Part 1 of this guide establishes the physical foundations and core equation families behind evaporation calculations. Once you see how energy balance, vapor pressure gradients, and boundary-layer transport fit together, choosing—and deriving—the right formula becomes straightforward. In Part 2, we will apply these foundations to concrete worked examples across hydrology, process engineering, and indoor water surfaces.
1. Defining what "Evaporation rate" means
Before plugging numbers into an equation, you have to lock down units. In technical literature, evaporation rate is reported in one of two ways:
Mass flux (JJJ or EmE_mEm): Mass of liquid converted to vapor per unit surface area per unit time, typically in kg m−2s−1 ext{kg m}^{-2} ext{s}^{-1}kg m−2s−1, g m−2h−1 ext{g m}^{-2} ext{h}^{-1}g m−2h−1, or lb ft−2h−1 ext{lb ft}^{-2} ext{h}^{-1}lb ft−2h−1. Process engineers and HVAC designers almost always prefer mass flux because it feeds directly into mass and enthalpy balances.
Volumetric depth rate (EEE): Volume of liquid lost per unit surface area per unit time, which simplifies geometrically to a linear depth change per unit time, such as mm day−1 ext{mm day}^{-1}mm day−1, cm h−1 ext{cm h}^{-1}cm h−1, or in month−1 ext{in month}^{-1}in month−1. Hydrologists, meteorologists, and irrigation engineers favor depth rates because they match rainfall measurements.
Converting between the two requires only the liquid density ρw ho_wρw (approximately 998 kg m−3998 ext{ kg m}^{-3}998 kg m−3 for water at 20∘C20^\circ ext{C}20∘C):
E=JρwE = \frac{J}{ ho_w}E=ρwJ
For example, a mass flux of 0.0115 kg m−2s−10.0115 ext{ kg m}^{-2} ext{s}^{-1}0.0115 kg m−2s−1 of water corresponds to:
E=0.0115998≈1.15×10−5 m s−1≈0.996 m day−1=996 mm day−1E = \frac{0.0115}{998} \approx 1.15 imes 10^{-5} ext{ m s}^{-1} \approx 0.996 ext{ m day}^{-1} = 996 ext{ mm day}^{-1}E=9980.0115≈1.15×10−5 m s−1≈0.996 m day−1=996 mm day−1
Notice how critical unit discipline is here: a factor-of-1,000 slip between kilograms and grams or meters and millimeters is the most common error in hand calculations.
2. The thermodynamic engine: Vapor Pressure and latent heat
Evaporation occurs at any temperature below boiling because molecules in a liquid do not all share the same kinetic energy. Their speeds follow a Maxwell–Boltzmann distribution; a high-energy tail of molecules near the surface overcomes intermolecular attractive forces and escapes into the gas phase. Meanwhile, vapor molecules in the air above the surface collide with the liquid and recondense.
Net evaporation is simply the difference between the escaping flux and the returning flux. Three thermodynamic quantities govern this balance.
Latent heat of vaporization (LvL_vLv)
Every kilogram of liquid that escapes into the vapor phase carries away energy equal to the latent heat of vaporization, LvL_vLv. For water, LvL_vLv decreases slightly as temperature rises:
Lv(T)≈2.501×106−2370 TL_v(T) \approx 2.501 imes 10^6 - 2370\,TLv(T)≈2.501×106−2370T
where TTT is in degrees Celsius (∘C^\circ ext{C}∘C) and LvL_vLv is in J kg−1 ext{J kg}^{-1}J kg−1. At 20∘C20^\circ ext{C}20∘C, Lv≈2.454×106 J kg−1L_v \approx 2.454 imes 10^6 ext{ J kg}^{-1}Lv≈2.454×106 J kg−1 (about 2.45 MJ kg−12.45 ext{ MJ kg}^{-1}2.45 MJ kg−1). Because this energy must come from somewhere, evaporation cools the remaining liquid unless external heat replaces it.
Saturation Vapor Pressure (ese_ses) and actual Vapor Pressure (eae_aea)
In a closed system at temperature TTT, evaporation continues until the partial pressure of the vapor reaches an equilibrium ceiling called the saturation vapor pressure, es(T)e_s(T)es(T). Warmer air holds vastly more water vapor at saturation, growing nonlinearly with temperature.
The relationship is governed by the Clausius–Clapeyron equation:
desdT=Lv esRv T2\frac{de_s}{dT} = \frac{L_v \, e_s}{R_v \, T^2}dTdes=RvT2Lves
where Rv=461.5 J kg−1K−1R_v = 461.5 ext{ J kg}^{-1} ext{K}^{-1}Rv=461.5 J kg−1K−1 is the specific gas constant for water vapor and TTT is absolute temperature in Kelvin.
