# The Hydrostatic Equation

Save this PDF as:

Size: px
Start display at page:

## Transcription

1 The Hydrostatic Equation Air pressure at any height in the atmosphere is due to the force per unit area exerted by the weight of all of the air lying above that height. Consequently, atmospheric pressure decreases with increasing height above the ground.

2 The Hydrostatic Equation Air pressure at any height in the atmosphere is due to the force per unit area exerted by the weight of all of the air lying above that height. Consequently, atmospheric pressure decreases with increasing height above the ground. The net upward force acting on a thin horizontal slab of air, due to the decrease in atmospheric pressure with height, is generally very closely in balance with the downward force due to gravitational attraction that acts on the slab.

3 The Hydrostatic Equation Air pressure at any height in the atmosphere is due to the force per unit area exerted by the weight of all of the air lying above that height. Consequently, atmospheric pressure decreases with increasing height above the ground. The net upward force acting on a thin horizontal slab of air, due to the decrease in atmospheric pressure with height, is generally very closely in balance with the downward force due to gravitational attraction that acts on the slab. If the net upward pressure force on the slab is equal to the downward force of gravity on the slab, the atmosphere is said to be in hydrostatic balance.

4 Balance of vertical forces in an atmosphere in hydrostatic balance. 2

5 The mass of air between heights z and z + δz in the column of air is ρ δz. 3

6 The mass of air between heights z and z + δz in the column of air is ρ δz. The downward gravitational force acting on this slab of air, due to the weight of the air, is gρδz. 3

7 The mass of air between heights z and z + δz in the column of air is ρ δz. The downward gravitational force acting on this slab of air, due to the weight of the air, is gρδz. Let the change in pressure in going from height z to height z + δz be δp. Since we know that pressure decreases with height, δp must be a negative quantity. 3

8 The mass of air between heights z and z + δz in the column of air is ρ δz. The downward gravitational force acting on this slab of air, due to the weight of the air, is gρδz. Let the change in pressure in going from height z to height z + δz be δp. Since we know that pressure decreases with height, δp must be a negative quantity. The upward pressure on the lower face of the shaded block must be slightly greater than the downward pressure on the upper face of the block. 3

9 The mass of air between heights z and z + δz in the column of air is ρ δz. The downward gravitational force acting on this slab of air, due to the weight of the air, is gρδz. Let the change in pressure in going from height z to height z + δz be δp. Since we know that pressure decreases with height, δp must be a negative quantity. The upward pressure on the lower face of the shaded block must be slightly greater than the downward pressure on the upper face of the block. Therefore, the net vertical force on the block due to the vertical gradient of pressure is upward and given by the positive quantity δp as indicated in the figure. 3

10 For an atmosphere in hydrostatic equilibrium, the balance of forces in the vertical requires that δp = gρ δz 4

11 For an atmosphere in hydrostatic equilibrium, the balance of forces in the vertical requires that In the limit as δz 0, δp = gρ δz p z = gρ. This is the hydrostatic equation. 4

12 For an atmosphere in hydrostatic equilibrium, the balance of forces in the vertical requires that In the limit as δz 0, δp = gρ δz p z = gρ. This is the hydrostatic equation. The negative sign ensures that the pressure decreases with increasing height. 4

13 For an atmosphere in hydrostatic equilibrium, the balance of forces in the vertical requires that In the limit as δz 0, δp = gρ δz p z = gρ. This is the hydrostatic equation. The negative sign ensures that the pressure decreases with increasing height. Since ρ = 1/α, the equation can be rearranged to give g dz = α dp 4

14 Integrating the hydrostatic equation from height z (and pressure p(z)) to an infinite height: p( ) p(z) dp = z gρ dz 5

15 Integrating the hydrostatic equation from height z (and pressure p(z)) to an infinite height: Since p( ) = 0, p( ) p(z) p(z) = dp = z z gρ dz gρ dz That is, the pressure at height z is equal to the weight of the air in the vertical column of unit cross-sectional area lying above that level. 5

16 Integrating the hydrostatic equation from height z (and pressure p(z)) to an infinite height: Since p( ) = 0, p( ) p(z) p(z) = dp = z z gρ dz gρ dz That is, the pressure at height z is equal to the weight of the air in the vertical column of unit cross-sectional area lying above that level. If the mass of the Earth s atmosphere were uniformly distributed over the globe, the pressure at sea level would be 1013 hpa, or Pa, which is referred to as 1 atmosphere (or 1 atm). 5

17 Geopotential The geopotential Φ at any point in the Earth s atmosphere is defined as the work that must be done against the Earth s gravitational field to raise a mass of 1 kg from sea level to that point. 6

18 Geopotential The geopotential Φ at any point in the Earth s atmosphere is defined as the work that must be done against the Earth s gravitational field to raise a mass of 1 kg from sea level to that point. In other words, Φ is the gravitational potential energy per unit mass. The units of geopotential are J kg 1 or m 2 s 2. 6

19 Geopotential The geopotential Φ at any point in the Earth s atmosphere is defined as the work that must be done against the Earth s gravitational field to raise a mass of 1 kg from sea level to that point. In other words, Φ is the gravitational potential energy per unit mass. The units of geopotential are J kg 1 or m 2 s 2. The work (in joules) in raising 1 kg from z to z + dz is g dz. Therefore dφ = g dz or, using the hydrostatic equation, dφ = g dz = α dp 6

20 The geopotential Φ(z) at height z is thus given by Φ(z) = z 0 g dz. where the geopotential Φ(0) at sea level (z = 0) has been taken as zero. 7

21 The geopotential Φ(z) at height z is thus given by Φ(z) = z 0 g dz. where the geopotential Φ(0) at sea level (z = 0) has been taken as zero. The geopotential at a particular point in the atmosphere depends only on the height of that point and not on the path through which the unit mass is taken in reaching that point. 7

22 The geopotential Φ(z) at height z is thus given by Φ(z) = z 0 g dz. where the geopotential Φ(0) at sea level (z = 0) has been taken as zero. The geopotential at a particular point in the atmosphere depends only on the height of that point and not on the path through which the unit mass is taken in reaching that point. We define the geopotential height Z as Z = Φ(z) = 1 g dz g 0 g 0 0 where g 0 is the globally averaged acceleration due to gravity at the Earth s surface. z 7

23 Geopotential height is used as the vertical coordinate in most atmospheric applications in which energy plays an important role. It can be seen from the Table below that the values of Z and z are almost the same in the lower atmosphere where g g 0. 8

24 The Hypsometric Equation In meteorological practice it is not convenient to deal with the density of a gas, ρ, the value of which is generally not measured. By making use of the hydrostatic equation and the gas law, we can eliminate ρ: p z = pg RT = pg R d T v 9

