HW5_solution - Homework 5 solution 4.1 All options are...

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1 Homework 5 solution 4.1 All options are calculated using the following formulas: Turbine: interpolation is needed at the outlet, the notation (lower) and (upper) denote the lower and upper limits for the interpolation. If the inlet entripy is less than the saturated vapor entropy at the outlet pressure, then the reversible outlet state will be wet steam, and the lower and upper limits for interpolation are the saturation values. The variable x denotes the fractional distance of the interval from the lower limit. If the outlet is wet steam, it is equivalent to the quality. x 4 ’=(S 3 -S lower )/(S upper -S lower ), H 4 ’=H lower +x 4 ’(H upper -H lower ) Δ H=Ws= η (H 4 ’-H 3 ), H 4 =H 3 + Δ H Condenser: outlet is assumed saturated liquid, H5 is the saturated liquid value at the condenser pressure, Q L =H 5 -H 4 Pump: Volume is saturated liquid volume at the condenser pressure. H 6 ’=H 5 +V L (P boiler -P condenser )*(1000 kJ/m 3 MPa), where V in m3/kg and P in MPa. H6=H5+ Δ H, where Δ H=Ws=(H 6 ’-H 5 )/ η pump Boiler: QH=H3-H6, where H3 and H6have already been determined. Flowrate=-(net output in MW)/(W s tubine +W s pump ), Q H =flowrate*Q H , Thermal efficiency = (net output)/Q H Option(i) Rankine Cycle Turbine inlet Condenser Overall parameters H 3 (kJ/kg) 3625.80 H 4 (kJ/kg) 2474.75 Net output (MW) 80.00 S 3 (kJ/kgK) 6.9045 H 5 (kJ/kg) 191.81 Flowrate (kg/s) 70.32 P 3 (Mpa) 10.00 P 5 (MPa) 0.01 Total Q H (kJ/s) 240543.79 Q L (kJ/kg) -2282.94 Thermal efficiency 0.33 Turbine outlet S(lower) 0.6492 Pump S(upper) 8.1488 V L (m 3 /kg) 0.001010 x 4 0.83 H 5 (kJ/kg) 191.81 H(lower) 191.81 H 6 ’ (kJ/kg) 201.90 H(upper) 2583.86 Η (pump) 0.75 H 4 2186.98
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HW5_solution - Homework 5 solution 4.1 All options are...

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