How much gunpowder does a rocket need?

  • Thread starter Thread starter 168918791999
  • Start date Start date
  • Tags Tags
    Rocket
AI Thread Summary
The discussion centers on calculating the amount of gunpowder needed for a rocket to reach a height of 75 meters, with initial calculations suggesting only 0.2 grams is required. However, participants highlight that real rockets require more gunpowder due to inefficiencies, drag, and the need for kinetic energy to overcome gravity. The average rocket operates at about 10% efficiency, meaning much of the energy from the gunpowder is lost to exhaust and drag forces. Additionally, the conversation emphasizes the importance of burn patterns and optimizing fuel usage to improve performance. Ultimately, the calculations must account for momentum and energy losses to provide a more accurate estimate of the required gunpowder.
168918791999
Messages
23
Reaction score
0

Homework Statement


m=100gr
g=9.81 m/s2
H=75m
Combustion of gunpowder = 2.7x10^6 J/Kg

This would mean we only need 0.2gr of gunpowder to get 75m high.
Why is there in a normal rocket more gundpowder?
Our physics and chemistry teacher checked my calculations but couldn't find a mistake, so what is the problem?

Homework Equations


U=m*g*h
---------
U=Joules
M=Kg
g=gravitational acceleration
H=hight in meters

Combustion of gunpowder = 2.7x10^6 J/Kg

Average rocket is 10% efficient

The Attempt at a Solution


U=0.1*9.81*75
U=73.57500J
U*10=735.8J

1000gr | 2.7*10^6
X | 735.8

X=Gunpowder needed
X=1000*735.8/2.7x10^6
X=0.27251851851gr
 
Physics news on Phys.org
168918791999 said:

Homework Statement


m=100gr
g=9.81 m/s2
H=75m
Combustion of gunpowder = 2.7x10^6 J/Kg

This would mean we only need 0.2gr of gunpowder to get 75m high.
Why is there in a normal rocket more gundpowder?
Our physics and chemistry teacher checked my calculations but couldn't find a mistake, so what is the problem?

Homework Equations


U=m*g*h
---------
U=Joules
M=Kg
g=gravitational acceleration
H=hight in meters

Combustion of gunpowder = 2.7x10^6 J/Kg

Average rocket is 10% efficient

The Attempt at a Solution


U=0.1*9.81*75
U=73.57500J
U*10=735.8J

1000gr | 2.7*10^6
X | 735.8

X=Gunpowder needed
X=1000*735.8/2.7x10^6
X=0.27251851851gr
You don't just pick up a rocket and put it on a shelf 75 m above the ground.

What about the KE of the rocket as it shoots upward, accelerating against the gravitational pull of the earth?
 
The work the rocket engine does on the rocket (and thus its efficiency) is highly dependent on the flight characteristics. One extreme: a burning rocket engine that fails to lift off does zero work on the rocket and has zero efficiency.
 
Last edited:
A rocket works on the principle of conservation of momentum. The rocket engine has nothing solid to push against.
Most of the energy goes into the exhaust fuel. How much goes into lifting the rocket depends on how quickly the fuel burns. You could have a vast supply of fuel, yet it burn it so slowly the rocket never lifts off.
You could take the optimum, that all the fuel burns instantly, and use momentum to figure out the speeds of the exhaust gas and the rocket lift-off.
 
I'm sorry but how can i calculate the momentum of the rocket if the fuel of the rocket is gundpowder?
 
Kinetic energy?
Also I don't think that is how a gun powder driven rocket operates.
The gp is used to bring it up to a max speed in a very short time period, or distance, and then it operates like a "thrown" projectile the rest of the distance.
 
Last edited:
168918791999 said:
I'm sorry but how can i calculate the momentum of the rocket if the fuel of the rocket is gundpowder?
You have the mass of the rocket and the mass of the gunpowder. The combusted gunpowder will have a greater mass.
Suppose the rocket takes off at speed u and the gunpowder exhaust gases exit at speed v. What equations can you write down?
 
Also a lot of the energy made availble by the gp will be used to do work against drag.
A typical drag coefficient seems to be 0.75.
 
andrevdh said:
Kinetic energy?
Also I don't think that is how a gun powder driven rocket operates.
The gp is used to bring it up to a max speed in a very short time period, or distance, and then it operates like a "thrown" projectile the rest of the distance.
Well, it takes energy to get the rocket moving at a non-zero velocity. The gunpowder is not like a compressed spring which throws the rocket upward when it is tripped.

