Evaporation and transpiration

Dr. Huidae Cho

Department of Civil Engineering...New Mexico State University

1 Definitions

Evaporation is

the process by which water is transferred from the land and water masses of the earth to the atmosphere Viessman and Lewis (2003)

Significant over large water bodies
Important for the design and operation of reservoirs

Transpiration is

the process by which soil moisture taken up by vegetation is eventually evaporated as it exits at plant pores Viessman and Lewis (2003)

Significant over heavily vegetated areas
Particular concern in agricultural production

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2 Statistics

Both combined are referred to as evapotranspiration, which is the largest component of losses in hydrology.

On average, about 40,000 billion gallons per day (bgd) of water vapor across the continental United States (CONUS).

Approximately 10 percent (about 4,200 bgd) is precipitated.

About two-thirds (2,750 bgd) of the precipitated amount is evapotranspirated.

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3 Evaporation

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3.1 Methods for estimating evaporation

Water budget: Inexpensive, common
Energy budget, Mass transfer: Quite expensive, costly to maintain observations
Pan: Least expensive, common, good for estimating annual evaporation

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3.2 Water budget method

\begin{equation} E_s=P+R_1-R_2+R_g-T_s-I-\Delta S_s \end{equation}

Defining the net seepage $O_s=R_g-I$ and assuming $T_s=0$, \begin{equation} E_s=P+R_1-R_2+O_s-\Delta S_s \label{eq:E_s} \end{equation}

$E_s$: Surface evaporation
$P$: Precipitation
$R_1$: Surface inflow
$R_2$: Surface outflow
$R_g$: Surface water returning from groundwater
$T_s$: Surface transpiration
$I$: Infiltration
$\Delta S_s$: Change in surface water storage

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3.2.1 Mass transfer product

Assume that evaporation is proportional to the mass transfer product $u(e_0-e_a)$.

$u$: Wind velocity
$e_0$: Saturated vapor pressure (vapor pressure at an equilibrium state; read more) at the water surface temperature (Table A.2)
$e_a$: Vapor pressure of the air (Table A.2)

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3.2.2 Net seepage determination

During no surface inflow and outflow ($R_1=R_2=0$), evaporation ($E_s$) and net seepage ($O_s$) are the only losses from Eq. (\ref{eq:E_s}).

Under these conditions during dry days ($P=0$), whenever the mass transfer product is 0 ($E_s=0$), the change in storage is equivalent to the net seepage ($O_s=\Delta S_s$).Figure 6.7 or Table A.2

Daily plot of change in elevation vs. mass transfer product ⇒ Intercept of the best-fit line ($E_s\propto u(e_0-e_a)=0$) ⇒ Net seepage $O_s$

However, the net seepage is a function of the reservoir stage and the season, and these effects need to be considered as well in many cases.

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3.3 Mass transfer method

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3.3.1 Meyer’s equation

Meyer (1944) developed a commonly used empirical equation: \begin{equation} E=C(e_0-e_a)\left(1+\frac{W}{10}\right) \end{equation}

$E$: Evaporation rate in in/day
$W$: Wind velocity in mph measured about 25 ft above the water surface
$C$: Pan empirical coefficientFigure 6.7 or Table A.2
- About 0.36 for an ordinary lake
- About 0.50 for wet soil surfaces, small puddles, and shallow pans

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3.3.2 Dunne’s equation

Dunne (1978) developed another equation: \begin{equation} E=(0.013+0.00016u_2)e_a\frac{100-R_h}{100} \end{equation}

$E$: Evaporation rate in cm/day
$u_2$: Wind velocity measured at 2 m above the surface in km/day
$e_a$: Vapor pressure of the air in millibars
$R_h$: Relative humidity in percent

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3.3.3 Example 6.1

Using the Meyer and Dunne equations, find the daily evaporation rate for a lake given that the mean value for air temperature was 87$^\circ$F, the mean value for water temperature was 63$^\circ$F, the average wind speed was 10 mph, and the relative humidity was 20%. Refer to Table A.2 for vapor pressure values.

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3.3.4 Example 6.2

Consider that the lake of Example 6.1 has a surface area of 1.5 mi². (a) If the average annual evaporation rate is estimated to be $\frac{2}{3}$ of the average daily rate calculated in Example 6.1, what volume in million gallons per day (mgd) and cubic meters per day would be lost to evaporation? (b) If the average water use of an urban community is 180 gallons per capita per day (gpcd), how large a community would the daily evaporation rate sustain?

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3.3.5 USGS (1954) equation

\begin{equation} E=Nu(e_0-e_a) \end{equation}

$E$: Evaporation rate in cm/day
$N$: Coefficient
$u$: Wind velocity at 2 m above the water surface in m/sec
$e_0$, $e_a$: In millibars

$N$ can be approximated using \begin{equation} N=\frac{0.0291}{A^{0.05}} \end{equation}

$A$: Surface area of the water in m²

For $A<4\times 10^6\text{m}^2$, the above equation should be used with caution.

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3.4 Evaporation pan method

Most widely used.

National Weather Service Class A Evaporation Pan

Image source

Assuming that air and water temperatures are the same, the daily pan evaporation in mm/day is \begin{equation} E_a=(e_0-e_a)^{0.88}(0.42+0.0029u_p) \end{equation} where $u_p$ is the wind movement 150 mm above the rim of the pan in km/day.

