Class exercises Chapter 3. Elementary Applications of the Basic Equations Section 3.1 Basic Equations in Isobaric Coordinates 3.1 For some (in fact many) applications we assume that the change of the Coriolis parameter with latitude can be approximated by: where = + =2Ω and y=0 at ϕ 0. This is called the midlatitude β-plane approximation (see Holton p.160). a. Derive the second equality in the definition of β as given above. b. Derive an expression for the divergence of the geostrophic wind on a pressure surface with this approximation. 3.2 A simplified vertical profile of the divergence in a cyclone is given in Figure 3.1. Figure 3.1 Simplified vertical profile of the divergence δ in a cyclone (p is in kpa). In the lower troposphere we have convergence, no divergence near 50 kpa and divergence in the upper troposphere. If the minimum divergence is given by the value δ = -X, and the maximum divergence by δ = +X, then the linear profile is given by =1 2 where p 0 = 100 kpa is the surface pressure. 1
a. From Holton Eq. (3.5) derive the profile of the pressure vertical velocity ω(p) assuming that ω(0)=0. b. Where do we find the maximum value of ω and what is its value? c. Make a sketch of the ω(p) profile between p=0 and p=p 0. 3.3 a. The temperature profile in (a part of) the atmosphere is dry adiabatic, i.e. we have: R c p p T ( p) = θ 0 p s where p s is the pressure of a standard level (usually 100 kpa) and θ 0 is the constant value of the potential temperature. Use Holton Eq. (3.7) to show that in that case the static stability parameter S P = 0. b. Derive an expression for S P (p) for an isothermal atmosphere. Section 3.2 Balanced Flow 3.4 a. Make a drawing of a circular (NH) cyclone with the geostrophic wind and the direction of n in the natural coordinate system. Next determine the sign of both where r is the vector pointing from the centre of the cyclone outward. The distinction between the two derivatives is important for many applications. b. Repeat the above but for an anticyclone. 3.5 a. For so-called balanced flow we have =0 What does this equality exactly imply using Holton Eq. (3.9)? b. And what does it also imply using Holton Eq. (3.10)? 3.6 An object is given a speed of 10 m/s, as a result its trajectory will be an inertial circle. If the initial position of the object was at 45 o N, what latitudes (northernmost and southernmost) will it reach when the initial direction was to the: a. west b. east c. north d. northwest (now a little geometry is involved). Neglect the dependence of the Coriolis parameter (f) on latitude. 3.7 Solve Holton problem 3.3. Hint: assume cyclostrophic balance and the gas law and integrate from the centre of the tornado to radius r 0. 2
3.8 Solve Holton problem 3.9. Hint: rewrite Holton Eq. (3.15b) and use 1+ 1+ 1 0 2 3.9 The trajectory of an air parcel in the Northern Hemisphere at 40 o N is shown in Figure 3.2. a. What is the direction of the ageostrophic wind? b. What is the speed of the ageostrophic wind (in m s -1 )? c. Calculate the speed and direction of the geostrophic wind. d. In Figure 3.1 sketch the height contours determined by the geostrophic wind and calculate the slope of the pressure surface e. If the motion is on a given pressure level, calculate the vertical velocity of the air parcel and explain why it is decelerating. (after Bluestein, 1992) Figure 3.2 Trajectory of an air parcel at 40 o N. (after Bluestein, 1992). Section 3.4 The Thermal Wind 3.10 At 50 N the geostrophic wind speed at 850 hpa is 15 m/s from the southeast and at 500 hpa it is 25 m/s from the southwest. a. Make a construction of this situation and calculate the thermal wind speed. b. Using the construction, calculate the two components u T and v T. c. Calculate the horizontal temperature gradient in the 850-500 hpa layer. d. Calculate the temperature advection in the 850-500 hpa layer, where this quantity is defined as: = e. Show that the above definition gives identical results for both the 850 and 500 hpa geostrophic wind. 3.11 Solve Holton Problem 3.12. 3
Section 3.5 Vertical Motion 3.12 The divergence of the horizontal wind at various pressure levels above a certain station is given in the following table. Compute the pressure vertical velocity ω (in hpa h -1 ) at each level assuming ω = 0 at 1000 hpa. Pressure (hpa) V ( 10-5 s -1 ) 1000-1.05 850-0.75 700-0.25 500 +0.05 300 +0.85 3.13 Solve Holton Problem 3.23. Note that westerly means: to the west. Vertical velocity (w) in cm s -1. Section 3.6 Surface Pressure Tendency 3.14 Using the divergence data given in Holton Problem 3.22 calculate the surface pressure tendency assuming the surface pressure equals 1000 hpa Answers 3.1 b. = 3.2 b. maximum: ω = -¼Xp 0 at p = ½p 0 3.3 b. = 3.6 a. 45 <ϕ<46.74 ; b. 43.26 <ϕ<45 ; c. 44.13 <ϕ<45.87 d. 44.74 <ϕ<46.49 3.7 p = 941.3 hpa 3.9 b. V a = 16.46 m s -1 ; c. Vg= 30 m s -1 direction = 237 ; d. 2.867 10-4 m m -1 e. w = 0.39 cm s -1 3.10 a. V T = 29.15 m s -1 ; b. u T = 28.3 m s -1 v T = 7.1 m s -1 ; c. 2.14 10-5 K m -1 d. 2.75 10-4 K s -1 3.11 900-700 hpa: 1.96 10-5 K m -1 and 0.5 K h -1 700-500 hpa: 1.04 10-5 K m -1 and 0 K h -1 δt = 45 h 4
3.12 Pressure (hpa) 1000 850 700 500 300 ω (hpa h -1 ) 0-4.86-7.56-8.28-5.04 3.13 w = 18.3 cm s -1 3.14 p s / t = +0.36 hpa h -1 5