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Unformatted text preview: 1 ME 201 Thermodynamics Old Exam #3 Solutions Directions: Work all three (3) problems. The exam is open notes and open text book. All problems have equal weight. Problem 1 An inventor claims to have designed a heat exchanger system that operates with air entering at 350 kPa, -17C, and 0.05 kg/s and is heated to 70C. Refrigerant-134a enters the other side as saturated vapor at 1.2 MPa at a rate of 0.25 kg/s. Determine the validity of this claim. Solution: This claim will be assessed by calculating the rate of entropy production for the system and comparing it to zero. The rate of entropy production will be given by T Q- s m- s m = S surr sys inlets i i outlets e e prod & & & & & & or writing it specifically for this heat exchanger ( ) ( ) T Q- s- s m s- s m = S surr HE a 134 R in out R134a air in out air prod & & & & + Now setting up to work a first law problem Substance #1: Air (Ideal Gas) Substance #2: R-134a (Compressible) System: Control Volume Device: Heat Exchanger kW Q = & kW W = & 1 st law: ( ) ( ) h- h m h- h m = a 134 R out in R134a air out in air & & + State Air in Air out R134a in R134a out T( C) -17 (256 K) 70 (343 K) 46.32 (sat.vap.) 46.32 (x=0.8874) P 350 kPa 350 kPa 1.2 MPa 1.2 MPa 1....
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- Spring '06