1. Lecture 6. EPS 5: 23 Sep 2010 1.Review the concept of the barometric law (hydrostatic balance: each layer of atmosphere must support the weight of the overlying column mass of atmosphere). Discuss the distribution of pressure with altitude in the atmosphere, or depth in the ocean. 2.Introduce buoyancy. Pressure force "upwards" on an object immersed in a fluid. 3.Archimedes principle: the buoyancy force on an object is equal to the weight of the fluid displaced by the object. Role of gravity. 4.The buoyancy of warm air. 5.A brief look at global weather patterns—sea surface temperature and buoyancy. 6.Introducing the properties of water.

2. Z 2 Z 1 P 1 P 2 By how much is P1 > P2? The weight of the slab of fluid between Z1 and Z2 is given by the density, , multiplied by volume of the slab) and g weight of slab =  (area  height)  g. Set the area of the column to 1 m 2, the weight is  g  (Z2 -Z1): If the atmosphere is not being accelerated, there must be a difference in pressure (P2 - P1) across the slab that exactly balances the force of gravity (weight of the slab). Relationship between density, pressure and altitude

3. oceanatmosphere 1 bar = 10 5 N/m 2

4. Buoyancy Buoyancy is the tendency for less dense fluids to be forced upwards by more dense fluids under the influence of gravity. Buoyancy arises when the pressure forces on an object are not perfectly balanced. Buoyancy is extremely significant as a driving force for motions in the atmosphere and oceans, and hence we will examine the concept very carefully here. The mass density of air  is given by mn, where m is the mean mass of an air molecule (4.81  10 -26 kg molecule -1 for dry air), and n is the number density of air (n =2.69  10 25 molecules m -3 at T=0 o C, or 273.15 K). Therefore the density of dry air at 0 C is  = 1.29 kg m -3. If we raise the temperature to 10  C (283.15 K), the density is about 4% less, or 1.24 kg m -3. This seemingly small difference in density would cause air to move in the atmosphere, i.e. to cause winds.

5. Buoyancy force: Forces on a solid body immersed in a tank of water. The solid is assumed less dense than water and to have area A (e.g. 1m 2 ) on all sides. P1 is the fluid pressure at level 1, and P1x is the downward pressure exerted by the weight of overlying atmosphere, plus fluid between the top of the tank and level 2, plus the object. The buoyancy force is P1 – P1x (up ↑) per unit area of the submerged block. P1x Net Force (Net pressure forces – Gravity)

6. The buoyancy force and Archimedes principle. 1. Force on the top of the block: P2  A =  water D 2 A g (A = area of top) weight of the water in the volume above the block 2. Upward force on the bottom of the block = P1  A =  water D 1 A g 3. Downward force on the bottom of the block = weight of the water in the volume above block + weight of block =  water D 2 A g +  block (D 1 - D 2 ) A g Unbalanced, Upward force on the block ( [2] – [3] ): F b =  water D 1 A g – [  water D 2 A +  block (D 1 - D 2 ) A ] g =  water g V block –  block g V block = (  water –  block ) (D 1 – D 2 ) A g weight of block BUOYANCY FORCE = weight of the water (fluid) displaced by the block Volume of the block = (D 1 – D 2 ) A

7. Archimedes principle: the buoyancy force on an object is equal to the weight of the fluid displaced by the object object immersed in a fluid weight of fluid displaced for the fluid itself, there will be a net upward force (buoyancy force exceeds object weight) on parcels less dense than the surrounding fluid, a net downward force on a parcel that is more dense. buoyancy can accelerate parcels in the vertical direction (unbalanced force). the derivation of the barometric law assumed that every air parcel experienced completely balanced forces, thus didn't accelerate. Buoyancy exactly balanced the weight of the parcel (“neutrally buoyant”) – this is approximately true even if the acceleration due to unbalanced forces is quite noticeable, because the total forces on an air parcel are really huge (100,000 N/m 2 ), and thus only small imbalances are needed to produce significant accelerations.

8. This experiment, done in this class, shows Archimedes principle. In frame A, the block is displacing water, and air. When we add oil, it displaces oil and water. Since oil has a higher density than air the buoyancy force increases, forcing the block upwards. It stops moving upwards when the weight of (oil + water) that it displaces equals the weight of the block. Density Data: water = 1000; HDPE=941; Veg oil =.894

9. What will happen when I add the oil on top of the water? 1. Block will move down; 2. Block will sink to the bottom; 3. Block will rise, but will remain submerged in the oil; 4 block will float to the top of the oil.. 9

10. A closer look at the U-tube experiment… compute the density of the paint thinner : U h1h1 h2h2 h3h3  w h 1 =  w h 3 +  p h 2  w (h 1 – h 3 ) =  p h 2 buoyancy force:  w h i g –  p h i g = (  w –  p )h i g Looks a lot like Archimedes' principle 2 h i = h 1 + h 2 + h 3

11. Lecture 6. EPS 5 1.Review the concept of the barometric law (hydrostatic balance: each layer of atmosphere must support the weight of the overlying column mass of atmosphere). Discuss the distribution of pressure with altitude in the atmosphere, or depth in the ocean. 2.Introduce buoyancy. Pressure force "upwards" on an object immersed in a fluid. 3.Archimedes principle: the buoyancy force on an object is equal to the weight of the fluid displaced by the object. Role of gravity. 4.The buoyancy of warm air.