25 The Hypsometric Equation In meteorological practice it is not convenient to deal with the density of a gas, ρ, the value of which is generally not measured. By making use of the hydrostatic equation and the gas law, we can eliminate ρ: p z = pg RT = pg R d T v Rearranging the last expression and using dφ = g dz yields dφ = g dz = RT dp p = R dp dt v p 9

26 The Hypsometric Equation In meteorological practice it is not convenient to deal with the density of a gas, ρ, the value of which is generally not measured. By making use of the hydrostatic equation and the gas law, we can eliminate ρ: p z = pg RT = pg R d T v Rearranging the last expression and using dφ = g dz yields dφ = g dz = RT dp p = R dp dt v p Integrating between pressure levels p 1 and p 2, with geopotentials Z 1 and Z 1 respectively, Φ2 p2 dp dφ = R d T v Φ 1 p 1 p or p2 dp Φ 2 Φ 1 = R d T v p p 1 9

27 Dividing both sides of the last equation by g 0 and reversing the limits of integration yields Z 2 Z 1 = R d g 0 p1 p 2 T v dp p The difference Z 2 Z 1 is called the geopotential thickness of the layer. 10

28 Dividing both sides of the last equation by g 0 and reversing the limits of integration yields Z 2 Z 1 = R d g 0 p1 p 2 T v dp p The difference Z 2 Z 1 is called the geopotential thickness of the layer. If the virtual temperature is constant with height, we get p1 ( ) dp p1 Z 2 Z 1 = H p 2 p = H log p 2 or [ p 2 = p 1 exp Z ] 2 Z 1 H where H = R d T v /g 0 is the scale height. Since R d = 287 J K 1 kg 1 and g 0 = 9.81 m s 2 we have, approximately, H = 29.3 T v. 10

29 Dividing both sides of the last equation by g 0 and reversing the limits of integration yields Z 2 Z 1 = R d g 0 p1 p 2 T v dp p The difference Z 2 Z 1 is called the geopotential thickness of the layer. If the virtual temperature is constant with height, we get p1 ( ) dp p1 Z 2 Z 1 = H p 2 p = H log p 2 or [ p 2 = p 1 exp Z ] 2 Z 1 H where H = R d T v /g 0 is the scale height. Since R d = 287 J K 1 kg 1 and g 0 = 9.81 m s 2 we have, approximately, H = 29.3 T v. If we take a mean value for virtual temperature of T v = 255 K, the scale height H for air in the atmosphere is found to be about 7.5 km. 10

30 Dividing both sides of the last equation by g 0 and reversing the limits of integration yields Z 2 Z 1 = R p1 d dp T v g 0 p 2 p The difference Z 2 Z 1 is called the geopotential thickness of the layer. If the virtual temperature is constant with height, we get p1 ( ) dp p1 Z 2 Z 1 = H p 2 p = H log p 2 or [ p 2 = p 1 exp Z ] 2 Z 1 H where H = R d T v /g 0 is the scale height. Since R d = 287 J K 1 kg 1 and g 0 = 9.81 m s 2 we have, approximately, H = 29.3 T v. If we take a mean value for virtual temperature of T v = 255 K, the scale height H for air in the atmosphere is found to be about 7.5 km. Exercise: Check these statements. 10

31 The temperature and vapour pressure of the atmosphere generally vary with height. In this case we can define a mean virtual temperature T v (see following Figure): log p1 log p1 log p T T v = 2 v d log p log p1 log p d log p = log p T 2 v d log p log(p 1 /p 2 ) 2 11

32 The temperature and vapour pressure of the atmosphere generally vary with height. In this case we can define a mean virtual temperature T v (see following Figure): log p1 log p1 log p T T v = 2 v d log p log p1 log p d log p = log p T 2 v d log p log(p 1 /p 2 ) 2 Using this in the thickness equation we get Z 2 Z 1 = R d g 0 p1 p 2 T v d log p = R d T v g 0 log p 1 p 2 11

33 The temperature and vapour pressure of the atmosphere generally vary with height. In this case we can define a mean virtual temperature T v (see following Figure): log p1 log p1 log p T T v = 2 v d log p log p1 log p d log p = log p T 2 v d log p log(p 1 /p 2 ) 2 Using this in the thickness equation we get Z 2 Z 1 = R d g 0 p1 p 2 T v d log p = R d T v g 0 log p 1 p 2 This is called the hypsometric equation: Z 2 Z 1 = R d T v g 0 log p 1 p 2 11

34 Figure 3.2. Vertical profile, or sounding, of virtual temperature. If area ABC is equal to area CDE, then T v is the mean virtual temperature with respect to log p between the pressure levels p 1 and p 2. 12

35 Constant Pressure Surfaces Since pressure decreases monotonically with height, pressure surfaces never intersect. It follows from the hypsometric equation that that the thickness of the layer between any two pressure surfaces p 2 and p 1 is proportional to the mean virtual temperature of the layer, T v. 13

36 Constant Pressure Surfaces Since pressure decreases monotonically with height, pressure surfaces never intersect. It follows from the hypsometric equation that that the thickness of the layer between any two pressure surfaces p 2 and p 1 is proportional to the mean virtual temperature of the layer, T v. Essentially, the air between the two pressure levels expands and the layer becomes thicker as the temperature increases. 13

37 Constant Pressure Surfaces Since pressure decreases monotonically with height, pressure surfaces never intersect. It follows from the hypsometric equation that that the thickness of the layer between any two pressure surfaces p 2 and p 1 is proportional to the mean virtual temperature of the layer, T v. Essentially, the air between the two pressure levels expands and the layer becomes thicker as the temperature increases. Exercise: Calculate the thickness of the layer between the 1000 hpa and 500 hpa pressure surfaces, (a) at a point in the tropics where the mean virtual temperature of the layer is 15 C, and (b) at a point in the polar regions where the mean virtual temperature is 40 C. 13

38 Solution: From the hypsometric equation, Z = Z 500 Z 1000 = R d T ( ) v 1000 ln = 20.3 g T v metres Therefore, for the tropics with virtual temperature T v = 288 K we get Z = 5846 m. 14

39 Solution: From the hypsometric equation, Z = Z 500 Z 1000 = R d T ( ) v 1000 ln = 20.3 g T v metres Therefore, for the tropics with virtual temperature T v = 288 K we get Z = 5846 m. For polar regions with virtual temperature T v = 233 K, we get Z = 4730 m. 14

