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How much gunpowder does a rocket need?

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant 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

    3. 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
     
  2. jcsd
  3. Dec 9, 2015 #2

    SteamKing

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    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?
     
  4. Dec 9, 2015 #3
    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: Dec 9, 2015
  5. Dec 10, 2015 #4

    haruspex

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    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.
     
  6. Dec 10, 2015 #5
    I'm sorry but how can i calculate the momentum of the rocket if the fuel of the rocket is gundpowder?
     
  7. Dec 10, 2015 #6

    andrevdh

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    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: Dec 10, 2015
  8. Dec 10, 2015 #7

    haruspex

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    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?
     
  9. Dec 10, 2015 #8

    andrevdh

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    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.
     
  10. Dec 10, 2015 #9

    SteamKing

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    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.
     
  11. Dec 10, 2015 #10

    jbriggs444

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    One useful parameter is exhaust velocity. Can you use the figure above to determine the exhaust velocity?
     
  12. Dec 10, 2015 #11

    andrevdh

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    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?
     
  13. Dec 10, 2015 #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: Dec 10, 2015
  14. Dec 10, 2015 #13

    jbriggs444

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    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.

    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.
     
  15. Dec 10, 2015 #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.
     
  16. Dec 10, 2015 #15

    haruspex

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    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.
     
  17. Dec 10, 2015 #16

    SteamKing

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    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.
     
  18. Dec 10, 2015 #17

    jbriggs444

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    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.
     
  19. Dec 10, 2015 #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]
     
  20. Dec 10, 2015 #19

    haruspex

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    [/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.
     
  21. Dec 11, 2015 #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?
     
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