Converting motor N-S to Joules

[les]

Not certain if this is the right forum - maybe it actually belongs in the restricted area.

Someone at work asked me this question knowing I'm involved in rocketry....

Our company is involved with aircraft electrification.
There is a requirement to use an enclosure with appropriate coatings to withstand a large lithium-ion battery pack cooking off.

A "pod" was developed to perform testing outside the building for safety purposes.
They remotely induce an overload into some batteries with an appropriately coated plate.

They asked me how many joules there are in a rocket motor. The thought is to use rocket motor(s) to emulate the batteries.
The motor would be secured with the nozzle aimed at the plate.

I know I've seen blast deflector plates get a hole burned through them if the rocket hangs on the rod.

 

[rcktnut]

Would it not be easier to use a propane torch (small 14oz cylinder) and determine the amount of propane expended then convert to btu's and then convert to Joules? You know how rocket motors perform, no 2 are exactly the same.

 

[Last Ox]

This is a little like asking the horsepower of a jet engine - rocket fuels are usually rated by specific impulse, which doesn't translate directly to delivered energy. You can work out an orders of magnitude answer, which may be good enough.

The combustion energy of propellants might run something like 3-10kj/gm. Motors range from something like 0.7 to 2.5 Nsec per gram of propellant. By definition, much of the energy (as much as possible!) developed by a rocket engine is as kinetic energy of the exhaust stream vice heat energy equivalent to I squared R discharge of a battery.

 

[r1dermon]

.

 

[aerostadt]

N-s is a measurement of total impulse. Joules is a unit of energy.

Impulse is force integrated over time. Joule is force integrated over distance.

 

[cls]

Exactly, these are "incommensurate units"

Look at the SI definitions of the units:

Ns = kg * m * s^-1. Kilogram neters per second

Joule = kg * m^2 * s^-2. So, like newtons times meter per second.

Not comparable.

 

[Last Ox]

That said, because Ns/g propellant is well behaved (lower for motors with lower expected efficiency, higher for higher energy propellants, but generally consistent), Ns of a motor is broadly proportional to its propellant energy.

There are bigger issues. A torching battery delivers the vast majority of its energy as heat, a rocket taps off a significant amount as kinetic energy of reaction products. I suspect the question came up because motors deliver a lot of energy very quickly, as opposed to a torch or similar simulant. But the amount of energy is also an issue. I came across a published figure of about 350kj/Ah capacity for complete combustion of a Li battery (or more than a kilo of black powder to replicate burning of a 10,000mAh cell phone recharging brick - my last one weighs 180gms). Not being a monopropellant, the combustion energy per gram is much higher than for a rocket motor, and at the temperatures of a battery fire, most of the battery's weight is combustible. To simulate a fire in a single segment of a battery big enough to power an aircraft would require quite a rocket motor.

There is probably no need to deal with expensive engineering (and significant thrust) of a large motor - what they probably really need is just the propellant. I'll bet with a bit of homework they could work up a test fixture with commercial smokeless powder that would be repeatable, describable numerically, and deliver the required heat power. A non-propulsive grain could be ginned up with the usual recipes, but may not be as safe or easy to quantify as a commercial propellant (some smokeless powder manufacturers publish heat of combustion, and in the interest of accuracy all produce a highly consistent product, and commercial powder reliably deflagrates rather than detonates).

 

[jderimig]

There isn't direct conversion but a rocket motor sitting there has potential energy that can be specified in Joules. But not all of that potential energy can be converted into work. The N-s specification can through simulation get this answer for you. For example do a flight simulation of your rocket in-flight in a vacuum with in zero gravity. At motor burnout determine the kinetic energy of the rocket. That is how many Joules your rocket motor delivered to the system in the conversion of its potential energy to kinetic energy.

 

[aerostadt]

There isn't direct conversion but a rocket motor sitting there has potential energy that can be specified in Joules. But not all of that potential energy can be converted into work. The N-s specification can through simulation get this answer for you. For example do a flight simulation of your rocket in-flight in a vacuum with in zero gravity. At motor burnout determine the kinetic energy of the rocket. That is how many Joules your rocket motor delivered to the system in the conversion of its potential energy to kinetic energy.

