Applied Thermodynamics - Solutions to Homework

8.12 A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor to develop net cycle power of 3 MW. Steam exits the reactor core at 100 bar, 520oC and expands through the turbine to the condenser pressure of 1 bar. Saturated liquid exits the condenser and is pumped to the reactor pressure of 100 bar. Isentropic efficiencies of the turbine and pump are 81% and 78%, respectively. Cooling water enters the condenser at 15oC with a mass flow rate of 114.79 kg/s. Determine (a) the thermal efficiency. (b) the temperature of the cooling water exiting the condenser, in oC. KNOWN: A nuclear power plant based on the Rankine cycle operates with a boiling-water reactor between specified pressures of 100 bar and 1 bar to produce 3 MW of net power. FIND: Determine (a) the thermal efficiency and (b) the temperature of the cooling water exiting the condenser, in oC. SCHEMATIC AND GIVEN DATA: p1 = 100 bar T1 = 520oC 1 Qin Turbine Reactor Core Wt 2 p = 1 bar 2 Wcycle 3 MW Condenser 4 p4 = p1 = 100 bar Q out 6 Cooling water 5 T5 = 15oC mcooling water 114 .79 kg/s Pump Wp 3 p3 = p2 = 1 bar x3 = 0 (saturated liquid) 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. Flow through the reactor core and condenser occurs at constant pressure. 3. Stray heat transfer in the turbine, condenser, and pump is ignored. 4. Kinetic and potential energy effects are negligible. 5. Condensate exits the condenser as saturated liquid. 6. For the cooling water, h ≈ hf(T). 1 ANALYSIS: First fix each principal state. State 1: p1 = 100 bar, T1 = 520oC → h1 = 3425.1 kJ/kg, s1 = 6.6622 kJ/kg∙K State 2s: p2s = p2 = 1 bar, s2s = s1 = 6.6622 kJ/kg∙K → x2s = 0.8849, h2s = 2415.6 kJ/kg State 2: p2 = 1 bar, h2 = 2607.4 kJ/kg (see below) ht  h1 h2 h1 h2 s h h1 ht (h1  h2 s)  3425.1 2 kJ kJ (0.81)(3425.1 2415.6) = 2607.4 kJ/kg kg kg State 3: p3 = 1 bar, saturated liquid → h3 = hf3 = 417.46 kJ/kg, v3 = vf3 = 0.0010432 m3/kg State 4: p4 = 100 bar, h4 = 430.70 kJ/kg (see below) hp  4h 417.46 kJ  kg (0.0010432 v3 ( p4  p3 ) v ( p  p3 ) h4 h3  3 4 h4  h3 hp m3 N )(100 1) bar 1000 2 kg 100 kPa 1 kJ m = 430.70 kJ/kg 0.78 1 bar 1 kPa 1000 N m State 5: T5 = 15oC, liquid → h5 ≈ hf5 at T5 = 62.99 kJ/kg (a) The thermal efficiency is h  Wt / m Wp / m Qin / m h ) (h4 h3) (h   1 2 (h1 h4 ) Substituting enthalpy values and solving yield h  (3425.1  2607.4) kJ/kg (430.70 417.46) kJ/kg = 0.2687 (26.87%) (3425.1 430.70) kJ/kg (b) The temperature of the cooling water exiting the condenser is determined by writing an energy balance for the condenser, solving for the specific enthalpy of the cooling water exiting the condenser, and finding the corresponding temperature for the specific enthalpy of the liquid in Table 2. With no stray heat transfer with the surroundings and no work, the energy balance for the condenser reduces to 0  m( h2  h3 ) mcw (h5 h6 ) 2 where m is the mass flow rate of the steam and mcw is the mass flow rate of the cooling water. Rearranging to solve for the specific enthalpy of the cooling water exiting the condenser, h6, gives m (h2  h3 ) h6 h5  mcw The mass flow rate of steam, m, is determined from the net cycle power m m Wcycle Wcycle  W h2 ) ( h4 h3 ) t / m Wp / m ( h1  3000 kW 1 kJ/s = 3.73 kg/s (3425.1 2607.4) kJ/kg (430.70  417.46) kJ/kg 1 kW Solving for the specific enthalpy at the condenser exit gives h6 62.99 kJ/kg  (3.73 kg/s) (2607.4 417.46) kJ/kg = 134.15 kJ/kg 114.79 kg/s Since the cooling water is liquid, h6 ≈ hf6 at T6. From Table A-2, T6 ≈ 32oC. 3 8.33 Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520oC. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500oC. 