Calculator
Example Data
| Method | Area (m²) | Water (°C) | Air (°C) | RH (%) | Wind (m/s) | hm (m/s) | Qnet (W/m²) |
|---|---|---|---|---|---|---|---|
| Mass transfer | 1.0 | 25 | 25 | 50 | 1.0 | 0.005 | - |
| Energy balance | 2.0 | 30 | 28 | 45 | 2.5 | - | 250 |
Formula Used
- Saturation vapor pressure (Tetens): \( e_s(T)=6.1078\times 10^{\frac{7.5T}{T+237.3}}\,\text{hPa} \) and \(1\,\text{hPa}=100\,\text{Pa}\).
- Actual vapor pressure: \( e_a = (RH/100)\,e_s(T_{air}) \).
- Vapor density (ideal gas): \( \rho_v = \frac{e}{R_v\,T} \), with \(R_v=461.5\,\text{J/(kg·K)}\).
- Mass-transfer evaporation: \( \dot m = h_m^{eff}\,A\,(\rho_{vs}-\rho_{va}) \), limited to nonnegative.
- Energy-balance evaporation: \( \dot m = \frac{Q_{net}\,A}{L_v(T)} \), with \(L_v\approx (2501-2.361T)\,\text{kJ/kg}\).
- Depth equivalent: \( \dot z = \frac{\dot m}{\rho_l A} \) and \(\text{mm/day}=\dot z\times 1000\times 86400\).
How to Use This Calculator
- Select a method: mass transfer or energy balance.
- Enter surface area, temperatures, humidity, and wind speed.
- For mass transfer, set the mass transfer coefficient \(h_m\).
- For energy balance, enter net heat flux \(Q_{net}\).
- Press Calculate to display results above the form.
- Use Download CSV or Download PDF for exports.
Technical Article
1) Overview of evaporation flux
Evaporation is a phase-change mass flux from liquid to vapor driven by a vapor-pressure deficit and enabled by heat supply. In environmental conditions, rates commonly range from about 1 to 10 mm/day for open water, but can exceed that under hot, dry, windy weather and strong radiation.
2) Two modeling paths in this calculator
This calculator offers a mass-transfer route and an energy-balance route. The mass-transfer route converts vapor-pressure conditions into vapor densities and applies a transport coefficient.
3) Temperature and saturation vapor pressure
Water temperature strongly increases the saturation vapor pressure at the surface. For example, saturation vapor pressure is about 3.17 kPa at 25°C and about 4.24 kPa at 30°C, meaning warmer water raises the maximum possible vapor density at the interface and therefore increases evaporation potential when the air is not saturated.
4) Humidity deficit as the driving gradient
Relative humidity controls the actual vapor pressure of air. If air at 25°C is 50% RH, its vapor pressure is near half the saturation value, leaving a substantial deficit compared with the water surface. As RH approaches 100%, the gradient collapses and evaporation slows.
5) Wind and boundary-layer transport
Wind reduces the thickness of the near-surface boundary layer and enhances turbulent exchange, increasing the effective transfer of vapor away from the surface. In the mass-transfer option, the tool applies a mild wind scaling to the base coefficient.
6) Energy inputs and latent heat
Evaporation requires energy to supply the latent heat of vaporization. Net heat flux represents the combined effect of solar input, longwave exchange, convection, and conduction. For example, a net 200 W/m² over 1 m² corresponds to a mass loss on the order of 0.00008 kg/s at typical latent heat, translating to several mm/day.
7) Interpreting units and depth rates
Results are reported as mass rate (kg/s and kg/hr) and as an equivalent depth rate (mm/day). The depth rate is obtained by dividing the mass flux by water density (about 997 kg/m³ near room temperature).
8) Practical data checks and use cases
Use realistic inputs: 0–100% RH, nonnegative wind speed, and plausible heat fluxes (often 0–400 W/m² in many daytime cases). Typical applications include estimating reservoir losses, cooling-pond performance, drying potential, or laboratory dish evaporation. For engineering reporting, export CSV or PDF to preserve assumptions and outputs.
FAQs
1) Which method should I choose?
Use mass transfer when you trust temperature, humidity, and a calibrated coefficient. Use energy balance when you have a credible net heat flux estimate or measured surface energy budget.
2) What does the mass transfer coefficient represent?
It captures boundary-layer transport and turbulence effects near the surface. It is site- and geometry-dependent, so calibration against measurements usually improves accuracy more than default values.
3) Why can energy-balance results be negative?
A negative net heat flux means the surface is losing energy. In that case, the phase-change term can reverse, implying condensation rather than evaporation under the chosen assumptions.
4) How sensitive is evaporation to humidity?
Very sensitive. Lower relative humidity reduces air vapor pressure, enlarging the vapor-density gradient. Near saturation (close to 100% RH), the driving gradient becomes small and evaporation drops sharply.
5) Do I need ambient pressure?