40 Solution: From the hypsometric equation, Z = Z 500 Z 1000 = R d T ( ) v 1000 ln = 20.3 g T v metres Therefore, for the tropics with virtual temperature T v = 288 K we get Z = 5846 m. For polar regions with virtual temperature T v = 233 K, we get Z = 4730 m. In operational practice, thickness is rounded to the nearest 10 m and expressed in decameters (dam). Hence, answers for this exercise would normally be expressed as 585 dam and 473 dam, respectively. 14

41 Using soundings from a network of stations, it is possible to construct topographical maps of the distribution of geopotential height on selected pressure surfaces. 15

42 Using soundings from a network of stations, it is possible to construct topographical maps of the distribution of geopotential height on selected pressure surfaces. If the three-dimensional distribution of virtual temperature is known, together with the distribution of geopotential height on one pressure surface, it is possible to infer the distribution of geopotential height of any other pressure surface. 15

43 Using soundings from a network of stations, it is possible to construct topographical maps of the distribution of geopotential height on selected pressure surfaces. If the three-dimensional distribution of virtual temperature is known, together with the distribution of geopotential height on one pressure surface, it is possible to infer the distribution of geopotential height of any other pressure surface. The same hypsometric relationship between the three-dimensional temperature field and the shape of pressure surface can be used in a qualitative way to gain some useful insights into the three-dimensional structure of atmospheric disturbances, as illustrated by the following examples: 15

44 Using soundings from a network of stations, it is possible to construct topographical maps of the distribution of geopotential height on selected pressure surfaces. If the three-dimensional distribution of virtual temperature is known, together with the distribution of geopotential height on one pressure surface, it is possible to infer the distribution of geopotential height of any other pressure surface. The same hypsometric relationship between the three-dimensional temperature field and the shape of pressure surface can be used in a qualitative way to gain some useful insights into the three-dimensional structure of atmospheric disturbances, as illustrated by the following examples: Warm-core hurricane 15

45 Using soundings from a network of stations, it is possible to construct topographical maps of the distribution of geopotential height on selected pressure surfaces. If the three-dimensional distribution of virtual temperature is known, together with the distribution of geopotential height on one pressure surface, it is possible to infer the distribution of geopotential height of any other pressure surface. The same hypsometric relationship between the three-dimensional temperature field and the shape of pressure surface can be used in a qualitative way to gain some useful insights into the three-dimensional structure of atmospheric disturbances, as illustrated by the following examples: Warm-core hurricane Cold-core upper low 15

46 Using soundings from a network of stations, it is possible to construct topographical maps of the distribution of geopotential height on selected pressure surfaces. If the three-dimensional distribution of virtual temperature is known, together with the distribution of geopotential height on one pressure surface, it is possible to infer the distribution of geopotential height of any other pressure surface. The same hypsometric relationship between the three-dimensional temperature field and the shape of pressure surface can be used in a qualitative way to gain some useful insights into the three-dimensional structure of atmospheric disturbances, as illustrated by the following examples: Warm-core hurricane Cold-core upper low Extratropical cyclone 15

47 Figure 3.3. Vertical cross-sections through (a) a hurricane, (b) a cold-core upper tropospheric low, and (c) a middle-latitude disturbance that tilts westward with height. 16

48 Reduction of Pressure to Sea Level In mountainous regions the difference in surface pressure from one measuring station to another is largely due to differences in elevation. 17

49 Reduction of Pressure to Sea Level In mountainous regions the difference in surface pressure from one measuring station to another is largely due to differences in elevation. To isolate that part of the pressure field that is due to the passage of weather systems, it is necessary to reduce the pressures to a common reference level. For this purpose, sea level is normally used. 17

50 Reduction of Pressure to Sea Level In mountainous regions the difference in surface pressure from one measuring station to another is largely due to differences in elevation. To isolate that part of the pressure field that is due to the passage of weather systems, it is necessary to reduce the pressures to a common reference level. For this purpose, sea level is normally used. Let Z g and p g be the geopotential and pressure at ground level and Z 0 and p 0 the geopotential and pressure at sea level (Z 0 = 0). 17

51 Reduction of Pressure to Sea Level In mountainous regions the difference in surface pressure from one measuring station to another is largely due to differences in elevation. To isolate that part of the pressure field that is due to the passage of weather systems, it is necessary to reduce the pressures to a common reference level. For this purpose, sea level is normally used. Let Z g and p g be the geopotential and pressure at ground level and Z 0 and p 0 the geopotential and pressure at sea level (Z 0 = 0). Then, for the layer between the Earth s surface and sea level, the hypsometric equation becomes where H = R d Tv /g 0. (Z g Z 0 ) = Z g = H ln p o p g 17

52 Once again, where H = R d Tv /g 0. Z g = H ln p o p g 18

53 Once again, where H = R d Tv /g 0. Z g = H ln p o p g This can be solved to obtain the sea-level pressure ( ) ( ) Zg g0 Z g p 0 = p g exp = p g exp H R d Tv 18

54 Once again, where H = R d Tv /g 0. Z g = H ln p o p g This can be solved to obtain the sea-level pressure ( ) ( ) Zg g0 Z g p 0 = p g exp = p g exp H R d Tv The last expression shows how the sea-level pressure depends on the mean virtual temperature between ground and sea level. 18

55 If Z g is small, the scale height H can be evaluated from the ground temperature. Also, if Z g H, the exponential can be approximated by ( ) Zg exp 1 + Z g H H. 19

56 If Z g is small, the scale height H can be evaluated from the ground temperature. Also, if Z g H, the exponential can be approximated by ( ) Zg exp 1 + Z g H H. Since H is about 8 km for the observed range of ground temperatures on Earth, this approximation is satisfactory provided that Z g is less than a few hundred meters. 19

57 If Z g is small, the scale height H can be evaluated from the ground temperature. Also, if Z g H, the exponential can be approximated by ( ) Zg exp 1 + Z g H H. Since H is about 8 km for the observed range of ground temperatures on Earth, this approximation is satisfactory provided that Z g is less than a few hundred meters. With this approximation, we get ( p 0 p g 1 + Z ) g or H p 0 p g ( pḡ H ) Z g 19

58 If Z g is small, the scale height H can be evaluated from the ground temperature. Also, if Z g H, the exponential can be approximated by ( ) Zg exp 1 + Z g H H. Since H is about 8 km for the observed range of ground temperatures on Earth, this approximation is satisfactory provided that Z g is less than a few hundred meters. With this approximation, we get ( p 0 p g 1 + Z ) g or H p 0 p g ( pḡ Since p g 1000 hpa and H 8 km, the pressure correction (in hpa) is roughly equal to Z g (in meters) divided by 8. p 0 p g 1 8 Z g H ) Z g 19

