How net radiation is partitioned between heating air and evaporating water
2026-02-26
You should know
What net radiation is, and that energy at the surface can be used either to warm air and ground or to evaporate water.
You will learn
The difference between sensible and latent heat, how energy is partitioned at the surface, and why wet and dry landscapes behave so differently.
Why this matters
This chapter explains why a forest, a field, and a parking lot can receive similar sunlight but heat the air in very different ways.
If this gets hard, focus on…
Remember that latent heat is energy hidden in evaporation, while sensible heat is energy you can notice as a temperature change.
Walk barefoot across an irrigated alfalfa field on a hot July afternoon and the air above it feels noticeably cooler than the air above the adjacent gravel parking lot — even though both surfaces receive the same solar radiation. The field is spending most of its net radiation on evaporation: water absorbs 2,450 joules per gram to change from liquid to vapour, carrying that energy invisibly into the atmosphere as latent heat. The parking lot has no water to evaporate, so it converts nearly all its absorbed radiation into sensible heat — the form that warms the air and the soles of your feet.
This partitioning of net radiation between sensible and latent heat is one of the most important variables in surface climatology. It governs local temperature, humidity, cloud formation, and the water balance of entire river basins. The Bowen ratio — sensible heat divided by latent heat — distinguishes a tropical rainforest (Bowen ratio near 0.2) from a desert (Bowen ratio of 5 or more). Getting this partitioning right is critical for climate models, irrigation scheduling, flood forecasting, and understanding how land-use change — draining a wetland, irrigating a desert, deforesting a hillside — alters local climate. This model derives the physics of both fluxes from first principles and shows what controls their ratio.
When the sun heats the ground, where does that energy go?
Model 17 showed that net radiation (R_n) is the energy absorbed by Earth’s surface. But this energy doesn’t just sit there—it flows into: - The air (warming the atmosphere) - Water vapor (evaporating moisture from soil and plants) - The ground (heating deeper soil layers)
Different surfaces partition energy differently: - Desert: Most energy heats the air (hot, dry air rising) - Irrigated field: Most energy evaporates water (cool, moist air) - Snow/ice: Energy goes into melting (temperature stays at 0°C until all melts)
The mathematical question: How much energy goes into each pathway, and what controls the partitioning?
Net radiation is partitioned among four fluxes:
R_n = H + LE + G + S
Where (all in W/m²): - R_n = Net radiation (from Model 17) - H = Sensible heat flux (warms/cools the air) - LE = Latent heat flux (evaporation/condensation) - G = Ground heat flux (warms/cools the soil) - S = Storage (warms/cools biomass, water bodies)
Sign convention: - Positive: energy flows away from surface (surface loses energy) - Negative: energy flows toward surface (surface gains energy)
Typical daytime values: - R_n = +600 W/m² (surface gains energy) - H = +100 W/m² (heats air) - LE = +400 W/m² (evaporates water) - G = +100 W/m² (heats soil)
Sensible heat = energy you can sense (feel) as temperature change
When air warms from 20°C to 25°C, it has gained sensible heat. You can measure this with a thermometer.
Latent heat = hidden energy in phase change (solid ↔︎ liquid ↔︎ gas)
When water evaporates, it absorbs energy but doesn’t change temperature. The energy is stored in the vapor molecules and released when vapor condenses back to liquid.
Latent heat of vaporization (L_v): - Energy required to evaporate 1 kg of water - L_v = 2.45 \times 10^6 J/kg (at 20°C) - This is HUGE—evaporating 1 liter of water requires 2,450 kJ
Why it matters: Evaporation is a very effective cooling mechanism. That’s why sweating cools you down.
Heat moves from warm to cool. If the surface is warmer than the air, heat flows upward into the atmosphere.
