Geothermal Power
Analysis of a Dual-Fluid Organic Rankine Cycle
CEEN 501 Energy System Fundamentals
Nick Mangione
Charlotte Xu
Mark de Lucovich
Agenda

Background

System Diagram

Methodology

Results

T-s Diagrams

Analysis


Exergy Destruction

Efficiencies

Comparisons
Conclusions
Background
Journal: Applied Thermal Engineering, March 2017
Title: Conventional and advanced exergy analyses of a
geothermal driven dual fluid organic Rankine cycle (ORC)
Authors Hossein Nami, Arash Nemati, Farshad Jabbari Fard
Faculty of Mechanical Engineering, University of Tabriz, Iran
Objectives

Reproduce the paper’s results and solution approach for
the “Ideal” case using data from NIST’s Chemistry Web
Book

Determine the energy and exergy efficiency, and exergy
destruction for each component

Analyze the effect of at least two operating parameters
on cycle performance to compare to the combined
effect of the 3 non-idealities in the “real” and
“unavoidable” cases

Identify the largest effects on cycle performance
System
Dual-Fluid
Organic Rankine Cycle
System
Adiabatic
High pressure loop
Isopentane
Low pressure loop
Isobutane
Geothermal loop
Water
System
System
The pre-heater output is always
saturated liquid at high pressure
The evaporator output is
always saturated vapor at
high pressure
The input to the pump is always
saturated liquid at low pressure
HPT high and low pressures are
1300 kPa and 400 kPa
1. Saturated vapor
P=1300
System
1. Saturated vapor
P=1300
System
1. Saturated vapor
P=1300
2. Super heated
s1=s2
P=400
System
1. Saturated vapor
P=1300
2. Super heated
s1=s2
P=400
3. Sat.Liq.
P2=P3
X3=0
System
1. Saturated vapor
P=1300
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
3. Sat.Liq.
P2=P3
X3=0
1. Saturated vapor
P=1300
System
5. Sat.Liq
P4=P5
X5=0
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
3. Sat.Liq.
P2=P3
X3=0
1. Saturated vapor
P=1300
System
The evaporator output is always
saturated vapor at high
pressure, so x6 = 1
In the ideal case, ∆T in the HPE,
LPE, and Condenser = 0, so
T6=T3
This determines P6
5. Sat.Liq
P4=P5
X5=0
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
3. Sat.Liq.
P2=P3
X3=0
6. Sat.Liq.
T6=T3
X6=1
1. Saturated vapor
P=1300
System
Determine State 8, before 7:
We don’t know P7 (=P8)!
The input to the pump is always
saturated liquid at low pressure
5. Sat.Liq
P4=P5
X5=0
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
3. Sat.Liq.
P2=P3
X3=0
T8 = Tcw2 (given): 35C
This is enough to determine P8.
8. Sat.Liq
T= Given, 35C
X8=0
6. Sat.Liq.
T6=T3
X6=1
1. Saturated vapor
P=1300
System
Condenser is isobaric
P8=P7
Low pressure turbine
Isentropic
s6=s7
5. Sat.Liq
P4=P5
X5=0
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
3. Sat.Liq.
P2=P3
X3=0
8. Sat.Liq
T= Given, 35C
X8=0
6. Sat.Liq.
T6=T3
X6=1
7. Superheatd
P8=P7
s6=s7
1. Saturated vapor
P=1300
System
5. Sat.Liq
P4=P5
X5=0
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
3. Sat.Liq.
P2=P3
X3=0
9.Comp.Liq.
S8=s9
P9=P6
8. Sat.Liq
T= Given, 35C
X8=0
6. Sat.Liq.
T6=T3
X6=1
7. Superheatd
P8=P7
s6=s7
1. Saturated vapor
P=1300
System
Given system Wnet = 3,443 kW
Now have all enthalpies, no
mass flows to calculate Wnet
E-bal on LPE:
mHP(h2-h3) = mLP(h6-h10)
Solve simultaneous equations to
get mHP and mLP
5. Sat.Liq
P4=P5
X5=0
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
6. Sat.Liq.
T6=T3
X6=1
3. Sat.Liq.
P2=P3
X3=0
9.Comp.Liq.
S8=s9
P9=P6
10. Sat.Liq
P9=P10
X10=0
8. Sat.Liq
T= Given, 35C
X8=0
7. Superheatd
P8=P7
s6=s7
System
1. Saturated vapor
P=1300
1gw. Comp.Liq
Given P=7000
