3
comparison of rankine with single regeneration
basic reference problem ...
Rankine_class_example.mcd
state 1: condenser outlet
kPa := 10 ⋅Pa
6
MPa := 10 Pa
3
kJ := 10 ⋅J bar := 0.1MPa
3
kN := 10 ⋅N
same as reference
3
T1 := 40
Table 1 or Table A.1.1
kJ
sf_1 := 0.5725
kg⋅ K
s1 := sf_1
p 1 := 7.384kPa
kJ
sfg_1 := 7.6845
kg⋅ K
v f_1 := 0.0010078
h f_1 := 167.57
kJ
m
v 1 := v f_1
kg
h fg_1 := 2406.7
kg
kJ
h 1 := h f_1
kg
assume vf = v1 constant, isentropic, ds = 0 =>T*ds= 0 => h2 = h1+v1*dp from
state 2: pump outlet
relationships Tds = dh + v*dp integrated with constant v and Tds = 0
p 2 := 30bar
s2 := s1
wp := h 1 − h 2
Cp := 4.184
using Cp
kJ
wp = −3.016
kg
and ... eqn 5.18
state 3: boiler outlet
p 3 := p 2
(
h 2 := h 1 + v 1 ⋅ p 2 − p 1
kJ
kJ
kg
kg⋅ K
h 2 = 170.586
(
)
T2 := T1 +
h2 − h1
Cp
interpolation
state 4: turbine outlet
h 3 = 3366.5
h 4 = 2216
kJ
η th =
(h3 − h4) − (h2 − h1)
h 3 − h 2
QH := h 3 − h 2
work_net
QH
=
QH + QL
QH
=
wt + wp
QH
η th = 0.359
=
kJ
s3 = 7.113
kg⋅ K
kJ
wt = 1150
kg
(h3 − h4) + (h1 − h2)
h3 − h2
η th :=
QL := h 1 − h 4
η th_1 :=
10/17/2005
kg
wt := h 3 − h 4
kg
thermal efficiency
η th :=
kJ
isentropic expansion to 40 deg C
s4 := s3
determine h4 from x s4 − sf_1
=>
s4 = sf_1 + x ⋅ sfg_1
x :=
x = 0.851
sfg_1
h 4 := h f_1 + h fg_1⋅x
1
kg
T2 = 40.721
p 3 = 3 MPa
from interpolation Table A.1.3 P=3MPa page 622
kJ
actual units
h 2 − h 1 = Cp⋅ T2 − T1
T3 := 460
)
wt + wp
h3 − h2
QH + QL
QH
η th = 0.359
η th_1 = 0.359
same cycle with regeneration
extract steam at 400kPa (4 bar) from turbine to mix
with condensate to become saturated liquid at 400kPa
1 - vacuum; (1-m 1 ) saturated liquid
5
T=40 C
Wt
Turbine
2 - sub cooled liquid at feed heater
pressure P=400 kPa
Boiler
Q_dotH
6
3 - saturated liquid at 400 kPa
7
4 - sub cooled liquid at boiler
pressure P=3 MPa
4
Feed
Pump
Condenser
Regenerative
3 feed heater 2
5 - superheated vapor T=460 C
Q_dotL
6 - (m1 ) superheated vapor @ 400
kPa or ... vapor + liquid @ saturation
temperature and pressure tbd
1
Wfp
Wcp
Condensate Pump
7 - (1-m1 ) vapor + liquid @ saturation
temperature and pressure
the state values are identical to the reference, however the fraction
of total mass flow is less = 1-m 1
state 1: condenser outlet (1-m1)
3
T1 := 40
Table 1 or Table A.1.1
kJ
sf_1 := 0.5725
kg⋅ K
s1 := sf_1
s2 := s1
(
h 2 := h f_1 + v f_1⋅ p 2 − p 1
kJ
wcp = −0.396
kg
state 3: regeneration out
kJ
kg
h fg_1 := 2406.7
kg
kJ
kg
h 1 := h f_1
)
T2 := T1 +
h 2 = 167.966
h2 − h1
Cp
kJ
kg
T2 = 40.095
mass rate = 1; saturated liquid at p_feed = 4bar
T3 := 143.63
p 2 = 400 kPa
v 3 := 1.0836⋅ 10
10/17/2005
h f_1 := 167.57
m
relationships Tds = dh + v*dp integrated with constant v and Tds = 0
wcp := h f_1 − h 2
p 3 := p 2
kJ
sfg_1 := 7.6845
kg⋅ K
v f_1 := 0.0010078
assume vf constant, isentropic, ds = 0 =>T*ds= 0 => h 2 = h1 +v1 *dp from
state 2: condensate
pump outlet (1-m1)
p 2 := 4bar
p 1 := 7.384kPa
h 3 := 604.74
3
−3m
kJ
kg
h fg_3:= 2133.9
kg
2
kJ
s3 := 1.7766
kg⋅ K
kJ
kg
kJ
sfg_3 := 5.1193
kg⋅ K
state 4: feed pump out
h 4 := h 3 + v 3 ⋅ p 4 − p 3
kJ
wfp = −2.817
kg
wfp := h 3 − h 4
