HOMEWORK #2 PROBLEM Advertised is a small toy that will send up a signal flare w

Question

HOMEWORK #2 PROBLEM Advertised is a small toy that will send up a signal flare with a parachute. Upon examination, we found that it is simply a metal tube 1.0 m long and 600 mm2 in cross-sectional ar plu g, which contains the parachute, fits into the tube and a trigger holds it in place. At this ition, the bottom edge of the plug is 0.5 m from the bottom. The plug weighs 1.0 kg. To operate the toy, the air below the plug is pumped up to a pressure of about 3.0 bar with a hand pump, and then the trigger is released, allowing the plug to fly out the top. The ambient temperature is 298 K, and the constant-volume heat capacity of air is 2.5 R, where R is the gas constant. (a) Assuming the entire device operates isothermally.Calculate the maximum height that the plug can go and the velocity of the plug at the top of the tube. Hint: To get the maximum possible height, you may assume that the gas expansion is reversible. x constant temperature (b) Now suppose the expansion is adiabatic, rather than isothermal. Calculate the maximum height that the plug can go and the velocity of the plug at the top of the tube. No heat trans fer (c) As an engineer, you are interested in making the toy work even better. Suppose that for safety reasons, the metal tube cannot withstand any pressure higher than 3.0 bar. But you can easily change where the trigger is positioned in the tube. Where would you place the trigger such that the maximum upward distance traveled by the plug is the greatest possible for the isothermal case? Also, rationalize why there exists an optimal placement of the trigger: what causes the plug to go less far if one places the trigger too high or too low? Now FOR oNLy CV ROCKY JONESTOYS COMPANY

Explanation / Answer

given: L=1m. ,Area A=600mm or 600x10-6 m. ,Mass M=1kg. Cv=2.5R & T=298k

Consider the procees is isothermal reversible process. for which the work done is given by;

W=nRT ln(P1/P2).

P1=3bar or 3x105 pas.

P2= 1atm or 1.01325x105 pas.

now work done in this process is equal to potentila energy of system.

by ideal gas law, PV=nRTformula will tranform.

W=3x105x 1 x 600x10-6 x ln(3x105/1.01325x105)               bcz (V=3x105x600x10-6)

W=180J

now, PE= mgz

so, 180=1x9.81xH

H=18.34m

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