Consider an Ideal steam regenerative cycle in which steam enters the turbine at 3 MPa, 400 C and exhausts to the condenser at 10 kPa. Steam is extracted from the turbine at 0.8 MPa for the open feedwater heater (FWH). The feedwater leaves the heater as saturated liquid. The appropriate pumps are used for the water leaving the condenser and FWH. Sketch and calculate the following: (Dead state P.-100 kPa and T.-25 C) (SHOW ALL WORK, INCLUDE TABLES, AND EQUATIONS) (a) Sketch the Cycle and T-s diagram (b) Thermal Efficiency of cycle (e) Carnot Eficiency of cycle (d) Second Law Efficiency of cycle Assume TH- Tro- 1000 "C Write Flow Exergy at states 5 (entrance to turbine), 6 (extraction form the turbine), and 7 (exhaust from the turbine to the condenser) FLOW EXERGY FOR THE STATES) (e) (DO NOT SOLVE THE
solve only the last part. the part E
solutions of part A to D are:
part A is solved in figure.
Cv for pump P1 ;
Wp2 = h2 - h1 = V1 (P2 - P1) = 0.00101 (800-10) = 0.798 KJ/KG
h2 = h1 + Wp1 = 191.81 + 0.798 = 192.61 KJ/KG
Cv for feedwater heater ;
x = m6 / mtotal ( extraction factor )
(1-x)h2 + xh6 = 1 h3
x = h3 - h2 / h6 - h2 = (721.1 - 192.61)/(2891.6 - 192.61) = 0.1958 ;
Cv for pump P2 ;
Wp2 = h4 - h3 = V3 (P4 - P3) ;
= 0.001115 (3000 - 800) = 2.45 KJ/KG
h4 = h3 + Wp2 = 721.1 + 2.45 = 723.55 KJ/KG
Cv for Boiler ;
qH = h5 - h4 = 3230.82 - 723.55 = 2507.3 KJ/KG
CV for turbine ;
2nd law, S7 = S6 = S5 = 6.9211 KJ/KG
P6, S6 = h6 = 2891.6 KJ/KG (super heated steam)
S7 = S6 = S5 = 6.9211
x7 = (6.9211 - 0.6492)/7.501 = 0.83614
h7 = 191.81 + x7 (2392.82) = 2192.55 KJ/KG
Turbine has full flow in HP section and fraction 1-x in LP section;
Wt / m5 = h5 - h6 + (1-x)(h6-h7)
Wt = 899.3 KJ/KG
P2 has the full flow and P1 has the 1-x fraction of the flow;
Wnet = Wt - (1-x)Wp1 - Wp2
Wnet = 899.3 - (1-0.1988) 0.798 - 2.45 = 896.2 KJ/KG
η = Wnet / qH = 896.2 / 2507.3 = 0.357 = 35.7 %
ηc = 1 - (TC / TH) ;
ηc = 1 - ( 45.817 / 1000 ) ; at 10 kpa, TC = 45.817. C ( From steam table ) , TH = 1000. C ( given );
ηc = 0.954 = 95.4 %
Second Law Efficiency = ηth / ηc ;
= 0.357 / 0.954
Second Law Efficiency = 0.3742 = 37.42%
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
