Please Show Your Work for Calculations

ATS 351, Spring 2009 Week 10, Lab #8 – due April 2nd What Makes the Wind Blow? - 60 points Please show your work for calculations.

Question #1: Sea Level Pressure (10 points) a) (4 points) How does sea-level pressure differ from station pressure? Can the two ever be the same? Explain.

Station pressure is the atmospheric pressure at the elevation of the weather station. Sea-level pressure is the station pressure the weather station would have if the station were located at sea level. The two would be the same if the station were located at sea level.

b) (3 points) Using the ideal gas law, please calculate what is referred to as the “standard” sea level pressure of the Earth. Values needed: average global temperature = 288K (15C); constant (R) = 287 J/kg/K; average sea level air density = 1.226 kg/m3. Please give pressure in millibars (what you calculate will be in Pascals).

Ideal Gas Law: p =  RT p = 1.226 kg/m3 * 287 J/kg/K * 288 K p = 101,336.26 Pa p = 101,336.26 Pa * (1 mb/ 100 Pa) p = 1013.36 mb

c) (3points) If the average global temperature increased to 295K, what would the new “standard” sea level pressure be?

p = 1.226 kg/m3 * 287 J/kg/K * 295 K p = 103,799.29 Pa p = 103,799.29 Pa * (1 mb/100 Pa) p = 1037.99 mb would be the new standard sea level pressure.

Question #2: Forces (5 points) a) (4 points) Identify the four forces that affect the movement of air.

Pressure gradient force, coriolis force, centripetal force, and friction

b) (1 point) Which force only influences winds at the surface and not aloft?

Friction Question #3: Coriolis Force (8 points) a) The equation for the Coriolis Force is: C o r i o l i s = 2 W s i n f The Earth’s angular rate of spin is 7.27x10-5 s-1.

By solving this equation,ﾠ please demonstrate how the Coriolis force changes with latitude (). Note: You need to solve this equation using radians, not degrees. To convert from degrees to radians, simply multiply your degree by pi and then divide by 180.

1 = 10 = 0.174 radians; 2 = 70 = 1.22 radians

-5 -1 CF1 = 2 * 0.14 kg * 7.27x10 s * 17 m/s * sin(0.174 rads) -5 -1 CF1 = 5.99 x 10 s

-5 -1 CF2 = 2 * 0.14 kg * 7.27x10 s * 17 m/s * sin(1.22 rads) -1 CF2 = 0.000325 s

By solving for both 10 and 70, it can be shown that the Coriolis force increases as you move further away from the equator. It is greatest at the poles.

Question #4: Drawing Forces using Vectors (21 points) a) (4 points) Assuming geostrophic and frictionless flow, please draw vectors for the pressure gradient and coriolis forces for the Northern Hemisphere.

1000 mb PGF 1004 mb Coriolis 1008 mb H

b) (2 points) Are geostrophic winds a good approximation for surface winds? Please explain why or why not.

No, geostrophic winds are not valid at the surface because isobars are rarely straight there and wind is not at a constant speed. Also, friction has a strong influence on the winds at the surface. Geostrophic wind is just the balance between the PGF and Coriolis, with no effect of friction included.

c) (10 points) Given the pictures below of low and high pressure, please draw the direction of the flow and the force vectors associated with areas of low and high pressure. Please give a brief discussion on your picture. (Hint: you will only be utilizing 3 out of the 4 forces at this point)

Cor Cor PGF LPGF H

Low pressure: the PGF vector is pointing towards the center of the low and the Coriolis force is directed to the right of the wind movement. Since centripetal force is pulling inward, the net force is going to be directed inwards towards the low and the PGF vector is therefore larger. High pressure: the PGF vector is pointing from high to low pressure (outward) while the Coriolis force is directed to the right of the wind movement (inward). Since centripetal force is again pulling inward, the net force will be towards the center of the high and the Coriolis vector is larger in this case.

d) (5 points) Please draw the vectors for PGF, Coriolis force, and friction that are associated with surface wind.

1000 mb PGF Sfc

Friction Cor 1004 mb H

Question #5: Pressure Gradient Force and Wind Movement (8 points) A sea-level pressure is shown below where isobars are drawn every 4mb (Northern Hemisphere). Answer the following questions.

a) Place a dashed line through the ridge and a solid line through the trough. (shown below) b) What would be the wind direction at point A and at point B?

A: winds from the southeast B: winds from the west c) Where would the stronger wind be blowing, at point A or B? Explain.

B: the pressure gradient is tighter at point B which indicates stronger winds. d) Compute the pressure gradient between points 1 and 2 and between points 3 and 4.

PG between 1 and 2: (1004-996mb)/1000km = 8mb/1000km = 0.008mb/km PG between 3 and 4: (1012-1000mb)/1000km = 12mb/1000km =0.012mb/km Question #6: Winds and Vertical Air Motions (8 points) a) (4 points) Draw and describe the type of vertical air motions associated with high and low pressure areas.

Above a surface low where there is convergence at the surface, there will be divergence aloft. This creates rising motion above a surface low.

Above a surface high where there is divergence at the surface, there will be convergence aloft. This creates sinking motion above a surface high.

b) (4 points) Since there is always an upward-directed pressure gradient force, why doesn’t the air rush off into space?

Even though a consistent PGF exists between the surface air and air aloft, this PGF is balanced by gravity, which keeps the air from rushing off into space. When gravity and PGF are in balance, they are said to be in hydrostatic equilibrium.