Using Stored heat (with Monte Carlo) and Numerical reservoir simulation. Lecture given at the 2017 LaGeo-UNUGTP Geothermal Diplomma Course.

1. Geothermal Reservoir Engineering
1. Geothermal Reserves Estimation
Manuel Rivera
LaGeo – Geothermal Studies and Assessment Dept.
cudus79@Gmail.com
October 2017

2. You need a concrete number for the
size of your resource
Ussher y
Hochwimmer, 2015

3. The solution
80
MWe
Data collection,
analysis
Conceptual
model
?

4. The solution
80
MWe
Data collection,
analysis
Conceptual
model
Reserves
Estimation
methods

5. ¿What’s in this lecture for you?
• You might be part of the team that eventually has to
transform the huge volumen of data into a number
(reserves size)
• You might have to assess the work and results of a
third party, understand the reasons for her proposals
• Understand the overall goal of your work in geotermal
exploration. How will my work be used to make
decisions?, how can I facilitate effective use of
information?
• Understand uncertainty magnitude in this kind of
assessment. Error bars in results

6. About me…

7. • General concepts
• Geology
• Geochemistry
• Geophysics
• Conceptual models
• Drilling
Reservoir
engineering
• Conversion technologies
• Environmental, social aspects

8. Logistics
• Theory and method
• Example and simple exercise
• Example use of spreadsheet (Monte Carlo
simulation)
• “complex” exercise

9. Reserves is the usable part of a
resource
Clotworthy et al.,
2006

10. The stored heat method

11. The underlying concept is the specific
heat capacity
𝑐 =
∆𝑄
𝑚∆𝑇
∆𝑇 =
∆𝑄
𝑚𝑐
Scienceblogs.scom

12. The energy we use comes mainly from
the hot rock
Example
• Volume 5 km3 (5x109 m3)
• Saturated with liquid water
• Porosity 0.1
• Initial temperature 250 °C
• Final temperature 180 °C

13. The energy we use comes mainly from
the hot rock
Qw = mw cw (Tr – T0)
= Vw φ ρw cw (Tr – T0)
= (5x109 m3)(0.1)(800 kg/m3)(4.82 kJ/kg°C)(250-180 °C)
= 1.35x1014 kJ
QR = VR (1-φ) ρR cR (Tr – T0)
= (5x109 m3)(1-0.1)(2700 kg/m3)(0.9 kJ/kg°C)(250-180 °C)
= 7.65x1014 kJ
Heat from water
Heat from fock

14. The energy we use comes mainly from
the hot rock
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
Heat from water Heat from rock
Storedenergy

15. A 100 MWe plant will "empty" the
reservoir 2+ times
Si h=1085 kJ/kg (liquido), Psep = 7 bar-g, xsep=0.17:
Vfluid = (5x109 m3)(0.1) = 5x108 m3
Required steam (2 kg/s-MWe)(100 MWe) = 200 kg/s
Required extraction (200 kg/s)/0.17/(800 kg/m3) = 1.47
m3/s
In 25 years operation, will be required:
Vreq = (1.47 m3/s)(25a)(365d)(24h)(3600s) = 11.6x108 m3
Vreq/Vfluid = 2.32
(Assuming reservoir remains liquid)

16. It is not technically feasible to extract
reservoir energy "to the last drop"
• Imperfect heat sweep, non-uniform, not all
the hot reservoir is permeable/accessible
• Numerical simulations in ideal conditions
(intergranular heat sweep) estimate
theoretical maximum close to 50%
• Qwh = Qr+w Rf Recovery factor
• In practice significantly lower (<20%)

17. It is not technically possible to convert
all thermal energy to electrical
• Maximum theoretical efficiency in any thermal
machine given by Carnot. Real efficiencies
significantly lower (f(h), 5~15%).
η = 1 −
𝑇 𝑐𝑜𝑙𝑑
𝑇ℎ𝑜𝑡
1 −
40 + 273
250 + 274
= 0.4

18. La energía aprovechable se “dosed” en
un período (años) arbitrario
𝑀𝑊𝑒 =
𝑄 𝑟𝑤 𝑅𝑓 η
𝐹 𝑡
F: Capacity factor
t: time [seconds]
Effective operational
time of power plant
All geoscientific data reduced to
2 parameters!

19. Assume hypothetical reservoir and
utilization with these characteristics:
Assume Carnot efficiency
What power plant size would you
recommend?
Area A = 8 km2 Thickness h = 1
km
Tr = 280 °C
T0 = 180 °C Rho rock = 2600
kg/m3
C rock= 0.9
kJ/kg°C
Rho water= 750
kg/m3
C water = 4.8
kJ/kg°C
Rf = 0.15
Fp = 0.9 t = 30 year Tcold = 40 °C
Φ = 0.1
Formulas:
Qw = VR φ ρw cw (Tr – T0)
QR = VR (1-φ) ρR cR (Tr – T0)
Qrw = Qr + Qw
η = 1 −
𝑇 𝑐𝑜𝑙𝑑
𝑇ℎ𝑜𝑡
𝑀𝑊𝑒 =
𝑄 𝑟𝑤
𝑅𝑓 η
1000 𝐹 𝑝
𝑡

