Math 5490 10/22/2014 Heat Imbalance Math 5490 October 22, 2014 - PDF document

Math 5490 10/22/2014 Heat Imbalance Math 5490 October 22, 2014 Topics in Applied Mathematics: Introduction to the Mathematics of Climate Monday’s mistakes corrected. Mondays and Wednesdays 2:30 – 3:45 http://www.math.umn.edu/~mcgehee/teaching/Math5490-2014-2Fall/ Streaming video is available at http://www.ima.umn.edu/videos/ Click on the link: "Live Streaming from 305 Lind Hall". Participation: https://umconnect.umn.edu/mathclimate Math 5490 10/22/2014 Heat Imbalance Heat Imbalance 70 60 sea level rise (meters) 50 Suppose that all the heat imbalance 40 0.1 went to melting the glaciers. 30 0.85 20 It takes 9.3 Wyr/m 2 to turn glaciers into 1.7 10 1 meter of ocean. If the heat imbalance is w W/m 2 , the sea level 0 0 1 2 3 4 5 6 7 8 9 10 would rise at the rate of w / 9.3 meters centuries 20 per year. At the current imbalance of 0.85 W/m 2 , the rate is about 0.091 sea level rise (meters) 15 meters per year, or 9.1 meters per century. 10 0.1 0.85 Melting all the glaciers would cause a 1.7 5 sea level rise of about 70 meters and would take about 760 years at the 0 current imbalance. 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 centuries James Hansen, et al, Earth’s Energy Imbalance: Confirmation and Implications , S CIENCE 308 (2005), p. 1431 Math 5490 10/22/2014 Math 5490 10/22/2014 Heat Imbalance Heat Imbalance 70 70 60 60 sea level rise (meters) sea level rise (meters) 50 50 Suppose now that all the heat Suppose instead that all the heat 40 40 imbalance first goes to raising the top imbalance first goes to raising the top 0.1 0.1 30 30 0.85 kilometer of ocean by 0.5 ° C, and then 0.85 kilometer of ocean by 1 ° C, and then 20 1.7 20 1.7 goes to melting the glaciers. goes to melting the glaciers. 10 10 It takes 46.5 Wyr/m 2 to raise the It takes 93 Wyr/m 2 to raise the 0 0 0 1 2 3 4 5 6 7 8 9 10 temperature of a kilometer of ocean by 0 1 2 3 4 5 6 7 8 9 10 temperature of a kilometer of ocean by centuries centuries 1 ° C. If the heat imbalance is w W/m 2 , 0.5 ° C. If the heat imbalance is 20 20 w W/m 2 , the increase would be the increase would be achieved in 93/ w achieved in 46.5/ w years, after which sea level rise (meters) 15 years, after which the sea level would sea level rise (meters) 15 the sea level would rise at w /9.3 rise at w /9.3 meters per year. 10 0.1 10 0.1 meters per year. At the current imbalance of 0.85 W/m 2 , 0.85 0.85 At the current imbalance of 0.85 W/m 2 , 1.7 5 the ocean temperature increase would 1.7 5 the ocean temperature increase would delay the sea level rise by about 112 delay the sea level rise by about 56 0 years. 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 years. centuries centuries Math 5490 10/22/2014 Math 5490 10/22/2014 Richard McGehee, University of Minnesota 1

Math 5490 10/22/2014 Heat Imbalance Heat Imbalance What’s Happening Now? Summary Currently, it appears that the heat imbalance is mostly going to heating the ocean, not to melting ice. If this pattern continues, the hiatus danger for this century is more likely to come from weather changes than from sea level rise. The current heat imbalance has the potential to raise the sea level by almost a meter per decade, a major threat to coastal cities worldwide. hiatus IPCC AR5 (2013) Figure 2.20 Math 5490 10/22/2014 Math 5490 10/22/2014 Heat Imbalance Heat Imbalance What’s Happening Now? What’s Happening Now? The Good News The heat imbalance is being absorbed by the ocean (at 1000 meters, not the surface). The surface temperatures are remaining fairly constant, so the perceived warming is small. This hiatus gives us an opportunity to address the basic problem (mitigation). The Bad News The surface temperatures are remaining fairly constant, so the perceived warming is small. This hiatus could lull us into complacency so that we do not address the basic problem. There is evidence that hiatuses (hiati?) correspond to 60 year cycles of the AMOC. Will we experience another strong warming period in 30 years? Chou & Tung, Science 345 (2014) p 897 Chou & Tung, Science 345 (2014) p 897 Math 5490 10/22/2014 Math 5490 10/22/2014 Ocean Circulation Ocean Circulation Atmospheric Circulation Drives the Surface Ocean Circulation Atlantic Meridional Overturning Circulation (AMOC) Cold salty water is dense, so it descends in the North Atlantic. The rate of circulation is a function of the temperature and salinity and can change over time. Math 5490 10/22/2014 Math 5490 10/22/2014 Richard McGehee, University of Minnesota 2

