Rankine cycle at various steam pressure and temperature
3
3
6
kN := 10 ⋅N kPa := 10 ⋅Pa
define some units
3
MPa := 10 Pa
kJ := 10 ⋅J bar := 0.1MPa
We have seen the calculation of a Rankine steam cycle in rankine_class_example.mcd Repeating these
calculations for various combinations of boiler pressure and temperature allows us to investigate pressure and
temperature on ideal thermal efficiency. the calculations are done in the area by doing a matrix of combinations:
 30 
60 
p 2 := 
bar
 90 
 120


 233.90 
275.64 
Tsat := 
 303.4 
 324.75


 Tsat 
T3 :=  460 
 540


(
Tsat
since Tsat varies with pressure this was
accomplished in a 4 x 3 array of temperatures
460 560
 233.90
275.64
TT3 := 
 303.4
 324.75

efficiency calculations
)
460 560 
 30 
460 560   60 
bar
460 560 
 90 

460 560 
 120 
0.45
0.44
efficiency (reversible)
0.42
0.41
0.39
0.38
0.36
0.35
0.33
0.32
0.3
30
39
48
57
66
75
84
pressure bars
93
102
111
saturated
460 deg C superheat
560 deg C superheat
This plot shows the ideal efficiency at various combinations of pressure and temperature.
data for saturation curve
10/14/2005
1
120
select_pressure :=
this plot for
ip := select_pressure − 1
p2
select_temperature :=
30
60
90
120
ip
= 6 MPa
saturated
superheat to 460 deg C
superheat to 560 deg C
iT := select_temperature − 1
TT3
ip , iT
data for T s and H s plots
T-s Plot
500
400
T degrees C
300
200
100
0
0
1
2
3
4
5
6
s entropy kJ/(kg*K)
7
8
9
h-s Plot
4000
h enthalpy kJ/kg
3000
2000
1000
0
10/14/2005
0
2
4
6
s entropy kJ/(kg*K)
2
8
10
10
= 460
close up of points 1 and 2
T - s plot
46
T degrees C
44
42
40
38
0.5
0.55
0.6
s entropy kJ/(kg*K)
0.65
180
175
h enthalpy kJ/kg
170
165
160
155
150
0.54
10/14/2005
0.56
0.58
0.6
s entropy kJ/(kg*K)
3
0.62
0.64
enthalpy average temperature approach for efficiency
we defined the entropy average temperatures in class and showed that the thermal efficiency (ideal) could be
calculated ...
T_bar_L
η=1−
using this approach let's calculate the thermal efficiency and compare with above
T_bar_H
 2804.2
2784.3
h3 = 
 2742.1
 2684.9

3366.6 3591.7 
3326.1 3564.2  kJ
3283.1 3535.7 
kg
3237.2 3506.2 
i := 0 .. 3
 170.586 
173.609 
kJ
h2 = 
K
 176.633 
kg⋅ K
 179.656


 6.187
5.889
s3 = 
 5.677
 5.492

7.114 7.402 
6.753 7.057  kJ
6.521 6.844 
kg⋅ K
6.34
6.684 
kJ
s2 = 0.572
kg⋅ K
j := 0 .. 2
calculate entropy average TH
QH = h 3 − h 2
or ... in indicial
notation
QL = h 1 − h 4
QH = h 3 − h 2
i, j
i, j
i
QL = h 1 − h 4
entropy average high temperature ...
T_bar_H
i, j
:=
h3
i, j
 469.082
491.036
T_bar_H = 
 502.57
 509.206

− h2
i
s3 − s2
i, j
488.545 500.917 
510.095 522.852 
K
522.253 535.608 
530.17
544.309 
entropy average low temperature ... is the constant condenser pressure - convert to K
(
)
T_bar_L := T1 + 273.15 ⋅K
η i , j := 1 −
(
Tsat
T_bar_L
T_bar_H
from above; identical !! as
expected
i, j
460 560
)
 0.332 0.359 0.375 
0.362 0.386 0.401 
η=
 0.377 0.4 0.415 
 0.385 0.409 0.425


10/14/2005
T1 = 40
(
Tsat
 30 
 60  bar
 90 
 120
 
460 560
)
 0.332 0.359 0.375 
0.362 0.386 0.401 
η th = 
 0.377 0.4 0.415 
 0.385 0.409 0.425


4
 30 
 60  bar
 90 
 120
 