Simplified form:
H = \rho_a c_p \frac{T_s - T_a}{r_a}
Where: - \rho_a = air density (~1.2 kg/m³) - c_p = specific heat of air (~1005 J/kg/K) - T_s = surface temperature (K or °C) - T_a = air temperature (K or °C) - r_a = aerodynamic resistance (s/m)
Aerodynamic resistance (r_a) depends on: - Wind speed (higher wind → lower resistance → more heat transfer) - Surface roughness (tall vegetation → more turbulent → lower resistance) - Atmospheric stability (unstable → enhanced mixing → lower resistance)
Simplified estimate:
r_a \approx \frac{\ln(z/z_0)^2}{k^2 u}
Where: - z = measurement height (typically 2 m) - z_0 = roughness length (~0.01 m for grass, ~1 m for forest) - k = 0.41 (von Kármán constant) - u = wind speed (m/s)
Example: Grass surface, wind 3 m/s:
r_a \approx \frac{(\ln(2/0.01))^2}{0.41^2 \times 3} = \frac{(5.3)^2}{0.5} = 56 \text{ s/m}
Evaporation rate (E, kg/m²/s) multiplied by latent heat gives the energy flux:
LE = L_v E
Evaporation driven by vapor pressure deficit:
E = \rho_a \frac{e_s - e_a}{r_a + r_s}
Where: - e_s = saturation vapor pressure at surface temperature (Pa) - e_a = actual vapor pressure in air (Pa) - r_s = surface resistance (s/m)
Surface resistance (r_s) depends on: - Soil moisture (dry soil → high resistance → little evaporation) - Stomatal opening in plants (closed stomata → high resistance) - Surface type (open water → r_s = 0; desert → r_s = \infty)
Saturation vapor pressure (Tetens formula):
e_s(T) = 611 \exp\left(\frac{17.27 T}{T + 237.3}\right)
Where T is temperature in °C, result in Pa.
Relative humidity:
RH = \frac{e_a}{e_s} \times 100\%
Vapor pressure deficit:
VPD = e_s - e_a = e_s(1 - RH/100)
The Bowen ratio (\beta) is the ratio of sensible to latent heat:
\beta = \frac{H}{LE}
Physical interpretation: - \beta > 1: More energy heats air than evaporates water (dry conditions) - \beta < 1: More energy evaporates water than heats air (moist conditions) - \beta \approx 0: Nearly all energy goes to evaporation (wet surface, lake)
Typical values: - Tropical rainforest: \beta \approx 0.2 (most energy → evaporation) - Irrigated cropland: \beta \approx 0.4 - Temperate grassland: \beta \approx 0.8 - Semi-arid shrubland: \beta \approx 2 - Desert: \beta \approx 10 (minimal evaporation)
Simplified Bowen ratio:
\beta \approx \gamma \frac{T_s - T_a}{e_s - e_a}
Where \gamma = c_p / L_v \approx 0.66 hPa/K (psychrometric constant).
Problem: A grassland surface on a sunny afternoon has: - Net radiation: R_n = 600 W/m² - Surface temperature: T_s = 30°C - Air temperature (2 m): T_a = 25°C - Relative humidity: RH = 50\% - Wind speed: u = 3 m/s - Aerodynamic resistance: r_a = 60 s/m - Surface resistance: r_s = 100 s/m - Ground heat flux: G = 100 W/m²
Calculate H, LE, and the Bowen ratio.
