How much gunpowder does a rocket need?

[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.