59 If Z g is small, the scale height H can be evaluated from the ground temperature. Also, if Z g H, the exponential can be approximated by ( ) Zg exp 1 + Z g H H. Since H is about 8 km for the observed range of ground temperatures on Earth, this approximation is satisfactory provided that Z g is less than a few hundred meters. With this approximation, we get ( p 0 p g 1 + Z ) g or H p 0 p g ( pḡ Since p g 1000 hpa and H 8 km, the pressure correction (in hpa) is roughly equal to Z g (in meters) divided by 8. p 0 p g 1 8 Z g In other words, near sea level the pressure decreases by about 1 hpa for every 8 m of vertical ascent. 19 H ) Z g

60 Exercise: Calculate the geopotential height of the 1000 hpa pressure surface when the pressure at sea level is 1014 hpa. The scale height of the atmosphere may be taken as 8 km. 20

61 Exercise: Calculate the geopotential height of the 1000 hpa pressure surface when the pressure at sea level is 1014 hpa. The scale height of the atmosphere may be taken as 8 km. Solution: From the hypsometric equation, Z 1000 = H ( p0 ) ( ln = 1000 H ln 1 + p ) ( ) H p where p 0 is the sea level pressure and the approximation for x 1 has been used. ln(1 + x) x 20

62 Exercise: Calculate the geopotential height of the 1000 hpa pressure surface when the pressure at sea level is 1014 hpa. The scale height of the atmosphere may be taken as 8 km. Solution: From the hypsometric equation, Z 1000 = H ( p0 ) ( ln = 1000 H ln 1 + p ) ( ) H p where p 0 is the sea level pressure and the approximation for x 1 has been used. ln(1 + x) x Substituting H 8000 m into this expression gives Z (p ) 20

63 Exercise: Calculate the geopotential height of the 1000 hpa pressure surface when the pressure at sea level is 1014 hpa. The scale height of the atmosphere may be taken as 8 km. Solution: From the hypsometric equation, Z 1000 = H ( p0 ) ( ln = 1000 H ln 1 + p ) ( ) H p where p 0 is the sea level pressure and the approximation for x 1 has been used. ln(1 + x) x Substituting H 8000 m into this expression gives Z (p ) Therefore, with p 0 = 1014 hpa, the geopotential height Z 1000 of the 1000 hpa pressure surface is found to be 112 m above sea level. 20

64 Exercise: Derive a relationship for the height of a given pressure surface p in terms of the pressure p 0 and temperature T 0 at sea level assuming that the temperature decreases uniformly with height at a rate Γ K km 1. 21

65 Exercise: Derive a relationship for the height of a given pressure surface p in terms of the pressure p 0 and temperature T 0 at sea level assuming that the temperature decreases uniformly with height at a rate Γ K km 1. Solution: Let the height of the pressure surface be z; then its temperature T is given by T = T 0 Γz 21

66 Exercise: Derive a relationship for the height of a given pressure surface p in terms of the pressure p 0 and temperature T 0 at sea level assuming that the temperature decreases uniformly with height at a rate Γ K km 1. Solution: Let the height of the pressure surface be z; then its temperature T is given by T = T 0 Γz Combining the hydrostatic equation with the ideal gas equation gives dp p = g RT dz 21

67 Exercise: Derive a relationship for the height of a given pressure surface p in terms of the pressure p 0 and temperature T 0 at sea level assuming that the temperature decreases uniformly with height at a rate Γ K km 1. Solution: Let the height of the pressure surface be z; then its temperature T is given by T = T 0 Γz Combining the hydrostatic equation with the ideal gas equation gives dp p = g RT dz From these equations it follows that dp p = g R(T 0 Γz) dz 21

68 Again: dp p = g R(T 0 Γz) dz 22

69 Again: dp p = g R(T 0 Γz) dz Integrating this equation between pressure levels p 0 and p and corresponding heights 0 and z, and neglecting the variation of g with z, we obtain p p o dp p = g R z 0 dz T 0 Γz. 22

70 Again: dp p = g R(T 0 Γz) dz Integrating this equation between pressure levels p 0 and p and corresponding heights 0 and z, and neglecting the variation of g with z, we obtain p p o dp p = g R z 0 dz T 0 Γz. Aside: Thus: dx ax + b = 1 a log(ax + b). 22

71 Again: dp p = g R(T 0 Γz) dz Integrating this equation between pressure levels p 0 and p and corresponding heights 0 and z, and neglecting the variation of g with z, we obtain p p o dp p = g R z 0 dz T 0 Γz. Aside: dx ax + b = 1 a log(ax + b). Thus: log p p 0 = g RΓ log ( ) T0 Γz T 0. 22

72 Again: dp p = g R(T 0 Γz) dz Integrating this equation between pressure levels p 0 and p and corresponding heights 0 and z, and neglecting the variation of g with z, we obtain p p o dp p = g R Aside: dx ax + b = 1 a z 0 dz T 0 Γz. log(ax + b). Thus: Therefore log p p 0 = g RΓ log z = T 0 Γ [ 1 ( ) T0 Γz ( p p o T 0 ) RΓ/g ]. 22

73 Altimetry The altimetry equation z = T 0 Γ [ 1 ( p p o ) RΓ/g ] forms the basis for the calibration of altimeters on aircraft. An altimeter is simply an aneroid barometer that measures the air pressure p. 23

74 Altimetry The altimetry equation z = T 0 Γ [ 1 ( p p o ) RΓ/g ] forms the basis for the calibration of altimeters on aircraft. An altimeter is simply an aneroid barometer that measures the air pressure p. However, the scale of the altimeter is expressed as the height above sea level where z is related to p by the above equation with values for the parameters in accordance with the U.S. Standard Atmosphere: T 0 = 288 K p 0 = hpa Γ = 6.5 K km 1 23

75 Exercise (Hard!): Show that, in the limit Γ 0, the altimetry equation is consistent with the relationship ( p = p 0 exp z ) H already obtained for an isothermal atmosphere. 24

76 Exercise (Hard!): Show that, in the limit Γ 0, the altimetry equation is consistent with the relationship ( p = p 0 exp z ) H already obtained for an isothermal atmosphere. Solution (Easy!): Use l Hôpital s Rule. Note: If you are unfamiliar with l Hôpital s Rule, either ignore this exercise or, better still, try it using more elementary means. 24

### Chapter 4 Atmospheric Pressure and Wind

Chapter 4 Atmospheric Pressure and Wind Understanding Weather and Climate Aguado and Burt Pressure Pressure amount of force exerted per unit of surface area. Pressure always decreases vertically with height

### When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid.

Fluid Statics When the fluid velocity is zero, called the hydrostatic condition, the pressure variation is due only to the weight of the fluid. Consider a small wedge of fluid at rest of size Δx, Δz, Δs

