advertisement

Evaporation ) The evaporation process ) Estimation of evaporation from water surface Water balance method Mass transfer method Energy budget method Combined method ) Potential evapotranspiration 1 Evaporation in hydrology Water supply reservoir – Loss of resources Soil moisture condition – Affect runoff conditions Continuous watershed simulation models – Overall water balance Atmospheric science Agricultural science - generation of precipitation - soil moisture available for plants 2 e Vapor pressure gradient es Saturated air Lake surface 3 Factors affecting evaporation 1. Availability of energy (latent heat) • • • Qe: Energy available for evaporation [cal/cm2-day] E : evaporation [cm/day] Le : latent heat of evaporation [cal/g] Qe = MLe [M : mass] = E × (1 cm2 )ρ Le E= Qe ρLe [cm / day] 2. Saturation deficit, es-e • Evaporation rate proportional to es-e E = C(es-e) 4 3. Temperature • Warm water will evaporate faster (less latent heat required) Le = 597.3 − 0.57 T • [cal/g], T in o C Warm air can hold more vapor 4. Wind • Removes saturated air and maintains vapor pressure gradient. Transpiration • Vegetation 5 Potential evapotranspiration Potential evapotranspiration from land surface ≈ Evaporation from open water surface Actual evapotranspiration depends on the dryness of the soil. 6 Method 1: Water budget Applicable to lake evaporation ∆storage = input − output ∆S = (I + P) − (O + E + GW) Or E = -∆S+I+P-O-GW I : inflow [cm] P : precipitation [cm] O: outflow [cm] E : Evaporation [cm] GW: Groundwater seepage [cm] Limitations: – Estimation of seepage (GW) – Estimation of precipitation 7 Mass transfer method Evaporation driven by – Vapor pressure gradient – Wind speed E = f(u)(es − ea ) = ( a + b u)(es − ea ) es : saturation vapor pressure at temperature above surface ea : vapor pressure at some level above surface u : wind speed at some level above surface a,b : empirical constants 8 Example Using Meyer's formula u E = 0.0269 1 + (e − e ) [cm / day] 16 [u in km / h, e in mb] 8 s a 8 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% 9 Solution Saturated vapor pressure above water surface Air temperature above surface ≈ water temperature = 15 °C 4278.6 es = 2.7489 × 108 exp − (T = 25) + 242.79 = 17.0 mb Vapor pressure at height of 8 m es(20 °C) = 23.3 mb e = RH es = 0.5×23.3 = 11.7 mb 10 u E = 0.0269 1 + (e − e ) 16 25 = 0.0269 1 + (17.0 − 11.7 ) 16 = 0.37 cm / day 8 s a = 11.1 cm / month 11 Energy budget method QN Qe Qh Qθ Qv QN : net radiation [cal/cm2-day] (solar radiation – reflection – radiation from lake) Qe : evaporation energy Qh : sensible heat transfer (water heats the air) Qv : advected energy Qθ : change in stored energy QN = Qe + Qh - Qv + Qθ 12 Energy budget method Sensible heat transfer difficult to measure Qh ≈ R × Qe where R=γ Ts − Ta es − ea [Eq. 1.16] Ta : air temperature [°C] Ts : water surface temperature [°C] ea : vapor pressure of the air [mb] es : saturation vapor pressure at water surface temp. [mb] γ : psychrometric constant = 0.66 (P/1000), P in mb 13 Energy budget method Daily evaporation depth: E = Energy balance Qe ρLe [cm / day] QN = Qe (1 + R) − Qv + Qθ = EρLe (1 + R) − Qv + Qθ or E = QN + Qv − Qθ ρLe (1 + R) with [cm / day] Q in [cal/cm2-day] Le in [cal/g] ρ in [g/cm3] 14 Combined method (Penman) Combined 'mass transfer' and 'energy budget': γ ∆ EρL = Q + E ∆+γ ∆+γ e N a [units as before – Eq. 1.17] ∆ : slope of es vs t curve (at air temperature - equation 1.18) Ea = ρLe ( a + bu)(esa − ea ) [cal / cm2 − day] where a,b : empirical constants esa : saturation vapor pressure at air temp. ea : actual vapor pressure 15 Example (textbook Ex. 1.5B) Assume Meyer's formula applies to a lake: E = 0.0269 (1+0.1 u)(es-ea) [cm/day] u in mi/h, e in mb Given: Ta = 32.2°C u = 32 km/h = 20 mi/h RH = 30% QN = 400 cal/cm2-day estimate daily evaporation using Penman's formula. 16 Solution 2.7489 × 108 × 4278.6 4278.6 exp − ∆= 2 (T + 242.79) T + 242.79 [eq. 1.18] with T = 32.2oC, ∆ = 2.72 mb /oC Actual and saturation vapor pressure: esa (T = 32.2o C) = 48.1 mb [eq. 1.6] ea = RH × esa = 0.3 × 48.1 = 14.4 mb Latent heat of evaporation at air temperature: Le = 597.3 − 0.57 × 32.2 = 579 cal / g [eq. 1.7] 17 Ea = ρLe 0.0269(1 + 0.1 u)(esa − ea ) = 1 × 579 × 0.0269(1 − 0.1 × 20)( 48.1 − 14.4) = 1590 cal/cm2 −day Penman's equation: ∆ γ EρLe = QN + Ea ∆+γ ∆+γ 0.66 2.72 = 1590 400 + 2.72 + 0.66 2.72 + 0.66 = 632 cal/cm2 − day Or: E= 632 = 1.1 cm/day 1 × 579 18