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# Chap4prac - Chapter 4 Practice Problems(P4.1 COP = coef of...

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Chapter 4 Practice Problems To accompany Introductory Chemical Engineering Thermodynamics © J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/24/2001) 1 of 2 (P4.1) COP = coef. of performance = net S C W Q , Using state numbers of Fig 4.9-4.10. P-H plot will look like Fig 4.10 of pg 151. Use P-H chart pg 653 and table pg 654: state 2 is satV at –40C ! H 2 = 372 kJ/kg (chart) state 3, outlet of the reversible compressor is found by following the isentropic line to 40C, where H 3 ’ = 438 kJ/kg. state 4 is satL at H 4 = 256 kJ/kg (table) state 1, H 1 = H 4 ( ) kg kJ H H Q C / 116 256 372 1 2 = = = W S = ( H 3 H 2 ) = 438 – 372 = 66 kJ/kg 76 . 1 66 116 = = = S C W Q COP & & Q C = COP (W S ) = 1.76 (9000 J/day) = 16 kJ/day (P4.2) Methane,. See Fig 4.9-4.10 pg 150-1.Methane chart pg 651. F lb Btu S lbm Btu H MPa P F T O satV satV O evap . / 35 . 2 / 339 032 . 0 , 280 2 2 = = = = Pressure of 3’ isn’t given, however, it will be the same as the condenser outlet (P 4 ) since the condenser will be considered to be isobaric. State 4 will be satL at 40F, so P4 = P3’ = 2 MPa. Follow isentropic line from the compressor inlet state to 2MPa. At P3’=2MPa,

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