state 5: boiler outlet
p 5 := p 4
(
p 4 := 3MPa
T5 := 460
state 6: turbine outlet
#1, partial flow (m1)
h 5 := 3366.6
h4 − h3
x 6 :=
s6 − s3
kJ
s4 := s3
kg
kJ
s5 := 7.1144
kg⋅ K
kg
x 6 = 1.043
sfg_3
T6 := T3
1 kJ
s6 = 7.114
K kg
kJ
T4 = 144.303
Cp
expansion to p3 with same ∆s as regeneration heating
determine h4 from x
=>
p = 4bar = 400kPa
h 4 = 607.557
Table 2 or interpolate as above
p 5 = 3 MPa
s6 = s3 + x 6 ⋅sfg_3
T4 := T3 +
)
p 6 := p 3
s6 := s5
new!! ... mass fraction has to be > 1
for heat balance to work. this says
extraction steam is superheated. x
relationship doesn't apply. need to
interpolate in superheated region
find T and h by interpolation
interpolation
h 6 := 2834.491
kJ
T6 := 187.702
kg
state 7: turbine outlet
#2, partial flow (1-m1)
expansion to p3 with same ∆s as regeneration
heating; determine h7 from x
s7 = s1 + x ⋅ sfg_1
x 7 :=
=>
h 7 := h 1 + h fg_1⋅x 7
s7 − s1
T7 := T1
x 7 = 0.851
sfg_1
h 7 = 2216.42
s7 := s5
kJ
kg
now we can do the turbine flow through 6 = m1 treat like x; combination of m 1 at h6 and (1-m1) at h2 = h3
for heat balance out of feed heater
(
)
m1 ⋅ h 6 + 1 − m1 ⋅ h 2 = h 3
h3 − h2
m1 =
h6 − h2
(
h3 − h2
m1 :=
h6 − h2
)(
wt := h 5 − h 6 + 1 − m1 ⋅ h 6 − h 7
thermal efficiency
η th :=
kJ
wt = 1048.94
kg
)
(
)
wt + wcp⋅ 1 − m1 + wfp
h 5 − h 4
wfp = −2817.36 Sv
η th = 0.379
η th_1 = 0.359
data for plots
10/17/2005
m1 = 0.164
3
Thermal efficiency using entropy average temperature approach
T5 = 460
h 5 = 3366.6
kJ
h 4 = 607.557
kg
kJ
s6 = 7.114
kg⋅ K
kJ
T1 = 40
kg
kJ
s4 = 1.777
kg⋅ K
1 kJ
s1 = 0.572
K
kg
we need to redefine efficiency for t average calculations
η th =
QH − QL
QH
=1−
⌠
QH = mH⋅  T ds
⌡
QL
QH
T_barH =
and ...
h5 − h4
T_barH :=
s5 − s4
⌠
QL = mL⋅ T ds
⌡
(
T constant
( ) = ∆hH
mH⋅ ( ∆sH)
mH⋅ ( ∆sH)
∆sH
T_barH⋅mH⋅ ( ∆sH) = QH
QH
=
mH⋅ ∆h H
T_barH = 516.888 K
)
T_barL := T1 + 273.15 ⋅K
mH := 1
T_barL = 313.15 K
( )
T_barL⋅mL⋅ ∆sL = QL
(∆sH = s5 − s4) ≠ (∆sL = s7 − s1)
but ...
as ...
and when inserting into Q η relationship have to put in m 1 values ...
η t_avg_1 := 1 −
(
)(
T_barL⋅ 1 − m1 ⋅ s7 −
s1
(
T_barH⋅ s6 − s3
)
)
but ...
s5 = s6 = s7
η th = 1 −
η t_avg_1 = 0.379
QL
QH
=1−
T_barL
T_barH
η t_avg_2 = 0.394
precise T avg
estimated T avg
η t_avg_1 = 0.379
η t_avg_2 = 0.394
actual regenerative
η th = 0.379
"perfect regeneration"
data for saturation curve
10/17/2005
4
( )
T_barH⋅mH⋅ ( ∆sH)
T_barL⋅mL⋅ ∆sL
matches as expected
suppose we didn't allow for the difference in ∆s and mass flow and just said
η t_avg_2 := 1 −
s4 ≠ s1
reference
η th_1 = 0.359
T-s Plot
500
T degrees C
400
300
200
100
0
0
1
2
3
4
5
6
s entropy kJ/(kg*K)
7
8
9
10
h-s Plot
4000
h enthalpy kJ/kg
3000
2000
1000
0
10/17/2005
0
2
4
6
s entropy kJ/(kg*K)
8
5
10
close up of points 1 and 2
T - s plot 40
39
0.56
T - s plot
146
T degrees C
T degrees C
41
close up of points 3 and 4
144
142
1.7
0.565
0.57
0.575
s entropy kJ/(kg*K)
170
1.75
1.8
s entropy kJ/(kg*K)
620
h enthalpy kJ/kg
h enthalpy kJ/kg
615
168
605
166
0.56
600
0.57
0.58
0.59
s entropy kJ/(kg*K)
N.B. these scales are very exaggerated !!
10/17/2005
610
6
1.76
1.78
s entropy kJ/(kg*K)