SME
Meteorology
Atmospheric thermodynamics
Use this for “Air at 263.15 K sits at 700 hPa. What is its potential temperature?”
Diagram
θ
=
T
(
1,000
p
)
κ
θ
=
263.15
×
(
1,000
700
)
0.286
=
291.4
K
Brought down dry to 1,000 hPa, the parcel warms along its adiabat to θ; every parcel on one adiabat has the same θ.
This is an example. Type your own values to start a new problem.
θ (K)
=
T (K)
×
(1,000
hPa
÷
p (hPa)
)
0.286
Formulas
θ
=
T
×
(
1,000
p
)
0.286
291.4106
=
263.15
×
(
1,000
700
)
0.286
Assumptions
•
θ is the temperature the air would have if brought dry and adiabatically to 1,000 hPa.
•
It stays the same for a parcel rising or sinking without heat or condensation.
•
κ = Rd ÷ cₚ = 287 ÷ 1,004 = 0.286 for dry air.
Step-by-step
WE KNOW
T = 263.15 K, p = 700 hPa
FIND
potential temperature (θ)
Step 1 · Find potential temperature (θ)
Use θ = T × (1,000 ÷ p)0.286
Bring the parcel down dry to 1,000 hPa: it warms by the pressure ratio to the power κ.
θ
=
263.15
×
(
1,000
700
)
0.286
1,000
700
=
1.429
1.429
0.286
=
1.107
θ
=
291.4106
K
Check
Put the numbers back in. Both sides should match.
291.4106
=
263.15
×
(
1,000
700
)
0.286
✓
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