T=175C
5. Sat.Liq
P4=P5
X5=0
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
6. Sat.Liq.
T6=T3
X6=1
3. Sat.Liq.
P2=P3
X3=0
9.Comp.Liq.
S8=s9
P9=P6
10. Sat.Liq
P9=P10
X10=0
8. Sat.Liq
T= Given, 35C
X8=0
7. Superheatd
P8=P7
s6=s7
System
1. Saturated vapor
P=1300
1gw. Comp.Liq
Given P=7000
T=175C
2gw. Comp.Liq
T2gw = T5
5. Sat.Liq
P4=P5
X5=0
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
6. Sat.Liq.
T6=T3
X6=1
3. Sat.Liq.
P2=P3
X3=0
9.Comp.Liq.
S8=s9
P9=P6
10. Sat.Liq
P9=P10
X10=0
8. Sat.Liq
T= Given, 35C
X8=0
7. Superheatd
P8=P7
s6=s7
System
1. Saturated vapor
P=1300
1gw. Comp.Liq
Given P=7000
T=175C
2gw. Comp.Liq
T2gw = T5
5. Sat.Liq
P4=P5
X5=0
Energy balance
high pressure
pre-heater
3gw. Comp.Liq
h3gw=h2gw(h5-h4)*(mHP/mgw)
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
6. Sat.Liq.
T6=T3
X6=1
3. Sat.Liq.
P2=P3
X3=0
9.Comp.Liq.
S8=s9
P9=P6
10. Sat.Liq
P9=P10
X10=0
8. Sat.Liq
T= Given, 35C
X8=0
7. Superheatd
P8=P7
s6=s7
System
1. Saturated vapor
P=1300
1gw. Comp.Liq
Given P=7000
T=175C
2gw. Comp.Liq
T2gw = T5
5. Sat.Liq
P4=P5
X5=0
3gw. Comp.Liq
h3gw=h2gw(h5-h4)*(mHP/mgw)
Energy balance
low pressure
pre-heater
4gw. Comp.Liq
h4gw= h3gw(h10-h9)*(mLP/mgw)
2. Super heated
s1=s2
P=400
4. Comp.Liq.
S3= S4
P=1300
6. Sat.Liq.
T6=T3
X6=1
3. Sat.Liq.
P2=P3
X3=0
9.Comp.Liq.
S8=s9
P9=P6
10. Sat.Liq
P9=P10
X10=0
8. Sat.Liq
T= Given, 35C
X8=0
7. Superheatd
P8=P7
s6=s7
Results
Loop
Fluid
stream
State
P (kPa)
T ( C)
quality
ψ (kJ/kg)
X_dot (kW)
none
0
dead state
100
25
–
High Pressure
Isopentane
1
Sat vapor
1300
129.49
1
500.13
1.3317
38.61
103.32
3989
High Pressure
Isopentane
2
super heated
400
94.28
–
456.47
1.3317
38.61
59.66
2303
High Pressure
Isopentane
3
sat liquid
400
74.54
0
113.31
0.3479
38.61
9.66
373.0
High Pressure
Isopentane
4
compressed
1300
75.01
–
114.91
0.3479
38.61
11.26
434.7
High Pressure
Isopentane
5
sat liquid
1300
129.49
0
265.82
0.7498
38.61
42.42
1638
Low Pressure
Isobutane
6
sat vapor
1199
74.54
1
651.11
2.3568
50.54
99.41
5025
Low Pressure
Isobutane
7
super heated
465.2
41.68
–
613.76
2.3568
50.54
62.06
3137
Low Pressure
Isobutane
8
sat liquid
465.2
35
0
283.69
1.2861
50.54
51.04
2580
Low Pressure
Isobutane
9
compressed
1199
35.42
–
285.05
1.2861
50.54
52.40
2649
Low Pressure
Isobutane
10
sat liquid
1199
74.54
0
388.98
1.6028
50.54
61.95
3131
geothermal
water
1gw
compressed
7000
175
–
744.27
2.0826
46.28
geothermal
geothermal
water
water
2gw
3gw
compressed
compressed
7000
7000
129.49
99.65
–
–
h (kJ/kg) s (kJ/kg.K) m_dot (kg/s)
–
548.79
422.88
1.6228
1.2979
46.28
46.28
geothermal
water
4gw
compressed
7000
72.54
–
309.36
0.9817
46.28
cooling
water
1cw
compressed
100
25
–
104.92
0.3672
399.11
cooling
water
2cw
compressed
100
35
–
146.72
0.5051
399.11
128.16
69.70
40.61
21.34
0.00
0.71
5931
3225
1879
987.3
0
281.7
Exergy Destruction
equipment
HP-turbine
LP-Turbine
inlet
stream
1
6
outlet
stream
2
7
work (kW)
1685.6
1887.8
X_d (kW)
0
0
HPVG (HPE+HPPH)
497
LPVG (LPE+LPPH)
0
61.7
68.7
446
275
0
0
0
0
0
0
1218.82
143.20
354.07
37.037
409.39
Condenser
HPF-Pump
LPF-Pump
1cw,7
3
8
2cw,8
4
9
TOTAL
HP-PreHeater
HP-Evaporator
LP-Evaporator
LP-PreHeater
2gw,4
1gw,5
2,10
3gw,9
3gw,5
2gw,1
6,3
4gw,10
T-s Diagrams
Isopentane T-S diagram
200,00
200,00
180,00
180,00
160,00
160,00
140,00
120,00
140,00
1
5
120,00
60,00
100,00
4
2
T [C˚]
T [C˚]
100,00
80,00
3
60,00
40,00
20,00
20,00
0,5000
1,0000
1,5000
s [kJ/kg-K]
2,0000
2,5000
6
10
80,00
40,00
0,00
0,0000
Isobutane T-S diagram
0,00
0,0000
9
7
8
0,5000
1,0000
1,5000
S [kJ/kg-K]
2,0000
2,5000
Energy and Exergy Efficiency

