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8 Pages

### Problem05

Course: NE 400, Spring 2010
School: N.C. State
Rating:

Word Count: 732

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proposed A steam cycle for a liquid metal reactor is illustrated in Figure 1. The reactor is to operate at 3600 MW. Compute the cycle efficiency. Assume the high pressure turbine has an efficiency of 90 % and the low pressure turbine has an efficiency of 85 %. You may assume the pumps to be ideal and the pressure change across the sodium pump is 125 psi. 6 4 21 mp Steam Loop Sodium Loop Turbine Superheater...

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proposed A steam cycle for a liquid metal reactor is illustrated in Figure 1. The reactor is to operate at 3600 MW. Compute the cycle efficiency. Assume the high pressure turbine has an efficiency of 90 % and the low pressure turbine has an efficiency of 85 %. You may assume the pumps to be ideal and the pressure change across the sodium pump is 125 psi. 6 4 21 mp Steam Loop Sodium Loop Turbine Superheater Turbine 9 m1 22 3 5 m3 8 m2 2 7 Steam Drum Steam Generator m4 Open Heater 18 17 13 Condenser 10 Core 19 1 11 15 14 m5 20 Sodium Pump 16 12 Figure 1. Steam Cycle for a LMR Advanced 18 Problem Data Point 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 Pressure (psia) 1650 1600 1600 1600 560 160 140 20 1 Temperature (F) Quality 50 % 100% 900 224 476 1600 950 885 Advanced 18 SOLUTION High Pressure Turbine P4 = 1600 psia T4 = 900 F h4 = 1425.2 Btu/lbm s4 = 1.5478 High Pressure Tap At P5 = 560 psia and an entropy of 1.5478, the steam is superheated. From the superheat tables h5s = 1293.9 + (1.5478 1.5431) (1324.9 1293.9) = 1298.99 (1.5717 1.5431) t = h4 h5 h5 = h4 t (h4 h5 s ) h4 h5 s h5 = 1425.2 - (0.90)(1425.2 1298.99) = 1322.61 Btu/lbm High Pressure Turbine Exhaust x6s = s4 s f s fg @ P6 = 160 psia sf = 0.5206 sfg = 1.0435 x6 s = 1.5478 0.5206 = 0.9844 1.0435 h6s = h f + x6s h fg @ P6 = 160 psia hf = 336.1 Btu/lbm hfg = 859 Btu/lbm h6 s = 336.1 + 0.9844 859 = 1181.7 t = h4 h6 h6 = h4 t ( h4 h6s ) h4 h6s h6 = 1425.2 (0.9)(1425.2 1181.7) = 1206.05 Low Pressure Turbine The entropy s6 at P6 = 160 psia and h6 = 1206.05 Btu/lbm is obtained by interpolating from the superheat tables Advanced 18 s6 = 1.5641 + (1206.05 1195.1) = 1.5771 (1.5906 1.5641) (1217.4 1195.1) High Pressure Tap At a pressure of 140 psia and an entropy of s6=1.5771 the steam is super heated and the ideal enthalpy at 7 is obtained by interpolating from the superheat tables h7 s = 1193.0 + (1.5771 1.5752) = 1194.59 (1220.9 1193.0) (1.6085 1.5752) t = h6 h7 h7 = h6 t ( h6 h7 s ) h6 h7 s h7 = 1206.05 (0.85)(1206.05 1194.59) = 1196.76 Low Pressure Turbine Tap x8 s = s6 s f s fg @ P8 = 20 psia sf = 0.3358 sfg = 1.3962 x8 s = 1.5771 0.3358 = 0.8891 1.3962 h8s = h f + x8s h fg @ P8 = 20 psia hf = 196.27 Btu/lbm hfg = 960.1 Btu/lbm h8 s = 196.27 + 0.8891 960.1 = 1049.85 t = h6 h8 h8 = h6 t ( h6 h8 s ) h6 h8 s h8 = 1206.05 (0.85)(1206.05 1049.85) = 1073.28 Low Pressure Turbine Exhaust x9 s = s6 s f s fg @ P9 = 1 psia sf = 0.1326 sfg = 1.8455 Advanced 18 x9 s = 1.5771 0.1326 = 0.7827 1.8455 h9s = h f + x9s h fg @ P9 = 1 psia hf = 69.73 Btu/lbm hfg = 1036.1 Btu/lbm h9 s = 69.73 + 0.7827 1036.1 = 880.7 t = h6 h9 h9 = h6 t ( h6 h9s ) h6 h9s h9 = 1206.05 (0.85)(1206.05 880.7) = 929.5 Condensate Pump wcp = ( P P ) = (0.0161)(140 1)(144 / 778) = 0.414 11 10 h10 = h f @ P = 1 psia = 69.73 Btu / lbm 10 h11 = h10 wcp h11 = 69.73 + 0.414 = 70.14 h12 = h f @ P = 20 psia = 196.27 Btu / lbm 12 h13 h f @ T13 = 224 F = 192.27 Btu / lbm h14 = h f @ P = 140 psia = 325 Btu / lbm 14 Low Pressure Feed Pump wlpfp = ( P P ) = (0.018)(1600 140)(144 / 778) = 4.86 15 14 h15 = h14 wlpfp h15 = 325 + 4.86 = 329.86 h16 = h f @ P = psia 560 = 460.9 + (560 550) 16 h17 h f @ T17 = 476 F = 459.9 Btu / lbm (471.7 460.9) = 463.06 (600 550) h18 = h f @ P = 1600 psia = 624.2 18 Boiler Feed Pump Advanced 18 w fp = ( P P ) = (0.02387)(1650 1600)(144 / 778) = 0.221 1 18 h1 = h18 w fp h1 = 624.2 + 0.221 = 624.42 h2 = h f + x2 h fg @ P2 = 1600 psia hf = 624.2 Btu/lbm hfg = 540.3 Btu/lbm h2 = 642.2 + (0.5)540.3 = 894.35 h3 = hg @ P3 = 1600 psia = 1164.5 Steam Drum & & & & m1h17 + m5h2 = m5h18 + m1h3 & m5 h3 h17 1164.5 459.9 = = = 2.608 & m1 h2 h18 894.35 624.2 High Pressure Heater & & & & m1h17 + m2 h16 = m1h15 + m2 h5 & m2 h17 h15 459.9 329.86 = = = 0.1513 & m1 h5 h16 1322.61 463.06 Open Feed Heater & & & & & & m2 h16 + m3 h7 + (m1 m2 m3 )h13 = m1h14 && && & m3 h14 (m2 /m1 )h16 (1 m2 /m1 )h13 = & m1 h7 h13 & m3 325 (0.1513) 463.06 (1 0.1513) 192.27 = = 0.0913 & m1 1196.76 192.27 Low Pressure Feed Heater & & & & m4 (h8 h12 ) = (m1 m2 m3 )(h13 h11 ) & && && m4 (1 m2 /m1 m3 /m1 )(h13 h11 ) = & m1 h8 h12 & m4 (1 0.1513 0.0913)(192.27 70.14) = = 0.1055 & m1 1073.28 196.27 Advanced 18 Superheater & & m p C p (T21 T22 ) = m1 (h4 h3 ) & mp h4 h3 = & m1 C p (T21 T22 ) & mp 1425.2 1164.5 = = 13.28 & m1 0.302 (950 885) Steam Generator & & m p C p (T22 T19 ) = m5 (h2 h1 ) T19 = T22 = 885 = 710 && m5 m1 (h2 h1 ) && m1 m p C p 2.608 (894.35 624.42) 13.28 .303 Sodium Pump 125 wsp = (P) = (144 / 778) = 0.431 53.7 C p (T20 T19 ) = wsp T20 = T19 wsp Cp 0.431 = 711.4 0.307 T20 = 710 + Cycle Efficiency = & & & & & & & & Wnet (Wthp + Wtlp + Wcp + Wlpfp + W fp + Wsp ) / m1 = & & /m Q Q &1 & & & & & Wthp = m1h4 m2 h5 (m1 m2 )h6 & Wthp && && = h4 (m2 / m1 )h5 (1 m2 / m1 )h6 & m1 & Wthp = 1425.2 (0.1513) 1322.61 (1 0.1513) 1206.05 & m1 = 201.51 Advanced 18 & & & & & & & & & Wtlp = (m1 m2 )h6 m3 h7 m4 h8 (m1 m2 m3 m4 )h9 & Wtlp && && && && && && = (1 m2 / m1 )h6 (m3 / m1 )h7 (m4 / m1 )h8 (1 m2 / m1 m3 / m1 m4 / m1 )h9 & m1 = (1 0.1513) 1206.05 (0.0913) 1196.76 (0.1055) 1073.28 (1 0.1513 0.0913 0.1055) 929.5 = 195.14 & & & & Wcp = (m1 m2 m3 ) wcp & Wcp && && = (1 m2 / m1 m3 / m1 ) wcp & m1 = (1 0.1513 0.0913)(0.414) = 0.314 & & Wlpfp = m1wlpfp & Wlpfp = wlpfp & m1 = 4.86 & & W fp = m5 w fp & W fp m5 & w fp = & & m1 m1 = 2.608 (0.221) = 0.576 & & Wsp = m p wsp & & Wsp m p w = & & sp m1 m1 = 13.28 (.431) = 5.72 && Q = m p C p (T21 T20 ) & & Q mp = C p (T21 T20 ) & & m1 m1 = 13.28 0.303 (950 711.4) = 960.1 = = & & & & & & & & Wnet (Wthp + Wtlp + Wcp + Wlpfp + W fp + Wsp ) / m1 = & & /m Q Q &1 (201.51 + 195.14 0.314 4.86 0.576 5.72) 960.1 = 40.1% Advanced 18
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