In practice, engineers use integrated empirical approximations rather than integrating Clausius–Clapeyron directly. A widely used form for water over liquid surfaces is the Magnus–Tetens equation:
es(T)=0.6108exp(17.27 TT+237.3)e_s(T) = 0.6108 \exp\left(\frac{17.27\,T}{T + 237.3} ight)es(T)=0.6108exp(T+237.317.27T)
where TTT is in ∘C^\circ ext{C}∘C and ese_ses is in kilopascals (kPa ext{kPa}kPa). Let us check two reference points:
At T=0∘CT = 0^\circ ext{C}T=0∘C: es=0.6108exp(0)=0.6108 kPae_s = 0.6108 \exp(0) = 0.6108 ext{ kPa}es=0.6108exp(0)=0.6108 kPa.
At T=20∘CT = 20^\circ ext{C}T=20∘C: es=0.6108exp(345.4257.3)≈0.6108exp(1.3424)≈2.338 kPae_s = 0.6108 \exp\left(\frac{345.4}{257.3} ight) \approx 0.6108 \exp(1.3424) \approx 2.338 ext{ kPa}es=0.6108exp(257.3345.4)≈0.6108exp(1.3424)≈2.338 kPa.
The actual partial pressure of vapor present in the ambient air is eae_aea, related to relative humidity (RH ext{RH}RH, expressed as a fraction between 000 and 111) by:
ea=RH⋅es(Tair)e_a = ext{RH} \cdot e_s(T_{ ext{air}})ea=RH⋅es(Tair)
Vapor Pressure deficit (VPD ext{VPD}VPD)
The driving potential for evaporation is the difference between the saturation vapor pressure at the evaporating surface and the actual vapor pressure in the surrounding air:
VPD=es(Tsurface)−ea ext{VPD} = e_s(T_{ ext{surface}}) - e_aVPD=es(Tsurface)−ea
When VPD ext{VPD}VPD is zero, the air adjacent to the surface is saturated and net evaporation stops. When VPD ext{VPD}VPD is large, hot surface, dry air, or both, the thermodynamic pull toward evaporation is strong.
3. The two fundamental modeling frameworks
Once you understand VPD ext{VPD}VPD and LvL_vLv, almost every evaporation formula falls into one of two physical frameworks, or combines both.
Framework a: Aerodynamic / Mass-Transfer Approach
Think of the air right above a water surface as having a thin boundary layer where vapor concentration transitions from es(Ts)e_s(T_s)es(Ts) at the interface to eae_aea in the free stream. Molecules cross this layer by molecular diffusion and turbulent eddy transport.
In its classic form (rooted in John Dalton's 1802 work), the mass-transfer equation writes:
E=f(u) (es−ea)E = f(u)\,(e_s - e_a)E=f(u)(es−ea)
Here, f(u)f(u)f(u) is an empirical or semi-empirical wind function that captures how wind speed uuu thins the boundary layer and boosts turbulent mixing. A typical linear wind function looks like:
f(u)=a+b uf(u) = a + b\,uf(u)=a+bu
where aaa represents free convection (buoyancy-driven motion when no wind blows) and b ub\,ubu represents forced convection.
Why mass transfer alone can be incomplete: Notice that ese_ses depends on the surface temperature TsT_sTs, not necessarily the bulk air temperature TairT_{ ext{air}}Tair. If you do not measure TsT_sTs directly, you cannot evaluate es(Ts)e_s(T_s)es(Ts) without knowing how heat flows into and out of the surface.
Framework b: Surface Energy-Balance Approach
Every joule leaving a surface as latent heat of evaporation (LvEL_v ELvE) must be balanced by energy arriving at or leaving that surface:
Rn−G−H−LvE=ΔSR_n - G - H - L_v E = \Delta SRn−G−H−LvE=ΔS
where:
RnR_nRn = net radiation (W m−2 ext{W m}^{-2}W m−2), incoming shortwave and longwave minus outgoing.
GGG = ground or sub-surface heat flux (W m−2 ext{W m}^{-2}W m−2).
HHH = sensible heat flux (W m−2 ext{W m}^{-2}W m−2), convective heat transfer to or from the air.
ΔS\Delta SΔS = change in heat storage within the water body or substrate (W m−2 ext{W m}^{-2}W m−2).
