Unit 7_heatflow_in_permafrost_nonneg
d28d1146c71542fdb241eb39279fc912
isee systems, inc.
Stella Professional
0
50000
1
m
meter
kg
kilogram
yr
year
degk
degree k
j
joule
s
second
Energy in layer 2 (100200 m): Units = joules (J)
The maximum number of iterations that can be run with STELLA is 30000. This does not afford a long enough time for this system to reach a final steady state when the reservoirs are initialized with 0 starting energy. To get around this problem, the model should first be run with all reservoirs having an inital value of 0, with a DT=1 year, and a run length of 030000 years. A table pad that keeps track of the layer energy should be created and the final values from the first 30000 year run used as initial values for a second 30000 year run.
Initial value for 2nd 30000 year run = 58140120105.89
60e9
flow_to_2
flow_to_1
j
Energy in layer 3 (200300 m): Units = joules (J)
The maximum number of iterations that can be run with STELLA is 30000. This does not afford a long enough time for this system to reach a final steady state when the reservoirs are initialized with 0 starting energy. To get around this problem, the model should first be run with all reservoirs having an inital value of 0, with a DT=1 year, and a run length of 030000 years. A table pad that keeps track of the layer energy should be created and the final values from the first 30000 year run used as initial values for a second 30000 year run.
Initial value for 2nd 30000 year run = 58364485003.41
60e9
flow_to_3
flow_to_2
j
Crosssectional area of each permafrost layer.
Units = meters squared (m^2)
1.0
meters^2
Number of seconds in a year. 1 yr = (60 s/min)*(60 min/hr)*(24 hr/day)*(365.25 days) = 3.15e7 s
Units = Seconds per Year (s/yr)
60.0*60.0*24.0*365.25
seconds/years
The temperature of each layer is calculated by dividing its energy content by its heat capacity (which is identical for all layers).
Units = joules per degrees Kelvin (J/K)
layer_thickness*unit_area*rock_density*rock_specific_heat
joules/degrees K
Average density of continental, crustal rock.
Units = kilograms per meters cubed (kg/m^3)
2700.0
kilograms/meters^3
Specific heat of continental, crustal rock. This value was taken from the value reported for granite at http://www.engineeringtoolbox.com/24_154.html
Units = joules per kilograms degrees Kelvin (J/(kg*K))
800.0
Energy in layer 4 (300400 m): Units = joules (J)
The maximum number of iterations that can be run with STELLA is 30000. This does not afford a long enough time for this system to reach a final steady state when the reservoirs are initialized with 0 starting energy. To get around this problem, the model should first be run with all reservoirs having an inital value of 0, with a DT=1 year, and a run length of 030000 years. A table pad that keeps track of the layer energy should be created and the final values from the first 30000 year run used as initial values for a second 30000 year run.
Initial value for 2nd 30000 year run = 58598012786.36
60e9
flow_to_4
flow_to_3
j
Energy in layer 5 (400500 m): Units = joules (J)
The maximum number of iterations that can be run with STELLA is 30000. This does not afford a long enough time for this system to reach a final steady state when the reservoirs are initialized with 0 starting energy. To get around this problem, the model should first be run with all reservoirs having an inital value of 0, with a DT=1 year, and a run length of 030000 years. A table pad that keeps track of the layer energy should be created and the final values from the first 30000 year run used as initial values for a second 30000 year run.
Initial value for 2nd 30000 year run = 58844971608.95
60e9
flow_to_5
flow_to_4
j
Heat flow from layer 3 to layer 2
Units = joules per years (J/yr)
k*(T_z300T_z200)/layer_thickness
j/yr
Heat flow from layer 1 to layer 0
Units = joules per years (J/yr)
k*(T_z100T_z0)/layer_thickness
j/yr
Heat flow from layer 2 to layer 1
Units = joules per years (J/yr)
k*(T_z200T_z100)/layer_thickness
j/yr
Heat flow from layer 4 to layer 3
Units = joules per years (J/yr)
k*(T_z400T_z300)/layer_thickness
j/yr
Energy in layer 6 (500600 m): Units = joules (J)
The maximum number of iterations that can be run with STELLA is 30000. This does not afford a long enough time for this system to reach a final steady state when the reservoirs are initialized with 0 starting energy. To get around this problem, the model should first be run with all reservoirs having an inital value of 0, with a DT=1 year, and a run length of 030000 years. A table pad that keeps track of the layer energy should be created and the final values from the first 30000 year run used as initial values for a second 30000 year run.
