The buoyancy acceleration is critical to the formation of thunderstorms and their associated hazards. The maximum vertical velocity attained depends on the vertical distribution (magnitude and depth) of positive buoyancy experienced by a rising air parcelwithin the storm. Consider two environments, one with positive buoyancy in a layer that is 6 km deep and the other with positive buoyancy in a layer 12 km deep. The 6-km deep layer has an average parcel-environment temperature difference of 4 K, while the 12-km deep layer has an average parcel-environment temperature difference of 2 K. The temperature at the bottom of the layer in both environments is 300 K and the average environmental lapse rate is 6 K km-1. a. Determine the maximum vertical velocity experienced by a rising air parcel in each environment. Which environment has the greatest w? b. Show that the maximum vertical velocity from a can also be found using a relationship that depends on convective available potential energy (CAPE): w = 2 CAPE

Explanation:

Below  are the given quantities:

Environmental lapse rate (\Gamma ) = 6 K/km

Dry adiabatic lapse rate (\Gammad) = 4 K/km for 12-km atmosphere, 2 K/km for 6-km atmosphere

Surface temperature (T) = 300 K

When we integrating the vertical momentum equation and assume that density is compensated by temperature, we have;

w.(dw/dz) = - \rho ' / \rho g = -[T ' -  \Gammadz - (T ' - \Gamma d dz)] g

w2 / 2 = ( \Gamma - \Gamma d / T ) gdz

For 6-km layer atmosphere,

w= (2 [(6 - 2) / 300] (10 / 6000) )1/2 = 0.0066 m/s

For 12-km atmosphere,

w= (2 [(6 - 4) / 300] (10 / 12000) )1/2 = 0.0033 m/s

Thus, we can say that the 6-km deep environment is having a greater w than 12-km deep environment.

b. The Convective Available Potential Energy (CAPE) will be computed as

((T - Te ) / Te ) gdz

We have this has the potential energy which is also equivalent to kinetic energy of the parcel, meaning. 1 / 2 w2,  maximum vertical velocity can be computed like;

1 / 2 w2 = CAPE

Thus, w = (2 CAPE) 1/2

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