In another sense that is not exactly correct, because some of the chemical energy goes into the exhaust, which is different from the mechanical energy. I reminded of the energy equation of the Navier-Stokes equations. It contains both the mechanical and the thermodynamic energy of the fluid. In principle momentum (F=ma) and mechanical energy methods are the same. Just remember not to include thermodynamic energy. One has to be careful on these matters on keeping things straight. It depends on what the user wants. If the user wants the total energy of the propellant, they could weigh the propellant that gives a certain total impulse and then multiply that value by the chemical energy per unit weight. Finding the mechanical or kinetic energy of the rocket will be something less, because not all the chemical energy is converted into the kinetic energy of the rocket.

 

[jderimig]

In another sense that is not exactly correct, because some of the chemical energy goes into the exhaust, which is different from the mechanical energy. I reminded of the energy equation of the Navier-Stokes equations. It contains both the mechanical and the thermodynamic energy of the fluid. In principle momentum (F=ma) and mechanical energy methods are the same. Just remember not to include thermodynamic energy. One has to be careful on these matters on keeping things straight. It depends on what the user wants. If the user wants the total energy of the propellant, they could weigh the propellant that gives a certain total impulse and then multiply that value by the chemical energy per unit weight. Finding the mechanical or kinetic energy of the rocket will be something less, because not all the chemical energy is converted into the kinetic energy of the rocket.

The motor's impulse rating if derived from test stand data already has losses subtracted out.

It is also not exactly correct for another reason.

Total impulse is equivalent to the change in momentum of the body (in a reversible system), So you can convert impulse to delta mv. So you have a motor with a N-S of 10, and a rocket with a mass of 1, this rocket will have a max potential velocity of 10. Kinetic energy = 1/2(1)100 = 50 Joules

But the same motor with rocket of mass 10, will have a max velocity of 1. Kinetic energy = 1/2(10)(1)= 5 Joules.

So you are right. Total impulse and Joules of energy are apples-carburetors.

 

[Tom Flint]

You can calculate the kinetic power of the rocket's exhaust using the formula 1/2 times the mass flow rate times the square of the change in velocity of the propellant as is exits the rocket. For a rocket that is not moving like a motor being static fired, this calculation is straightforward. For a moving rocket, things get a bit more interesting. Since power in watts equals joules per second, you can then calculate the total energy that the motor imparted to its exhaust jet over the entire burn time of the motor.

 

[rocket_troy]

You can calculate the kinetic power of the rocket's exhaust using the formula 1/2 times the mass flow rate times the square of the change in velocity of the propellant as is exits the rocket. For a rocket that is not moving like a motor being static fired, this calculation is straightforward. For a moving rocket, things get a bit more interesting. Since power in watts equals joules per second, you can then calculate the total energy that the motor imparted to its exhaust jet over the entire burn time of the motor.

Well, really, that's what you should be calculating anyway.

Isp = total impulse / fuel weight
effective exhaust velocity (c) = Isp * g
Use the kinetic energy equation to convert c (m/s) into joules
KE = (c^2)/2
Multiply by mass flow or total mass to gather total energy. Ball park stuff, but close enough for purpose.

TP

 

[jqavins]

There's a lot of overthinking going on here.

This is a little like asking the horsepower of a jet engine

No, it's more like trying to convert minutes to meters, or lumens to pounds, for the reason others have stated. Trying to convert Ns to J is just nonsense, and not worth considering.

The combustion energy of propellants might run something like 3-10kj/gm.

And stop right there. Look up or ask the manufacturer what that number is for a particular propellent, multiply by the propellent mass, and there's your answer. For rocketry, we may care about how that relates to the impulse generated (not converts directly, but relates) but for using a motor to simulate a battery failure none of that matters.

As an alternative, you might consider getting a can of BP and weighing out a pile of the appropriate size.

 

[jsdemar]

Another interesting measure of combustion energy capacity is called impetus. It can be theoretically predicted for a solid mixture from thermodynamic data, evolved gas constants, etc. But, it's most useful as a measured value which includes practical limitation and losses such as particle size, and heat transferred to the enclosure. The (mixed) units for impetus best explain what it is: PSI-in^3/gram. It's the solid's ability to pressurize a volume when burned within an enclosed space.