20. Qw = (8x1x109 m3)(0.1)(750 kg/m3)(4.82 kJ/kg°C)(280-180 °C)
= 2.892 x 1014 kJ
Qr = (8x1x109 m3)(1-0.1)(2600 kg/m3)(0.9 kJ/kg°C)(280-180 °C)
= 16.85 x 1014 kJ
Qrw = 2.892 x 1014 kJ + 16.85 x 1014 kJ
= 19.74 x 1014 kJ
η = 1 −
40+273
280+273
= 0.43
𝑀𝑊𝑒 =
(19.74 𝑥 1014) (0.15) (0.43)
1000(0.9)(30)(365)(24)(3600)
= 149 MWe

21. Stored heat method assumes no heat
recharge
• Can be valid, but not always. Think of heat
extraction rate vs. natural heat recharge rate.
• Conservative assumption

22. Guide for parameters selection

23. Area: surface features, geophysical
anomalies, well temperature contours
Fumarole
Contour 5 ohm-mChloride
spring

24. Thickness: conductive layer (MT),
measured well temperatures, drilling
cost
50 °C
220-260 °C
Caprock
Reservoir
Expensive
drilling!
2500 m

25. Thickness: conductive layer (MT),
measured well temperatures, drilling cost
Reservoir
Reservoir
T < 180 °C???
T < 180 °C

26. Recovery factor is (perhaps) the most
uncertain parameter
Porosity φ
Rf Linear
Interpolation
Rf = 2.5φ
Nathenson, 1975. Numerical
simulation heat sweep.
Φ=0.2, Rf~0.5
No porosity:
No intergranular sweep
Starting point.
May be
modified at
discretion
Φ=f(depth)

27. Reservoir temperature: geochemistry,
well measurements
Tenorio y D’Amore, 1996

28. T0: reservoir temperature, conversion
technology
NetMW
180 220 250
Temperature °C
Sanyal et al., 2007
100 160 220
Temperature °C
NetMW
0816
NetMW
059
Self-flowing (flash) Pumped (binary)
180 °C 125-140 °C

29. Conversion efficiency: resource
enthalpy, conversion technology
Zarrouk and Moon, 2014

30. Rock properties: measurements or
literature
Eoas.ubc.ca
Jones, 2015

31. Water thermophysical properties:
IAPWS tables
• Density, specific heat
• F(T,[P,x])

32. Capacity factor: statistics
90 – 95%

33. Utilization time: standard 25 or 30
years

34. Monte Carlo Simulation

35. It is practically impossible to choose
"the right value" for variables
rfdoil.com

36. But we can bound a range...
Tenorio y D’Amore, 1996

37. ... and perhaps propose a more likely
value…
Tenorio y D’Amore, 1996
295 °C

38. And build probability distributions for
the variables
Lognormal
Triangular
Uniforme

39. Using random sampling, repeat
calculation many times
MWe
𝑄 𝑟𝑤 𝑅𝑓 η
𝐹 𝑡
= 𝑀𝑊𝑒
Deterministic model

40. Uncertainty of inputs propagates
forward to output variable (MWe)
The result is not a
single value, but a
probability
distribution

41. Result expressed in probability terms
(percentiles, P90, P50, P10)
0.00%
20.00%
40.00%
60.00%
80.00%
100.00%
120.00%
0
20
40
60
80
100
120
140
160
180
200
220
240
260
280
300
320
340
360
380
400
420
440
460
480
500
ymayor...
MWe
Cumulative frequency
P90 P50 P10
P90: 90% of simulation results greater tan this value

42. It takes lots of samples to properly
reproduce the underlying pdf
Underlying PDF
N=10
N=100
N=1000
N=10000
Random
sampling

43. Example
Answer in Excel sheet

44. Numerical simulation method

45. Uses much more detailed physical
models
Vs.𝑐 =
∆𝑄
𝑚∆𝑇

46. Allows spatial variation of parameters
and thermodynamic variables
Software.dicam.unibo.it O’Sullivan y Yeh, 2010

47. First, calibrate using natural
state temperatures and
pressures

48. Second, calibrate model using
production data
Availability of this data
Is the main limitation for
the method

49. Then, make forecasts using calibrated
model

50. Reserves estimation (using any method)
linked to quality of conceptual model

51. Sometimes, estimates can be
conservative or realistic...
Berlin
1982:
55 MWe (numérical)
150 MWe (volum.)
1993:
Proven 55~65 MWe
Probable 100 MWe
2003:
90 MWe OK
(numérical)

52. … But some challenges are difficult to
anticipate…
P90: 87 MWe
P50: 140 MWe
105 MWe
(flow tests)
Bjornsson, 2008
U1
U2
U3B
Reality:
• Scaling problems
• Reservoir cooling
Momotombo

53. … Therefore, staged development has
advantages (and disadvantages)
Steingrimsson et al., 2005

54. Other reserves estimation methods
exist, but not standard
• Heat flow method correlates reserves with natural surface heat
discharges.
• Lumped parameter model are a super-simplified version of numerical
simulation, needs production data as well, but less inputs
• Decline curves have been used mainly in vapor-dominated reservoirs
• Areal analogy method gives an interesting (realistic) reference point.
• Plane fracture method (Bodvarsson, 1974) useful in reservoirs mostly
confined to large fractures/faults

55. Summary
• 2 standard methods for reserves estimation area stored heat (with Monte Carlo)
and numerical reservoir simulation
• Stored heat is quick, simplified (over-simplified?), all exploration data reduced to 2
parameters
• Numerical simulation more detailed, more parameters, requires production data
• Other methods available, application not standard
• The base for most methods is the conceptual model
• All methods have significant uncertainties >> staged development