SME
Meteorology · Problem type

The lifting condensation level

Use this for “Air at the ground is 30 °C with a dew point of 14 °C. How high is the cloud base, and how warm is it there?”

Diagram

parcel’s temperature: −9.8 °C per kmits dew point: −1.8 °C per km0123010203040Temperature, °CAltitude, kmcloud basezLCL = 2,000 m30 °C14 °C
The
parcel
cools
9.8
°C
per
km
and
its
dew
point
falls
1.8
°C
per
km,
so
they
close
8
°C
per
km:
h
=
30 − 14
8
=
2
km
=
2,000
m.
There the air is 10.4 °C, saturated, and a cloud forms; above it the parcel cools about 6 °C per km.
This is an example. Type your own values to start a new problem.
zLCL (m)
=
125
×
(
T (°C)
−
Td (°C)
)
ΔT
Spread
calculated · °C
Tbase
Temperature at the base
calculated · °C

Formulas

ΔT
=
T
−
Td
16 = 30 − 14
zLCL = 125 × ΔT
2,000 = 125 × 16
Tbase
=
T
−
9.8
×
zLCL
1,000
10.4
=
30
−
9.8
×
2,000
1,000

Assumptions

•
A rising parcel cools 9.8 °C per km (dry adiabatic) and its dew point falls 1.8 °C per km.
•
They close 8 °C per km, so they meet, and a cloud forms, 125 m up for each °C of spread.
•
The dew point is never above the temperature.

Step-by-step

WE KNOW
T = 30 °C, Td = 14 °C
FIND
spread (ΔT), cloud base (zLCL), temperature at the base (Tbase)
Step 1 · Find spread (ΔT)
Use ΔT = T − Td
How far the air is from saturation.
ΔT = 30 − 14
ΔT = 16 °C
Step 2 · Find cloud base (zLCL)
Use zLCL = 125 × ΔT
The
two
lines
close
9.8
−
1.8
=
8
°C
per
km,
so
they
meet
1,000
8
=
125
m
up
per
°C.
zLCL = 125 × 16
zLCL = 2,000 m
Step 3 · Find temperature at the base (Tbase)
Use Tbase = T − 9.8 × zLCL ÷ 1,000
The parcel cools 9.8 °C for each km it rises, dry.
Tbase
=
30
−
9.8
×
2,000
1,000
Tbase
=
30
−
19,600
1,000
Tbase = 30 − 19.6
Tbase = 10.4 °C
Check
Put the numbers back in. Both sides should match.
16 = 30 − 14 ✓
2,000 = 125 × 16 ✓
10.4
=
30
−
9.8
×
2,000
1,000
✓
Topic check
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