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Test_2_Sp96 - ME 326 THERMODYNAMICS I Test 2 1(33 Argon gas...

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ME 326: THERMODYNAMICS I Test 2, March 29, 1995 1) (33%)  Argon gas passes through a well-insulated valve in a supply line at a rate of  0.5 kg/s. Upstream of the valve, the pressure is 10 atm and the temperature is 127 ° C.  Downstream of the valve, the pressure is 1 atm. What is the entropy change of the argon across the valve? Note any assumptions that are required to get your answer. SOLUTION: Because argon is a monatomic gas, c P  is constant. First Law for this case, assuming  valve is adiabatic and  KE and  PE across the valve are negligable, is 0 ( ) ( ) out in P out in Q W m h h mc T T & & & & so T out =T in . The specific entropy change across the valve for a adiabatic steady-state  steady flow is  s out - s in = c P ln T out T in - R ln P out P in = 0 - 0.20813 kJ / kg K ( 29 ln 1 10 = 0.4792 kJ / kg K or, if in terms of a rate of entropy change, 0.5( / ) 0.4792( / ) 0.2396 / S m s kg s kJ kg
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