Rearranging for latent heat flux gives:
LvE=Rn−G−H−ΔSL_v E = R_n - G - H - \Delta SLvE=Rn−G−H−ΔS
This equation establishes a hard physical ceiling. Even if the air above a pond is bone-dry, water cannot evaporate faster than the energy supply allows, unless the water body is actively cooling down (ΔS<0\Delta S < 0ΔS<0) and sacrificing its own stored thermal energy to pay the latent heat cost.
4. Bridging both worlds: The Combination principle
In 1948, Howard Penman showed that you can eliminate the unknown surface temperature TsT_sTs by combining the energy-balance equation with the aerodynamic mass-transfer equation.
The key mathematical trick uses the slope of the saturation vapor pressure curve, Δ\DeltaΔ:
Δ=desdT∣T=Tair\Delta = \left.\frac{de_s}{dT} ight\vert{}_{T = T_{ ext{air}}}Δ=dTdesT=Tair
Using the Magnus–Tetens derivative, Δ\DeltaΔ (in kPa ∘C−1 ext{kPa }^\circ ext{C}^{-1}kPa ∘C−1) is:
Δ=4098[0.6108exp(17.27 TT+237.3)](T+237.3)2\Delta = \frac{4098 \left[0.6108 \exp\left(\frac{17.27\,T}{T + 237.3} ight) ight]}{(T + 237.3)^2}Δ=(T+237.3)24098[0.6108exp(T+237.317.27T)]
By linearizing es(Ts)e_s(T_s)es(Ts) around TairT_{ ext{air}}Tair via Δ\DeltaΔ and pairing it with the psychrometric constant γ\gammaγ:
γ=cp Pε Lv≈0.665×10−3 P\gamma = \frac{c_p \, P}{\varepsilon \, L_v} \approx 0.665 imes 10^{-3}\,Pγ=εLvcpP≈0.665×10−3P
(where cp≈1013 J kg−1K−1c_p \approx 1013 ext{ J kg}^{-1} ext{K}^{-1}cp≈1013 J kg−1K−1 is specific heat of moist air at constant pressure, ε=0.622\varepsilon = 0.622ε=0.622 is the ratio of molecular weights of water vapor to dry air, and PPP is atmospheric pressure in kPa ext{kPa}kPa), Penman derived the famous combination equation:
E=ΔΔ+γRn−GLv+γΔ+γEaE = \frac{\Delta}{\Delta + \gamma}\frac{R_n - G}{L_v} + \frac{\gamma}{\Delta + \gamma} E_aE=Δ+γΔLvRn−G+Δ+γγEa
Look at the two terms on the right-hand side:
The radiation term ΔΔ+γRn−GLv\frac{\Delta}{\Delta + \gamma}\frac{R_n - G}{L_v}Δ+γΔLvRn−G scales with available net energy.
The aerodynamic term γΔ+γEa\frac{\gamma}{\Delta + \gamma} E_aΔ+γγEa scales with the vapor pressure deficit (es(Tair)−ea)(e_s(T_{ ext{air}}) - e_a)(es(Tair)−ea) and wind speed via an aerodynamic term EaE_aEa.
The weighting factors ΔΔ+γ\frac{\Delta}{\Delta + \gamma}Δ+γΔ and γΔ+γ\frac{\gamma}{\Delta + \gamma}Δ+γγ sum to 111, automatically shifting emphasis from radiation-dominated evaporation in cool, humid climates to vapor-deficit-dominated evaporation in hot, arid conditions.
5. How specialized formulas branch from these foundations
Depending on your domain, practical formulas simplify or extend the equations above:
| Domain | Typical Governing Approach | Key Specific Variables |
|---|---|---|
| Open water & reservoirs | Penman combination, Rohwer, or Meyer equations | Net radiation, fetch/basin area, wind at 2 m height |
| Crops & soil (evapotranspiration) | FAO-56 Penman–Monteith | Surface (canopy) resistance rs, aerodynamic resistance ra, crop height |
| Indoor pools & wet surfaces | Empirical mass-transfer correlations | Water-to-air temperature difference, surface area, air velocity |
| Chemical solvents & industrial tanks | Diffusion / convective mass transfer | Vapor molecular weight, Antoine equation vapor pressure, Schmidt number |
In Part 2, we will walk through the step-by-step calculation procedures, unit conversions, and worked numerical examples for each of these major regimes so you can pick the exact equation for your problem and solve it cleanly without unit traps.
Advanced Aerodynamic and Combination models: Beyond basic Dalton
While Dalton’s foundational mass-transfer approach provides an intuitive entry point, modern hydrologists, meteorologists, and environmental engineers rarely rely on vapor pressure and wind speed alone for precision work. In natural or controlled settings, the temperature of the evaporating surface (TsT_sTs) is rarely identical to the ambient air temperature (TaT_aTa). Solar radiation constantly heats the surface, while evaporation simultaneously cools it through latent heat loss.
