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SECTION IV – FIRST LAW APPLIED TO OPEN SYSTEMS Refrigerant-12 vapour enters a steady-flow compressor as a saturated vapour at 12C. The outlet conditions are 0.6 MPa and 50C, and the process is assumed to be adiabatic. Calculate the power in Kilowatts, required if the refrigerant flow rate is 20 kg/min. Determine the diameter of the inlet tubing to the compressor if the inlet velocity is not to exceed 3 m/s. Schematic of compressor system boundary ie Inlet conditions TI = 12C, saturated vapour Exit Conditions Pe = 0.6 MPa, Te = 50C Pe < Osat = 1.2193 MPa corresponding to Te = 50C Refrigerant-12 is superheated at the exit. e e 0 Process is adiabetic Q ie i m e m Flow through compressor is steady m 20 kg / min 0.333 kg / sec m Assumptions - Properties are uniformly distributed at the inlet and exit. - epot and ekin are negligible compared to h. Steady-state, steady-flow form of the conservation of energy for the compressor as an open system is then: m (he hi ) -W ie hi h ig h ig 191.602 T 12C 12 10 h e 216.141 kJ / kg 193.644 191.602 h ig 192.419 kJ / kg 15 10 m (h i h e ) 0.333(192.419 216.141) 7.90 KW W ie Power Input D i2 1 4(0.333) i i Vi A i Vi m D i2 vi vi 4 3 Section IV – First Law Applied to Open Systems Page 19 v i 0.040914 0.035413 0.040914 v i 0.0387136 m 3 / kg 12 10 15 10 D i2 4(0.333)(0.0387136) Di 0.074 m 74 mm 3 Steam enters a turbine at 600C and 6 MPa with a velocity of 300 m/s. The mass flow rate entering the turbine is 400 kg/min. Assume the turbine is well-insulated and the exhaust steam leaves the turbine at a low velocity. The steam exits the turbine at 200 KPa and 260C. Calculate the power developed by the turbine and the inlet-duct area. Schematic of turbine system boundary Inlet Conditions: TI = 600C, PI = 6 MPa, i 400 kg / min 6.67 kg / s, Vi 300 m / s m Exit Conditions: Te = 260C, Pe = 200 KPa ie e e 0 Turbine is well-insulated Q ie Assumptions - flow through turbine is steady i.e. no mass build-up in turbine - flow properties are uniformly distributed at the inlet and exit to turbine - changes in potential energy of working fluid are negligible compared to change in enthalpy - neglect kinetic energy at the exit since the velocity will be low Steady-state, steady-flow form the conservation of energy for the turbine as an open system is then: W ie m(he hi ekini ) steam is superheated at the inlet since T > Tsat = 275.64C corresponding to P = 6 MPa hI = 3658.4 kJ kg e kini 1 (300) 2 45 kJ kg 2 1000 steam is also superheated at the outlet since P < Psat = 4.688 Mpa corresponding to P = 200 KPa Section IV – First Law Applied to Open Systems Page 20 he 2971.0 3071.8 2971.0 kJ he 2991.2 260 250 300 250 kg Wie m(hi ekini he ) 6.67(3658.4 45 2991.2) Wie 4750.4 KW 4.75MW VA mv 6.67 x0.06525 m i iVi Ai i i Ai i i vi Vi 300 A i 0.00145m 2 14.5cm 2 4-64 Boles & Cengel Refrigerant –12 at 1MPa and 80C is cooled to 1MPa and 30C in a condenser by air. The air enters at 100KPa and 27C with a volume flow rate of 800 m3/min and leaves at 95KPa and 60C. Determine the mass flow rate of the refrigerant. Air 1 = 800 3/min R-12 P3 = 100 kPa T3 = 27C system boundary P1 = 1 MPa T1 = 80C P4 = 95 kPa T4 = 60C P2 = 1 MPa T2 = 30C Section IV – First Law Applied to Open Systems Page 21 Can air be treated as an ideal gas under the given conditions? 