### One Atmospheric Pressure. Measurement of Atmos. Pressure. Units of Atmospheric Pressure. Chapter 4: Pressure and Wind

Chapter 4: Pressure and Wind Pressure, Measurement, Distribution Hydrostatic Balance Pressure Gradient and Coriolis Force Geostrophic Balance Upper and Near-Surface Winds One Atmospheric Pressure (from

### Lecture 4: Pressure and Wind

Lecture 4: Pressure and Wind Pressure, Measurement, Distribution Forces Affect Wind Geostrophic Balance Winds in Upper Atmosphere Near-Surface Winds Hydrostatic Balance (why the sky isn t falling!) Thermal

### V. Water Vapour in Air

V. Water Vapour in Air V. Water Vapour in Air So far we have indicated the presence of water vapour in the air through the vapour pressure e that it exerts. V. Water Vapour in Air So far we have indicated

### Synoptic Meteorology I: Hydrostatic Balance, the Hypsometric Equation, and Thickness. 23, 25 September 2014

Synoptic Meteorology I: Hydrostatic Balance, the Hypsometric Equation, and Thickness Deriation of the Hydrostatic Equation 23, 25 September 2014 Consider a unit olume (V = 1 m 3, with sides of 1 m each)

### Chapter 3: Weather Map. Weather Maps. The Station Model. Weather Map on 7/7/2005 4/29/2011

Chapter 3: Weather Map Weather Maps Many variables are needed to described weather conditions. Local weathers are affected by weather pattern. We need to see all the numbers describing weathers at many

### Air Pressure and Winds-I. GEOL 1350: Introduction To Meteorology

Air Pressure and Winds-I GEOL 1350: Introduction To Meteorology 1 2 Pressure gradient force is in balance with gravity Hydrostatic relations Means no vertical motion initially 3 How does atmospheric pressure

### 39th International Physics Olympiad - Hanoi - Vietnam - 2008. Theoretical Problem No. 3

CHANGE OF AIR TEMPERATURE WITH ALTITUDE, ATMOSPHERIC STABILITY AND AIR POLLUTION Vertical motion of air governs many atmospheric processes, such as the formation of clouds and precipitation and the dispersal

### ATM 316: Dynamic Meteorology I Final Review, December 2014

ATM 316: Dynamic Meteorology I Final Review, December 2014 Scalars and Vectors Scalar: magnitude, without reference to coordinate system Vector: magnitude + direction, with reference to coordinate system

### Physics of the Atmosphere I

Physics of the Atmosphere I WS 2008/09 Ulrich Platt Institut f. Umweltphysik R. 424 Ulrich.Platt@iup.uni-heidelberg.de heidelberg.de Last week The conservation of mass implies the continuity equation:

### Chapter 3: Weather Map. Station Model and Weather Maps Pressure as a Vertical Coordinate Constant Pressure Maps Cross Sections

Chapter 3: Weather Map Station Model and Weather Maps Pressure as a Vertical Coordinate Constant Pressure Maps Cross Sections Weather Maps Many variables are needed to described dweather conditions. Local

### The vertical structure of the atmosphere

Chapter 3 The vertical structure of the atmosphere In this chapter we discuss the observed vertical distribution of temperature, water vapor and greenhouse gases in the atmosphere. The observed temperature

### 2 Atmospheric Pressure

2 Atmospheric Pressure meteorology 2.1 Definition and Pressure Measurement 2.1.1 Definition Pressure acts in all directions, up and sideways as well as down, but it is convenient in meteorology to regard

### The Ideal Gas Law. Gas Constant. Applications of the Gas law. P = ρ R T. Lecture 2: Atmospheric Thermodynamics

Lecture 2: Atmospheric Thermodynamics Ideal Gas Law (Equation of State) Hydrostatic Balance Heat and Temperature Conduction, Convection, Radiation Latent Heating Adiabatic Process Lapse Rate and Stability

### Pressure, Forces and Motion

Pressure, Forces and Motion Readings A&B: Ch. 4 (p. 93-114) CD Tutorials: Pressure Gradients, Coriolis, Forces & Winds Topics 1. Review: What is Pressure? 2. Horizontal Pressure Gradients 3. Depicting

### Chapter 27 Static Fluids

Chapter 27 Static Fluids 27.1 Introduction... 1 27.2 Density... 1 27.3 Pressure in a Fluid... 2 27.4 Pascal s Law: Pressure as a Function of Depth in a Fluid of Uniform Density in a Uniform Gravitational

### 2.016 Hydrodynamics Reading #2. 2.016 Hydrodynamics Prof. A.H. Techet

Pressure effects 2.016 Hydrodynamics Prof. A.H. Techet Fluid forces can arise due to flow stresses (pressure and viscous shear), gravity forces, fluid acceleration, or other body forces. For now, let us

### AP2 Fluids. Kinetic Energy (A) stays the same stays the same (B) increases increases (C) stays the same increases (D) increases stays the same

A cart full of water travels horizontally on a frictionless track with initial velocity v. As shown in the diagram, in the back wall of the cart there is a small opening near the bottom of the wall that

### Course 2 Mathematical Tools and Unit Conversion Used in Thermodynamic Problem Solving

Course Mathematical Tools and Unit Conversion Used in Thermodynamic Problem Solving 1 Basic Algebra Computations 1st degree equations - =0 Collect numerical values on one side and unknown to the otherside

### For Water to Move a driving force is needed

RECALL FIRST CLASS: Q K Head Difference Area Distance between Heads Q 0.01 cm 0.19 m 6cm 0.75cm 1 liter 86400sec 1.17 liter ~ 1 liter sec 0.63 m 1000cm 3 day day day constant head 0.4 m 0.1 m FINE SAND

### Description of zero-buoyancy entraining plume model

Influence of entrainment on the thermal stratification in simulations of radiative-convective equilibrium Supplementary information Martin S. Singh & Paul A. O Gorman S1 CRM simulations Here we give more

### Dynamics IV: Geostrophy SIO 210 Fall, 2014

Dynamics IV: Geostrophy SIO 210 Fall, 2014 Geostrophic balance Thermal wind Dynamic height READING: DPO: Chapter (S)7.6.1 to (S)7.6.3 Stewart chapter 10.3, 10.5, 10.6 (other sections are useful for those

### Physics 181- Summer 2011 - Experiment #8 1 Experiment #8, Measurement of Density and Archimedes' Principle

Physics 181- Summer 2011 - Experiment #8 1 Experiment #8, Measurement of Density and Archimedes' Principle 1 Purpose 1. To determine the density of a fluid, such as water, by measurement of its mass when

### Problem Set 4 Solutions

Chemistry 360 Dr Jean M Standard Problem Set 4 Solutions 1 Two moles of an ideal gas are compressed isothermally and reversibly at 98 K from 1 atm to 00 atm Calculate q, w, ΔU, and ΔH For an isothermal

### Figure 1.6 shows a column of mercury held in place by a sealing plate. Recall from physics that at the bottom of the column of mercury

Lecture no. 7 1.6 PRESSURE Pressures, like temperatures, can be expressed by either absolute or relative scales. Pressure is defined as "normal force per unit area." Figure 1.5 shows a cylinder of water.

