A Climate Senior Project Presentation

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Simulating Climate Using A Simple Model Jonathan Fivelsdal May 12, 2010 Jonathan Fivelsdal Simulating Climate Using A Simple Model 1/44

Transcript of A Climate Senior Project Presentation

Page 1: A Climate Senior Project Presentation

Simulating Climate UsingA Simple Model

Jonathan Fivelsdal

May 12, 2010

Jonathan Fivelsdal Simulating Climate Using A Simple Model 1/44

Page 2: A Climate Senior Project Presentation

General Outline

• Introduction

• Basic Facts About Climate

• Climate Model Equation

• Results

• Future Work

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Introduction

• My Research Focuses on Global Climate

• I use a Computer Model of a 1-D EBM

• I Investigate How Land Configs Affect TemperatureChange Across the Globe

• Simulation Runs Over 1-Year Period

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Introduction

• My Research Focuses on Global Climate

• I use a Computer Model of a 1-D EBM

• I Investigate How Land Configs Affect TemperatureChange Across the Globe

• Simulation Runs Over 1-Year Period

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Introduction

• My Research Focuses on Global Climate

• I use a Computer Model of a 1-D EBM

• I Investigate How Land Configs Affect TemperatureChange Across the Globe

• Simulation Runs Over 1-Year Period

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Introduction

• My Research Focuses on Global Climate

• I use a Computer Model of a 1-D EBM

• I Investigate How Land Configs Affect TemperatureChange Across the Globe

• Simulation Runs Over 1-Year Period

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Introduction

• My Research Focuses on Global Climate

• I use a Computer Model of a 1-D EBM

• I Investigate How Land Configs Affect TemperatureChange Across the Globe

• Simulation Runs Over 1-Year Period

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1-D Energy Balance Model

1-D Energy Balance Model

• Used to Examine Properties of Earth’s Climate System

• Based on Equilibrium Condition

• Considers Temperature Variations Across Lines of LatitudeOnly

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Physics of Heat Exchange

• Energy is the ability to do work

• Heat flux is the process of energy transfer from one body toanother

• Example: Earth Receives Energy From the Sun and inResponse to the Energy Exchange Heats Up

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Physics of Heat Exchange

• Energy is the ability to do work

• Heat flux is the process of energy transfer from one body toanother

• Example: Earth Receives Energy From the Sun and inResponse to the Energy Exchange Heats Up

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Physics of Heat Exchange

• Energy is the ability to do work

• Heat flux is the process of energy transfer from one body toanother

• Example: Earth Receives Energy From the Sun and inResponse to the Energy Exchange Heats Up

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Specific Heat

C =ΔQ

mΔT

• Specific heat capacity: heat energy required to raise thetemperature of a substance

Terms DescriptionΔQ Heat required to raise substance to a given temperaturem Mass of a given substanceΔT Amount of temperature change in a given substanceC Specific Heat Capacity

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Specific Heat (Examples)

• specific heat of rock is 850 J (kg K)−1

• specific heat of water 4186 J (kg K)−1

• Since the specific heat of water > specific heat of rock;more energy to increase temp of oceans vs. land

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Specific Heat (Examples)

• specific heat of rock is 850 J (kg K)−1

• specific heat of water 4186 J (kg K)−1

• Since the specific heat of water > specific heat of rock;more energy to increase temp of oceans vs. land

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Albedo

Albedo (�)

• ratio of radiation reflected by a surface

• 0 ≤ � ≤ 1

• Net Albedo of Earth is determined by factors such as % ofland vs. ocean and glacier coverage

• Earth’s Average Albedo is .3

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Blackbody Radiation

• A blackbody is a surface that completely absorbs allincident radiation

• Earth absorbs most of the radiation it receives (about 70%)

• Earth can be approximately described as a blackbody

• The amount of energy released by a blackbody over agiven area is obtained by the Stefan-Boltzmann law

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Blackbody Radiation

• A blackbody is a surface that completely absorbs allincident radiation

• Earth absorbs most of the radiation it receives (about 70%)

