SME
Meteorology
Severe weather (thunderstorms, tornadoes, hurricanes)
Use this for “A parcel is on average 4 K warmer than its surroundings (260 K) through 10 km. Estimate the CAPE and the updraft it could reach.”
Chart
wₘₐₓ(1,509) ≈ 54.94
This is an example. Type your own values to start a new problem.
CAPE (J/kg)
=
g
×
(
ΔT (K)
÷
T̄ (K)
)
×
D (km)
wₘₐₓ
Strongest updraft
worked out (not typed) · m/s
Formulas
CAPE
=
9.81
×
(
ΔT
T̄
)
×
D
1,509.2308
=
9.81
×
(
4
260
)
×
10,000
wₘₐₓ
=
√
2 × CAPE
54.9405
=
√
2 × 1,509.2308
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
The buoyancy g × ΔT ÷ T̄ is averaged over the layer, with g = 9.81 m/s².
•
wₘₐₓ = √(2CAPE) ignores mixing and the weight of water, so real updrafts are about half of it.
•
CAPE above 2,500 J/kg is very unstable: strong storms can grow.
Step-by-step
WE KNOW
ΔT = 4 K, T̄ = 260 K, D = 10 km
FIND
convective energy (CAPE), strongest updraft (wₘₐₓ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
D = 10 km = 10,000 m (1 km = 1000 m)
Step 2 · Find convective energy (CAPE)
Use CAPE = 9.81 × (ΔT ÷ T̄) × D
The
parcel’s
upward
push
per
kilogram
is
g
×
ΔT
T̄
;
times
the
depth
it
acts
over,
that
is
the
energy.
CAPE
=
9.81
×
(
4
260
)
×
10,000
9.81
×
4
260
=
0.15092
m/
s
2
0.15092 × 10,000 = 1,509.2
CAPE = 1,509.231 J/kg
Step 3 · Find strongest updraft (wₘₐₓ)
Use wₘₐₓ = √(2 × CAPE)
All
the
energy
turned
into
motion:
1
2
w
2
=
CAPE,
so
w
is
the
square
root
of
twice
it.
wₘₐₓ
=
√
2 × 1,509.231
wₘₐₓ
=
√
3
,018.462
wₘₐₓ = 54.9405 m/s
Check
Put the numbers back in. Both sides should match.
1,509.231
=
9.81
×
(
4
260
)
×
10,000
✓
54.9405
=
√
2 × 1,509.231
✓
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A front from the weather it brings
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The updraft that holds up a hailstone
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