Initial value for 2nd 30000 year run = 59109263261.61
60e9
flow_to_6
flow_to_5
j
Energy in layer 7 (600700 m): Units = joules (J)
The maximum number of iterations that can be run with STELLA is 30000. This does not afford a long enough time for this system to reach a final steady state when the reservoirs are initialized with 0 starting energy. To get around this problem, the model should first be run with all reservoirs having an inital value of 0, with a DT=1 year, and a run length of 030000 years. A table pad that keeps track of the layer energy should be created and the final values from the first 30000 year run used as initial values for a second 30000 year run.
Initial value for 2nd 30000 year run = 59394316740.35
60e9
flow_to_7
flow_to_6
j
Energy in layer 8 (700800 m): Units = joules (J)
The maximum number of iterations that can be run with STELLA is 30000. This does not afford a long enough time for this system to reach a final steady state when the reservoirs are initialized with 0 starting energy. To get around this problem, the model should first be run with all reservoirs having an inital value of 0, with a DT=1 year, and a run length of 030000 years. A table pad that keeps track of the layer energy should be created and the final values from the first 30000 year run used as initial values for a second 30000 year run.
Initial value for 2nd 30000 year run = 59702994712.71
60e9
flow_to_8
flow_to_7
j
Heat flow from layer 7 to layer 6
Units = joules per years (J/yr)
k*(T_z700T_z600)/layer_thickness
j/yr
Heat flow from layer 6 to layer 5
Units = joules per years (J/yr)
k*(T_z600T_z500)/layer_thickness
j/yr
Heat flow from layer 5 to layer 4
Units = joules per years (J/yr)
k*(T_z500T_z400)/layer_thickness
j/yr
Energy in layer 9 (800900 m): Units = joules (J)
The maximum number of iterations that can be run with STELLA is 30000. This does not afford a long enough time for this system to reach a final steady state when the reservoirs are initialized with 0 starting energy. To get around this problem, the model should first be run with all reservoirs having an inital value of 0, with a DT=1 year, and a run length of 030000 years. A table pad that keeps track of the layer energy should be created and the final values from the first 30000 year run used as initial values for a second 30000 year run.
Initial value for 2nd 30000 year run = 60037515431.60
60e9
flow_to_9
flow_to_8
j
Energy in layer 10 (9001000 m): Units = joules (J)
The maximum number of iterations that can be run with STELLA is 30000. This does not afford a long enough time for this system to reach a final steady state when the reservoirs are initialized with 0 starting energy. To get around this problem, the model should first be run with all reservoirs having an inital value of 0, with a DT=1 year, and a run length of 030000 years. A table pad that keeps track of the layer energy should be created and the final values from the first 30000 year run used as initial values for a second 30000 year run.
Initial value for 2nd 30000 year run = 60399392227.20
60e9
Geothermal_Inflow
flow_to_9
j
Air temperature in degrees Kelvin (K). 5 deg. C = 268.15 K.
Oscillation temperature equation: 268.15+5*(sin(2*pi*time/1000))
Step change in temperature logical statement: if(time<10000)then(268.15)else(273.15)
273.15
Heat flow from layer 8 to layer 7
Units = joules per years (J/yr)
k*(T_z800T_z700)/layer_thickness
j/yr
Heat flow from layer 9 to layer 8
Units = joules per years (J/yr)
k*(T_z900T_z800)/layer_thickness
j/yr
Heat flow from earth to layer 10 (earth to 1000 m depth).