### Fluid Mechanics. Fluid Statics [3-1] Dr. Mohammad N. Almasri. [3] Fall 2010 Fluid Mechanics Dr. Mohammad N. Almasri [3-1] Fluid Statics

1 Fluid Mechanics Fluid Statics [3-1] Dr. Mohammad N. Almasri Fluid Pressure Fluid pressure is the normal force exerted by the fluid per unit area at some location within the fluid Fluid pressure has the

### oil liquid water water liquid Answer, Key Homework 2 David McIntyre 1

Answer, Key Homework 2 David McIntyre 1 This print-out should have 14 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making

### STRAIGHT LINE MOTION. 1 Equations of Straight Line Motion (Constant Acceleration)

1 CHAPTER. STRAIGHT LINE MOTION (CONSTANT ACCELERATION) 1 INSTITIÚID TEICNEOLAÍOCHTA CHEATHARLACH INSTITUTE OF TECHNOLOGY CARLOW STRAIGHT LINE MOTION By definition, mechanics is the study of bodies at

### = 800 kg/m 3 (note that old units cancel out) 4.184 J 1000 g = 4184 J/kg o C

Units and Dimensions Basic properties such as length, mass, time and temperature that can be measured are called dimensions. Any quantity that can be measured has a value and a unit associated with it.

### Figure 2.1: Warm air rising from SAPREF stacks, October 2002 Source: GroundWork

13 CHAPTER TWO SOME CONCEPTS IN CLIMATOLOGY 2.1 The Adiabatic Process An important principle to remember is that, in the troposphere, the first layer of the atmosphere, temperature decreases with altitude.

### δy θ Pressure is used to indicate the normal force per unit area at a given point acting on a given plane.

2 FLUID PRESSURES By definition, a fluid must deform continuously when a shear stress of any magnitude is applied. Therefore when a fluid is either at rest or moving in such a manner that there is no relative

### PRESSURE Gas Pressure

PRESSURE Gas Pressure 1 Gas molecules inside a volume (e.g. a balloon) are constantly moving around freely. They frequently collide with each other and with the surface of any enclosure. Figure 1: The

### Performance 4. Fluid Statics, Dynamics, and Airspeed Indicators

Performance 4. Fluid Statics, Dynamics, and Airspeed Indicators From our previous brief encounter with fluid mechanics we developed two equations: the one-dimensional continuity equation, and the differential

### CHAPTER 3: FORCES AND PRESSURE

CHAPTER 3: FORCES AND PRESSURE 3.1 UNDERSTANDING PRESSURE 1. The pressure acting on a surface is defined as.. force per unit. area on the surface. 2. Pressure, P = F A 3. Unit for pressure is. Nm -2 or

### Chapter 13 - Solutions

= Chapter 13 - Solutions Description: Find the weight of a cylindrical iron rod given its area and length and the density of iron. Part A On a part-time job you are asked to bring a cylindrical iron rod

### Physics Principles of Physics

Physics 1408-002 Principles of Physics Lecture 21 Chapter 13 April 2, 2009 Sung-Won Lee Sungwon.Lee@ttu.edu Announcement I Lecture note is on the web Handout (6 slides/page) http://highenergy.phys.ttu.edu/~slee/1408/

### Physical and chemical properties of water Gas Laws Chemical Potential of Water Rainfall/Drought

Lecture 14, Water, Humidity, Pressure and Trace Gases, Part 1 Physical and chemical properties of water Gas Laws Chemical Potential of Water Rainfall/Drought Atmospheric Gas Composition constitue nt percent

### (a) Calculate the voidage (volume fraction occupied by voids) of the bed.

SOLUTIONS TO CAPTER 6: FLOW TROUG A PACKED BED OF PARTICLES EXERCISE 6.1: A packed bed of solid particles of density 500 kg/m 3, occupies a depth of 1 m in a vessel of cross-sectional area 0.04 m. The

### CHAPTER 13. Definite Integrals. Since integration can be used in a practical sense in many applications it is often

7 CHAPTER Definite Integrals Since integration can be used in a practical sense in many applications it is often useful to have integrals evaluated for different values of the variable of integration.

### Boyle s Law and the Law of Atmospheres

previous index next Boyle s Law and the Law of Atmospheres Michael Fowler, UVa 6/14/06 Introduction We ve discussed the concept of pressure in the previous lecture, introduced units of pressure (Newtons

### Gauge Pressure, Absolute Pressure, and Pressure Measurement

Gauge Pressure, Absolute Pressure, and Pressure Measurement Bởi: OpenStaxCollege If you limp into a gas station with a nearly flat tire, you will notice the tire gauge on the airline reads nearly zero

### Performance. 12. Gliding Flight (Steady State)

Performance 12. Gliding Flight (Steady State) If the engine is turned off, (T = 0), and one desires to maintain airspeed, it is necessary to put the vehicle at such an attitude that the component of the

### PSS 17.1: The Bermuda Triangle

Assignment 6 Consider 6.0 g of helium at 40_C in the form of a cube 40 cm. on each side. Suppose 2000 J of energy are transferred to this gas. (i) Determine the final pressure if the process is at constant

### Physics 1114: Unit 6 Homework: Answers

Physics 1114: Unit 6 Homework: Answers Problem set 1 1. A rod 4.2 m long and 0.50 cm 2 in cross-sectional area is stretched 0.20 cm under a tension of 12,000 N. a) The stress is the Force (1.2 10 4 N)

### Gas Properties and Balloons & Buoyancy SI M Homework Answer K ey

Gas Properties and Balloons & Buoyancy SI M Homework Answer K ey 1) In class, we have been discussing how gases behave and how we observe this behavior in our daily lives. In this homework assignment,

### CH-205: Fluid Dynamics

CH-05: Fluid Dynamics nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering) Solutions of Mid Semester Examination Data Given: Density of water, ρ = 1000 kg/m 3, gravitational acceleration, g

### Physics Honors Page 1

1. An ideal standard of measurement should be. variable, but not accessible variable and accessible accessible, but not variable neither variable nor accessible 2. The approximate height of a 12-ounce

### I SOIL WATER POTENTIAL

I SOIL WATER POTENTIAL 1.1. Introduction Soil water content is not sufficient to specify the entire status of water in soil. For example, if soils with a same water content but with different particle

### Serway_ISM_V1 1 Chapter 4

Serway_ISM_V1 1 Chapter 4 ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. Newton s second law gives the net force acting on the crate as This gives the kinetic friction force as, so choice (a) is correct. 2. As

### 12 BAROMETRIC PRESSURE

12 BAROMETRIC PRESSURE OBJECTIVE To observe and measure the change in atmospheric pressure with height. ACTIVITY 1 PREDICTIONS 1. What is the standard barometric pressure at sea level in atmospheres (atm)?

