The temperature 50 kin north of a weather station is 3degreeC cooler than at the

Question

The temperature 50 kin north of a weather station is 3degreeC cooler than at the station. a) If the wind is blowing from the north at 20 m/s and there is no external heating, what is the local rate of temperature change at the station? b) If the wind is blowing from the northeast at 20 m/s and there is no external heating, what is the local rate of temperature change at the station? c) If the wind is blowing from the northeast at 20 m/s and the air is being heated by the sun's radiation at the rate of 1degreeC/hour, what is the local rate of temperature change at the station?

Explanation / Answer

For advective cooling
dT/dt = -(udT/dx + vdT/dy + wdT/dz)
a) dT/dt = -(20*3/50000) = -0.0012 deg C/ s
b) dT/dt = -(20*3 + 20*0)/sqroot(2)*50,000 = -8.4852*10^-4 C/s
c) dT/dt = 1 - 8.4852*10^-4 = 0.99915 C/s

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