Problem 1: What is the direction of the geostrophic wind 100 a) From the northwe
ID: 121040 • Letter: P
Question
Problem 1: What is the direction of the geostrophic wind 100 a) From the northwest b) From the northeast c) From the southwest d) From the southeast e) None of the previous 40 30 -60 20 20 NGM 961019/0000012 1000MB TEMP. DEPT. SFC PRES 60 80 100 120 Problem 2: Calculate the corresponding potential temperature of an air parcel at height 100 m and with temperature 5 °C. Make the calculations using theKelvin scale Problem 3: Calculate the Coriolis acceleration on an air parcel at latitude 30 °N if its horizontal wind components are u = 1 m/s, v = 1 m/sExplanation / Answer
1. Winds flow from high pressure to low pressure due to pressure gradient force. Corriolis force play an important role in the direction of winds and air parcel deflected right in the northern hemisphere and left in the southern hemisphre due to this coriolis force. Due to this phenomena, when winds starts blowing parallel to the isobars, that winds referred to as geostropic winds.
Direction of geostropic wind (Considering northern hemisphere):
a) From the northwest- northeast
b) From the northeast- southeast
c) From the Southwest- northwest
d)From the southeast- south west
e) None of the previous- Parallel to the isobars.
2) 0 degree celsius= 273.15 kelvin
Givent height= 100m.
In terms of mathematics, drop in temperature above the surface 9.8 deg per km. This is due to decrease in air pressure. As pressure decreases, air expands and temperature also decreases.
Since here height is only 100 m, the effect will be very less.
So for 100 m , lose in temperature will be = 0.98 degree C
Since the given surface temperature is 5 deg C, so temp. at the height of 100 m will be 5-0.98=4.02 deg C
The corresponding temp will be 277.17 K.
3) In atmospheric dynamics, the Coriolis acceleration appears only in the horizontal equations due to the neglect of products of small quantities and other approximations. Hence, the formula for coriolis acceleration is
-fk*(u,v) , where , k is a unit local vertical, f = 2sin(latitude) is Coriolis parameter and (u,v) are the horizontal components of the velocity.
(earth's angular rotation) = 7.292 x 10-5sec-1
Using the coriolis parameter, the equations for the coriolis acceleration in terms of the wind vector (u and v) are:
Fx = fv Fy = fu
where, the acceleration in the x direction is a function of v, and the acceleration in the y direction is a function of u.
Put the value in the above equation to get the answer.
Hence,
Fx= 2sin(30).1 = 2* 7.292 x 10-5 * = -0.000144 ms-2 = 1.44*10-4 ms-2= Fy