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: Qin Reheat Section Steam Generator p3 = p2 = 6 MPa T3 = 500oC 3 2 p2 = 6 MPa 1 Turbine 1 p1 = 28 MPa T1 = 520oC Wt Turbine 2 h t1 = h t2 = 78% p6 = p1 = 28 MPa 4 p4 = p 6 Condenser h p = 82% 5 Pump Wp Qout p5 = p4 x5 = 0 (saturated liquid) 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: p1 = 28 MPa, T1 = 520oC → h1 = 3192.3 kJ/kg, s1 = 5.9566 kJ/kg∙K 1 State 2s: p2s = p2 = 6 MPa, s2s = s1 = 5.9566 kJ/kg∙K → h2s = 2822.2 kJ/kg State 2: p2 = 6 MPa, h2 = 2903.6 kJ/kg (see below) h t1  h1 h2 h1  h2s h2 h h2 s) 3192.3 1 ht1(h1  kJ kJ (0.78)(3192.3 2822.2 ) = 2903.6 kJ/kg kg kg State 3: p3 = 6 MPa, T3 = 500oC → h3 = 3422.2 kJ/kg, s3 = 6.8803 kJ/kg∙K State 4s: p4s = p4 = 6 kPa, s4s = s3 = 6.8803 kJ/kg∙K → x4s = 0.8143, h4s = 2118.8 kJ/kg State 4: p4 = 6 MPa, h4 = 2405.5 kJ/kg (see below) h t2  h3  h4 h3 h4s h4 h h4 s) 3422.2 3 ht2 (h3  kJ kJ = 2405.5 kJ/kg (0.78)(3422.2 2118.8) kg kg State 5: p5 = 6 kPa, saturated liquid → h5 = hf5 = 151.53 kJ/kg, v5 = vf5 = 0.0010064 m3/kg State 6: p6 = p1 = 28 MPa, h6 = 185.89 kJ/kg (see below) hp  v5 ( p6  p5 ) v ( p p5 ) h6 h5  5 6 h6  h5 hp 0.0010064 h6 151.53 kJ/kg  m3 N (28,000 6)kPa 1000 kg 1 kJ m2 = 185.89 kJ/kg 0.82 1 kPa 1000 Nm (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, hf4 = 151.53 kJ/kg and hfg4 = 2415.9 kJ/kg, as follows: h hf 4 (2405.5 151.53) kJ/kg = 0.9330 x4  4  hfg 4 2415.9 kJ/kg The cycle thermal efficiency is Wcycle / m h  Qin / m h  (h1  W t1 / m Wt2 / mWp / m Q61 / m Q23 / m ( h3 h4 ) (h6  h5 ) h 2)  h6) ( h3 h2) ( h1  2 Substituting enthalpy values and solving yield a thermal efficiency of  .6) kJ/kg ( (3192.3 2903 3422.2  2405.5) kJ/kg (185.89 151.53) kJ/kg (3192.3  185.89) kJ/kg (3422.2 2903.6) kJ/kg h  h = 0.3606 (36.06%) (b) For p ranging from 4 kPa to 70 kPa, IT gives the following results: IT Output for p4 = 6 kPa IT Code // Known Properties eff_thermal h1 h2 h2s h3 h4 h4s h5 h6 p2 p2s p4s p5 p6 s1 s2s s3 s4s v5 x4 eff_p eff_t1 eff_t2 p1 p3 p4 T1 T3 x5 p1 = 28000 // kPa T1 = 520 // oC p6 = p1 p3 = 6000 // kPa T3 = 500 // oC p2 = p3 p2s = p2 p4 = 6 // kPa p4s = p4 p5 = p4 x5 = 0 // Known Operating Parameters eff_t1 = 0.78 eff_t2 = 0.78 eff_p = 0.82 // Calculations for Quality at State 4 h3 = h_PT("Water/Steam", p3, T3) s3 = s_PT("Water/Steam", p3, T3) s4s = s3 h4s = h_Ps("Water/Steam", p4s, s4s) h4 = h3 - eff_t2*(h3 - h4s) x4 = x_hP("Water/Steam", h4, p4) // Calculations for Thermal Efficiency h1 = h_PT("Water/Steam", p1, T1) s1 = s_PT("Water/Steam", p1, T1) s2s = s1 h2s = h_Ps("Water/Steam", p2s, s2s) h2 = h1 - eff_t1*(h1 - h2s) h5 = hsat_Px("Water/Steam", p5, x5) v5 = vsat_Px("Water/Steam", p5, x5) h6 = h5 + (v5*(p6 - p5)/eff_p) eff thermal = ((h1 - h2) + (h3 - h4) - (h6 - h5))/((h1 - h6) + (h3 - h2)) 3 0.3606 3192 2903 2821 3422 2405 2118 151 185.4 6000 6000 6 6 2.8E4 5.956 5.956 6.879 6.879 0.001007 0.933 0.82 0.78 0.78 2.8E4 6000 6 520 500 0 Results from IT for p4 = 6 kPa correspond closely to the results obtained using steam tables and hand calculations. Graphical Results from IT are shown below: Quality versus Condenser Pressure 1 0.4 Thermal Efficiency versus Condenser Pressure 0.99 0.98 0.3 0.97 0.96 0.2 0.95 0.94 0.1 0.93 0.92 0 5 10 15 20 25 30 35 40 Pressure [kPa] 45 50 55 60 65 0 0 70 10 20 30 40 50 60 Condenser Pressure (kPa) In general, as the condenser pressure increases the quality of the steam increases and the thermal efficiency decreases since the average temperature of heat rejection is higher. As shown by the Quality versus Condenser Pressure Graph for the conditions of this problem, when the condenser pressure reaches approximately 70 kPa, the liquid-vapor mixture quality becomes 1. Above this pressure steam is superheated vapor and the quality is not defined. 4 70