ME300HW11Solution - Pump work consumption is small compared...

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1 8.33 Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520 o C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500 o C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure, p . (a) For p = 6 kPa, determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency. (b) Plot the quantities of part (a) versus p ranging from 4 kPa to 70 kPa. KNOWN: A supercritical reheat cycle operates with steam as the working fluid. FIND: (a) For p = 6 kPa, determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency, and (b) plot the quantities of part (a) versus p ranging from 4 kPa to 70 kPa. SCHEMATIC AND GIVEN DATA: ENGINEERING MODEL: 1. Each component of the cycle is analyzed as a control volume at steady state. The control volumes are shown on the accompanying sketch by dashed lines. 2. For all components stray heat transfer is ignored. 3. Flow through the steam generator, reheater, and condenser is at constant pressure. 4. Kinetic and potential energy effects are negligible. ANALYSIS: First fix each principal state. State 1: p 1 = 28 MPa, T 1 = 520 o C → h 1 = 3192.3 kJ/kg, s 1 = 5.9566 kJ/kg∙K t W 4 Condenser out Q Pump p W 5 6 1 Steam Generator in Q p 5 = p 4 x 5 = 0 (saturated liquid) p 1 = 28 MPa T 1 = 520 o C p 4 = p p 6 = p 1 = 28 MPa Turbine 2 Turbine 1 2 3 p 2 = 6 MPa p 3 = p 2 = 6 MPa T 3 = 500 o C Reheat Section h t1 = h t2 = 78% h p = 82%
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2 State 2s : p 2s = p 2 = 6 MPa, s 2s = s 1 = 5.9566 kJ/ kg∙K h 2s = 2822.2 kJ/kg State 2 : p 2 = 6 MPa, h 2 = 2903.6 kJ/kg (see below) kg kJ ) 2 . 2822 3 . 3192 )( 78 . 0 ( kg kJ 3 . 3192 ) ( 2 1 t1 1 2 2 1 2 1 t1 s s h h h h h h h h h h = 2903.6 kJ/kg State 3: p 3 = 6 MPa, T 3 = 500 o C → h 3 = 3422.2 kJ/kg, s 3 = 6.8803 kJ/kg∙K State 4s : p 4s = p 4 = 6 kPa, s 4s = s 3 = 6.8803 kJ/kg∙K → x 4s = 0.8143, h 4s = 2118.8 kJ/kg State 4 : p 4 = 6 MPa, h 4 = 2405.5 kJ/kg (see below) kg kJ ) 8 . 2118 2 . 3422 )( 78 . 0 ( kg kJ 2 . 3422 ) ( 4 3 t2 3 4 4 3 4 3 t2 s s h h h h h h h h h h = 2405.5 kJ/kg State 5 : p 5 = 6 kPa , saturated liquid → h 5 = h f5 = 151.53 kJ/kg, v 5 = v f5 = 0.0010064 m 3 /kg State 6 : p 6 = p 1 = 28 MPa, h 6 = 185.89 kJ/kg (see below) p 5 6 5 5 6 5 6 5 6 5 p ) ( ) ( h h p p h h h h p p v v m N 1000 kJ 1 kPa 1 m N 1000 82 . 0 kPa ) 6 000 , 28 ( kg m 0010064 . 0 kJ/kg 53 . 151 2 3 6 h = 185.89 kJ/kg (a) The quality of the steam at the exit of the second stage of the turbine (state 4) is determined using values from Table A-4, h f4 = 151.53 kJ/kg and h fg4 = 2415.9 kJ/kg, as follows: kJ/kg 9 . 2415 kJ/kg ) 53 . 151 5 . 2405 ( 4 fg 4 f 4 4 h
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