Solution set 9

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Jan 9, 2024

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EMAE 251: Thermodynamics Solution Set 9 Problem 1 [Chapter 8] 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, 520°C 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 15°C with a mass flow rate of 114.79 kg/s. Determine: a.) the thermal efficiency. *Isentropic efficiencies = n State 1- P1= 100 bar, T1= 520 C => Refprop h1=3425.1 kJ/kg, s1=6.66 kJ/(kgK) State 2- P2s= 1 bar, s2s=s1=6.66 kJ/(kgK) => x2s= (s2s-sf)/(sg-sf)=0.8849, h2s=hf+(x2s)(hfg)=2415.6 kJ/kg, h2=h1-nt(h1-h2s)=3426.4-.81(3426.4-2415.6)=2607.4 kJ/kg State 3- P3=1 bar, x3=0 => h3=417.46kJ/kg, v1=0.0010432 m^3/kg State 4- P4=100 bar => h4=h3 + (v3(P4-P3))/(np)=430.7 kJ/kg Cooling Water- Tin=15 C => hin=62.99kJ/kg Solving for Thermal Efficiency- n= ((h1-h2)-(h4-h3))/(h1-h4)= 26.9%/.

b.) the temperature of the cooling water exiting the condenser, in °C. Wnet= Wt-Wp = m[(h1-h2)-(h4-h3)] m= (3000 kJ/s)/[(3426.4-2607.4)-(430.7-417.46)kJ/kg]= 3.73 kg/s Qout= m(h2-h3) = m(cooling water)(hcoolingwater,out - hcoolingwater,in) hcoolingwater,out = ((3.73kg/s)(2607.4-417.46)kJ/kg)/114.79kg/s + 62.99 kJ/kg= 134.15 kJ/kg hcoolingwater,out= hf(Tcoolingwater,out) Tcoolingwater, out = 32 C Problem 2 [Chapter 8] Water is the working fluid in an ideal Rankine cycle with reheat. Superheated vapor enters the turbine at 10 MPa, 320°C, and the condenser pressure is 8 kPa. Steam expands through the first-stage turbine to 1 MPa and then is reheated to 320°C. Determine for the cycle a.) the heat addition, in kJ per kg of steam entering the first-stage turbine. State 1: P1=10 MPa, T1=320°C, h1=2780 kJ, s1=5.71kJ/kg-k State 2: P2= 1 MPa, s1=s2=5.71 KJ/kg-k, h2=2380 kJ/kg-k State 3: P3=1 MPa , T3=320°C h3=3090 kJ/kg, s3= 7.21 kJ/kg-k State 4: P4=8 kPa, s3=s4=7.21 kJ/kg-k, h4=2250 kJ/kg State 5: P5= 8 kPa, x5=0 h5=174 kJ/kg v5=0.00101m^3 State 6: P6= 10x10^3 h6=h5+v5(P6-P5)=174+0.00101(10x10^3-8) → h6=184.09 kJ/kg Qin = (h1-h6)+(h3-h2) → (2780-184.09) + (3090-2380) → Qin = 3305.9 kJ/kg Qout=(h4-h5) → 2250-174 → Qout = 2076 kJ/kg b.) the thermal efficiency. = 1 - (Qout/Qin) = 1 - (2076/3305.9) → =37.2% η ?ℎ η ?ℎ

c.) the heat transfer from the working fluid passing through the condenser to the cooling water, in kJ per kg of steam entering the first-stage turbine. Qout=(h4-h5) → 2250-174 → Qout = 2076 kJ/kg

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Problem 3 For the cycle in the previous problem, reconsider the analysis assuming the pump and each turbine stage have an isentropic efficiency of 80%. Determine for the cycle: [Chapter 8] Water is the working fluid in an ideal Rankine cycle with reheat. Superheated vapor enters the turbine at 10 MPa, 320°C, and the condenser pressure is 8 kPa. Steam expands through the first-stage turbine to 1 MPa and then is reheated to 320°C. Determine for the cycle State 1: P1=10 MPa, T1=320°C, h1=2780 kJ, s1=5.71kJ/kg-k State 2: P2= 1 MPa, s2s=s1=5.71 KJ/kg-k, h2s=2380 kJ/kg-k State 3: P3=1 MPa , T3=320°C h3=3090 kJ/kg, s3= 7.21 kJ/kg-k

State 4: P4=8 kPa, s3=s4s=7.21 kJ/kg-k, h4s=2250 kJ/kg State 5: P5= 8 kPa, x5=0 h5=174 kJ/kg v5=0.00101m^3 State 6: P6= 10x10^3 a.) the heat addition, in kJ per kg of steam entering the first-stage turbine. η ? = 0. 8 = ℎ 1 −ℎ 2 ℎ 1 −ℎ 2? ℎ 2 = ℎ 1 − η ? (ℎ 1 − ℎ 2? ) = 2780 − 0. 8(2780 − 2380) = 2461 𝑘𝐽/𝑘? − 𝑘 η ? = 0. 8 = ℎ 3 −ℎ 4 ℎ 3 −ℎ 4? ℎ 4 = ℎ 3 − η ? (ℎ 3 − ℎ 4? ) = 3090 − 0. 8(3090 − 2250) = 2040 𝑘𝐽/𝑘? − 𝑘 η ? = 0. 8 = ? 5 (? 6 −? 5 ) ℎ 6 −ℎ 5 ℎ 6 = ℎ 5 + ? 5 (? 6 −? 5 ) η ? = 184 𝑘𝐽/𝑘? − 𝑘 heat addition = ? 𝑖? ' ? ' = (ℎ 1 − ℎ 6 ) + (ℎ 3 − ℎ 2 ) = (2780 − 184) + (3090 − 2461) = 3227 𝑘𝐽/𝑘? b.) the thermal efficiency. ? ' ? ' = (ℎ 1 −ℎ 2 )+(ℎ 3 −ℎ 4 )−(ℎ 6 −ℎ 5 ) ? ' ? ' = (2780−2461)+(3090−2418)−(184−174) 3227 = 0. 3038 = 30. 38% c.) the heat transfer from the working fluid passing through the condenser to the cooling water, in kJ per kg of steam entering the first-stage turbine. ? ??? ' ? ' = (ℎ 4 − ℎ 5 ) = (2420. 7 − 173. 8) = 2246. 9 𝑘𝐽/𝑘?

Here is an accurately-scaled T-s diagram for problems 2 and 3. Note that states 5, 6, and 6s are nearly indistinguishable at this scale since the temperature change across the pump is so small.

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Points earned Problem Name Points Contribution (initial solution, improvement, question, answer) 1 Owen Myer 4 Initial Solution 2 Jack Bryer 5 Initial Solution 3 Joseph Ahmed 5 Initial Solution Brock Joyce 2 Improvement