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3.4.1 Daily evaporation

Penman derived the daily evaporation equation: \begin{equation} E=\frac{\Delta Q_n+\gamma E_a}{\Delta+\gamma} \end{equation}

$\Delta$: Slope of the saturated vapor pressure vs. temperature curve at the air temperature $T_a$
$Q_n$: Net radiant energy in the same units as those of $E$
$\gamma=\frac{0.61p}{1000}$
$p$: Atmospheric pressure in mb

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3.4.2 Average daily lake evaporation

Assumptions:

Any net advection* into the lake is balanced by the change in energy storage
The net transfer of sensible heat through the pan is negligible
The pan exposure is representative

\begin{equation} E_L=0.70\frac{\Delta Q_n+\gamma E_a}{\Delta+\gamma} \end{equation} or \begin{equation} E_L=0.70\left[E_a+0.00051P\alpha_p(0.37+0.0041u_p)(T_0-T_a)^{0.88}\right] \end{equation}

$\alpha_p$: Proportion of advected energy used for evaporation
$T_0$: Outerface temperature of the pan in degrees Fahrenheit
$T_a$: Air temperature in degrees Fahrenheit
$u_p$: Wind velocity in mi/day
$P$: Atmospheric pressure in in Hg

\begin{equation} \alpha_p=0.13+0.0065T_0-(6.0\times 10^{-8}T_0^3)+0.016u_p^{0.36} \end{equation} or graphical method using Figures 6.3 and 6.4

*Advection: Horizontal transfer of any atmospheric property by the wind

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4 Transpiration

Seasonal consumptive use for a particular crop can be estimated using \begin{equation} U=k_sB \end{equation}

$U$: Consumptive use of water during the growing season in in
$k_s$: Seasonal consumptive use coefficient (Table 6.3)
$B=\sum\frac{tp}{100}$: Summation of monthly consumptive use factors for a given season
$t$: Mean monthly temperature in degrees Fahrenheit
$p$: Monthly daytime hours given as percentage of the year (Table 6.4)

Given the monthly consumptive use coefficient $k$, the monthly consumptive use can be estimated using \begin{equation} u=\frac{ktp}{100} \end{equation}

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4.1 Example 6.3

Determine the monthly consumptive use of an alfalfa crop grown in southern California for the month of July if the average monthly temperature is 72$^\circ$F, the average value of daytime hours in percentage of the year is 9.88, and the mean monthly consumptive use coefficient for alfalfa is 0.85.

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4.2 Example 6.4

Determine the seasonal consumptive use of a tomato crop grown in New Jersey (a humid area) if the mean monthly temperatures for May, June, July, and August are 61.6, 70.3, 75.1, and 73.4$^\circ$F, respectively, and the percent daylight hours for the given months are 10.02, 10.08, 10.22, and 9.54 as percent of the year, respectively.

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5 Evapotranspiration

Potential evapotranspiration is

the water loss which will occur if at no time there is a deficiency of water in the soil for the use of vegetation Thornthwaite

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5.1 ARS equation

The Agricultural Research Service (ARS) estimates potential evapotranspiration using \begin{equation} \text{ET}=\text{GI}\times k\times E_p\times\left(\frac{S-\text{SA}}{\text{AWC}}\right)^\frac{\text{AWC}}{G} \end{equation}

$\text{ET}$: Potential evapotranspiration in in/day
$\text{GI}=\frac{\text{ET}}{\text{ET}_\text{max}}$: Growth index of crop in percentage of maturity
$k$: Ratio of $\text{GI}$ to pan evaporation; 1.0–1.2 for short grasses, 1.2–1.6 for crops up to shoulder height, 1.6–2.0 for forest
$E_p$: Pan evaporation in in/day
$S$: Total porosity
$\text{SA}$: Available porosity unfilled by water
$G$: Moisture freely drained by gravity
$\text{AWC}=S-G$: Available water capacity

Use Table 6.5 for $S$, $G$, and $\text{AWC}$.

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5.2 Penman method

Widely used method that combines the mass transport and energy budget theories.

\begin{equation} \text{ET}=\frac{\Delta H+0.27E}{\Delta+0.27} \end{equation}

$\text{ET}$: Evapotranspiration or consumptive use for a given period in mm/day
$\Delta$: Slope of the saturated vapor pressure vs. temperature curve at the absolute temperature in mm Hg/$^\circ$F (Figure 6.8)
$H$: Daily heat budget at the surface (estimate of net radiation) in mm/day
$E$: Daily evaporation in mm

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5.2.1 Estimation of $E$ and $H$

\begin{equation} E=0.35(e_a-e_d)(1+0.0098u_2) \end{equation}

$e_a$: Saturated vapor pressure at the mean air temperature in mm Hg (Figure 6.7 or Table A.2)
$e_d$: Saturated vapor pressure at the mean dew point (actual vapor pressure in the air) in mm Hg ($e_a\times\text{Relative Humidity}$)
$u_s$: Mean wind speed at 2 m above the ground in mi/day

\begin{equation} H=R(1-r)(0.18+0.55S)-B(0.56-0.092e_d^{0.5})(0.10+0.90S) \end{equation}

$R$: Mean monthly extraterrestrial radiation in mm H₂O evaporated per day (Table 6.6)
$r$: Estimated percentage of reflecting surface
$B$: Temperature-dependent coefficient (Table 6.7)
$S$: Estimated ratio of actual duration of bright sunshine to maximum possible duration of bright sunshine

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5.3 Example 6.5

Using the Penman method, estimate ET given the following data: temperature at water surface is 22$^\circ$C, temperature of air is 33$^\circ$C, relative humidity is 45%, wind velocity is 1.5 mph. The month is June at latitude 33$^\circ$ north, $r$ is given as 0.07 and $S$ is found to be 0.70.

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6 Homework: Evaporation and transpiration

Problem 6.2
Problem 6.3
Problem 6.4

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