To account for this coupled energy-mass dynamic, Howard Penman formulated the landmark Penman Combination Equation in 1948. By merging the surface energy balance with aerodynamic mass transfer, Penman eliminated the need to directly measure elusive surface temperatures.
The standard open-water Penman formulation is expressed as:
E0=Δ(Rn−G)+γEaΔ+γE_0 = \frac{\Delta (R_n - G) + \gamma E_a}{\Delta + \gamma}E0=Δ+γΔ(Rn−G)+γEa
Where:
E0E_0E0 = Open-water evaporation rate (typically expressed in mm d−1 ext{mm d}^{-1}mm d−1 or kg m−2s−1 ext{kg m}^{-2} ext{s}^{-1}kg m−2s−1)
Δ\DeltaΔ = Slope of the saturation vapor pressure versus temperature curve at air temperature (kPa ∘C−1 ext{kPa }^\circ ext{C}^{-1}kPa ∘C−1)
RnR_nRn = Net radiation flux density at the surface (MJ m−2d−1 ext{MJ m}^{-2} ext{d}^{-1}MJ m−2d−1)
GGG = Soil or water heat flux density (MJ m−2d−1 ext{MJ m}^{-2} ext{d}^{-1}MJ m−2d−1) (often assumed negligible or zero over daily cycles for deep open water bodies)
γ\gammaγ = Psychrometric constant (kPa ∘C−1 ext{kPa }^\circ ext{C}^{-1}kPa ∘C−1)
EaE_aEa = Aerodynamic mass-transfer term, defined by a wind function f(u)f(u)f(u) multiplied by the vapor pressure deficit ( es−ea )(\,e_s - e_a\,)(es−ea)
The beauty of the Penman approach lies in its division into two distinct drivers: the radiation term (driven by net available energy) and the aerodynamic term (driven by wind speed and vapor pressure deficit). In humid environments, the radiation term dominates, whereas in arid, windy environments, the aerodynamic term dictates the bulk of the evaporation rate.
Empirical Corrections: Rohwer and Meyer formulations
For localized field evaluations where radiation data is missing or difficult to procure, engineers frequently adapt empirical mass-transfer models developed through large-scale pan and reservoir experiments. Two notable historical adaptations remain widely referenced:
1. The Rohwer Formula
Rohwer expanded upon Dalton’s equation by conducting controlled experiments that explicitly accounted for barometric pressure variations (elevation) and ground-level wind speeds:
Ev=(1.465−0.0186B)(0.44+0.118W)(es−ed)E_v = (1.465 - 0.0186 B)(0.44 + 0.118 W)(e_s - e_d)Ev=(1.465−0.0186B)(0.44+0.118W)(es−ed)
Where:
EvE_vEv = Evaporation rate in inches per day
BBB = Mean barometric pressure in inches of Hg
WWW = Mean wind velocity at ground level in miles per hour
es,ede_s, e_des,ed = Saturation vapor pressure of the water surface and dew-point vapor pressure, respectively (in inches of Hg)
2. The Meyer Formula
For deeper natural lakes and reservoirs, Meyer introduced a simplified velocity coefficient that accounts for wind measured at higher elevations (typically 9 meters above ground level):
EL=CM(es−ea)(1+W916)E_L = C_M (e_s - e_a)\left(1 + \frac{W_9}{16} ight)EL=CM(es−ea)(1+16W9)
Where CMC_MCM is an empirical coefficient (typically 0.360.360.36 for large deep lakes and 0.500.500.50 for small, shallow water bodies), and W9W_9W9 is the wind velocity in kilometers per hour at a 9-meter height. These equations underline a vital engineering principle: wind speed scaling depends heavily on the measurement height and boundary layer roughness.
Step-by-Step practical engineering calculation
To see how these formulas function in practice, let us calculate the daily open-water evaporation rate for an industrial reservoir using a modified aerodynamic mass-transfer and vapor pressure deficit approach.