### Unit 4: Science and Materials in Construction and the Built Environment. Chapter 14. Understand how Forces act on Structures

Chapter 14 Understand how Forces act on Structures 14.1 Introduction The analysis of structures considered here will be based on a number of fundamental concepts which follow from simple Newtonian mechanics;

### Chapter 8 Circulation of the Atmosphere

Chapter 8 Circulation of the Atmosphere The Atmosphere Is Composed Mainly of Nitrogen, Oxygen, and Water Vapor What are some properties of the atmosphere? Solar Radiation - initial source of energy to

### 04-1. Newton s First Law Newton s first law states: Sections Covered in the Text: Chapters 4 and 8 F = ( F 1 ) 2 + ( F 2 ) 2.

Force and Motion Sections Covered in the Text: Chapters 4 and 8 Thus far we have studied some attributes of motion. But the cause of the motion, namely force, we have essentially ignored. It is true that

### Thermal Properties of Matter

Chapter 18 Thermal Properties of Matter PowerPoint Lectures for University Physics, Thirteenth Edition Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Goals for Chapter 18 To relate the

### Pressure in Fluids Ans. Ans. Ans. 4. (i) (ii) (iii) Ans. (i)

Pressure in Fluids 1. Derive an expression for pressure in a liquid at a point which is at a depth h units and density of liquid is ρ units. Ans. Consider a point x, at a depth h in a liquid of density

### Homework 9. Problems: 12.31, 12.32, 14.4, 14.21

Homework 9 Problems: 1.31, 1.3, 14.4, 14.1 Problem 1.31 Assume that if the shear stress exceeds about 4 10 N/m steel ruptures. Determine the shearing force necessary (a) to shear a steel bolt 1.00 cm in

### THE KINETIC THEORY OF GASES

Chapter 19: THE KINETIC THEORY OF GASES 1. Evidence that a gas consists mostly of empty space is the fact that: A. the density of a gas becomes much greater when it is liquefied B. gases exert pressure

### Lecture L2 - Degrees of Freedom and Constraints, Rectilinear Motion

S. Widnall 6.07 Dynamics Fall 009 Version.0 Lecture L - Degrees of Freedom and Constraints, Rectilinear Motion Degrees of Freedom Degrees of freedom refers to the number of independent spatial coordinates

### Newton s Third Law. object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1

Newton s Third Law! If two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1!! Note on notation: is

### RECOMMENDATION ITU-R P THE RADIO REFRACTIVE INDEX: ITS FORMULA AND REFRACTIVITY DATA (Question ITU-R 201/3) ANNEX 1. n = 1 + N 10 6 (1)

Rec. ITU-R P.453-7 1 RECOMMENDATION ITU-R P.453-7 THE RADIO REFRACTIVE INDEX: ITS FORMULA AND REFRACTIVITY DATA (Question ITU-R 201/3) Rec. ITU-R P.453-7 (1970-1986-1990-1992-1994-1995-1997-1999) The ITU

### Min-218 Fundamentals of Fluid Flow

Excerpt from "Chap 3: Principles of Airflow," Practical Mine Ventilation Engineerg to be Pubished by Intertec Micromedia Publishing Company, Chicago, IL in March 1999. 1. Definition of A Fluid A fluid

### Fluid Statics. [Ans.(c)] (iv) How is the metacentric height, MG expressed? I I

Fluid Statics Q1. Coose te crect answer (i) Te nmal stress is te same in all directions at a point in a fluid (a) only wen te fluid is frictionless (b) only wen te fluid is frictionless and incompressible

### Mass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati

Mass Transfer Operations I Prof. Bishnupada Mandal Department of Chemical Engineering Indian Institute of Technology, Guwahati Module - 1 Diffusion Mass Transfer Lecture - 6 Diffusion coefficient: Measurement

### The Equipartition Theorem

The Equipartition Theorem Degrees of freedom are associated with the kinetic energy of translations, rotation, vibration and the potential energy of vibrations. A result from classical statistical mechanics

### 1.4 Review. 1.5 Thermodynamic Properties. CEE 3310 Thermodynamic Properties, Aug. 26,

CEE 3310 Thermodynamic Properties, Aug. 26, 2011 11 1.4 Review A fluid is a substance that can not support a shear stress. Liquids differ from gasses in that liquids that do not completely fill a container

### WORK DONE BY A CONSTANT FORCE

WORK DONE BY A CONSTANT FORCE The definition of work, W, when a constant force (F) is in the direction of displacement (d) is W = Fd SI unit is the Newton-meter (Nm) = Joule, J If you exert a force of

### Fluid Mechanics: Static s Kinematics Dynamics Fluid

Fluid Mechanics: Fluid mechanics may be defined as that branch of engineering science that deals with the behavior of fluid under the condition of rest and motion Fluid mechanics may be divided into three

### Answer, Key Homework 6 David McIntyre 1

Answer, Key Homework 6 David McIntyre 1 This print-out should have 0 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making

### Applied Fluid Mechanics

Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and

### Exponential Functions and Their Graphs

Exponential Functions and Their Graphs Definition of Exponential Function f ( x) a x where a > 0, a 1, and x is any real number Parent function for transformations: f(x) = b a x h + k The graph of f moves

### Activity 5a Potential and Kinetic Energy PHYS 010. To investigate the relationship between potential energy and kinetic energy.

Name: Date: Partners: Purpose: To investigate the relationship between potential energy and kinetic energy. Materials: 1. Super-balls, or hard bouncy rubber balls. Metre stick and tape 3. calculator 4.

### Pressure. Pressure is one of those words we frequently use, perhaps knowing intuitively what it means. In science, we define pressure as follows:

Weather reports in the media provide information on variables such as temperature, precipitation and wind speed. In this chapter, we discuss three physical quantities that help determine weather: (1) Temperature,

### List the Oceans that hurricanes can form in. (There should be 4) Describe the shape of the winds and the direction that they spin.

