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SCIENCE · INVESTIGATE AND EXPLAIN

Estimate a lifting condensation level

Find where simplified parcel temperature and dew-point paths meet, then apply a separate saturated cooling rule.

Learning outcome: Derive the approximate LCL and retain continuity across the change in cooling rate.

Starting recap

The LCL is the saturation height of a hypothetically lifted parcel. Here its pre-saturation temperature falls 9.8°C/km and dew point falls an assumed 1.8°C/km.

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Try

An initial 8°C separation closes at 8°C/km, giving an approximate LCL of 1 km and parcel temperature 20.2°C there.

Open the experiment

Predict

If the initial temperature–dew-point separation doubles, how does this model’s LCL change?

Experiment

Compare surface differences 0, 8 and 16°C. Compute each LCL, check the parcel temperature there and explain the slope change above it.

Lift a parcel through a temperature profile

Worked teaching model · synthetic values · Synthetic parcel and environment both start at 30 °C. Prescribed cooling: unsaturated parcel 9.8 °C/km, its dew point approximately 1.8 °C/km, saturated parcel a fixed teaching approximation of 6 °C/km. Actual moist cooling varies with pressure and temperature. Height spans 0–3 km above the starting level. No mixing, virtual-temperature correction, condensate loading, pressure solver, lift force, time evolution or precipitation. Temperature differences are only a buoyancy proxy, not CAPE or an actual forecast.

Reset keeps your prediction, captures and practice history.

Rising parcel and surroundings · height above the starting levelSolid: parcel · dashed: environment · dotted: parcel dew point before saturationDry cooling 9.8 °C/km · saturated teaching approximation 6 °C/km00.511.522.53km-10010203035Temperature (°C) · circle: approximate condensation level 1.00 kmThe dotted dew-point approximation stops at saturation; these are not measured soundings.

Conditionally unstable lapse-rate comparison. This compares assumed lapse rates, not the weather outcome. The post-saturation parcel/environment temperature equality is 3.80 km (beyond this 3 km plot). The temperature equality is not a virtual-temperature-corrected level of free convection.

Compare parcel and environment at the same height
Height above startParcelEnvironmentParcel minus environmentParcel stage
0.00 km30.00 °C30.00 °C0.00 °CUnsaturated approximation
0.50 km25.10 °C26.50 °C-1.40 °CUnsaturated approximation
1.00 km · condensation level20.20 °C23.00 °C-2.80 °CSaturated approximation
1.50 km17.20 °C19.50 °C-2.30 °CSaturated approximation
2.00 km14.20 °C16.00 °C-1.80 °CSaturated approximation
2.50 km11.20 °C12.50 °C-1.30 °CSaturated approximation
3.00 km8.20 °C9.00 °C-0.80 °CSaturated approximation

Positive temperature difference means warmer, negative means colder. At equal pressure and comparable composition this suggests a buoyancy tendency; actual density also depends on moisture and suspended condensate. This model supplies neither vertical speed nor the force needed to lift a colder parcel.

The initial gap is 8 °C. Its approximate closing rate is 9.8 − 1.8 = 8 °C/km, giving 1.00 km (1000 m). Rising air expands as surrounding pressure falls. Unlike the preceding humidity state comparison, pressure is not held fixed along this conceptual ascent. Cloud formation and precipitation require additional microphysics and observations.

Approximate lifting condensation level 1.00 km. At 3 km parcel minus environment = -0.80 °C.

Conditionally unstable lapse-rate comparison. Condensation level comes from the closing temperature–dew-point gap: 8/(9.8 − 1.8) km. Saturation does not guarantee that the parcel is warmer than its environment or that rain develops.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Find where two approaching temperatures meet

The surface parcel starts at 30°C. If the selected surface temperature–dew-point difference is a, its starting dew point is 30 − a. Before saturation, this model cools the parcel at 9.8°C/km and lowers its dew point at an approximate 1.8°C/km. The two temperatures approach each other because the parcel temperature falls faster.

The lifting condensation level, or LCL, describes where an initially unsaturated parcel becomes saturated under a lifting construction. NWS teaching materials illustrate it by the intersection of a dry-adiabatic path and a constant-mixing-ratio path on a thermodynamic diagram. NWS explanation of the LCL. The browser replaces those curved thermodynamic paths with explicit fixed-rate approximations.

Derive the gap-closing rate

Before saturation, parcel temperature is 30 − 9.8z and dew point is 30 − a − 1.8z, with z in kilometers. Their difference is a − 8z. Setting the difference to zero gives z_LCL = a/8 km, because 9.8 − 1.8 = 8°C/km.

At a = 8°C, the LCL is 1 km. At a = 16°C, it is 2 km. At a = 0, the parcel starts saturated at the surface, so there is no initial unsaturated segment. These are consequences of the selected rates; they are not measured cloud-base altitudes.

Create a paper table for a values 0, 8 and 16°C. Divide each gap by the closing rate and check the units: degrees Celsius divided by degrees Celsius per kilometer leaves kilometers. A larger gap requires more lifting before the two prescribed temperatures meet.

Keep the temperature path continuous

At the LCL, parcel temperature is T_LCL = 30 − 9.8z_LCL. Above it, the model uses T_parcel = T_LCL − 6(z − z_LCL). The second expression starts from the temperature already reached; it does not restart at 30°C.

For a = 8°C, T_LCL is 20.2°C. At 2 km, the parcel is 14.2°C, and at 3 km it is 8.2°C. Draw both line segments and check that they meet at 1 km with no temperature jump. Only the slope changes.

A real saturated ascent is not described by one universal 6°C/km rate. Moist adiabatic cooling depends on state, including temperature and pressure. The constant rate here makes the change in slope inspectable while leaving the detailed thermodynamics outside the model. NWS lapse-rate definitions.

Stop extending the unsaturated dew-point rule

The 1.8°C/km dew-point decrease applies only to this model’s pre-saturation construction. Continuing that separate line above the LCL would imply the parcel temperature and dew point diverge in the wrong way for the assumed saturated branch. It would mix incompatible rules.

The saturated branch assumes saturation but does not calculate condensate mass, droplet number or precipitation. The model also does not compute the lifting force or prove that the parcel can traverse an unfavorable layer below saturation. A displayed LCL is a conditional threshold, not a promise that a cloud forms there in actual weather.

This differs from the earlier humidity-state experiment, where total pressure and vapor pressure were held fixed while temperature changed. During hypothetical ascent, environmental pressure decreases and the parcel expands. Its dew point therefore need not remain constant even before condensation. The two experiments hold different quantities fixed.

Use the approximation as an evidence question

If comparing with a sounding or a cloud-base observation, record the starting parcel definition, temperature and humidity uncertainty, and the limitations of the fixed-rate construction. A mismatch could reflect those assumptions or a different air sample; it does not automatically invalidate the observation.

Transfer to a new surface gap

Use a = 12°C. Calculate the LCL, parcel temperature there, and parcel temperature at 3 km using the two-branch rule.

Check the lifting-condensation transfer

The approximate LCL is 12/8 = 1.5 km. Parcel temperature there is 30 − 9.8 × 1.5 = 15.3°C. Another 1.5 km at 6°C/km gives 6.3°C at 3 km. These values use the prescribed approximation and do not establish an observed cloud or rain event.

Apply

With an initial 12°C temperature–dew-point separation and a closing rate of 8°C/km, the approximate LCL is:

Public practice checks a specific claim. It does not establish independent mastery.

Show the gap-closing calculation, name the fixed-rate assumptions and explain why this LCL does not guarantee a cloud or rain event.

Continue

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