12. Cold, relatively dense air has higher density than adjacent warm air, the warm air is buoyant (the cold air is "negatively buoyant"). The "warm air rises" (is buoyant!). Buoyancy and air temperature. Consider two air parcels at the same pressure, but different temperatures. P =  1 (k/m) T 1 =  2 (k/m) T 2 Then  1 /  2 = T 2 /T 1 ; if T 1 > T 2,  1 <  2. Warmer air, lower density!

13. Lecture 6. EPS 5 1.Review the concept of the barometric law (hydrostatic balance: each layer of atmosphere must support the weight of the overlying column mass of atmosphere). Discuss the distribution of pressure with altitude in the atmosphere, or depth in the ocean. 2.Introduce buoyancy. Pressure force "upwards" on an object immersed in a fluid. 3.Archimedes principle: the buoyancy force on an object is equal to the weight of the fluid displaced by the object. Role of gravity. 4.The buoyancy of warm air. 5.Introduce properties of water vapor.

14. Lecture 6. EPS 5: 23 Sep 2010 Review the concept of the barometric law (hydrostatic balance: each layer of atmosphere must support the weight of the overlying column mass of atmosphere). Discuss the distribution of pressure with altitude in the atmosphere, or depth in the ocean. 1.Introduce buoyancy. Pressure force "upwards" on an object immersed in a fluid. 2.Archimedes principle: the buoyancy force on an object is equal to the weight of the fluid displaced by the object. Role of gravity. 3.The buoyancy of warm air. 4.A brief look at global weather patterns—sea surface temperature and buoyancy. 5.Introducing the properties of water.

15. Global Sea Surface Temperatures February 2002

16. Global Sea Surface Temperature Anomalies, December 2001

17. 10 Feb 2002 GOES ir image

18. 10 Feb 2002 GOES ir image

19. http://www.cira.colostate.edu/Special/CurrWx/g8full40.asp 10 Feb 2003 GOES ir image

20. Global Sea Surface Temperature Anomalies, Dec. 2001, Jan 2003 12-2001 La Niña 01-2003 El Niño

21. Lecture 6. EPS 5: 23 Feb. 2010 Review the concept of the barometric law (hydrostatic balance: each layer of atmosphere must support the weight of the overlying column mass of atmosphere). Discuss the distribution of pressure with altitude in the atmosphere, or depth in the ocean. 1.Introduce buoyancy. Pressure force "upwards" on an object immersed in a fluid. 2.Archimedes principle: the buoyancy force on an object is equal to the weight of the fluid displaced by the object. Role of gravity. 3.The buoyancy of warm air. 4.A brief look at global weather patterns—sea surface temperature and buoyancy. 5.Introducing the properties of water.

22. 22 Net Exchange of CO 2 in a Mid-Latitude Forest S. C. Wofsy 1, M. L. Goulden 1, J. W. Munger 1, S.-M. Fan 1, P. S. Bakwin 1, B. C. Daube 1, S. L. Bassow 2, and F. A. Bazzaz 2 1 Division of Applied Science and Department of Earth and Planetary Science 2 Department of Organismic and Evolutionary Biology, Harvard University, Cambridge, MA 02138 The eddy correlation method was used to measure the net ecosystem exchange of carbon dioxide continuously from April 1990 to December 1991 in a deciduous forest in central Massachusetts. The annual net uptake was 3.7 +/- 0.7 metric tons of carbon per hectare per year. Ecosystem respiration, calculated from the relation between nighttime exchange and soil temperature, was 7.4 metric tons of carbon per hectare per year, implying gross ecosystem production of 11.1 metric tons of carbon per hectare per year. The observed rate of accumulation of carbon reflects recovery from agricultural development in the 1800s. Carbon uptake rates were notably larger than those assumed for temperate forests in global carbon studies. Carbon storage in temperate forests can play an important role in determining future concentrations of atmospheric carbon dioxide. Submitted on December 4, 1992 Accepted on March 23, 1993 Science 28 May 1993: Vol. 260. no. 5112, pp. 1314 - 1317