SI units are: Pa-m^3/kg. Or: (N/m^2)*m^3/kg = N-m/kg = Joules/kilogram. In other words, the potential energy in the form of heated pressurization gasses per mass of the burned solid.

One might think of this as a black powder ejection charge pressurizing a parachute compartment. After converting a solid's chemical energy into gas energy, there's an impetus to do work. No work has been done yet until the pressurization force moves the parachute and nosecone. If the gas is left enclosed, the energy will eventually transfer as heat to what it's held in (energy conserved, no work is done).

Applying this to a composite propellant, the solid mixture produces combustion gasses within an enclosed volume and has the potential (impetus) to do work. There are heat transfer losses and work done to move the gas products (and remaining condensed products) out the nozzle. In a static test, no additional mechanical work is done. The energy is transferred to the expanding gasses and as heat. In a rocket on the pad, work is done by accelerating the mass of the rocket and some energy is transferred (lost) by the aerodynamic losses and as the expanding exhaust gasses cool. This is a simplification, but that gives the major energy transfer modes.

I used the experimental measure of impetus to compare pyrogens in my igniter study. Combined with the combustion temperature and time-to-peak-pressure, the impetus value (pressurization per gram) shows the pyrogens' relative capabilities as a "motor starter".

If someone wishes to play loosely with units, total impulse (N-sec) could be substituted into the Impetus measure: Impetus = N-m/kg = (N-sec)*(m/sec)/kg = TotalImpulse * m/sec/kg. With m/sec the same units as cstar (characteristic velocity of exhaust gasses), more hand waving produces:
Impetus = Joules/kilogram = TotalImpulse * cstar / kilogram.
Which gives:
Joules = Total Impulse * cstar / kilogram of propellant.

That will give a rough relative idea of the energy in the propellant given the total impulse. You need to know the characteristic velocity and the mass of the propellant.

 

[rocket_troy]

No, it's more like trying to convert minutes to meters, or lumens to pounds, for the reason others have stated. Trying to convert Ns to J is just nonsense, and not worth considering.

Getting back to the OP: Les was basically asking how much energy is there within a given rocket motor. If we take the assumption he's happy enough to settle on *useable* energy for propulsion, the calculation is rather straightforward as per my previous post.
As rocket motors essentially work on energy to ultimately provide the work they produce (yes, like batteries), it's not quite the apple-oranges different universes you're making it out as. A chemical rocket propellant fundamentally acts like a chemical battery. They both use the chemical energy within the storage - the potential energy - to provide the impetus to do the work. The efficiency to which that conversion takes place is obviously never 100% for either.
If you wanted to know the theoretical *total* chemical energy available within the propellant ie. both usable, unused and unusable for propulsion, that, whilst computationally complex to calculate, is available from my JeffPep spreadsheet (probably the only tool in the world available to report that). However, for the sake of this exercise, you shouldn't need that.
The whole Ns diversion was an unfortunate side-track of the narrative by... perhaps... naivety or preconceptions IMHO.

TP

 

[cls]

They both use the chemical energy within the storage - the potential energy - to provide the impetus to do the work.

Show me units equations for these things: chemical energy, potential energy, impetus, work.

 

[jsdemar]

Show me units equations for these things: chemical energy, potential energy, impetus, work.

A battery's stores energy in Joules = C x V = charge in coulombs times volts. The electrolyte chemistry provides a band gap voltage and capacity to hold a charge per kg. So, the impetus analogy here is C V/mass (coulomb-volts/kg which is joules/kg). The battery can deliver this energy as heat to a resistor at some level of efficiency. Internal resistance of the battery causes the main loss, but there are also temperature dependencies in the ev chemistry. But, no work is done (loosely related to a rocket motor in a test stand). To do work, the electrical energy has to be converted to mechanical energy... motor, pump, etc., with additional losses in transferring the energy and losses in the device (resistive loss, mechanical friction, fan for cooling, and so on).

I'm guessing my posts are falling into the /TLDR/ category.

