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Help needed

  1. Apr 8, 2007 #1
    1. "w" grams of a specific fuel is burnt in a steel box of volume "V". the calorific value of fuel is given , suppose "c", now we have to calculate the force exerted on the steel walls of the box due to expansion of air inside the box due to burning of fuel. what will b the presure change? if the volume of the box could be changed .. to what volume will the inside air expand upto?
    Part 2: consider the figure; Supposing the box has an opening into a pipe of cross-section area, a1 , and pipe is attached to this opening with a massless piston(as shown in the figure). When the air in the box would expand, it pushes the piston to the other end. Now, supposing the pipe to be of infinite length, to what distance will the massless piston travel, till it comes to a stop? do not consider the loss of heat from any surface, assume the piston to be air-tight.

    2. Relevant equations: no idea .. am pretty confused .. anyone mind tellin me??

    3. My trial:
    For the second part, the teacher gave the hint: the piston must develop some momentum due to which it will keep moving untill there is development of vaccum in the box and the vaccum force will cause the piston to stop.
    Now what is vaccum force??:confused: :eek:

    Attached Files:

  2. jcsd
  3. Apr 8, 2007 #2
    If you know how much fuel is combusted, you can know how much energy is released. This energy, assuming 100% efficiency, will be all converted to heat energy. You can then find the temperature change of the air if you know the volume of air inside the chamber.

    Then use PV = nRT to solve for P and V respectively. (n and R are constant)
  4. Apr 8, 2007 #3
    Is there any other equation for calculating the volume change of air? or is this sufficient?
  5. Apr 8, 2007 #4
    well since it isn't stated that gas is being generated, which would really be sneaking up a rocket engine, I think that is all you can use.
  6. Apr 9, 2007 #5
    The problem states that the "fuel" is burnt in the steel box, and becoz of the heat developed the air inside it must expand ....
    what part of "thermodynamics" would possibly be used to solve this problem?
    Rocket engines must be different ....
  7. Apr 9, 2007 #6
    Rocket engines are different because they depend on genarating lots of gas from combustibles which is then expelled at the rear as rapidly as possible given constraints of physical containment. I think sapplingg is right, just use the ideal gas law where T is raised due to combustion of fuel. The second part of the problem I was a little confused by--at least your profs hint about momentum and vacuum force. I would think it would expand until the pressures are equal on both sides. I suspect it will be a good distance and since physical dimensions are not given, frankly, not sure how to tackle this.
  8. Apr 10, 2007 #7
    well .. I can try to solve the second part .. but I am confused by "vaccum force" .. what will it be equal to??
  9. Apr 10, 2007 #8
    Assuming you more or less quoted him accurately, I'm mystified by your profs explanation as well as how to solve this except in the most general terms. First he suggests that the piston has momentum, which I find a little surprising since its explicitly states that the piston has no mass. Then he uses the term Vacuum force to descibe the conditions under which it will stop. What I think you can say is that the pressures from both sides of the membrane (piston) act on the same area and therefore any difference in pressure will cause the membrane to move until there is no net force on it, ie when the pressures are the same. This would imply a change in volume that the gas occupies.
    Last edited: Apr 10, 2007
  10. Apr 10, 2007 #9


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    Burning the fuel adds heat to the system (gas in the box). In part 1 the gas does no work (no expansion or compression of the gas). This means that the heat causes an increase in the internal energy of the gas, which will manifest as a temperature increase. We assume that the amount of gas stays unchanged. This means that

    [tex]\frac{p}{T} = constant[/tex]
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