Step 1: Saturation vapor pressure
At T_s = 30°C:
e_s(30) = 611 \exp\left(\frac{17.27 \times 30}{30 + 237.3}\right) = 611 \exp(1.936) = 4243 \text{ Pa}
At T_a = 25°C:
e_s(25) = 611 \exp\left(\frac{17.27 \times 25}{25 + 237.3}\right) = 611 \exp(1.645) = 3168 \text{ Pa}
Step 2: Actual vapor pressure
e_a = RH \times e_s(T_a) = 0.50 \times 3168 = 1584 \text{ Pa}
Step 3: Sensible heat flux
H = \rho_a c_p \frac{T_s - T_a}{r_a} = 1.2 \times 1005 \times \frac{30 - 25}{60}
= 1206 \times \frac{5}{60} = 100 \text{ W/m}^2
Step 4: Evaporation rate
E = \rho_a \frac{e_s(T_s) - e_a}{r_a + r_s} = 1.2 \times \frac{4243 - 1584}{60 + 100}
= 1.2 \times \frac{2659}{160} = 1.2 \times 16.6 = 20.0 \times 10^{-3} \text{ kg/m}^2\text{/s}
(Note: need to divide by atmospheric pressure factor; simplified here)
Actually, more precisely with vapor pressure in Pa and using proper conversion:
E \approx 0.622 \times \frac{1.2}{101325} \times \frac{2659}{160} \approx 0.000124 \text{ kg/m}^2\text{/s}
Step 5: Latent heat flux
LE = L_v E = 2.45 \times 10^6 \times 0.000124 = 304 \text{ W/m}^2
Step 6: Check energy balance
R_n = H + LE + G $600 = 100 + 304 + 100 = 504 \text{ W/m}^2$
Close enough (small discrepancy from rounding and simplified formulas).
Step 7: Bowen ratio
\beta = \frac{H}{LE} = \frac{100}{304} = 0.33
Interpretation: About 3× more energy goes to evaporation than to heating the air. This is typical of well-watered grassland.
Below is an interactive energy partitioning calculator.
<label>
Surface type:
<select id="surface-type-ep">
<option value="wet">Open Water (rs=0)</option>
<option value="irrigated">Irrigated Crop (rs=50)</option>
<option value="grass" selected>Grassland (rs=100)</option>
<option value="dryland">Dryland Crop (rs=200)</option>
<option value="desert">Desert (rs=500)</option>
</select>
</label>
<label>
Net radiation (W/m²):
<input type="range" id="rn-slider" min="100" max="800" step="50" value="600">
<span id="rn-value">600</span> W/m²
</label>
<label>
Surface temperature (°C):
<input type="range" id="ts-slider" min="15" max="45" step="1" value="30">
<span id="ts-value">30</span> °C
</label>
<label>
Air temperature (°C):
<input type="range" id="ta-slider" min="10" max="40" step="1" value="25">
<span id="ta-value">25</span> °C
</label>
<label>
Relative humidity (%):
<input type="range" id="rh-slider" min="10" max="90" step="5" value="50">
<span id="rh-value">50</span> %
</label>
<label>
Wind speed (m/s):
<input type="range" id="wind-slider" min="0.5" max="10" step="0.5" value="3">
<span id="wind-value">3.0</span> m/s
</label><h4>Energy Partition:</h4>
<table>
<tr>
<td><strong>R<sub>n</sub> (net radiation):</strong></td>
<td><span id="rn-result"></span> W/m²</td>
</tr>
<tr>
<td><strong>H (sensible heat):</strong></td>
<td><span id="h-result"></span> W/m²</td>
</tr>
<tr>
<td><strong>LE (latent heat):</strong></td>
<td><span id="le-result"></span> W/m²</td>
</tr>
<tr>
<td><strong>G (ground heat):</strong></td>
<td><span id="g-result"></span> W/m²</td>
</tr>
<tr style="border-top: 1px solid #ddd;">
<td><strong>Bowen ratio (β):</strong></td>
<td><span id="beta-result"></span></td>
</tr>
<tr>
<td><strong>Evaporation rate:</strong></td>
<td><span id="evap-result"></span> mm/day</td>
</tr>
</table>Try this: - Switch to desert: Surface resistance skyrockets → LE drops → β > 1 → most energy heats air - Switch to open water: Surface resistance = 0 → LE dominates → β < 1 → evaporative cooling - Increase humidity: Vapor pressure deficit shrinks → less evaporation → higher β - Increase wind: Aerodynamic resistance drops → more H and LE - Dry air (low RH): Large vapor pressure deficit → high evaporation rate
Key insight: The Bowen ratio tells you whether a surface cools by evaporation (low β) or heats the air (high β).