• Earth can be approximately described as a blackbody

• The amount of energy released by a blackbody over agiven area is obtained by the Stefan-Boltzmann law

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Blackbody Radiation

• A blackbody is a surface that completely absorbs allincident radiation

• Earth absorbs most of the radiation it receives (about 70%)

• Earth can be approximately described as a blackbody

• The amount of energy released by a blackbody over agiven area is obtained by the Stefan-Boltzmann law

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Blackbody Radiation

• A blackbody is a surface that completely absorbs allincident radiation

• Earth absorbs most of the radiation it receives (about 70%)

• Earth can be approximately described as a blackbody

• The amount of energy released by a blackbody over agiven area is obtained by the Stefan-Boltzmann law

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Stefan Boltzmann LawAmount of Energy Emitted by a Blackbody

E = ��T 4

• E: Flux Density• �: Emissivity (Ability to radiate as a blackbody (�bb = 1)• �: Stefan-Boltzmann Constant *• T: Temperature

* � = 5.67 ⋅ 10−8 Wm−2 K−4

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Greenhouse Effect

• Ability of Atmosphere to Absorb and Reflect Heat RadiatedFrom Earth’s Surface

• Atmosphere is a layer of gas

• (ie. 78% Nitrogen,21% Oxygen, and < 1% Carbon Dioxide)

• Thicker Atmosphere Absorbs Heat More Effectively

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Greenhouse Effect

• Ability of Atmosphere to Absorb and Reflect Heat RadiatedFrom Earth’s Surface

• Atmosphere is a layer of gas

• (ie. 78% Nitrogen,21% Oxygen, and < 1% Carbon Dioxide)

• Thicker Atmosphere Absorbs Heat More Effectively

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Greenhouse Effect

• Ability of Atmosphere to Absorb and Reflect Heat RadiatedFrom Earth’s Surface

• Atmosphere is a layer of gas

• (ie. 78% Nitrogen,21% Oxygen, and < 1% Carbon Dioxide)

• Thicker Atmosphere Absorbs Heat More Effectively

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Greenhouse Effect

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Facts About Earth’s Temperature

• Earth’s Effective Blackbody Temperature (No Atmosphere)is 255 K

• Freezing Temperature of Water 273 K

• Earth’s Temperature With Atmosphere (Actual) is 288 K

• Earth’s Temperature With Atmosphere (Model) is 303 K

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More Facts About Earth’sTemperature

• Difference Between Actual Mean Temperature (288 K) andEffective (255 K) is 30 K

• The Temp Difference is due to the Greenhouse Effect

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Incoming Solar Radiation

• Amount of Solar Radiation Earth Receives in Units Wm−2

• Solar Radiation is also referred to as Short Wave(SW)

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Incoming Solar Radiation

• Amount of SW that reaches Top of Atmosphere is 1380Wm−2

• Total Amount of SW Received by an Average Point onEarth is 343 Wm−2

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Incoming Solar Radiation

• Amount of SW that reaches Top of Atmosphere is 1380Wm−2

• Total Amount of SW Received by an Average Point onEarth is 343 Wm−2

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Incoming Solar Radiation

• Amount of SW that reaches Top of Atmosphere is 1380Wm−2

• Total Amount of SW Received by an Average Point onEarth is 343 Wm−2

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Incoming Solar Radiation

• Only 25 % of the SW at the Top of the Atmosphere(TOA)Reaches Earth’s Surface

• This is because Earth’s Sphericity and only half of theearth is lit by the sun

• (ie. day and night)

Solar Radiation ValuesRadiation at TOA 1380 Wm−2

Incoming Radiation 343 Wm−2

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Geographic Features and Climate

Geographic Features that Affect Climate are

• Distribution of Land and Sea• Ocean Currents• Orientation of Land Masses

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Climate Models

Advantages of Simple Climate Models VS Complex Models

• Easier to Interpret• Less Computationally Expensive• Easier to Implement

• Complex Models use Systems of PDE’s• My Model is Based on 1 Partial Differential Equation

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The Model Equation

Climate Model Equation

∂T

∂t=

1

�cpℎ[Qs(1− �)− I +D]

