. You have one pound of propellant, so your mass flow is 0.167 pounds per second or 0.167 lb/sec. So you take the (constant) thrust of 20 lb, divide this by the mass flow of 0.167 lb/sec, and you get:

20 lb / 0.167 lb/sec = 20 lb * 1 sec / 0.167 lb = 20 sec / 0.167 = 120 seconds

If you cannot assume constant thrust on your rocket, then you have to figure out the burn rate and thrust curve. You can do this theoretically if the coring is simple, like a cylinder, but for more complex coring you end up needing some more advanced math. Or you end up just doing it experimentally.

[PB666]

56 minutes ago, sevenperforce said:

Specific impulse is easier to calculate for liquid rockets because thrust and fuel flow can be controlled. Solid rockets are not nearly so refined.

Just an aside -- are you talking about a commercial solid rocket booster, a hobby/amateur solid rocket motor, or a homemade rocket?

If you can assume roughly constant thrust on your rocket, then you can simply take the total propellant and divide by burn time. For example, let's say you have a hobby rocket motor which produces 20 pounds of thrust, weighs 1.1 pounds "full", and weighs 0.1 pounds "empty". Let's say that it takes 6 seconds to burn out. You have one pound of propellant, so your mass flow is 0.167 pounds per second or 0.167 lb/sec. So you take the (constant) thrust of 20 lb, divide this by the mass flow of 0.167 lb/sec, and you get:

20 lb / 0.167 lb/sec = 20 lb * 1 sec / 0.167 lb = 20 sec / 0.167 = 120 seconds

If you cannot assume constant thrust on your rocket, then you have to figure out the burn rate and thrust curve. You can do this theoretically if the coring is simple, like a cylinder, but for more complex coring you end up needing some more advanced math. Or you end up just doing it experimentally.

Make a half or quarter scale length, some length in which the final speed is less than x.

Then measure h at apogee and make some assumption about burn time. From that you can determine dV by determining velocity at end burn based on h and final h (or better the velocity at end burn).

Vend burn = (a * burn time) - (g * burn time) = (a - g ) burn Time = V or a - g = V/burn time

T/m = a/burntime + g    m = average mass. If the PL of the rocket is sufficiently large then velocity will be low and the relative mass difference is small, you can calculate T * time pretty easily. 
 

A simpler and far less sophisticated way to do it is take 4 fishing scales (they need to be tuned to the rockets theoretical thrust) and mount them to the rocket in such a way that when it fires you can see the weight and multiply it be 4.

BTW, playing with SRBs is dangerous.

Edited March 15, 2018 by PB666

[sevenperforce]

P.S. Note that in the above example, you can convert the specific impulse of 120 seconds into the exhaust velocity by multiplying with the acceleration of gravity, g₀. You get 1,177 m/s, which is about the right exhaust velocity for a hobby rocket motor without a good nozzle.

[Cheif Operations Director]

5 hours ago, sevenperforce said:

Specific impulse is easier to calculate for liquid rockets because thrust and fuel flow can be controlled. Solid rockets are not nearly so refined.

Just an aside -- are you talking about a commercial solid rocket booster, a hobby/amateur solid rocket motor, or a homemade rocket?

If you can assume roughly constant thrust on your rocket, then you can simply take the total propellant and divide by burn time. For example, let's say you have a hobby rocket motor which produces 20 pounds of thrust, weighs 1.1 pounds "full", and weighs 0.1 pounds "empty". Let's say that it takes 6 seconds to burn out. You have one pound of propellant, so your mass flow is 0.167 pounds per second or 0.167 lb/sec. So you take the (constant) thrust of 20 lb, divide this by the mass flow of 0.167 lb/sec, and you get:

20 lb / 0.167 lb/sec = 20 lb * 1 sec / 0.167 lb = 20 sec / 0.167 = 120 seconds

If you cannot assume constant thrust on your rocket, then you have to figure out the burn rate and thrust curve. You can do this theoretically if the coring is simple, like a cylinder, but for more complex coring you end up needing some more advanced math. Or you end up just doing it experimentally.

I'm not understanding this simpily because of the lack of mathematical signs. 

20 lb ÷ 0.167 ÷ what exactly = 20 Ib and that's where you completely lost be I'm not following simpily because their are no math symbols. Can you please use words instead.

 

[1101]

17 minutes ago, Cheif Operations Director said:

I'm not understanding this simpily because of the lack of mathematical signs. 

20 lb ÷ 0.167 ÷ what exactly = 20 Ib and that's where you completely lost be I'm not following simpily because their are no math symbols. Can you please use words instead.

 

I think it is 20 pounds divided by 0.167 pounds burned per second is equal to a specific impulse of 120 seconds.

 

Or, 20 lb/0.167lb.s^-1 = 120s

[sevenperforce]

35 minutes ago, Cheif Operations Director said:

I'm not understanding this simpily because of the lack of mathematical signs. 

20 lb ÷ 0.167 ÷ what exactly = 20 Ib and that's where you completely lost be I'm not following simpily because their are no math symbols. Can you please use words instead.

Well, / is the same as ÷, haha.

But here, I wrote it all out in as plain of notation as I know how:

Spoiler

 

 

Edited March 15, 2018 by sevenperforce

[Cheif Operations Director]

On 3/15/2018 at 4:02 PM, sevenperforce said:

Well, / is the same as ÷, haha.

But here, I wrote it all out in as plain of notation as I know how:

  Hide contents

 

 

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? 

 

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)