Pressure has a small effect because vapor density uses the ideal-gas relationship with temperature. For most near-sea-level work, the default value is adequate unless you are at high altitude.
6) Why is depth rate reported in mm/day?
Hydrology and meteorology commonly report evaporation as an equivalent water depth removed per day. It lets you compare directly with pan evaporation, lake budgets, and climate datasets.
7) How can I improve accuracy?
Use measured water temperature, local humidity, and wind at the surface height. Calibrate the mass transfer coefficient or estimate net heat flux from observations. Keep units consistent and export results for traceability.
Technical Article
1) Overview of evaporation flux
Evaporation is a phase-change mass flux from liquid to vapor driven by a vapor pressure deficit and sustained by available heat. For open water under mild weather, depth-equivalent losses often sit around 1-10 mm/day. Hot, dry, windy afternoons can push higher, while humid, calm air can suppress losses.
2) Two modeling paths in this calculator
This tool provides a mass-transfer model and an energy-balance model. The mass-transfer route estimates a transport-limited flux from the difference between saturation vapor density at the water surface and vapor density in air, scaled by an effective coefficient. The energy route converts net heat flux (W/m^2) into a mass rate using latent heat.
3) Temperature and saturation vapor pressure
Surface temperature controls saturation vapor pressure strongly. Using the Tetens relation, saturation pressure is about 3.17 kPa at 25 C and about 5.62 kPa at 35 C. That increase raises saturation vapor density and typically increases evaporation even if humidity stays fixed. Accurate temperature inputs matter more than small pressure variations.
4) Humidity deficit as the driving gradient
Relative humidity sets the ambient vapor pressure as a fraction of saturation at air temperature. For example, at 25 C, 50% RH implies the air holds about half the saturation vapor pressure, creating a strong gradient from the surface. When RH rises toward 90%, the gradient shrinks and the mass-transfer estimate can drop sharply.
5) Wind and boundary-layer transport
Wind enhances ventilation by thinning the near-surface boundary layer and removing moist air. Many field correlations show approximately linear growth of transfer with wind speed over moderate ranges. In this calculator, the effective mass-transfer coefficient is gently increased with wind as a practical approximation. If you have a site-calibrated coefficient, enter it directly for best results.
6) Energy inputs and latent heat
Evaporation requires energy equal to latent heat of vaporization, roughly 2.44e6 J/kg near 25 C. If net heat flux is 200 W/m^2, the energy model gives m'' about 200/2.44e6 = 8.2e-5 kg/m^2/s, which corresponds to about 7.1 mm/day of water depth. This method is useful when radiation and heat budgets are known.
7) Interpreting units and depth rates
The calculator reports both mass rate (kg/s and kg/hr) and depth-equivalent rate (mm/day). Depth rate is often easier for water management because it is independent of area, while total mass rate scales with area. Converting depth loss to volume is straightforward: volume loss per day equals depth (m/day) times area (m^2).
8) Practical data checks and use cases
Use the two methods as a consistency check. Under steady daytime heating, both can produce similar orders of magnitude. If the energy method predicts strong evaporation but the mass-transfer method predicts near zero, humidity inputs may be too high or temperatures may be inconsistent. Typical applications include pools, cooling basins, reservoirs, and lab pans where you can track conditions and compare exported results over time.
FAQs
1) Which method should I choose?
Use mass transfer when you trust temperature, humidity, wind, and a reasonable transfer coefficient. Use energy balance when you have net heat flux or a heat-budget estimate. Compare both to spot input inconsistencies.
2) Why can the mass-transfer result become zero?
If the air vapor density equals or exceeds the surface saturation vapor density, the gradient reverses and evaporation stops. This can occur with very high humidity or cooler water than air.
3) What is a reasonable mass transfer coefficient?
For small water surfaces in light ventilation, values around 0.003 to 0.01 m/s are common starting points. Larger, rougher, windier settings can be higher. If you have measurements, calibrate hm to match observed losses.
4) Can net heat flux be negative?
Yes. At night or under strong cooling, Qnet can be negative, implying condensation or reduced evaporation. The energy model will reflect that by producing a smaller or negative mass rate, which you can interpret as deposition.
5) How accurate is the saturation vapor pressure formula?
The Tetens form is a practical approximation over typical ambient temperatures and is widely used in meteorology. For extreme temperatures or high-precision work, a more detailed formulation may be preferred, but trends remain consistent.
6) Why include ambient pressure?
Vapor density uses the ideal-gas relation and pressure appears implicitly through vapor pressure terms. Across typical sea-level to modest altitude ranges, the effect on evaporation estimates is usually smaller than temperature, humidity, and wind.
7) How do I use the result for a pool or reservoir?
Use the mm/day depth rate to estimate daily water-level drop. Multiply depth (m/day) by surface area (m^2) for volume loss (m^3/day). Use CSV/PDF exports to track changes by weather or season.