Model Parameters Description�cpℎ Thermal Capacity FactorQ Insolations = s(y) Latitude Distribution� AlbedoI OLRD Difffusion

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Model Parameters Explained

∂T

∂t=

1

�cpℎ[Qs(1− �)− I +D]

• Q = 343 Wm−2: (Incoming Solar Radiation)• s(y): Describes How Solar Radiation is Distributed Over

Latitudes• y = sin(�): where � is Latitude• (1-�): Ratio Solar Radiation Absorbed by the Earth• I: Outgoing Long Wave Radiation• D: Diffusion (Heat Transport Term)

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Outward Longwave Radiation

Since �rock ≈ 1 and �water ≈ 1 we represent I as

I = ��T 4

= (1)�T 4

= �T 4

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Thermal Capacity Factor

∂T

∂t=

1

�cpℎ[Qs(1− �)− I +D]

• �cpℎ is the thermal capacity factor

• Amount of Heat Per Unit Surface Area needed to Raise theSurface Temperature of the Earth by 1∘ C

• Units are in J K−1 m−2

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Thermal Capacity Factor

∂T

∂t=

1

�cpℎ[Qs(1− �)− I +D]

• �cpℎ is the thermal capacity factor

• Amount of Heat Per Unit Surface Area needed to Raise theSurface Temperature of the Earth by 1∘ C

• Units are in J K−1 m−2

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Thermal Capacity Factor

∂T

∂t=

1

�cpℎ[Qs(1− �)− I +D]

• �cpℎ is the thermal capacity factor

• Amount of Heat Per Unit Surface Area needed to Raise theSurface Temperature of the Earth by 1∘ C

• Units are in J K−1 m−2

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Properties of Rock and Water

• Rock radiates heat differently from water

• Specific Heat of Rock (cpr) is 850 J K−1 kg−1

• Specific Heat of Water (cpw) is 4198 J K−1 kg−1

• Specific Heat of Water is about 5 times larger than SpecificHeat of Rock

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Properties of Rock and Water

• Rock radiates heat differently from water

• Specific Heat of Rock (cpr) is 850 J K−1 kg−1

• Specific Heat of Water (cpw) is 4198 J K−1 kg−1

• Specific Heat of Water is about 5 times larger than SpecificHeat of Rock

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Properties of Rock and Water

• Rock radiates heat differently from water

• Specific Heat of Rock (cpr) is 850 J K−1 kg−1

• Specific Heat of Water (cpw) is 4198 J K−1 kg−1

• Specific Heat of Water is about 5 times larger than SpecificHeat of Rock

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Properties of Rock and Water

• Rock radiates heat differently from water

• Specific Heat of Rock (cpr) is 850 J K−1 kg−1

• Specific Heat of Water (cpw) is 4198 J K−1 kg−1

• Specific Heat of Water is about 5 times larger than SpecificHeat of Rock

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Capacity Factors: Rock and Water

∂T

∂t=

1

�cpℎ[Qs(1− �)− I +D]

Rock Thermal Capacity Factor

rc = �rcprℎr

Water Thermal Capacity Factor

wc = �wcpwℎw

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Radiative Imbalance

• Excess Radiation (Tropics) Deficit (Poles)

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Hadley Circulation

• A Physical Manifestation of Heat Transport

• Large Scale Circulations Develop in the Atmosphere due toUneven Heating

• 1. Air Rises Near The Equator

• 2. Then Air Flows Poleward and Cools

• 3. Air Sinks Near The Poles

• 4. Air Flows Close to Surface Back Towards Equator

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Gulf Stream

• Series of Fast Moving Ocean Currents in the Atlantic• Carry Warm Air Poleward• Results in Mild Climates in North America and Europe• The Diffusion term (D) Represents this Process

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Model Diffusion Compared toRealistic Diffusion

• Model Diffusion is Given by the equation D = C(T − T )

• Realistic Diffusion is Represented by

D ∝ ∂2T

∂x

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Model Diffusion Compared toRealistic Diffusion