Math 5490 10/22/2014 Ocean Circulation Ocean Circulation Surface Ocean Circulation Overturning Ocean Circulation Kaper & Engler Kaper & Engler Math 5490 10/22/2014 Math 5490 10/22/2014 Ocean Circulation Ocean Circulation Simplified Overturning Ocean Circulation Simplified Overturning Ocean Circulation “Conveyer Belt”, “Thermohaline Circulation. View Looking Down on Antarctic http://commons.wikimedia.org/wiki/File:Conveyor_belt.svg Kaper & Engler Math 5490 10/22/2014 Math 5490 10/22/2014 Ocean Circulation Ocean Box Models Thermohaline Circulation The circulation is driven by density differences, which are functions of temperature and salinity. Atlantic Meridional Overturning Circulation (AMOC) Part of the “thermohaline circulation”. thermo: heat haline: salt Cold salty water is dense, so it descends in the North Atlantic. The rate of circulation is a function of the temperature and salinity and can change over time. Kaper & Engler Math 5490 10/22/2014 Math 5490 10/22/2014 Richard McGehee, University of Minnesota 3

Math 5490 10/22/2014 Ocean Box Models Ocean Box Models Stommel Model Stommel Model Henry Stommel, Thermohaline Convection with Two Stable Regimes of Flow , T ELLUS XII (1961), 224-230. dT  c T   T ( ) dt * * dS    d S S ( ) dt * * T : temperature S : salinity Stars indicate constant bath temperature and salinity. Stommel, T ELLUS XII (1961) Kaper & Engler Math 5490 10/22/2014 Math 5490 10/22/2014 Ocean Box Models Ocean Box Models Dynamical Systems Approach Dynamical Systems Approach dT dx    c T T ( )    x dt dt x x dS    d S S ( ) dt     0 0 Both equation have the form: dx       x x x ( ), where and are constant. dt t t Consider first dx    x dt General solution x t  ce   t c ( ) , where is an arbitrary constant.   x x c Let (0) x x 0 x e   x t  t ( ) 0 Math 5490 10/22/2014 Math 5490 10/22/2014 Ocean Box Models Ocean Box Models Dynamical Systems Approach Dynamical Systems Approach dx    x Back to: x dt x dx       x x x ( ), where and are constant. dt     0 0 Equilibrium solution dx        x x x t x 0 ( ), so ( ) (constant). x* x* dt General solution t t x t  x   ce   t ( ) , where is an arbitrary constant.        x x x c c x x (0) , so . “Asymptotic stability” 0 0 of the equilibrium  x e    x t  x  x  t ( ) ( ) 0 x solution x x* x*   x t  x  t   If 0, ( ) as . stable unstable Math 5490 10/22/2014 Math 5490 10/22/2014 Richard McGehee, University of Minnesota 4

Math 5490 10/22/2014 Ocean Box Models Ocean Box Models Dynamical Systems Approach Dynamical Systems Approach dT dT    c T T ( )  c T   T ( ) dt       dt T t T T T e ct Solve each ( ) ( ) 0 dS dS  separately:  d S  S       dt ( )    S t S S S e d S S ( ) ( ) ( ) dt 0 dt “Non ‐ dimensionalize” S S We want to (1) take away as many units as possible, and  d c (2) get rid of as many constants as possible.  d c   T S T S ( , )     y x T S Let , ( , )   T S T T dy dT c  T  dy 1         T T c c y   ( )  1  (1 ) c y (1 )    dt T dt T  T  dt dx dx dS d  S  1           d x S S d d x (1 ) ( )  1  (1 )    dt dt S dt S  S  The new variables x and y are “dimensionless”. “Phase Portraits” Math 5490 10/22/2014 Math 5490 10/22/2014 Ocean Box Models Ocean Box Models Dynamical Systems Approach Dynamical Systems Approach dy dy   y  c  y (1 ) 1  dt d dx d dx     x d x (1 ) (1 )  d c dt “Time scaling” Finally, introduce the ratio of the rates at which the temperature and We are not wedded to seconds. Introduce a new time variable τ : salinity equilibrate. d d   cdt   c dy dy dy 1 1 (1    c  y   y ) 1 dT dy d  cdt c dt c  c T   T   ( ) y 1  dt d dx dx dx d 1 1    d  x   x dS (1 ) (1 ) dx        d cdt c dt c c d S S x ( ) (1 ) dt d  dy   y 1 d  By rescaling the temperature, salinity, and time, we have reduced the dx d   original system with four parameters, to one with just one parameter. x (1 )  d c Math 5490 10/22/2014 Math 5490 10/22/2014 Richard McGehee, University of Minnesota 5