Prelim_1_problem_1

# Prelim_1_problem_1 - left = 286.16 kJ/kg Thus u(T = 250.11...

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Prelim 1, Problem 1 Given: A rigid adiabatic container is divided into two sections containing equal amounts of air at different T and P. On the left is 10 kg at 2 bar and 400 K; on the right is 10 kg at 3 bar and 300 K. The partition breaks. The air behaves as an ideal gas. Find the final temperature and pressure? 2 bar 400 K 3bar 300 K T=? P=? Solution for the temperature: For a rigid, adiabatic container: Q = 0 & W = 0. Therefore U = 0 = 20 kg * u(T) – 10kg * (u(T right ) + u(T left )) = 0. This gives u(T) = (u(T right ) + u(T left ))/2. From Table A-22: u(T right ) = 214.07 kJ/kg and u(T
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Unformatted text preview: left ) = 286.16 kJ/kg. Thus, u(T) = 250.11 kJ/kg. Then from A-22, T = 350 K! To find pressure used the ideal gas law. This gives p = RT/v. By definition v = V/m where V = (V left + V right ). m is 20 kg, and the volumes V left and V right are found from the ideal gas law. V left = (m/2)RT left /p left and V right = (m/2)RT right /p right . Note that m/2 = 10 kg! Substitution gives for pressure bar p T p T R RT p right left 33 . 2 2 = + • =...
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## This note was uploaded on 03/15/2010 for the course ENGRD 221 taught by Professor Staff during the Fall '08 term at Cornell.

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