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

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Unformatted text preview: Chapter 4 Practice Problems (P4.1) COP = coef. of performance = QC WS ,net 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 ! H2 = 372 kJ/kg (chart) state 3, outlet of the reversible compressor is found by following the isentropic line to 40C, where H3’ = 438 kJ/kg. state 4 is satL at H4 = 256 kJ/kg (table) state 1, H1 = H4 QC = (H 2 − H 1 ) = 372 − 256 = 116kJ / kg WS = (H3 – H2) = 438 – 372 = 66 kJ/kg & Q 116 ⇒ COP = C = = 1.76 & 66 W S QC = COP (WS) = 1.76 (9000 J/day) = 16 kJ/day (P4.2) Methane,. See Fig 4.9-4.10 pg 150-1.Methane chart pg 651. Tevap = −280 O F , P = 0.032MPa ⇒ H 2satV = 339 Btu / lbm S 2satV = 2.35Btu / lb.O F Pressure of 3’ isn’t given, however, it will be the same as the condenser outlet (P4) 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, and S3’ = 2.35 Btu/lbmF, H3’ = 485 Btu/lbm, and T3’ = 40F. At the outlet of the condenser, the fluid will be saturated liquid at 2 MPa, T4 = −160 O F , P4 = 2MPa ⇒ H 4satL = 218Btu / lbm satliq S4 1.54 Btu / lb.O F State 1 2 Sat'd vap 3' 4 COP = O T( F) O P(MPa) H(Btu/lb) S(Btu/lb* F) -280 0.032 218 -280 0.032 339 2.35 40 485 2 2.35 -160 2 218 1.54 QC H 2 − H1 H − H 4 339 − 218 = =2 = ′ ′ H 3 − H 2 WS ,net H 3 − H 2 485 − 339 ⇒ COP = 0.83 (P4.3) Rankine Cycle, Fig 4.3 Pg. 143 temperatures 200 O C & 99.6 O C , Sat vapor from Turbine outlet can be found in the sat Pressure table p642. To accompany Introductory Chemical Engineering Thermodynamics © J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/24/2001) 1 of 2 Chapter 4 Practice Problems The turbine is assumed to be adiabatic and reversible since no other specifications are given. The inlet state will then have the same entropy as the outlet. Since the boiler/superheater outlet is to operate at T3 = 200 C, and S3 = 7.3592, we must hunt in the superheated tables to find this combination. This will occur between 0.2 and 0.3 MPa. Interpolating, the value of H can be found. Values are tabulated below: O State TC P(MPa) H(kJ/kg) S(kJ/kg*K) V(m3/kg) 4’ sat vap 99.6 0.098 2675 7.359 5 sat liq. 99.6 0.098 417 1.3028 0.001043 3 200 0.271 2867 7.3592 6 417.18 ′ WS′,turbine = (H 4 − H 3 ) = 2675 − 2867 = −192kJ / kg ∆H pump = WS , pump = ∫ VdP = V∆P 1043cm 3 * (0.271 − 0.098) 1J kJ *3 *3 kg cm * MPa 10 J = 0.180kJ / kg. ⇒ ∆H pump = ⇒ ∆H pump ⇒ WS′,net = −192 + 0.180 = −191.82kJ / kg &η = − WS′,net Qboiler = − (− 191.82 ) − (− 191.82 ) = (H 3 − H 6 ) (2867 − H 6 ) ⇒ H 6 = 417 + 0.18 = 417.18kJ / kg − (− 191.82 ) = 0.0783 (2867 − 417.18) ⇒ η = 7.83% ⇒η = To accompany Introductory Chemical Engineering Thermodynamics © J.R. Elliott, C.T. Lira, 2001, all rights reserved. (7/24/2001) 2 of 2 ...
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This note was uploaded on 01/31/2010 for the course CHEM 2106 taught by Professor Liz during the Spring '08 term at University of Minnesota Crookston.

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