0.1 0.027 3.76 300 Tr3 2.26 133 0.095 Pr4 0.025 3.76 333 Tr4 2.50 133 Pr3 Z3 = Z4 1 (See p. 807 of Text) Assumptions: Flow of Refrigerant –12 is steady m1 m2 m R12 Flow of air is steady m 3 m4 m air Heat lost by R-12 is gained by air Qie air = -Qie R-12 Properties are uniformly distributed at 1 & 2 and at 3 & 4. Conservation of energy applied to the flow of air: m air (h4 – h3) P 100 800 m air 3 1 3 1 15.486 kg / s RT3 0.287 300 60 h3 hT 300K 300.19 KJ Kg see p. 788 of text Q ie air W ie air = h4 hT 333k h4 330.34 333 330 340.42 330.34 340 330 h4 333.36 KJ Kg Q ie air 15.486(333.36 300.19) 513.7 KJ Conservation of energy applied to the flow of Refrigerant –12: Q ie R 12 Wie R 12 m R 12 ( h 2 h 1 ) Qie R 12 Q ie air 513.7 KJ m R 12 513.7 (h2 h1 ) (no work done on or by the Refrigerant –12) Section IV – First Law Applied to Open Systems Page 22 T1 = 80C Tsat = 41.64C corresponding to P1 = 1MPa R-12 is a superheated vapor at 1 h1 = 232.91 KJ/Kg (see p.784 of Text) T2 = 30C Tsat = 41.64C corresponding to P2 = 1MPa R-12 is a subcooled liquid at 2 h 2 h fT 30 C m vf (PCL PSat ) 64.59 0.0007739(1000 744.90) 64.79 KJ Kg T 30C (see pp. 780-781 of text) 513.7 Kg Kg 3.06 183.3 (64.79 232.91) S min Section IV – First Law Applied to Open Systems Page 23 Section IV – First Law Applied to Open Systems Page 24 Section IV – First Law Applied to Open Systems Page 25 The operating data from the simple, steady-flow, steam-power plant cycle shown in the accompanying figure are summarized in the table below. Determine the power output of the turbine, the heat-transfer rate in the steam generator, the power input to the pump, and the thermal efficiency of the cycle. Location Pressure, Mpa Quality or Temperature 1 0.01 0.0 2 48°C 3 5 450° C 4 0.95 Other data: Mass flow rate of steam = 20 kg/s Negligible pressure drops through steam generator and condenser Section IV – First Law Applied to Open Systems Location 1 2 3 4 Page 26 Pressure, Mpa 0.01 5 5 0.01 Quality or Temperature 0.0 48C 450 0.95 Assumptions: ep << h, ek << h for all units Properties are uniformly distributed at the inlet and exit of each unit of the cycle - Pressure drop through steam generator is negligible P2 P3 = 5Mpa - Pressure drop through condenser is negligible P4 P1 = 0.01 Mpa = 10KPa m 20 Kg S Flow through the steam generator: system boundary First law: W no work h 2 ) Q 23 23 m( h 3 done T2 = 48C < Tsat = 263.99C Corresponding to P2 = 5MPa water is a compressed liquid at state (2) h2 hf T 48 C vf h f 188.45 209.33 188.45 T 48 C PCL Psat (48 45) h f 200.98 KJ Kg (50 45) v f 0.001010 (0.001012 0.001010) (48 45) m3 v f 0.001011 (50 45) Kg ( Psat 9.593) (48 45) Psat 11.247 KPa (12.349 9.593) (50 45) h2 200.98 0.001011(5000 11.247) 206.02 KJ Kg Section IV – First Law Applied to Open Systems Page 27 T3 = 450C > Tsat = 263.99C corresponding to P3 = 5Mpa water is a superheated vapor at state(3). h3 = 3316.2 KJ/Kg Q 23 20(3316.2 206.02) 62203.6 KW Flow through the turbine: First Law: Q 34 W 34 m(h4 h3 ) turbine is adiabatic Steam turbine P 10 KPa x 4 h fg P 10 KPa W 34 m (h3 h4 ) system boundary h4 hf 191.83 (0.95)(2392.8) 2465 KJ Kg W34 20(3316.2 2465) 17024 KW Flow through the pump: First law: system boundary h4 hf P 10 KPa Q 12 W12 m(h2 h1 ) pump is adiabatic 191.83 kJ / kg W12 20(191.83 206.02) 283.8KW Power input W12 m(h 1 h 2 ) Section IV – First Law Applied to Open Systems Page 28 Flow through the condenser: system boundary First law: Q 21 W21 m(h 1 h 4 ) no work done in condenser Q12 20(191.83 2465) 45463.4 KW Thermal Efficiency: th th W net Q steam generator (17024 283.8) 0.269 (or 26.9%) (62203.6) Q Check: W net net Q W Q W 34 12 23 4l (17024 – 283.8) = (52203.6 – 45463.4) 16740.2 16740.2 Check!