Given Parameters
Mean Air Temperature (TaT_aTa): 25∘C25^\circ ext{C}25∘C
Relative Humidity (RHRHRH): 50%50\%50% (0.500.500.50)
Wind Speed at 2 meters height (u2u_2u2): 2.5 m s−12.5 ext{ m s}^{-1}2.5 m s−1
Atmospheric Pressure (PPP): 101.3 kPa101.3 ext{ kPa}101.3 kPa
Step 1: Calculate saturated Vapor Pressure (ese_ses)
Using the Tetens equation or standard saturation vapor pressure formula:
es(Ta)=0.6108exp(17.27⋅TaTa+237.3)e_s(T_a) = 0.6108 \exp\left(\frac{17.27 \cdot T_a}{T_a + 237.3} ight)es(Ta)=0.6108exp(Ta+237.317.27⋅Ta)
Substituting Ta=25∘CT_a = 25^\circ ext{C}Ta=25∘C:
es=0.6108exp(17.27⋅2525+237.3)≈3.167 kPae_s = 0.6108 \exp\left(\frac{17.27 \cdot 25}{25 + 237.3} ight) \approx 3.167 ext{ kPa}es=0.6108exp(25+237.317.27⋅25)≈3.167 kPa
Step 2: Calculate actual Vapor Pressure (eae_aea)
Using the relative humidity (RHRHRH):
ea=es⋅(RH100)=3.167⋅0.50=1.584 kPae_a = e_s \cdot \left(\frac{RH}{100} ight) = 3.167 \cdot 0.50 = 1.584 ext{ kPa}ea=es⋅(100RH)=3.167⋅0.50=1.584 kPa
Step 3: Compute the Vapor Pressure deficit (VPD)
VPD=es−ea=3.167−1.584=1.583 kPa ext{VPD} = e_s - e_a = 3.167 - 1.584 = 1.583 ext{ kPa}VPD=es−ea=3.167−1.584=1.583 kPa
Step 4: Apply the Aerodynamic Wind function
Using a standard FAO-56 mass transfer wind function coefficient for open water (f(u)=0.61+0.50 u2f(u) = 0.61 + 0.50 \, u_2f(u)=0.61+0.50u2):
f(u)=0.61+(0.50⋅2.5)=1.86 mm d−1 kPa−1f(u) = 0.61 + (0.50 \cdot 2.5) = 1.86 ext{ mm d}^{-1} ext{ kPa}^{-1}f(u)=0.61+(0.50⋅2.5)=1.86 mm d−1 kPa−1
Step 5: Compute final Evaporation rate (EEE)
E=f(u)⋅(VPD)=1.86⋅1.583≈2.94 mm d−1E = f(u) \cdot ( ext{VPD}) = 1.86 \cdot 1.583 \approx 2.94 ext{ mm d}^{-1}E=f(u)⋅(VPD)=1.86⋅1.583≈2.94 mm d−1
Thus, under these specific meteorological conditions, the reservoir experiences an approximate water column loss of 2.94 millimeters per day2.94 ext{ millimeters per day}2.94 millimeters per day.
Common pitfalls, edge cases, and error analysis
When implementing evaporation formulas in real-world engineering or scientific projects, several subtle physical phenomena can introduce major calculation errors if ignored:
Water Surface Temperature Discrepancies (Ts≠TaT_s eq T_aTs=Ta): Assuming the water surface temperature equals the air temperature is a frequent shortcut. In reality, solar radiation absorption can raise shallow water temperatures significantly above air temperature, exponentially boosting ese_ses and underestimating actual evaporation if air temperature is used instead.
Salinity and Total Dissolved Solids (TDS): Standard formulas assume pure water. According to Raoult's Law, dissolved salts lower the vapor pressure of a liquid. For hyper-saline ponds, brine pools, or industrial tailings basins, a salinity reduction factor (typically 0.900.900.90 to 0.950.950.95) must be applied to account for the suppressed evaporation rate.
Height Corrections for Wind Speed: Wind sensors are rarely mounted at the exact standard reference height of 2 meters. Engineers must convert wind measurements using the logarithmic wind profile law before inputting speeds into aerodynamic equations:
u2=uz⋅ln(2/z0)ln(z/z0)u_2 = u_z \cdot \frac{\ln(2 / z_0)}{\ln(z / z_0)}u2=uz⋅ln(z/z0)ln(2/z0)
where zzz is the measurement height and z0z_0z0 is the roughness length of the surrounding surface.
Conclusion and model selection summary
Calculating the evaporation rate accurately is a multi-disciplinary challenge balancing meteorology, fluid mechanics, and thermodynamics. Selecting the correct formula depends entirely on data availability and system constraints:
Use Dalton / Simple Mass Transfer when data is sparse and only rough baseline estimates are required.
Use Rohwer or Meyer Empirical Equations for regional hydrology assessments and open-air reservoirs where wind and barometric pressure records are well-established.
Use the Penman Combination Equation when high-precision scientific rigor, radiation data, and complex microclimate variables are accessible.
By matching your data constraints to the appropriate mathematical model, you ensure reliable, defensible water resource management and engineering design.