Consider an ideal steam regenerative cycle in which stearm oin 700 F, and exhaus
ID: 1718469 • Letter: C
Question
Consider an ideal steam regenerative cycle in which stearm oin 700 F, and exhausts to the condenser at 2 lbf/in.2 Steam is extracted from the steam regenerative cycle in which steam enters the turbine at 500 urbine at 120 Ibf/in.2 and also at 20 lbin2 for heating the boiler feedwater in two open Ieedwater heaters. The feedwater leaves each heater at the temperature of the condensing feedwater steam. The appropriate pumps are used for the water leaving the condenser and the two eedwater heaters. Calculate the cycle thermal efficiency and the net work per Ibm of stearn. ?- Mg 132 t0 10sExplanation / Answer
This is a standard Rankine cycle with an open FWH
C.V Pump P1
wP1 = h2 - h1 = v1(P2 - P1) = 0.00101(800 - 10) = 0.798 kJ/kg
=> h2 = h1 + wP1 = 191.81 + 0.798 = 192.61 kJ/kg
C.V. FWH Call m.
6 / m.
tot = x (the extraction fraction)
(1 - x) h2 + x h6 = 1 h3
x =
h3 - h2
h6 - h2
=
721.1 - 192.61
2891.6 - 192.61 = 0.1958
C.V 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 Boiler: qH = h5 - h4 = 3230.82 - 723.55 = 2507.3 kJ/kg
CV Turbine
2nd Law s7 = s6 = s5 = 6.9211 kJ/kg K
P6 , s6 => h6 = 2891.6 kJ/kg (superheated vapor)
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
W.
T / m.
5 = h5 - h6 + (1 - x) (h6 - h7)
wT = 3230.82 – 2891.6 + (1 - 0.1988) ( 2891.6 – 2192.55) = 899.3
P2 has the full flow and P1 has the fraction 1-x of the flow
wnet = wT - (1 - x) wP1 - wP2
= 899.3 - (1 - 0.1988)0.798 – 2.45 = 896.2 kJ/kg
Kcycle = wnet / qH = 896.2 / 2507.3 = 0.357
Solution:
Desired exit State 4: P4 = 500 kPa, sat. vap. => x4 = 1.0, T4 = 151.9qC
h4 = hg = 2748.7 kJ/kg, s4 = sg = 6.8212 kJ/kg-K
Inlet State: 20qC, 100 kPa h1 = hf = 83.94 kJ/kg, v1 = vf = 0.001002 m3
/kg
Without Cogeneration; The water is pumped up to 500 kPa and then heated in
the steam generator to the desired exit T.
C.V. Pump: wPw/o = v1( P4- P1) = 0.4 kJ/kg
h2 = h1 + wPw/o = 84.3 kJ/kg
C.V. Steam Generator: qw/o = h4 - h2 = 2664.4 kJ/kg
With Cogeneration; The water is pumped to 5 MPa, heated in the steam
generator to 400qC and then flows through the turbine with desired exit state.
C.V. Pump: wPw = ³ vdP = v1( P2- P1) = 4.91 kJ/kg
h2 = h1 + wPw = 88.85 kJ/kg
C.V. Steam Generator: Exit 400qC, 5 MPa => h3 = 3195.64 kJ/kg
qw = h3 - h2 = 3195.64 - 88.85 = 3106.8 kJ/kg
C.V.: Turbine, Inlet and exit states given
wt
= h3 - h4 = 3195.64 - 2748.7 = 446.94 kJ/kg
Comparison
Additional Heat Transfer: qw - qw/o = 3106.8 - 2664.4 = 442.4 kJ/kg
Q.
extra = m.
(qw - qw/o) = 4424 kW
Difference in Net Power: wdiff = (wt
- wPw) + wPw/o,
wdiff = 446.94 - 4.91 + 0.4 = 442.4 kJ/kg
W.
diff = m.
wdiff = 4424 kW
By adding the extra heat transfer at the higher pressure and a turbine all the
extra heat transfer can come out as work (it appears as a 100% efficiency)