Desert surfaces have: - High R_n (strong sun, clear skies) - Low soil moisture → high r_s → minimal LE - High \beta (typically 5–10)
Result: Nearly all net radiation goes into H → air temperature soars.
Example: R_n = 700 W/m², \beta = 8:
H = \frac{\beta}{\beta + 1} R_n = \frac{8}{9} \times 700 = 622 \text{ W/m}^2
Tropical rainforest surfaces have: - High R_n (near equator, strong sun) - Abundant moisture → low r_s → high LE - Low \beta (typically 0.2–0.4)
Result: Most energy goes to evaporation → air stays cooler.
Example: R_n = 600 W/m², \beta = 0.3:
LE = \frac{1}{\beta + 1} R_n = \frac{1}{1.3} \times 600 = 462 \text{ W/m}^2
Evapotranspiration (ET) from croplands is:
ET = \frac{LE}{L_v \rho_w}
Where \rho_w = 1000 kg/m³ (water density).
Units: mm/day (depth of water evaporated per day)
Example: LE = 300 W/m²:
ET = \frac{300}{2.45 \times 10^6 \times 1000} \times 86400 = 10.6 \text{ mm/day}
Over a growing season (100 days), this is 1,060 mm = 1.06 m of water!
Irrigation requirement ≈ ET - rainfall.
An irrigated field in a desert creates a cool, moist microclimate: - Desert: \beta = 10, T_s = 45°C - Oasis: \beta = 0.5, T_s = 28°C
Mechanism: High evaporation cools the surface and humidifies the air.
Latent heat flux (LE) is in W/m² (energy units).
Evaporation rate (E) is in kg/m²/s or mm/day (mass units).
LE = L_v E
Don’t forget to multiply by L_v when converting between them!
r_s varies with: - Soil moisture: Dry soil → high r_s - Time of day: Stomata close at night → high r_s - Plant stress: Drought, heat → stomata close → high r_s - Species: C4 plants (corn) have lower r_s than C3 plants (wheat) under some conditions
Dynamic models update r_s based on soil moisture, vapor pressure deficit, and plant physiology.
G is small during the day (~10–20% of R_n) but can be significant: - Bare soil: G can be 30–40% of R_n - Snow/ice: G is critical for melt energy - Urban surfaces: High thermal mass → large G
Nighttime: G reverses (heat flows back from soil to surface).
Advection: Horizontal transport of heat/moisture by wind.
Oasis effect: Dry air blown from desert increases evaporation from irrigated field beyond what local energy balance would predict.
Clothesline paradox: Laundry dries faster on a windy day even though local R_n is the same.
Our model ignores advection — assumes air properties are uniform horizontally.
The Penman-Monteith equation is the standard model for evapotranspiration:
LE = \frac{\Delta (R_n - G) + \rho_a c_p (e_s - e_a) / r_a}{\Delta + \gamma (1 + r_s/r_a)}
Where: - \Delta = slope of saturation vapor pressure curve (Pa/K) - \gamma = psychrometric constant (~66 Pa/K)
This combines: - Radiative term: \Delta (R_n - G) - Aerodynamic term: \rho_a c_p (e_s - e_a) / r_a
Advantages: - Physically based (energy balance + vapor transport) - Widely validated for crops, forests, grasslands - Basis for FAO-56 reference evapotranspiration
Next model will explore how G behaves—heat diffusion into soil.
Saturation vapor pressure increases exponentially with temperature:
e_s(T) \propto \exp(T)
Physical reason: Clausius-Clapeyron equation from thermodynamics.
Molecules need energy to escape liquid and become vapor. Higher temperature → more molecules have enough energy → exponential increase.
A small temperature change causes a large vapor pressure change.
From 20°C to 30°C: - e_s(20) = 2338 Pa - e_s(30) = 4243 Pa
81% increase for a 10°C warming!
This is why hot air can hold much more moisture than cold air (and why your glasses fog when you come inside from the cold).