Also, this rocket is only going 75 m. Drag is not a significant factor if the velocity is low.
 
  • #10
168918791999 said:
Combustion of gunpowder = 2.7x10^6 J/Kg
One useful parameter is exhaust velocity. Can you use the figure above to determine the exhaust velocity?
 
  • #11
SteamKing
For 75 J of kinetic energy the max speed needs to be about 40 m/s which is about 150 km/h,
but you are saying it burns throughout the whole stage? Shouldn't it stop burning way before
75 m in order to reach zero velocity at the top?
 
  • #12
So to calculate the kinetic energy I found the equation:

KE=½*m*v2
mass = 100gr
velocity = 40 m/s (I still don't know how you found this number)
KE=80000J
and the gunpowder you would need is

80000*1000/2700000 ≈ 29.63 gr
thank you for your response this number is more likely but I am still not sure if I did it right
 
Last edited:
  • #13
168918791999 said:
So to calculate the kinetic energy I found the equation:

KE=½*m*v2
mass = 100gr
velocity = 40 m/s (I still don't know how you found this number)

KE=80000J
Units matter. The formula for KE=½mv2 only works if your measurements are all in the same system of units (and assumes that it is not a gravitational system). To be consistent with Joules and meters per second, the mass of your object needs to be expressed in kilograms.

80000*1000/2700000 ≈ 29.63 gr
This is still incorrect. It assumes that all of the energy in the gunpowder is delivered to the payload and that none is delivered to the exhaust stream. The exhaust stream will be moving fast and will be carrying away a LOT of kinetic energy. That means that the result you have obtained (once corrected for the kilogram versus gram problem) is dramatically understated.

That is what Haruspex was getting at in #4 above. If you want to figure out how much energy goes into the exhaust stream, it helps to know the velocity of the exhaust stream. Haruspex in #7 above was suggesting that you start things off by treating it as an unknown named v. Post #10 above suggested that you could calculate the exhaust velocity from the information you already have.

Try looking at #7 again and see what problems you run into.
 
  • #14
If you find the amount of gunpowder needed is not intuitive you might consider an experiment done
by a group of high school students measuring the muzzle velocity, of a .223 caliber rifle, versus
the powder load.
To be brief they obtained velocities in the neighborhood of 2500 ft / sec using around 25 grains of
powder or something less than 2 grams.
Using the range formula R = V^2 / g you would get a range (maximum) of 37 mi.
In practice the maximum range would be much less than this because of air resistance.
 
  • #15
J Hann said:
If you find the amount of gunpowder needed is not intuitive you might consider an experiment done
by a group of high school students measuring the muzzle velocity, of a .223 caliber rifle, versus
the powder load.
To be brief they obtained velocities in the neighborhood of 2500 ft / sec using around 25 grains of
powder or something less than 2 grams.
Using the range formula R = V^2 / g you would get a range (maximum) of 37 mi.
In practice the maximum range would be much less than this because of air resistance.
A rifle might not be sufficiently like a rocket. A rifle traps the exhaust gases in a chamber, allowing more of the energy to be transferred to the bullet.
 
  • #16
J Hann said:
If you find the amount of gunpowder needed is not intuitive you might consider an experiment done
by a group of high school students measuring the muzzle velocity, of a .223 caliber rifle, versus
the powder load.
To be brief they obtained velocities in the neighborhood of 2500 ft / sec using around 25 grains of
powder or something less than 2 grams.
Using the range formula R = V^2 / g you would get a range (maximum) of 37 mi.
In practice the maximum range would be much less than this because of air resistance.
I wonder about that range formula. Sure, you get units of length out, but is this formula like an old Wives' Tale?

Range is dependent on initial firing angle, and there's nothing about firing angle in this formula.

Even firing a rifle perfectly horizontally, the bullet remains in the air for an amount of time equal to what it takes for a stationary bullet to free-fall to the ground. For a rifle held horizontally 1.5 m above the ground, that's a little over half a second. The range of the bullet therefore would be a little over 1250 feet or so, if it leaves the muzzle at 2500 fps, not 37 miles or some ridiculous distance.
 
  • #17
As Haruspex points out, we are not talking about a bullet fired from a rifle. Nor is the velocity of a bullet as it leaves a rifle particularly relevant to computing the exhaust velocity for gunpowder expelled from a rocket.

Perhaps we should be focusing more carefully on the original problem.
 
  • #18
The range formula is actually R = v^2 * sin (2 theta) / g.
The maximum range occurs at theta = 45 deg.
My example only referred to the maximum range.

Range is dependent on initial firing angle, and there's nothing about firing angle in this formula.

Even firing a rifle perfectly horizontally, the bullet remains in the air for an amount of time equal to what it takes for a stationary bullet to free-fall to the ground. For a rifle held horizontally 1.5 m above the ground, that's a little over half a second. The range of the bullet therefore would be a little over 1250 feet or so, if it leaves the muzzle at 2500 fps, not 37 miles or some ridiculous distance.[/QUOTE]
SteamKing said:
I wonder about that range formula. Sure, you get units of length out, but is this formula like an old Wives' Tale?

Range is dependent on initial firing angle, and there's nothing about firing angle in this formula.

Even firing a rifle perfectly horizontally, the bullet remains in the air for an amount of time equal to what it takes for a stationary bullet to free-fall to the ground. For a rifle held horizontally 1.5 m above the ground, that's a little over half a second. The range of the bullet therefore would be a little over 1250 feet or so, if it leaves the muzzle at 2500 fps, not 37 miles or some ridiculous distance.
 
  • #19
J Hann said:
The range formula is actually R = v^2 * sin (2 theta) / g.
The maximum range occurs at theta = 45 deg.
My example only referred to the maximum range.

Range is dependent on initial firing angle, and there's nothing about firing angle in this formula.

Even firing a rifle perfectly horizontally, the bullet remains in the air for an amount of time equal to what it takes for a stationary bullet to free-fall to the ground. For a rifle held horizontally 1.5 m above the ground, that's a little over half a second. The range of the bullet therefore would be a little over 1250 feet or so, if it leaves the muzzle at 2500 fps, not 37 miles or some ridiculous distance.
[/QUOTE]
Guys, please stop discussing rifles and bullets on this thread. They are really not relevant to the question at hand. By all means start another thread for it.
 
  • Like
Likes jbriggs444
  • #20
I found the equation:
P=m*v
So the momentum of our rocket would be 0.1*40 so 4 Newtons per second
So what now?
 
  • #21
What is your Physics background?
 
  • #22
168918791999 said:
I found the equation:
P=m*v
So the momentum of our rocket would be 0.1*40 so 4 Newtons per second
So what now?
That 40 meters per second comes out of the blue at post #11. It is not trustworthy. The point of these forums is not to give you the answers. It is to give you the ability to find the answers. So before we proceed forward, we need to see how to get this far. There are two steps.

First, we need to figure out the best "burn pattern" for our gunpowder rocket. This sort of optimization problem is pretty advanced subject matter. Stuff that you can handle with the "calculus of variations". But we can get by with some more simple-minded heuristics.

Suppose that we light off the motor and that it burns at a steady rate. The rocket rises at a constant acceleration and eventually passes 75 meters. Are there any problems with this acceleration profile? Can we do better?

Problem 1: The rocket passes the 75 meter mark with residual velocity. We could turn the motor off and coast partway through the trip and do better.
Problem 2: We are carrying fuel up partway into the trip and burning later. We've wasted energy lifting that fuel. We'd be better served burning it lower.
Problem 3: We are taking time in the flight. The more time we take, the longer gravity has to act. All other things being equal, we'd be better off if we could make the trip take less time -- by increasing our average velocity.

The bottom line is that we are better off burning our fuel off quickly at the bottom and then coasting to the top, crossing the 75 meter mark with no velocity remaining.

Haruspex mentioned this in #4. Andrevdh addressed it in #6.

Second, we need to figure out what velocity we need to have at the end of that short burn in order to coast 75 meters up.

Can you find the velocity that the rocket needs to rise 75 meters into the air so that it arrives at the top with no remaining speed? Alternately, can you find the speed it will have if it falls from 75 meters up and crashes into the ground? [the two answers should be identical].
 
Last edited:
  • #23
168918791999 said:
I found the equation:
P=m*v
So the momentum of our rocket would be 0.1*40 so 4 Newtons per second
So what now?
No, you don't have that yet since the velocity is what we're trying to find. (And it's Newton seconds, Ns, not Newtons per second, N/s.)
You know the mass of the rocket, but not the velocity.
You know the mass of the gunpowder, but not the velocity of the exhaust gases.
You know that momentum is conserved, ##\Sigma m_iv_i=0##
You know the energy released. If you assume all that gets converted to KE you have ##\Sigma \frac 12m_i{v_i}^2=E##.
Two equations, two unknowns. Solve.

Edit: Funny, I didn't see the other replies until I posted that.
Jbriggs gives a much more complete answer. Mine is just for the extreme (optimal) case that all the fuel is burnt instantly. Without that assumption, you need to know the burn time and pattern, and get tangled up with a varying mass payload.
 
  • #24
PE=m*g*h
KE=½*m*v2
m=100g
g=9.81 m/s2
h=75
PE=KE
m*g*h=½*m*v2
9.81*75=½*v2
735.75*2=v2
v=√1471.5
v=38.36 m/s

Is this correct?
 
  • #25
So now the momentum would be:
P=m*v
m=0.1kg
v=38.36 m/s
P=0.1*38.36
P=3.836 Ns
 
  • #26
168918791999 said:
So now the momentum would be:
P=m*v
m=0.1kg
v=38.36 m/s
P=0.1*38.36
P=3.836 Ns
Ok. So what does that tell you about the momentum of the exhaust gases?
 
  • #27
It must be the same because of conservation of momentum?
 
  • #28
168918791999 said:
It must be the same because of conservation of momentum?
Well, equal and opposite. So what KE do the exhaust gases have?
 
  • #29
it must be half of the momentum because toghether it must be 3.8Ns
so 1.9 Ns
 
  • #30
168918791999 said:
it must be half of the momentum because toghether it must be 3.8Ns
so 1.9 Ns
No, they have the same magnitude of momentum, but opposite signs. The total must be zero.
So you have three unknowns:
The mass of the gunpowder (which I presume is the same as the mass of the exhaust gases)
The energy in the gunpowder
The exhaust speed of the gases
These are related by three equations:
Energy density of the powder
Total momentum=0
Energy of the gunpowder = total KE (at least)
 
  • #31
the energy in the gunpowder is 2.7x106 J/Kg right?
So 2.7x106 J/m3
 
Last edited:
  • #32
168918791999 said:
the energy in the gunpowder is 2.7x106 J/Kg right?
So 2.7x106 J/m3
Gunpowder does not have a density of 1 kg per cubic meter. Nor does water for that matter. But that is irrelevant. The volume of the gunpowder does not enter in. Do not get ahead of yourself. What must the momentum of the burnt gunpowder be if it is equal and opposite to the upward momentum of the rocket?
 
  • #33
The momentum of the gunpowder must be the same as the momentum of the rocket but it is in the opposite direction.
So -3.8Ns
 
  • #34
168918791999 said:
The momentum of the gunpowder must be the same as the momentum of the rocket but it is in the opposite direction.
So -3.8Ns
Good. Now can you figure out the exhaust velocity of the burnt gunpowder?

Hint: Assume that all of the chemical energy in, for example, a kilogram of gunpowder goes into accelerating that kilogram of gunpowder into motion in a straight line. How much mass would the resulting gasses have? How fast would they need to be moving if energy is conserved?
 
  • #35
168918791999 said:
so what is the problem?
Efficiency. There is an upper limit to the efficiency of about 1/3.
 
  • #36
F=m*Δv
3.8=0.1*Δv
Δv=38m/s2
 
  • #37
168918791999 said:
F=m*Δv
3.8=0.1*Δv
Δv=38m/s2
I don't know which post this is a reply to, but it makes no sense.
If F is supposed to be a force then the first equation is wrong. The right hand side would yield a momentum.
Or maybe Δv is supposed to be an acceleration?
Secondly, you seem to be rediscovering the initial velocity of the rocket. You previously calculated 38m/s as the initial velocity to reach the desired height, then multiplied by the mass to get the momentum. Dividing that by the same mass will just give you back the initial velocity.
You need to write down two equations with two unknowns. The unknowns are the mass of gunpowder and the exhaust speed of the gases. Write down one equation for the momentum of the exhaust gases and another saying that total KE equals energy produced by combustion.
 
  • #38
KE=2,7x106*mg
mg = Mass of the gunpowder
P=mg*v
P=mg*38
 
  • #39
168918791999 said:
KE=2,7x106*mg
mg = Mass of the gunpowder
P=mg*v
P=mg*38
P=mgvg. vg is unknown.
 
  • #40
However can you say the momentum (p) is 3.8 Ns?
And that the kinetic energy is
KE=0.5*mg*V2
KE=0.5*mg*38.362
Ke=735.7448*mg
 
  • #41
Bystander said:
Efficiency. There is an upper limit to the efficiency of about 1/3.
168918791999 said:
However can you say the momentum (p) is 3.8 Ns?
And that the kinetic energy is
KE=0.5*mg*V2
KE=0.5*mg*38.362
Ke=735.7448*mg
That kinetic energy is irrelevant now. We used it to calculate the launch velocity of the rocket. We used the launch velocity of the rocket to calculate the launch momentum of the rocket. We used the launch momentum of the rocket to calculate the downward momentum of the exhaust gasses.

Back to post #34. Calculate the exhaust velocity.
 
  • #42
P=v/g
3.8=v/9.81
V=3.8*9.81
V=37.278 m/s
This is the only equation i could found
 
  • #43
168918791999 said:
However can you say the momentum (p) is 3.8 Ns?
Yes.
168918791999 said:
KE=0.5*mg*V2
KEg=0.5*mg*Vg2, and Vg is UNKNOWN.
Please try to follow the plan I laid out in the last paragraph of post #37.
 
  • #44
P=Mg*Vg
3.8=Mg*Vg
But how to calculate any of the unknowns?
 
  • #45
168918791999 said:
P=Mg*Vg
3.8=Mg*Vg
But how to calculate any of the unknowns?
The way I described in the last paragraph of post #37: if the mass of gunpowder is mg, you can write down an expression for the energy in the gunpowder. If its exhaust speed is vg, you can write down an expression for the KE of the exhaust. Finally, total KE equals total available energy (though in practice a bit less). This will leave you with two equations, an energy equation and a momentum equation, involving two unknowns, mg and vg. Solve.
 
  • #46
Eg=Mg*2.7x106
KEg=½*Mg*Vg
 
  • #47
KEg=P2/2mg
KEg=3.82/2mg
KEg=14.44/2mg
Can you say KEg=Eg?
Or is KEg+KEr=Eg?
 
Last edited:
  • #48
168918791999 said:
KEg=P2/2mg
KEg=3.82/2mg
KEg=14.44/2mg
Can you say KEg=Eg
Yes. It is a reasonable assumption that essentially all of the original chemical energy (Eg) in the gunpowder turns into kinetic energy (KEg) of the exhaust gasses.

[The assumption rests on the nozzle being ideal, the exhaust stream being much faster than the rocket among other idealizations]
 
  • #49
Keg = Eg
KEg = 0.5*Mg*Vg2
Eg = 2.7x106*mg
0.5*Vg2 * mg= 2.7x106*mg
0.5*Vg2 = 2.7x106
Vg2 = 5.4x106
Vg = 2323.79 m/s = 8365.644 km/h
Would this not be too fast?
 
Last edited:
  • #50
168918791999 said:
Vg = 2323.79 m/s = 8365.644 km/h
Would this not be too fast?
It is in the right ballpark. Liquid oxygen/liquid hydrogen manages 4462 m/s. Nitrogen tetroxide/hydrazine manages 3369.
 

Similar threads

Is this solution correct? (amount of fuel needed for orbital maneuvers)
Replies
12
Views
1K
How do I calculate the height this water rocket will rise?
Replies
3
Views
3K
Deflection of Fixed Rotated Beam
Replies
4
Views
2K
What Is the Blood Pressure in the Head When Standing?
Replies
1
Views
2K
Speed on an object given F and Ek....
Replies
3
Views
1K
Altitude of a Rocket Launched into Space
Replies
2
Views
5K
Find v_1 for projectile using start position, final position and angle
Replies
10
Views
2K
Max Height of Rocket A Above Earth's Surface
Replies
2
Views
4K
How Does a Spring Affect the Motion of a Rocket?
Replies
3
Views
4K
How to Calculate the Flight Time and Apex for a Model Rocket Launch in Excel?
Replies
2
Views
5K

Hot Threads

Recent Insights

Back
Top