Ordinary-cell convective storms - Severe Convective Storms and Tornadoes: Observations and Dynamics - page 148

Figure 3.43. As in Figure 3.37, but at the right edge of the domain the flow is uniform and from
right to left at speed c, above z ¼ h 0 , while the wind speed increases linearly by D u with
decreasing height until at the surface (z ¼ 0) c is c þD u from the right; at the left edge of
the domain, the wind u ¼ u c (from the right) above z ¼ h þ h b , while the wind speed increases
linearly with decreasing height to u b at the top of the cold pool, at z ¼ h. If there is a shear layer
on the right-hand side, there must also be one on the left-hand side because, following the
streamlines which do not pass through the cold pool, horizontal vorticity is conserved (based
on an illustration in Emanuel, 1994).

layer moves up and over the density current (in the reference frame of the density
current). Second, there is an airstream having no vertical shear that lies above the
shear layer. In the case of low-level vertical shear, the depth of the shear layer way
ahead of the cold pool is h 0 , which is not necessarily the same as h (the depth of
the cold pool) and the depth of the shear layer above the density current is h b ; the
flow relative to the density current is u b in the shear layer and u c above, in the
no-shear layer (u ¼ 0 in the cold pool, as before).
The basic equation is (3.26), the flux form of the vorticity equation. Integrated
over the entire domain of the density current, we find that:

2

2 c 2 2 u c ¼ g h ½ð 1 0 Þ= 0

1 1 1

2 ð c þD

u Þ þ ð 3 : 51 Þ

where w ¼ 0atz ¼ 0andH, @ w =@ x ¼ 0; and B ¼ 0atx ¼1 , u 1; H ¼ c,

u 1; 0 ¼ð c þD u Þ , and u 1; H ¼ u c , and u 1; 0 ¼ 0. Conservation of mass for the

no-shear layer is found again by integrating (3.9) over the x-z-plane:
u c ½ H ð h þ h b Þ¼ c ð H h 0 Þ 52 Þ

We could have used as an additional constraint the macroscale conservation of
mass in the shear layer, but it is not needed for our analysis of the special case of
a ''shallow'' density current that we are going to consider. Eliminating u c from
(3.51) and (3.52), we find that
c 2

ð 3 :

2 2

½ð H h 0 Þ=ð H h h b Þ þ 2c D u þðD u Þ ¼ 2 g h ½ð 1 0 Þ= 0 ð 3 : 53 Þ