EXAMPLE PROBLEM II-2-5

FIND:

The 10-m wind speed, the wind direction, and the coefficient of drag.

GIVEN:

A pressure gradient of 5 mb in 100 km,

an air-sea temperature difference of -5o C

(i.e. the water is warmer than the air, as is

typical in autumn months), the latitude of

the location of interest (equal to 45o N), and

the geographic orientation of the isobars.

SOLUTION:

Option 1 - From Equation II-2-10, wind speed is calculated (in cgs units) as

Ug = 1/(1.2x10-3 x 1.03x10-4) x (dp/dn)

(a)

= 1/1.236x10-7 x (5 x 1000 )/(100 x 100000) (b)

= 4045 cm/sec

(c)

= 40.45 m/sec

The 1.2x10-3 factor in step (a) is air density in g/cm3.

The underlined 1000 factor in step (b) converts mb to dynes/cm2. The 100000 factor in step (b)

converts km to cm. From Ug and ∆T and Figure II-2-13

U10/Ug = 0.68

and

U10 = Ug x U10/Ug = 40.45 x 0.68 = 27.5 m/s

From Figure II-2-5

CD = 0.0024

Wind Direction: Parallel to isobars, counterclockwise circulation around low, therefore the

direction is west

Option 2 - Use Figure II-2-12, though it requires pressure gradient information in a different

form than given in this example.

II-2-24

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