Introduction & Context

Evaporative salt production is the oldest industrial route to crystallise sodium chloride from brine. In solar ponds the rate-limiting step is the evaporation of water; once the brine reaches its saturation concentration the salt precipitates. Accurate prediction of the water evaporation flux therefore translates directly into a daily salt yield. This calculation is embedded in pond sizing, harvest scheduling, energy balances and economic feasibility studies for both artisanal and mechanised salt works.

Methodology & Formulas

Vapour driving force
The air above the pond is characterised by its relative humidity RH. The corresponding bulk vapour density is \[ \rho_{v,\infty}= \text{RH}\;\rho_{\text{sat}}(T) \] while the brine surface is assumed to be in equilibrium with saturated vapour at the pond temperature \(T\). The evaporation flux is driven by the difference \(\rho_{\text{sat}}-\rho_{v,\infty}\).
Reynolds number
Air flow over the pond is treated as forced convection across a flat plate whose characteristic length is the hydraulic gap \(H_{\text{gap}}\): \[ Re = \frac{\rho_{\text{air}}\,U_{\infty}\,H_{\text{gap}}}{\mu_{\text{air}}} \]

Sherwood correlation
Mass transfer is governed by the Sherwood number. The appropriate correlation depends on the flow regime:

Regime	Condition	Sherwood Number
Laminar	\(Re < 2300\)	\(Sh = 3.66\)
Turbulent	\(Re \ge 2300\)	\(Sh = 0.023\,Re^{0.8}\,Sc^{0.4}\)

The Schmidt number for water vapour in air is \[ Sc = \frac{\mu_{\text{air}}}{\rho_{\text{air}}\,D_{\text{AB}}} \] with \(D_{\text{AB}}\) the binary diffusion coefficient.

Mass-transfer coefficient
\[ h_{m}= \frac{Sh\;D_{\text{AB}}}{H_{\text{gap}}} \]
Evaporation flux
\[ N_{\text{v}} = h_{m}\left(\rho_{\text{sat}}-\rho_{v,\infty}\right) \] Expressed per hour: \(N_{\text{v,h}}=3600\,N_{\text{v}}\).
Salt production rate
When the brine reaches the saturation mass fraction \(C_{\text{sat}}\) (kg salt per kg solution) the precipitated salt is \[ \dot{m}_{\text{salt}}=N_{\text{v,h}}\;A_{\text{pond}}\;\frac{C_{\text{sat}}}{1-C_{\text{sat}}} \] with \(A_{\text{pond}}=L_{\text{pond}}\,W_{\text{pond}}\). Daily output is obtained with a factor of 24 h d^-1.

The fundamental relation is: P = E · A · C · η
Where:

P = salt production rate (kg h⁻¹)
E = net evaporation rate (m h⁻¹) from pond water balance
A = active pond surface area (m²)
C = brine density at harvest concentration (kg m⁻³)
η = fractional harvest efficiency (0–1) accounting for losses

Measure E with a Class-A pan corrected by a pan coefficient (≈0.7) and adjust for rainfall and seepage. Typical η for well-managed ponds is 0.85–0.90.

Maintain the pond at 25.5–26.0 °Bé (≈295–305 g L⁻¹ NaCl) at 25 °C. Above 26.5 °Bé, supersaturation spikes and nighttime cooling cause secondary nucleation on liners and pumps, cutting harvest efficiency by up to 15 %. Use in-line density meters with ±0.1 °Bé accuracy and automate harvest gates at the set-point.

In tropical sites, every 10 % rise in relative humidity above 60 % reduces evaporation by 0.7 mm day⁻¹. Build a humidity-adjusted evaporation model:

Collect 10-year NOAA or local RH, wind, and solar data
Apply Penman-Monteith to generate monthly E factors
Schedule pond fill and harvest so that brine reaches target 26 °Bé during the lowest-humidity quarter
Keep a 30-day buffer capacity to absorb monsoon delays

This typically advances harvest by 2–3 weeks in arid years and delays it by up to 5 weeks in humid years.