Figure:Dark Line ≡ Temp. DistributionA,B ≡ Specfic Temps.Dotted ≡ Mean Temperature

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Model Diffusion Compared toRealistic Diffusion

• With Model Diffusion A and B are Treated the Same• Due to (T − T ) for A equaling (T − T ) for B

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Model Diffusion Compared toRealistic Diffusion

• With Realistic Diffusion Negative Diffusion at A andPositive Diffusion at B

• A: Neg. Curvature and B: Pos. CurvatureJonathan Fivelsdal Simulating Climate Using A Simple Model 32/44

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Land Configurations

Land Types

• Type 1: Land in Poles Only (land > 55 S and land > 55 N)

• Type 2: Land in Poles and Tropics

• Type 3: Land in Tropics Only (23.5 S < land < 23.5 N)

• Type 4: Land mostly in Northern Hemisphere (MostlyAbove the Equator)

• Land types were run for each initial temperature

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Percentage Land Vs. Water

Land Type: Type 1 Type 2 Type 3 Type 4Land 39% 65% 26% 39%Water 61 % 35 % 74 % 61%

• Notice: Land Type 3 has the most water

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Initial Condition 1

Annual Average Temperatures With Initial Temperature 255 K

250

255

260

265

270

275

280

285

290

-90 -75 -60 -45 -30 -15 0 15 30 45 60 75 90

Latitude

Tem

pera

ture

(Kel

vin)

Land Type 1Land Type 2Land Type 3Land Type 4Initial Temp

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Results for Initial Condition 1

• Land Type 3 Changes the Least From the InitialTemperature of 255 K

• This is because Land Type 3 has the most water and thehigh heat capacity of water

• Land Type 2 Changes the Most Since it has the Most Land• Land Type 1 and 4 have similar Temp. Distributions since

same amount of land

Land Type: Type 1 Type 2 Type 3 Type 4Land 39% 65% 26% 39%Water 61 % 35 % 74 % 61%

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Initial Condition 2

Annual Average Temperatures With Initial Temperature 273 K

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Latitude

Tem

pera

ture

(Kel

vin)

Land Type 1Land Type 2Land Type 3Land Type 4Initial Temp

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Results for Initial Condition 2

Land Type 3 Changes the Least From the Initial Temp.

Again this is likely due to the higher heat capacity of water

Land Type: Type 1 Type 2 Type 3 Type 4Land 39% 65% 26% 39%Water 61 % 35 % 74 % 61%

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Initial Condition 3

Annual Average Temperatures With Initial Temperature 288 K

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-90 -75 -60 -45 -30 -15 0 15 30 45 60 75 90

Latitude

Tem

pera

ture

(Kel

vin)

Land Type 1Land Type 2Land Type 3Land Type 4Initial Temp

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Results for Initial Condition 3

Land Type: Type 1 Type 2 Type 3 Type 4Average Temps. 279.30 K 278.88 K 282.20 K 279.25 KΔ Initial Temp. -8.7 K -9.12 K -5.8 K -8.75 K

• Land types 1,2, and 4 have average temperatures about 10K less than 288 K.

• Land type 3 has an average temperature about 6 K lessthan 288 K

• Therefore Land Type 3 changes the least from the initialtemperature

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Initial Condition 4

Annual Average Temperatures With Initial Temperature 303 K

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-90 -75 -60 -45 -30 -15 0 15 30 45 60 75 90Latitude

Tem

pera

ture

(Kel

vin)

Land Type 1Land Type 2Land Type 3Land Type 4Initial Temp

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Results for Initial Condition 4

One difference from other graphs is all temp. distributions liebelow initial temp.

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Discussion

All temp. distributions have a similar shape (indicatesdominance of solar term Qs(y))

Every temp. distribution has cold temperatures in poles andwarm temperatures in tropics

Land Type 3 is the most distinct distribution

Land Type 3 Changes Least from Initial Temperature (has themost water)

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Future Work

Test if large percent of water coverage is sole reason for smalltemp. changes

Method 1: Keep ratio of land to water constant but changedistributions of land

Method 2: Keep land distribution constant but change ratio ofland to water

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