 

[rocket_troy]

Show me units equations for these things: chemical energy, potential energy, impetus, work.

you don't really need to worry about impetus or even work to gather the energy.
Working backwards:
Total impulse is the product of Isp and total mass flow (ie. propellant mass)
Isp *is* the effective exhaust velocity (just divided by a *constant* g)
note: effective exhaust velocity is the exhaust velocity with the velocity contribution of pressure thrust (negative for Pe<>Pa)
The Ns of total impulse is a product of propellant mass in N and the seconds of Isp
so, we can say It= propellant mass in N * m/s/g or propellant mass in Kg * m/s
"c" (effective exhaust velocity) in m/s
we can use the KE equation to gather total kinetic energy KE = 1/2 propellant mass in Kg * c^2 in m/s
That's for the entire propellant which is assumed to be averaged over the burn duration if we're starting with total impulse and propellant mass.
So, that total KE is equal to the amount of total chemical energy available for propulsion. In this instance, I'm saying for the sake of this exercise that usable chemical energy is your usable potential energy.
.
TP

 

[jsdemar]

So, that total KE is equal to the amount of total chemical energy available for propulsion. In this instance, I'm saying for the sake of this exercise that usable chemical energy is your usable potential energy.
.
TP

The total impulse is derived from a test firing and therefore includes the losses in the chemical process and the losses in the motor as heat. It also includes the gain derived form the nozzle's thrust coefficient. So, the energy calculated from a test firing is the useful potential energy to do mechanical work.

 

[rocket_troy]

The total impulse is derived from a test firing and therefore includes the losses in the chemical process and the losses in the motor as heat. It also includes the gain derived form the nozzle's thrust coefficient. So, the energy calculated from a test firing is the useful potential energy to do mechanical work.

Indeed. I think - again, in the context of the OP - those losses and gains should be included, otherwise, you find yourself heading towards a thought exercise in the theoryland of total available enthalpy of the propellant etc... which is never really a metric used for any such (macro perspective) system.

TP

 

[jderimig]

The total impulse is derived from a test firing and therefore includes the losses in the chemical process and the losses in the motor as heat. It also includes the gain derived form the nozzle's thrust coefficient. So, the energy calculated from a test firing is the useful potential energy to do mechanical work.

I forgot much of my physics and kinetics, but bear me out on this question.

On a test stand the rocket motor does no work, so all of its potential energy goes into heat.

Let the motor fly and some kinetic energy is produced.

Does that mean a flying motor heats the world less than a test stand motor?

 

[rocket_troy]

I forgot much of my physics and kinetics, but bear me out on this question.

On a test stand the rocket motor does no work, so all of its potential energy goes into heat.

Let the motor fly and some kinetic energy is produced.

Does that mean a flying motor heats the world less than a test stand motor?

Yes.

Well, I think so

Certainly true for the exhaust, but we also need to consider the heating the vehicle itself provides to the atmosphere, but that will be influenced by things like the cd and air density independently of exhaust interactions.

TP

 

[cls]

units are: Pa-m^3/kg. Or: (N/m^2)*m^3/kg = N-m/kg = Joules/kilogram. In other words, the potential energy in the form of heated pressurization gasses per mass of the burned solid.

Ummm... How did you get joules (N-m)

One might think of this as a black powder ejection charge pressurizing a parachute compartment. After converting a solid's chemical energy into gas energy, there's an impetus to do work. No work has been done yet

When no work has been done... Where did the extra m come from for the N-m above?

 

[jsdemar]

Ummm... How did you get joules (N-m)

When no work has been done... Where did the extra m come from for the N-m above?

Subtract 'm' unit powers in the numerator and denominator: (N/m^2)*m^3/kg = N-m/kg = Joules/kilogram
This leaves the 'N' and one 'm' in the numerator, with just kg in the denominator.

There doesn't have to be work done for the impetus units to reduce like this. There's a static pressure generated from the combustion gasses: Pounds per in^2 or Pascals per m^2. This is contained in a static volume: in^3 or m^3. So, all of the distance units come from static measures, thus no work is done.

 

[jqavins]

Getting back to the OP: If you wanted to know the theoretical *total* chemical energy available within the propellant ie. both usable, unused and unusable for propulsion, that, whilst computationally complex to calculate, is available from my JeffPep spreadsheet (probably the only tool in the world available to report that). However, for the sake of this exercise, you shouldn't need that.

