The diagram on the next page shows a hybrid single flash/binary geothermal power plant. The conditions at the wellhead are 28 kg/s, 2.5 MPa and 180oC. The pressure in the water cooled condenser is 10 kPa (at 4 and 5). The isobutane (at location 8) is heated to a temperature that is 6oC lower than the brine temperature coming from the separator (at 6). The isentropic efficiency of both turbines is 89%. The fluid is saturated liquid leaving both condensers. The isobutane must stay between 4 and 30 Bar pressures. Your task is to determine the optimal pressures in the system (the brine pressure after flash, and the two pressures in the binary portion). Compare the net power output and efficiency of the plant to determine the best pressures.- Write a summary report with an introduction, methodology, results and conclusions. Also, attach any hand calculations or spreadsheets to your work.- I have attached some work I have done, you can continue and finish the problem.Section 1
T1
P1_mpa
P1
m1
h1
180
2.5
25
28
763.05
C
MPa
bar
kg/s
kJ/kg
Section 2
P2
T2
m2
h2
x2
0.015
53.97
28
763.05
0.226
Mpa
celsius
kg/s
kJ/kg
Section 3
m3
T3
P3
h3
sat vap
6.328
53.97
0.015
2598.3
kg/s
celsius
Mpa
kJ/kg
x3
s3
1
8.0071
kj/kg*K
Section 4
T4
P4
h4a
h4s
x4
Efficiency of turbines
η
0.89
Sections 8-11
Pmax
Pmin
30
4
OPTIMIZE Choose pressure and temp. Find X with h2
T1>T2
(Table A-6)
T2P4
h4s=hf+X4s(hfg)
x4s=s4s-Sf/Sfg
h4a=h3-Nt(h3-h4s)
x4s=x4 (isentropic)
(h3-h4a)/(h3-h4s)
h4a=h3-n(h3-h4s)
bar
bar
s4s
Wout_h2o
8.0071
417.84
kj/kg*K
kw
s4s=s3=sg @ P4
sf
Section 5
T5
P5_kpa
P5
h5
m5
Qout
sat liq
45.81
10
0.1
191.81
6.328
14753.43
Celsius
kpa
bar
kJ/kg
kg/s
kw
Qout=m5(h4a-h5)
Section 6
m6
T6
P6
h6
s6
19.74
53.97
0.015
225.94
0.7549
sat liquid
kg/s
Celsius
Mpa
kJ/kg
kJ/Kg-K
Section 7
T7
P7
m7
19.74
kg/s
Section 8
T8
P8
h8
m8
47.97
6.7
640
19.74
Section 9
T9
P9
m9
h9s
h9a
Wout_iso
Section 10
Sat Liquid
T10
P10
h10
Qout
s10
T8+6
p6=p3
sat vpor
celsius
Bar
kJ/kg
Need to calculate between 4 and 30 bar
s8=s9
30
4
19.74
590
595.5
878.43
Celsius
Bar
kg/s
kJ/kg
kJ/kg
kw
4
2738.1
Celsius
Bar
kJ/kg
1
sg
Pumps and turbines are isentropic
X9
0.93
m9(h8-h9a)
s10 = s11
Qout=m10(h9a-h10)
No m10 yet
Pumps and turbines are isentropic
Section 11
T11
P11
s11
vocab
flash chamber
separator
turbine
water cooled condensor
air cooled condensor
pump
heat exchanger
W_out_tot
*not work total total
Flash chamber: sudden decrease in pressure causes t
Seperator: gravity is utilized in a vertical vessel to separat
produces work; reduces pressure. Use turb effici
heat out at condensor; saturated liquid?
Pumps: Make the p
1296.27
kw
nd 30 bar
crease in pressure causes the liquid water to “flash,” or vaporize, into steam.
a vertical vessel to separate the liquid and the vapor. The liquid settles to the bottom of the vessel, where it is withdrawn.
es pressure. Use turb efficiency to calc.
; saturated liquid?
Pumps: Make the pressure go up substantially. Generally Qdot = 0 unless energy is purposfully added or taken away. Isentropic pr
re it is withdrawn.
dded or taken away. Isentropic process.
x
Seperator
W
flash,geo
Water cooled
condenser
Steam
turbine
w
5
2
w.
binary.geo
Isobutane
turbine
Air cooled
condenser
Flash
chamber
Heat
Exchanger
10
Pump
11
15
W
elec.geo
000.00
2000
0260
9000
50,00
R600a W. Raya Thamodynamic Properties
DTU, Dietmast Energy Engisting
40,00 Vis al. The C
M S & K 06-04-14
30,00
2.000
000
-0.00
CI
0,015
100
0.020
20,00
2006
0.030
-0010
0.010
w of
00204
10.00-
9.00
8,00
7,00
6,00
5,00
4,00
0,060
0060
0.070
+0,060
090
0.00
ya 2030
0000
+40.15
Pressure Bar
3.00
0.30
0060
0:00
2,00
0.30
0,20
015
30
0.30
0.50
-0,20
0.30
000
1.0
.
1.00
U 90
0,80
0,70
0,60
0,50
0.40
0
040
060
-15
0,30
20
0,20
3.0
0.30
0.40
80
100
120
140
-0,10
3-0.80
0.20
1.00
0,50
1.40
0,60
1.50
0.70
1.80
0,80
2.00
0.90
2.20
1.20
50
100
150
200
250
300
350
350
600
650
700
750
800
850
400 450 500
Enthalpy (kJ/kg)
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