Could this question be calculated without knowing the change in length?

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Discussion Overview

The discussion revolves around the calculation of the final temperature of air in a test facility following the impact of a bullet embedded in a piston, focusing on the principles of momentum conservation and adiabatic processes. The participants explore the necessary equations and concepts from thermodynamics and kinematics to solve the problem without knowing the change in length of the chamber.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the initial volume of air and expresses the need for the final volume to determine the final temperature using the adiabatic relation.
  • Another participant emphasizes that the problem requires understanding both kinematics and thermodynamics, noting that momentum conservation is key to the calculations.
  • A different participant suggests that the kinetic energy of the piston and bullet must be converted into work done on the gas, hinting at the relationship between kinetic energy and internal energy changes.
  • One participant proposes using the change in internal energy and the relationship between specific heats to find the temperature change, indicating a method involving the first law of thermodynamics.
  • Another participant expresses uncertainty about the method being used, suggesting that the calculations may be correct but the approach could be flawed.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct method to solve the problem. There are multiple competing views on how to approach the calculations, with some suggesting different equations and methods without resolving which is the most appropriate.

Contextual Notes

Participants express uncertainty about the correct approach and calculations, indicating potential limitations in their understanding of the adiabatic process and energy conservation principles. There is also mention of missing steps in the calculations, which remain unresolved.

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In order to measure the energy of projectiles, a test facility consisting of a horizontal insulated tube, closed at one end and with a frictionless piston of mass 1.5 kg at the other is used. The tube has a cross sectional area of 0.05 m2 and contains 1.15 kg of air at 20 °C and 100 kPa. During a test the piston is struck by a bullet of mass 0.15 kg traveling at 800 m/s, which remains embedded in the piston. It rapidly moves and compresses the air bringing the piston and bullet to rest.
For air R = 0.287 kJ kg–1 K–1 ; Cp = 1.005 kJ kg–1 K–1 .

Assuming the bullet is embedded in the piston, and momentum is conserved, and reversible adiabatic compression of the air in the tube, calculate the temperature of air at the end of compression when the piston has stopped.

__________________________________________

P1 = 100 kPa; T1 = 293K and so using PV=mRT, I calculated V1 to be 0.967 (3dp) and so the length of the chamber would be the V/A => 19.34m and if I was given the length it was compressed to (or if someone could kindly point out how to calculate it) I could work out T2 using T2 = T1 (V1/V2)^k-1 as it's adiabatic.

Any help would be really appreciated!
 
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The solution to this problem is not a plug into one formula and find the answer. It involves kinematics and thermodynamics and tests how well you understand the concepts.

Problem states that momentum is conserved, so that is your first clue for a calculation.
What next might be important, well maybe the kinetic energy of the piston imbedded with the bullet at impact.
Then the problem states that the piston stops, so you would have kinetic energy go from some initial value at impact to zero.
Well, now there must be some force acting on the piston to slow it down, and we all know that F=ma and W=Fd, so that looks a little promising, but ...

Aha - all the kinetic energy must go into compressing the gas, and since it is an s adiabatic process then the work on the gas can be calculated from a formula,

And since PV^gamma=C, the temperature can be found also.

( Hope my post is not too lame or missing a step, as I haven't had my morning coffee yet )
 
Last edited by a moderator:
Your calculations look finebut I guess it is the incorrect method.

Try
Cv = R - Cp

For adiabatic
dU +dW = Q = 0 ==> dU = dW

\DeltaU = -P\DeltaV

\DeltaU = m Cp \DeltaT

And the change in internal energy would equal the change in kinetic energy

solve for \DeltaT
 
256bits said:
Your calculations look finebut I guess it is the incorrect method.

Try
Cv = R - Cp

I would't try that one!
I'll try to get involved here later if the right answer is still not forthcoming.
 

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