# hw07 - College of Engineering and Computer Science...

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College of Engineering and Computer Science Mechanical Engineering Department Mechanical Engineering 370 Thermodynamics Fall 2010 Course Number: 14319 Instructor: Larry Caretto Unit Six Homework Solutions, October 7, 2010 1 Consider an 8 L evacuated rigid bottle that is surrounded by the atmosphere at 100 kPa  and 17 o C.  A valve at the neck of the bottle is now opened and the atmospheric air is  allowed to flow into the bottle.  The air trapped in the bottle eventually reaches thermal  equilibrium with the atmosphere as a result of heat transfer through the wall of the bottle.  The valve remains open during the process so that the trapped air also reaches  mechanical equilibrium with the atmosphere.  Determine the net heat transfer through the  wall of the bottle during this filling process. If we define the bottle as our system, we see that we have an unsteady problem because mass enters through the one inlet and there are no outlets for the mass to exit. The general first law equation for unsteady open systems is shown below. + + + + + - - = + + - + + inlet i i i i outlet i i i i u system gz V h m gz V h m W Q gz V u m gz V u m 2 2 2 2 2 2 1 2 1 2 2 2 We see that there is no mechanism for useful work in this system so we set W u = 0 and make the usual assumption that kinetic and potential energy terms are zero. This gives the following expression for the first law. [ ] in in system h m Q u m u m u m u m + = - = - 1 1 2 2 1 1 2 2 Since the cylinder is initially evacuated, we have m 1 = 0, so that m 2 = m in = m. This gives the following result for the first law. ( 29 in in in in in h u m h m u m h m u m u m Q - = - - = - - = 2 2 2 1 1 2 2 0 We can compute the mass from the given data for the final state. The temperature and pressure in the bottle at the final state are the same as those of the atmosphere due to the thermal and mechanical equilibrium. Thus, P 2 = 100 kPa and T 2 = 17 o C = 290.15 K. We find the gas constant for air from Table A-1: R = 0.2870 kJ/kg K = 0.2870 kPa m 3 /kg K. We then find the mass as follows: kg K K kg m kPa L m L kPa RT V P m m m bottle in 0096 . 0 ) 15 . 290 ( 2870 . 0 1000 ) 8 )( 100 ( 3 3 2 2 2 = = = = = Jacaranda (Engineering) 3519 Mail Code Phone: 818.677.6448 E-mail: [email protected] 8348 Fax: 818.677.7062

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We can use the ideal gas properties for air in Table A-17. For a temperature of 290.15, which is both the temperature of the inlet air and the final temperature in the cylinder, we find u 2 = 206.91 kJ/kg and h in = 290.16 kJ/kg. We use these values and the mass to find the heat transfer. ( 29
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hw07 - College of Engineering and Computer Science...

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