How Does a Portable Device for Pressurizing Tires Impact Thermodynamics?

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SUMMARY

The discussion focuses on the thermodynamic analysis of a portable tire pressurizing device containing methane (CH4) at an initial pressure of 500 psia. The process is adiabatic and reversible, involving calculations of mass flow, entropy changes, and final states of the gas in both the container and the tire. Key calculations include determining the mass of gas in the container and tire, the final pressure in the tire, and the entropy change for the closed system. The ambiguity in the final temperature of the gas in the tire is highlighted as a critical point for accurate calculations.

PREREQUISITES
  • Understanding of adiabatic processes in thermodynamics
  • Familiarity with the ideal gas law and state equations
  • Knowledge of entropy and its calculation in closed systems
  • Experience with thermodynamic charts for gases
NEXT STEPS
  • Study the principles of adiabatic and isothermal processes in thermodynamics
  • Learn how to use thermodynamic charts for gases, focusing on methane
  • Research the application of the first law of thermodynamics in closed systems
  • Explore methods for calculating entropy changes in thermodynamic processes
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This discussion is beneficial for mechanical engineers, thermodynamics students, and professionals involved in the design and analysis of gas pressurization systems.

siddharth
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The question verbatim is:

"A portable device for pressurising tyres consists of a small container of 1.5 cft volume, initially filled with, say, CH4 at 500 psia. When a tyre (volume=0.5 cft) is to be filled, the gas is allowed to flow into it from the container through a valve till the pressure in the tyre reaches 50 psia. The process is essentially adiabatic and the initial pressure in the tyre may be taken as 15 psia. Let the subscripts 1 & 2 and i & f refer to the initial and final states of the gas in the container and tyre respectively. Assume that P1=500 psia, Pi=15, T1=Ti=80 F

a) Assume that the filling process is reversible. Use the chart to read v,s and h for the gas (i)initially in the container (ii) initially in the tyre and (iii) finally in the tyre."
b) Calculate m1, mi and mf
c) What's the value of s2? Read t2 and h2 from the chart
d) Use the first law for container + tyre to find m2(h2-p2v2)
e) Solve for P2 (:confused:) by trial and error so as to satisfy the first law
f) Calculate the change in entropy of
(i)The closed system (container + tyre)
(ii)The environmentFirst of all, I think there's some ambiguity in the statement of the question on the notations. Leaving that aside, I have a doubt regarding the final state of the air in the tyre. The thing is that T_f for the tyre isn't known. So how can I calculate T_f (so that I can then calculate h,v,s from the chart)?
Now the process is adiabatic and reversible, so the entropy change for the whole system (ie, container + tyre, neglect valve) is 0. But, that doesn't help me find T_f because the entropy change in the tyre alone wouldn't be 0. I think there's some information which needs to be given, or some assumption to make to calculate the final state. Once I do that, the rest of the problem seems easy.
 
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