EVAPORATION

Evaporation (E) is a type of vaporization of a liquid that occurs only on the surface of a liquid. The other type of vaporization is boiling, which, instead, occurs on the entire mass of the liquid. Evaporation is also part of the water cycle.

FACTORS INFLUENCING THE RATE OF EVAPORATION

• Concentration of the substance evaporating in the air
•  If the air already has a high concentration of the substance evaporating, then the given substance will evaporate more slowly.
• Flow rate of air
• This is in part related to the concentration points above. If fresh air is moving over the substance all the time, then the concentration of the substance in the air is less likely to go up with time, thus encouraging faster evaporation.
•  Pressure
• Evaporation happens faster if there is less exertion on the surface keeping the molecules from launching themselves.
• Surface area
• A substance that has a larger surface area will evaporate faster, as there are more surface molecules that are able to escape.
• Temperature of the substance
• If the substance is hotter, then its molecules have a higher average kinetic energy, and evaporation will be faster.
•  Density
• The higher the density the slower a liquid evaporates.

METHODS OF ESTIMATING EVAPORATION

•         Water budget methods

•         Energy budget methods

•         Mass transfer techniques (e.g., Meyer, Thornthwaile-Holzman)

•         Combination of energy budget and mass transfer methods (e.g.,Penman)

Method I: Water budget calculations

Approximation with Storage Equation Approach. Storage equation generally can be written as follow:

E = P + I – G – O ± ∆S

Where,

            ET – Evapotranspiration

            E – Evaporation

            R – Surface runoff

            G – Groundwater

P – Rainfall depth

I – Surface runoff that enter catchments area

O – Surface runoff out from catchments area

∆S – Change in stroge, above and below land surface

Example

The drainage area of the Sungai Limbang at Sarawak is 11839 km². If the mean annual runoff is determined to be 144.4 m³/s and the average annual rainfall is 1.08m, estimate the ET losses for the area. How does this compare with the lake evaporation of 1 m/year measured at Sungai Limbang, Sarawak.

Solution

Assuming that G = 0 and ∆S = 0, P = 1.08, R = 144.4 m³/s  

by using this equation:            ET = P – R

R = Runoff converted from m³/s à m/year

    = (144.4 x 86400 x 365) / (11839 x 10⁶)

    = 0.38 m

Therefore,        ET = 1.08 – 0.38

                              = 0.7 m < 1 m/year

Method II: Energy budget method

Q_N – Net radiation (cal/cm2-day)

Qe – Evaporation energy

Qh – sensible heat transfer (water heats the air)

Qv – Advected energy

Qo – Change in stored energy

Q_N = Qe + Qh – Qv + Q₀

Daily evaporation depth: E = Qe/pLe (cm/day)

Energy balance:           Q_N = Qe(1 + R) – Qv + Q₀

                                         = EpLe(1 + R) – Qv + Q₀

                                                                Or

                                      E = Q_N + Qv - Q₀ / pLe(1 + R) cm/day

With,

Q in [cal/cm²-day]

Le in [cal/g]

Ρ in [g/cm³]

Method III: Mass Transfer Technique

Evaporation driven by

•  Vapor pressure gradient
•  Wind speed

E = f(u)(e_s - e_a)

   = (a + bu)(e_s - e_a)

Where,

e_s: saturation vapor pressure at temperature above surface

e_a: vapor pressure at some level above surface

u: wind speed at some level above surface

a,b: empirical constants

Example

Using Meyer's formula

E = 0.0269 (1 + u_a/16)(e_s - e_a) cm/day

Determine lake evaporation for a month in which

·         average air temperature = 20°C,

·         average water temperature = 15 °C,

·         average wind speed at 8 m = 15 km/h

·         average relative humidity is 50%

Solution

Saturated vapor pressure above water surface

Air temperature above surface ≈ water temperature = 15 °C

Es = 2.7489 x 10⁸ (-4278.6 / (25 + 242.79))

     = 17.0 mb

Vapor pressure at height of 8m

es(20 °C) = 23.3 mb

e = RH es = 0.5 x 23,3 = 11.7 mb

by using this formula:

E = 0.0269 (1 + u_a/16)(e_s - e_a) cm/day

    = 0.0269 (1 + 25/16)(17.0 – 11.7) cm/day

    = 0.37 cm/day àcm/month

    = 11.1 cm/month

Method IV: Penman Method

Combined 'mass transfer' and 'energy budget':

EpLe = ∆ / ∆ + γ . Q_{N +}  γ / ∆ + γ . Ea

∆: slope of e_s vs t curve

Ea = pLe(a + bu)(e_sa - e_a) cal/cm²-day

where

a,b : empirical constants

e_sa: saturation vapor pressure at air temp.

e_a: actual vapor pressure

Example

Assume Meyer's formula applies to a lake:

E = 0.0269 (1 + 0.1u)(e_s - e_a) cm/day

Given:

Ta = 32.2°C

u = 32 km/h = 20 mi/h

RH = 30%

Q_N = 400 cal/cm²-day

Estimate daily evaporation using Penman's formula.

∆ = 2.7489 x 10⁸ x 4278.6 / (T + 242.79)² exp (-4278.6 / T + 242.79)

T = 32.2°C , ∆ = 2.72 mb/°C

Actual and saturation vapor pressure:

e_sa(T = 32.2°C) = 48.1 mb

ea = RH x e_sa = 0.3 x 48.1 = 14.4 mb

Latent heat of evaporation at air temperature:

Le = 597.3 - 0.57 x 579 cal/g

Ea = Ple 0.0269(1 + 0.1u)(e_sa - e_a)

     = 1 x 579 x 0.02691(1-0.1 x 20)(48.1 – 14.4)

     = 1590 cal/cm³-day

Penman’s Equation

EpLe = ∆ / ∆ + γ . Q_{N +}  γ / ∆ + γ . Ea

         = 2.72/(2.72 + 0.66) . 400 + 0.66/(2.72 + 0.66) . 1590

         = 632 cal/cm²-day

E = 632/(1 X 579)

   = 1.1 cm/day