Ordinary-cell convective storms - Severe Convective Storms and Tornadoes: Observations and Dynamics - page 141

The above is not a satisfactory equation for finding c because we have an
unknown constant p t . To eliminate p t , we must find an independent equation
in which p t appears. We consider the horizontal equation of motion again, but
this time we consider it everywhere—not just at z ¼ 0. The inviscid, steady-state,
two-dimensional equation of motion for this case is

p 0 u @ u =@ x þ w @ u =@ z þ 0 @ =@ x ¼ 0 ð 3 : 21 Þ

which is similar to (3.18), except that the vertical advection term is retained.
Combining (3.21) with the Boussinesq two-dimensional equation of continuity
@ u =@ x þ@ w =@ z ¼ 0 ð 3 : 22 Þ

we find that

u 2 p 0

@ =@ x þ@=@ z ð uw Þþ 0 @ =@ x ¼ 0 ð 3 : 23 Þ

We now integrate (3.23) over the x-z-plane from x ¼1 to x ¼1 and from
z ¼ 0toz ¼ H (i.e., over the entire domain), making use of (3.16) and (3.17), and
eliminate u using the mass conservation relation (3.13), to find that
c 2 Hh

2 g h 2

1

=ð H h Þ¼ 0 p t H ð 1 0 Þ= 0 ð 3 : 24 Þ

Eliminating p t from (3.24) and (3.20), we find that
c 2

¼ g h ½ð 1 0 Þ= 0 f 2 ½ð H h Þ=ð H þ h Þ½ð H

1

2 h Þ=

H g

ð 3 :

25 Þ

Alas, it is not obvious whether or not (3.25) will work for all values of h H,sowe

must find yet another constraint to nail down c. To do so, we now make use of
the Boussinesq vorticity (the y-component) equation (2.51) in flux form, subject to
the conditions of being steady state and two dimensional (

=

@=@ y ¼ 0), and to the

continuity equation (3.22):
@ð u Þ=@ x þ@ð w Þ=@ z ¼@ B =@ x ð 3 : 26 Þ

where x, the y-component of the vorticity vector. Integrating

(3.26) over the entire domain and using the upper and lower kinematic boundary
conditions w ¼ 0atz ¼ 0 and z ¼ H, and assuming that w ¼ 0atx ¼1
(altogether assuming that w ¼ 0 along all the edges of the domain), it is found
that

¼ @ u =@ z @ w =@

2 u 2 H ;1 ¼ g h ½ð 1 0 Þ= 0

1

ð 3 :

27 Þ

We need to find an expression now for c in terms of u H ;1 and eliminate the
latter. To do so, we use of the conservation of mass relation (3.13), noting that
u u H ;1 and find that

c 2

2

28 Þ
We now have two independent equations for c 2 . Setting (3.28) and (3.25) equal to
each other, we obtain the following relationship between H and h:

¼ g h ½ð 1 0 Þ= 0 2 ½ð H h Þ= H ð 3 :

2

H 2

1

ð H h Þ = ¼½ð H h Þ=ð H þ h Þ½ð H

2 h Þ=

H ð 3 : 29 Þ

This equation has solutions h ¼ H = 2, h ¼ H,andH ¼1 . The solution h ¼ H is