No, that is exactly what the OP does need. His coworkers want to use a burning motor as a simulant for a burning battery. The total energy released is the only thing he cares about. Which is why a naked pile of BP is probably a better choice for the application than a motor.

The opening question - how do you convert the (readily available) impulse value to the (desired) energy value - has no answer, and there's the end of it. All the rest is the sound and fury of a burning motor, signifying nothing.

 

[jsdemar]

No, that is exactly what the OP does need. His coworkers want to use a burning motor as a simulant for a burning battery. The total energy released is the only thing he cares about. Which is why a naked pile of BP is probably a better choice for the application than a motor.

The opening question - how do you convert the (readily available) impulse value to the (desired) energy value - has no answer, and there's the end of it. All the rest is the sound and fury of a burning motor, signifying nothing.

Hey, Joe: I've given the answer for both the energy released by the static firing of a rocket motor or the energy released by a pile of pyrogen.

Now @les needs to know the energy released by the burning Lithium battery to find the equivalent. That's not the electrical energy stored in the battery, but the chemical combustion energy released by the content of the battery once it it hits it's self-ignition temperature. Other factors are the combustion temperature and the combustion pressure of the lithium goop contained in the battery case. This will help choose an equivalent propellant/pyrogen composition and how it should be contained. All reasonable to calculate or find experimentally.

 

[jqavins]

That's all true. My point has been that calculating it by starting with the impulse and using the average exhaust speed, g, and other stuff is the long way around, and is probably constrained to particular situations.

Here's how the conversation should go:

Coworker: Hey Les, we need to simulate a battery fire by burning something to get X Joules. Can I use one of those rocket motors you use? Which one would be best?​

Les: Well, if you're going to use a black powder motor, the specific energy of the propellant is about [Les does a Google search] about 3 MJ/kg. So you'll need about X Joules over 3 MJ/kg, or Y grams. You could just get a can of BP and measure out what you need. But if you want to use a motor, let's see which one comes closest.​

[Les then spends a couple of minutes at thrustcurve.org]​

Scenario 1:​

Les: Here's you go, it looks like a <whatever motor> is your best bet.​

​

Scenario 2:​

Les: Oh, it looks like BP motors don't come that big. Let's look at composite motors. [Les does another Google search] It says here that APCP has a specific energy of about 31 MJ/kg. That's probably an average, because the motors don't all use exactly the same formula. Anyway, we need a tenth as much APCP as BP, let's see what motor is the best fit that way.​

[Les spends several minutes on thrustcurve.org, as there are many more options with composite motors]​

Scenario 2A:​

Here you go.​

Scenario 2B:​

Well, you could use one of these, but that's a damn expensive motor. This really doesn't look like a good idea. Sorry.​

 

[cls]

There doesn't have to be work done for the impetus units to reduce like this. There's a static pressure generated from the combustion gasses: Pounds per in^2 or Pascals per m^2. This is contained in a static volume: in^3 or m^3. So, all of the distance units come from static measures, thus no work is done.

Hmmm... Maybe... I'll get another cup of coffee and cogitate about this.

 

[cls]

Here's how the conversation should go:

Scenario 3:

Coworker: Hey Les, we need to simulate a battery fire by burning something to get X Joules.

Les: ok great, Let's get a Kill-a-Watt meter, an electric furnace, and one of those infrared thermometers.

 

[jsdemar]

You're all making this more difficult than it needs to be. But, of course it's the Internet. ;-)

Just replace the combustion energy of one solid compound (LiPF6, for example) with another one that is cheaper than the equivalent battery.

Use BP or APCP or whatever. I'd prefer something safer than BP, such as an APCP grain.

Contain it a similar way as the battery is designed. Check the combustion speed and temperature of the replacement compound to make sure it's close enough for test purposes.

Think of this as replacing black powder in an ejection charge with an equivalent energetic compound. I've done this with granular BKNO3+Viton. The impetus of BKNO3 is about twice that of BP, so you need half as much by mass. Containment is not as important because BKNO3 combustion is not sensitive to ambient pressure.