Sulfate enters via feed brine and gypsum dissolution. Keep SO₄²⁻ below 5 g L⁻¹ at harvest by:

Blending high-sulfate intake brine with low-sulfate sources to achieve ≤3 g L⁻¹ before pond entry
Operating a 0.5 m deep pre-treatment settling pond with 24 h residence to precipitate CaSO₄ at 30–35 °C
Seeding with 1 kg m⁻³ of 20 µm gypsum crystals to accelerate nucleation
Monitoring weekly with ion chromatography; bleed 5 % pond volume when SO₄²⁻ >4 g L⁻¹

These steps raise final NaCl purity from 96 % to >99 % on a dry basis.

Worked Example: Estimating Daily Salt Harvest from a Laboratory-Scale Evaporative Pond

A process-development engineer is designing a 1 m² pilot pond to evaluate solar salt recovery from brine. The pond is housed in a low-speed wind tunnel that maintains 25 °C, 1 atm, and a gentle 0.15 m s^-1 air flow. The brine surface is 15 mm below the rim of the pond, and the tunnel humidity is kept near zero to maximise evaporation. Using the Chilton–Colburn analogy, determine the mass of NaCl that can be harvested per day if the brine is always at saturation.

Knowns

Air temperature, \(T_{\text{air}}\) = 25 °C
Atmospheric pressure, \(P_{\text{atm}}\) = 1.013 bar
Air velocity, \(U_{\infty}\) = 0.150 m s^-1
Pond length, \(L_{\text{pond}}\) = 1.000 m
Pond width, \(W_{\text{pond}}\) = 1.000 m
Air gap above brine, \(H_{\text{gap}}\) = 0.015 m
Relative humidity, RH = 0 %
Diffusion coefficient water vapour in air, \(D_{\text{AB}}\) = 2.5×10^-5 m² s^-1
Kinematic viscosity air, \(\nu_{\text{air}}\) = 1.562×10^-5 m² s^-1
Schmidt number, Sc = 0.610
Saturation concentration water vapour, \(C_{\text{sat}}\) = 0.264 kg m^-3
Density NaCl, \(\rho_{\text{salt}}\) = 2165 kg m^-3

Step-by-Step Calculation

Compute the Reynolds number for the gap: \[ Re = \frac{U_{\infty} H_{\text{gap}}}{\nu_{\text{air}}} = \frac{0.150 \times 0.015}{1.562 \times 10^{-5}} = 144 \]
Because \(Re < 2300\), the flow is laminar; for fully developed laminar mass transfer between parallel plates, the asymptotic Sherwood number is: \[ Sh = 3.66 \]
Calculate the average convective mass-transfer coefficient: \[ h_m = Sh \frac{D_{\text{AB}}}{H_{\text{gap}}} = 3.66 \frac{2.5 \times 10^{-5}}{0.015} = 0.0061 \; \text{m s}^{-1} \]
Determine the molar (mass) flux of water vapour: \[ N_v = h_m (C_{\text{sat}} - C_{\infty}) = 0.0061 (0.264 - 0) = 0.00161 \; \text{kg m}^{-2} \text{s}^{-1} \]
Convert to hourly rate: \[ N_{v,\text{h}} = 0.00161 \times 3600 = 5.796 \; \text{kg m}^{-2} \text{h}^{-1} \]
Compute the equivalent salt mass flux. At 25 °C, 1 kg of evaporated water precipitates 0.359 kg of NaCl: \[ \dot{m}_{\text{salt}} = 5.796 \times 0.359 = 2.08 \; \text{kg m}^{-2} \text{h}^{-1} \]
Project the daily harvest for the 1 m² pond: \[ \dot{m}_{\text{salt,day}} = 2.08 \times 24 = 49.9 \; \text{kg day}^{-1} \]

Final Answer

The pilot pond can produce approximately 49.9 kg of salt per square metre per day under the specified wind-tunnel conditions.

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle