How much gunpowder does a rocket need?

[168918791999]

F=m*Δv
3.8=0.1*Δv
Δv=38m/s²

 

[haruspex]

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.

 

[168918791999]

KE=2,7x10⁶*m_g
m_g = Mass of the gunpowder
P=m_g*v
P=m_g*38

 

[haruspex]

KE=2,7x10⁶*m_g
m_g = Mass of the gunpowder
P=m_g*v
P=m_g*38

P=m_gv_g. v_g is unknown.

 

[168918791999]

However can you say the momentum (p) is 3.8 Ns?
And that the kinetic energy is
KE=0.5*m_g*V²
KE=0.5*m_g*38.36²
Ke=735.7448*m_g

 

[jbriggs444]

Efficiency. There is an upper limit to the efficiency of about 1/3.

However can you say the momentum (p) is 3.8 Ns?
And that the kinetic energy is
KE=0.5*m_g*V²
KE=0.5*m_g*38.36²
Ke=735.7448*m_g

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.

 

[168918791999]

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

 

[haruspex]

However can you say the momentum (p) is 3.8 Ns?

Yes.

KE=0.5*m_g*V²

KE_g=0.5*m_g*V_g², and V_g is UNKNOWN.
Please try to follow the plan I laid out in the last paragraph of post #37.

 

[168918791999]

P=M_g*V_g
3.8=M_g*V_g
But how to calculate any of the unknowns?

 

[haruspex]

P=M_g*V_g
3.8=M_g*V_g
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 m_g, you can write down an expression for the energy in the gunpowder. If its exhaust speed is v_g, 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, m_g and v_g. Solve.

 

[168918791999]

E_g=M_g*2.7x10⁶
KE_g=½*M_g*V_g

 

[168918791999]

KE_g=P²/2m_g
KE_g=3.8²/2m_g
KE_g=14.44/2m_g
Can you say KE_g=E_g?
Or is KE_g+KE_r=E_g?

 

[jbriggs444]

KE_g=P²/2m_g
KE_g=3.8²/2m_g
KE_g=14.44/2m_g
Can you say KE_g=E_g

Yes. It is a reasonable assumption that essentially all of the original chemical energy (E_g) in the gunpowder turns into kinetic energy (KE_g) of the exhaust gasses.

[The assumption rests on the nozzle being ideal, the exhaust stream being much faster than the rocket among other idealizations]

 

[168918791999]

Ke_g = E_g
KE_g = 0.5*M_g*V_g²
E_g = 2.7x10⁶*m_g
0.5*V_g² * m_g= 2.7x10⁶*m_g
0.5*V_g² = 2.7x10⁶
V_g² = 5.4x10⁶
V_g = 2323.79 m/s = 8365.644 km/h
Would this not be too fast?

 

[jbriggs444]

V_g = 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.

 

[168918791999]

p=m_g*v_g
3.8=m_g*2323.79
m_g=0.00163525964 Kg
m_g=1.63525964 gr
Do you still have to multiply it by 10 because a rocket is on average only 10% efficient?
Because then you get 16.35 gr and that would be more realistic than 1.6 gr

 

[insightful]

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.

What is the temperature of the exhaust gases? How much chemical energy goes into heating the exhaust gases?

 

[jbriggs444]

m_g=1.63525964 gr
Do you still have to multiply it by 10 because a rocket is on average only 10% efficient?

A figure of 10% for rocket efficiency is unfounded. Before such a figure could even make sense, one would need to come up with a measure of efficiency. One such measure is the fraction of chemical energy which goes into accelerating the payload. That measure depends on how fast the rocket is going already. When rocket velocity is equal to exhaust velocity, 100% efficiency can be approached. When rocket velocity is zero, 0% efficiency always results.

Edit: However, it scores few points with the instructor to object that a given of the problem is wrong. Let us try to make it meaningful. Assume that the 10% is a measure of how much of the chemical energy in the propellant is divvied up as kinetic energy going to the rocket and to the bulk of the exhaust stream. Then the remaining 90% would be lost as waste heat, sound, vibration and kinetic energy in the expanding exhaust cloud.

What is the temperature of the exhaust gases? How much chemical energy goes into heating the exhaust gases?

For a suitably ideal nozzle operating in vacuum, the exhaust gas temperature can become negligible. Adopting realistic assumptions for this seems to be out of scope for the problem at hand.

Edit: Possibly the "realistic assumption" is the 10% figure from the problem statement.

 

[168918791999]

However thank you for your help!

 

[insightful]

However thank you for your help!

Using your method from your original post and an efficiency of 0.6% will give you a reasonable answer.
This efficiency is derived from data on Estes model rocket performance.

 

[168918791999]

Using your method from your original post and an efficiency of 0.6% will give you a reasonable answer.
This efficiency is derived from data on Estes model rocket performance.

No, that would mean we needed:
1.63525964 / 0.6 *100 = 272.543273333 grams of gunpowder
That would mean that we needed more gunpowder than the rocket itself is

 

[insightful]

No, that would mean we needed:
1.63525964 / 0.6 *100 = 272.543273333 grams of gunpowder
That would mean that we needed more gunpowder than the rocket itself is

What is 1.63... and why multiply 0.6 times 100?

 

[168918791999]

1.6 is the mass of the gunpowder
so if it is 0.6% efficient that would mean:
1.6 gr | 0.6%
M_r | 100%

 

[insightful]

1.6 is the mass of the gunpowder

Do you mean "at 100% efficiency"? If so, redo your calculations.

 

[jbriggs444]

Do you mean "at 100% efficiency"? If so, redo your calculations.

First, define 100% efficiency.

 

[insightful]

100% efficiency = all gunpowder energy ends up as mgh energy.

 

[jbriggs444]

100% efficiency = all gunpowder energy ends up as mgh energy.

That is an incorrect definition of rocket efficiency.

 

[insightful]

That is an incorrect definition of rocket efficiency.

I believe this is the OP's definition and a useful one here if defined as such.

 

[168918791999]

okay to calculate the amount of gunpowder we need in total (at 100%)
i assumed that the 1.6 grams is the amount of gunpowder we needed only to fly up to 75 meters
because it is only 0.6% efficient it means that if i have an amount of gunpowder (100%) only 0.6% goes into flying the rocket and that is the 1.6 grams we needed.
so if 1.6 = 0.6%
272.5 grams = 100%

 

[insightful]

Show calculation to get 1.6 grams.

 

[jbriggs444]

I believe this is the OP's definition and a useful one here if defined as such.

Given the numbers we have come up with for exhaust velocity and rocket velocity at burnout, a figure of 10% as given in the problem statement cannot possibly be compatible with the definition that you propose.

The definition that you propose is extremely sensitive to rocket velocity at burnout. That makes it exeedingly unreliable to pick a figure up from the Internet and try to plug it into an arbitrary problem. Suppose, for instance, that we double the amount of gunpowder burned. We double velocity at burnout and we achieve a high point with roughly four times the energy according to E=mgh. We have doubled the efficiency of our rocket motor without doing anything to the rocket motor. That is not a good thing to see in an efficiency metric.

 

[insightful]

Given the numbers we have come up with for exhaust velocity and rocket velocity at burnout, a figure of 10% as given in the problem statement cannot possibly be compatible with the definition that you propose.

Where did I say a figure of 10% is compatible? I said 0.6%.
Anyway, so what answer did you get?

 

[jbriggs444]

Your figure of 0.6% is not compatible either. No number is.

Edit: The homework forums are not a place for answers to be posted. The calculation of the 1.6 gram figure has been shown. The calculation of the exhaust velocity has been shown.

 

[insightful]

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?

It is in the right ballpark. Liquid oxygen/liquid hydrogen manages 4462 m/s. Nitrogen tetroxide/hydrazine manages 3369.

Doesn't this calculation assume all the chemical energy ends up as mechanical energy of the exhaust?

Oops, just reread #48; never mind.

So, how much energy is being used to heat the exhaust and push back the atmosphere?
(Space shuttle exhaust from the main engines is at 1600^oC.)

 

[insightful]

A fascinating thread.
I would love to see the OP post the "official" correct answer.