Heat Transfer Tab: (Use the textbook or the ScienceSaurus book) Box 1: Radiation: (on the colored tab) Draw a picture (on the white tab directly below it) Box 2: Convection: (on the colored tab) Draw a

### SKEW-T, LOG-P DIAGRAM ANALYSIS PROCEDURES

SKEW-T, LOG-P DIAGRAM ANALYSIS PROCEDURES I. THE SKEW-T, LOG-P DIAGRAM The primary source for information contained in this appendix was taken from the Air Weather Service Technical Report TR-79/006. 1

### Units and Dimensions in Physical Chemistry

Units and Dimensions in Physical Chemistry Units and dimensions tend to cause untold amounts of grief to many chemists throughout the course of their degree. My hope is that by having a dedicated tutorial

### Synoptic Meteorology II: Isentropic Analysis. 31 March 2 April 2015

Synoptic Meteorology II: Isentropic Analysis 31 March 2 April 2015 Readings: Chapter 3 of Midlatitude Synoptic Meteorology. Introduction Before we can appropriately introduce and describe the concept of

### v v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )

Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution

### Statistical Mechanics, Kinetic Theory Ideal Gas. 8.01t Nov 22, 2004

Statistical Mechanics, Kinetic Theory Ideal Gas 8.01t Nov 22, 2004 Statistical Mechanics and Thermodynamics Thermodynamics Old & Fundamental Understanding of Heat (I.e. Steam) Engines Part of Physics Einstein

### State Newton's second law of motion for a particle, defining carefully each term used.

5 Question 1. [Marks 20] An unmarked police car P is, travelling at the legal speed limit, v P, on a straight section of highway. At time t = 0, the police car is overtaken by a car C, which is speeding

### 12.307. 1 Convection in water (an almost-incompressible fluid)

12.307 Convection in water (an almost-incompressible fluid) John Marshall, Lodovica Illari and Alan Plumb March, 2004 1 Convection in water (an almost-incompressible fluid) 1.1 Buoyancy Objects that are

### Differential Relations for Fluid Flow. Acceleration field of a fluid. The differential equation of mass conservation

Differential Relations for Fluid Flow In this approach, we apply our four basic conservation laws to an infinitesimally small control volume. The differential approach provides point by point details of

### Ocean Processes I Oceanography Department University of Cape Town South Africa

Ocean Processes I Isabelle.Ansorge@uct.ac.za Oceanography Department University of Cape Town South Africa 1 9/17/2009 Lecturer in Oceanography at UCT About me! BSc in England University of Plymouth MSc

### Weather Basics. Temperature Pressure Wind Humidity. Climate vs. Weather. Atmosphere and Weather

Climate vs. Weather Atmosphere and Weather Weather The physical condition of the atmosphere (moisture, temperature, pressure, wind) Climate Long term pattern of the weather in a particular area Weather

### Fall 2013 RED barcode here Physics 105, sections 1, 2 and 3 Exam 1 Please write your CID

Fall 03 RED barcode here Physics 05, sections, and 3 Exam Colton Please write your CID No time limit. No notes. No books. Student calculators only. All problems equal weight, 60 points total. Constants/Materials

### Physics Exam Q1 Exam, Part A Samples

Physics Exam Q1 Exam, Part A Samples 1. An object starts from rest and accelerates uniformly down an incline. If the object reaches a speed of 40 meters per second in 5 seconds, its average speed is (A)

### Problem Set 3 Solutions

Chemistry 360 Dr Jean M Standard Problem Set 3 Solutions 1 (a) One mole of an ideal gas at 98 K is expanded reversibly and isothermally from 10 L to 10 L Determine the amount of work in Joules We start

### This document is copyrighted EMD International A/S. Unauthorized use is prohibited.

WindPRO / ENERGY Modelling of the Variation of Air Density with Altitude through Pressure, Humidity and Temperature Height [m] Ratio of Moist Air Density to Dry Air Density 5000 4000 3000 2000 1000 0 Ratio

### 1. Model: A representation of a real object or event. Ex: Map Globe Graphs A. Models are useful because they allow an easier way to study objects or

1. Model: A representation of a real object or event. Ex: Map Globe Graphs A. Models are useful because they allow an easier way to study objects or events. 2. The shape of the Earth A. The earth is round.

### The Boltzmann distribution law and statistical thermodynamics

1 The Boltzmann distribution law and statistical thermodynamics 1.1 Nature and aims of statistical mechanics Statistical mechanics is the theoretical apparatus with which one studies the properties of

### Thermodynamic Systems

Student Academic Learning Services Page 1 of 18 Thermodynamic Systems And Performance Measurement Contents Thermodynamic Systems... 2 Defining the System... 2 The First Law... 2 Thermodynamic Efficiency...

### p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh

IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: P..Sleigh@leeds.ac.uk r J Noakes:.J.Noakes@leeds.ac.uk January 008 Module web site: www.efm.leeds.ac.uk/ive/fluidslevel1

### A satellite of mass 5.00x10² kg is in a circular orbit of radius 2r around Earth. Then it is moved to a circular orbit radius of 3r.

Supplemental Questions A satellite of mass 5.00x10² kg is in a circular orbit of radius 2r around Earth. Then it is moved to a circular orbit radius of 3r. (a) Determine the satellite s GPE in orbit. (b)

### Natural Convection. Buoyancy force

Natural Convection In natural convection, the fluid motion occurs by natural means such as buoyancy. Since the fluid velocity associated with natural convection is relatively low, the heat transfer coefficient

### Typhoon Haiyan 1. Force of wind blowing against vertical structure. 2. Destructive Pressure exerted on Buildings

Typhoon Haiyan 1. Force of wind blowing against vertical structure 2. Destructive Pressure exerted on Buildings 3. Atmospheric Pressure variation driving the Typhoon s winds 4. Energy of Typhoon 5. Height

### Basic Fluid Mechanics. Prof. Young I Cho

Basic Fluid Mechanics MEM 220 Prof. Young I Cho Summer 2009 Chapter 1 Introduction What is fluid? Give some examples of fluids. Examples of gases: Examples of liquids: What is fluid mechanics? Mechanics

### Wave Hydro Dynamics Prof. V. Sundar Department of Ocean Engineering Indian Institute of Technology, Madras

Wave Hydro Dynamics Prof. V. Sundar Department of Ocean Engineering Indian Institute of Technology, Madras Module No. # 02 Wave Motion and Linear Wave Theory Lecture No. # 03 Wave Motion II We will today

### PHYSICS MIDTERM REVIEW

1. The acceleration due to gravity on the surface of planet X is 19.6 m/s 2. If an object on the surface of this planet weighs 980. newtons, the mass of the object is 50.0 kg 490. N 100. kg 908 N 2. If