Chapter 8 Examples

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1 8.25 Superheated steam at 20 MPa, 560 o C enters the turbine of a vapor power plant. The pressure at the exit of the turbine is 0.5 bar, and liquid leaves the condenser at 0.4 bar at 75 o C. The pressure is increased to 20.1 MPa across the pump. The turbine and pump have isentropic efficiencies of 81 and 85%, respectively. Cooling water enters the condenser at 20 o C with a mass flow rate of 70.7 kg/s and exits the condenser at 38 o C. For the cycle, determine (a) the mass flow rate of steam, in kg/s. (b) the thermal efficiency. KNOWN: Water is the working fluid in a vapor power plant. Data are given at various states in the cycle. FIND: (a) the mass flow rate of steam, in kg/s and (b) the thermal efficiency. 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. Stray heat transfer in the turbine, condenser, and pump is ignored. 3. Kinetic and potential energy effects are negligible. ANALYSIS: First fix each principal state. State 1 : p 1 = 20 MPa (200 bar), T 1 = 560 o C → h 1 = 3423.0 kJ/kg, s 1 = 6.3705 kJ/kg∙K State 2s : p 2s = p 2 = 0.5 bar, s 2s = s 1 = 6.3705 kJ/kg∙K → x 2s = 0.8119, h 2s = 2212.2 kJ/kg State 2 : p 2 = 0.5 bar, h 2 = 2442.3 kJ/kg (see below) t W ± Turbine Cooling water 2 Condenser Pump p W ± 3 4 1 Boiler in Q ± p 3 = 0.4 bar T 3 = 75 o C p 1 = 20 MPa T 1 = 560 o C p 2 = 0.5 bar p 4 = 20.1 MPa out Q ± h t = 81% h p = 85% 6 5 T 5 = 20 o C T 6 = 38 o C kg/s 7 . 70 cw m ± T 4 3 2 1 20 MPa s 2s T 1 = 560 o C 0.5 bar 0.4 bar 20.1 MPa
2 kg kJ ) 2 . 2212 0 . 3423 )( 81 . 0 ( kg kJ 0 . 3423 ) ( 2 1 t 1 2 2 1 2 1 t ± ± ± ± o ± ± s s h h h h h h h h h h = 2442.3 kJ/kg State 3 : p 3 = 0.4 bar, T 3 = 75 o C From Table A-2 p 3 > p sat @ 75 o C. Thus, state 3 is a sub- cooled liquid state. Since the pressure is low, h 3 h f3 at 75 o C = 313.93 kJ/kg, v 3 v f3 at 75 o C = 0.0010259 m 3 /kg State 4 : p 4 = 20.1 MPa (201 bar), h 4 = 338.14 kJ/kg (see below) p 3 4 3 3 4 3 4 3 4 3 p ) ( ) ( h h p p h h h h p p ± ² o ± ± v v m N 1000 kJ 1 kPa 1 m N 1000 bar 1 kPa 100 85 . 0 bar ) 4 . 0 201 )( kg m 0010259 . 0 ( kg kJ 93 . 313 2 3 4 ± ² h = 338.14 kJ/kg State 5 : T 5 = 20 o C, liquid → h 5 h f5 at 20 o C = 83.96 kJ/kg State 6 : T 6 = 38 o C, liquid → h 6 h f6 at 38 o C = 159.21 kJ/kg (a) The mass flow rate of the steam can be determined by writing an energy balance for the condenser. With no stray heat transfer with the surroundings and no work, the energy balance for the condenser reduces to ) ( ) ( 0 6 5 cw 3 2 h h m h h m ± ² ± ± ± where m ± is the mass flow rate of the steam and cw m ± is the mass flow rate of the cooling water. Rearranging to solve for the mass flow rate of steam gives ) ( ) ( 3 2 5 6 cw h h h h m m ± ± ± ± Substituting values and solving give kJ/kg ) 93 . 313 .3 2442 ( kJ/kg ) 96 . 83 .21 159 ( kg/s) 0.7 7 ( ± ± m ± = 2.50 kg/s (b) The thermal efficiency is
3 ) ( ) ( ) ( 4 1 3 4 2 1 h h h h h h ± ± ± ± h Substituting enthalpy values and solving yield kJ/kg ) .14 338 3423.0 ( kJ/kg ) 93 . 313 .14 338 ( kJ/kg ) 3 . 2442 3423.0 ( ± ± ± ± h = 0.3101 (31.01%)
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 s . 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%
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 ± ± ± ± o ± ± 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 ± ± ± ± o ± ± 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 ± ² o ± ± 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 ± ² 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- 3 , 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 h h x = 0.9330 The cycle thermal efficiency is m Q m Q m W m W m W m Q m W ± ± ± ± ± ± ± ± ± ± ± ± ± ± / / / / / / / 23 61 p t2 t1 in cycle ² ± ² h ) ( ) ( ) ( ) ( ) ( 2 3 6 1 5 6 4 3 2 1 h h h h h h h h h h ± ² ± ± ± ± ² ± h
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