SME
Meteorology
Atmospheric structure
Use this for “How thick is the layer from 1,000 hPa up to 500 hPa when its mean temperature is 255 K?”
Diagram
H
=
RT̄
g
=
287
×
255
9.81
=
7,460
m.
Δz
=
H
ln(
p₁
p₂
)
=
7,460
×
ln(
1,000
500
)
=
5,171
m.
A warmer layer has a larger H, so the same two pressures lie farther apart.
This is an example. Type your own values to start a new problem.
Δz (m)
=
H (m)
×
ln(
p₁ (hPa)
÷
p₂ (hPa)
)
T̄
Mean temperature
example · K
Formulas
H
=
287
×
T̄
9.81
7,460
=
287
×
255
9.81
Δz
=
H
×
ln(
p₁
p₂
)
5,171.0475
=
7,460
×
ln(
1,000
500
)
Assumptions
•
Dry air in hydrostatic balance: Rd = 287 J/(kg·K) and g = 9.81 m/s².
•
One mean temperature T̄ for the whole layer stands in for adding up thin slices.
•
A warm layer is thicker than a cold one, which is why 500 hPa heights map the temperature below.
Step-by-step
WE KNOW
p₁ = 1,000 hPa, p₂ = 500 hPa, T̄ = 255 K
FIND
scale height (H), thickness (Δz)
Step 1 · Find scale height (H)
Use H = 287 × T̄ ÷ 9.81
The
scale
height
Rd
T̄
g
:
the
climb
over
which
the
pressure
falls
to
1
e
of
its
value.
H
=
287
×
255
9.81
H
=
73,185
9.81
H
=
7,460.245
m
Step 2 · Find thickness (Δz)
Use Δz = H × ln(p₁ ÷ p₂)
The hypsometric equation: the scale height times the log of the pressure ratio.
Δz
=
7,460.245
×
ln(
1,000
500
)
Δz = 7,460.245 × ln(2)
Δz = 7,460.245 × 0.6931
Δz = 5,171.048 m
Check
Put the numbers back in. Both sides should match.
7,460.245
=
287
×
255
9.81
✓
5,171.048
=
7,460.245
×
ln(
1,000
500
)
✓
Topic check
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