Analyze a Reheat-Regenerative Rankine cycle (with a closed heater). The cycle sp

Question

Analyze a Reheat-Regenerative Rankine cycle (with a closed heater). The cycle specifications are given. Use Qin @ TH = max. cycle T = T1 and Qout @ TC = min. cycle T = T5, and “dead state” at 300 K, 100 kPa.

B) Analyze a Reheat-Regenerative Rankine cycle (with a closed heater). The cycle specifications are given. Use Qin @ T-max. cycle T = T1 and Qout @ Tc = min. cycle T = T5, and ''dead state', at 300 K, 100 kPa. Find the following: State properties and mass flowrates at each section (clearly organize a table & plot T-s diagram) INCLUDE a “large" state table with room for all numbered states plus (2s), (4s), (6s) INCLUDE a "full-scale" T-s diagram clearly showing all states-use the class T-s chart for steam Cycle heat input (MW), heat rejected (MW), net power output (MW), thermal efficiency INCLUDE a table of specific work and specific heat transfer for each component except the heater With units of kJ/kq of main steam at (1) Exergy destruction rate (MW) for the steam gen., condenser & feed HTR (ho-113.3 kJkg, So-0.3954 kJ/kg-K Cycle Specifications Steam generator mass flowrate at (1) = 200 kg/s Superheat Temperature (1) Reheat Temperature @ (3) Primary steam generator pressure Reheater pressure Condenser pressure Reheat section 565 °C 510 °C 200 bar (20000 kP 10 bar (1000 kPa) 0.04 bar (4 kPa) Low-pressure turbine Closed feedwater heater pressure, p8 8 bar (800 kPa) Qin (Lower than p2 to allow friction & for control valves.) Boiler feedpump exit pressure 220 bar (22000 kPa Higher than p1 to allow friction &for; control valves.) High- Turbine isentropic efficiencies Pump isentropic efficiencies 0.90 (both) 0.70 (both) Unless otherwise specified Neglect external heat transfer, ke, & pe Condenser&Closed; Heater Condensate Drain Sat. Liquid Closed Heater Main Flow is not always Sat. Liquid Steam out Closed Condenser Set up Interactive Thermo IT V3 not required but useful ater 6 Pump (8) No mixing

Explanation / Answer

and the formulas for finding the heat rejected is Q = (1- m1-m2-m3) (h7-h8)

for finding out the m1 m2 m3 the equations are as fallows

m1h2 + ( 1- m1)h13 =h14

m2h5 + (1 - m2-m3) h11 = (1 - m1 ) h12

m3h6 = ( 1- m1 -m2 -m3)h9 = (1- m1 -m2 )h10

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