“Ideal” exergy efficiency from paper:
69.85%

Calculated exergy efficiency:
69.65%

Using exergy destroyed:
75.35%

Using ηrev between T1 and T8:
72.87%

Calculated energy efficiency: 17.11%
Real vs Ideal Case


The “Real” and “Unavoidable”
cases in the paper change 3
operating conditions at the same
time:

Turbines’ & pumps’ isentropic
efficiency

ΔP for all heat exchangers

ΔT in the HPE, LPE, and Condenser
Paper exergy efficiency results:

Ideal 69.85%

Real
47.03%
Effect of Changed Conditions

Varying the isentropic efficiency of
only the turbines had a large
negative effect


Varying that of the pumps would be
much smaller
Adding pressure drops had a small
negative effect on energy
efficiency

Higher P in HPE increased exergy
efficiency very slightly because it
reduced ∆T between the
geothermal water and HP loop
Ideal
ηT =
0.85
∆P = 3%
for HPE, HPPH,
LPPH
Real
Energy
Efficiency
17.11% 14.5%
17.07%
—
Exergy
Efficiency
69.9%
70.1%
47.0%
59.2%
Results

The remaining drop in exergy
efficiency in the “Real” case must
come from the 10°C drops in the
heat exchangers

This could be mitigated by larger
heat exchangers, but this adds cost

As the paper suggests,
improvements in turbine isentropic
efficiency would bring the largest
improvements in efficiency, even
though they are not the largest
sources of exergy destruction
Summary

We were able to closely duplicate the results of the “Ideal” analysis, and recreated the design logic used in the paper despite it not being stated

The isentropic efficiency of the turbines has a large effect on cycle energy
and exergy efficiency, whereas pressure drops do not, unlike gas power cycles

The low exergy efficiency of the “Real” case presented in the paper is due to
a combination of turbine and pump isentropic efficiency and temperature
drops through real heat exchangers

As the original authors concluded, most of the exergy destruction in the heat
exchangers is unavoidable given technical limitations, and so improvements
to turbine isentropic efficiency are the most promising route
Questions?
References
[1] Çengel Yunus A., & Boles, M. A. (2011). Thermodynamics: an engineering
approach. New York, NY: McGraw-Hill.
[2] Nami, H., Nemati, A., & Fard, F. J. (2017). Conventional and advanced exergy
analyses of a geothermal driven dual fluid organic Rankine cycle (ORC). Applied
Thermal Engineering, 122, 59–70. doi: 10.1016/j.applthermaleng.2017.05.011
[3] National Institute of Standards and Technology Chemistry WebBook, SRD 69.
(n.d.). Retrieved from https://webbook.nist.gov/chemistry/fluid/
[4] J. P. Roy & Ashok Misra (2017) Comparative performance study of
differentconfigurations of organic Rankine cycle using low-grade waste heat for
power generation,International Journal of Green Energy, 14:2, 212-228, DOI:
10.1080/15435075.2016.1253570