Units = joules per years (J/yr)
Mean_Heat_Flow*sec_per_year*unit_area
j/yr
Temperature at the top of layer 4 (300 m depth) in degrees Kelvin (K).
energy_in_layer_4/heat_capacity
degK
Temperature at the top of layer 5 (400 m depth) in degrees Kelvin (K).
energy_in_layer_5/heat_capacity
degK
Temperature at the top of layer 6 (500 m depth) in degrees Kelvin (K).
energy_in_layer_6/heat_capacity
degK
Temperature at the top of layer 3 (200 m depth) in degrees Kelvin (K).
energy_in_layer_3/heat_capacity
degK
Temperature at the top of layer 2 (100 m depth) in degrees Kelvin (K).
energy_in_layer_2/heat_capacity
degK
Temperature at the top of layer 1 (0 m depth) in degrees Kelvin (K). This temperature is prescribed by whatever the temperature of the overlying air is.
air_temperature
degK
Thermal conductivity is measured in W/(m*K). Since 1 W = 1 J/s, thermal conductivity also has units of J/(m*K*s). Turcotte and Shubert have a table on pg. 135 of their book that gives typical k values. The average seems to be about 3.6 and this is converted to J/mKyr by multiplying by the number of seconds in a year to allow the geothermal heat flux and this term to have the same units.
Salt conductivity = 6.1 W/(m*K), shale conductivity = 1.8 W/(m*K).
3.6*sec_per_year
j/(mdegKyr)
Temperature at the top of layer 7 (600 m depth) in degrees Kelvin (K).
energy_in_layer_7/heat_capacity
degK
Temperature at the top of layer 8 (700 m depth) in degrees Kelvin (K).
energy_in_layer_8/heat_capacity
degK
Temperature at the top of layer 9 (800 m depth) in degrees Kelvin (K).
energy_in_layer_9/heat_capacity
degK
Temperature at the top of layer 10 (900 m depth) in degrees Kelvin (K).
energy_in_layer_10/heat_capacity
degK
According to Turcotte and Shubert (pg. 135), the mean heat flow for all continents is 65 +/ 1.6 mW/(m^2). This is converted to 65.0/1000.0 W/(m^2).
The mean heat flow for the ocean basins is 101 +/ 2.2 mW/(m^2).
(65.0/1000.0)
joules/meters^2seconds
Thickness of each model layer.
Units = meters (m)
100.0
m
60e9
flow_to_1
to_atmosphere
j
energy_in_layer_1
j/yr
air_temperature
T_z0
T_z200
energy_in_layer_4
T_z300
T_z100
layer_thickness
heat_capacity
rock_specific_heat
heat_capacity
Geothermal_Inflow
unit_area
heat_capacity
rock_density
heat_capacity
Geothermal_Inflow
energy_in_layer_5
T_z400
T_z200
flow_to_3
flow_to_2
flow_to_3
energy_in_layer_3
T_z200
flow_to_4
T_z200
flow_to_2
energy_in_layer_2
T_z100
T_z100
flow_to_2
T_z100
flow_to_1
T_z0
flow_to_1
flow_to_3
flow_to_4
flow_to_5
flow_to_6
flow_to_1
flow_to_5
flow_to_6
flow_to_7
flow_to_8
flow_to_7
flow_to_8
flow_to_9
flow_to_9
T_z300
flow_to_3
T_z300
flow_to_4
T_z400
flow_to_4
T_z400
flow_to_5
T_z400
T_z300
T_z500
T_z500
flow_to_5
T_z500
flow_to_6
T_z600
flow_to_6
T_z600
flow_to_7
T_z700
flow_to_7
T_z700
flow_to_8
T_z800
flow_to_8
T_z800
flow_to_9
T_z700
T_z600
T_z800
T_z900
T_z900
flow_to_9
energy_in_layer_10
T_z900
energy_in_layer_9
T_z800
energy_in_layer_8
T_z700
energy_in_layer_7
T_z600
energy_in_layer_6
T_z500
sec_per_year
k
Mean_Heat_Flow
Geothermal_Inflow
flow_to_2
flow_to_1































energy_in_layer_1
to_atmosphere
heat_capacity
unit_area
sec_per_year
k
layer_thickness
k
layer_thickness
k
k
layer_thickness
k
layer_thickness
heat_capacity
heat_capacity
heat_capacity
layer_thickness