5. Determine the average daily evaporation from a lake using the following data

Question

5. Determine the average daily evaporation from a lake using the following data from sensor situated at 2 m
over the lake surface:
• Average wind speed: 2.16 ms 1
• Air pressure, P: 101,100 P a
• Density of air a: 1.24 kgm 3
• Vapor pressure at the sensor e: 2.58 kP a
• Air temperature at z0: 28 degC
hint: vapor pressure is at the saturation point at z0
Be careful with units, make sure they are all coherent. Also, the following information is made available to
you:
• Roughness height, z0: 0.0015 m
• Zero-plane displacement, zd: 0.0 m
Give the daily evaporation E in mm/day (20% of grade)
Note: we use the convention that fluxes are negative when they are pointed away from the surface (i.e. the
surface is losing water or energy)
What is the latent he at flux in the previous example? G ive the flux in Wm 2 (15% of grade)
Note: we use the convention that fluxes are negative when they are pointed away from the surface (i.e. the
surface is losing water or energy)

Explanation / Answer

(i) Evaporation is a type of vaporization of a liquid that occurs from the surface of a liquid into a gaseous phase that is not saturated with the evaporating substance. The other type of vaporization is boiling, which is characterized by bubbles of saturated vapor forming in the liquid phase. Steam produced in a boiler is another example of evaporation occurring in a saturated vapor phase. Evaporation that occurs directly from the solid phase below the melting point, as commonly observed with ice at or below freezing or moth crystals (napthalene or paradichlorobenzene), is called sublimation.

Given data is,

sensor situated at 2 m
over the lake surface:
• Average wind speed: 2.16 ms 1
• Air pressure, P: 101,100 P a
• Density of air a: 1.24 kgm 3
• Vapor pressure at the sensor e: 2.58 kP a
• Air temperature at z0: 28 degC
hint: vapor pressure is at the saturation point at z0 and

• Roughness height, z0: 0.0015 m
• Zero-plane displacement, zd: 0.0 m

Empirical evaporation Equations

Meyer’s Formula :

EL=KM (ew–ea)(1+u9/16)

In which,

u9=monthly Mean wind velocity in km/h at about 9m above ground

And

KM=coefficient accounting for various other factors with

a value of 0.36 for large deep and 0.50 For small shallow waters.

The wind velocity can be assumed to follow the 1/7 power law

Uh= C h 1/7

Where, Uh= wind velocity at a height h above the ground and

C = constant.

This equation can be used to determine the velocity at any desired level.

Where as here, EL=KM (ew–ea)(1+u9/16)

Km=0.36 for large deep

ew=2.58 Kpa and ea=101100 Pa=10.11 Kpa

Uh=Ch(1/7) here U1=2.16 m/s=7.76km/h

Uh=C(1)1/7

U9/U1=C(9)1/7/C(1)1/7=U9=U1*(9)1/7                         (1/7)=0.142

U9= (7.76) (9)0.142=(7.76) (1.36) = 10.55 km/h

Now substitute all values into eq EL=KM (ew–ea)(1+u9/16)

=0.36(10.11-2.58)(1+10.55/7.76)

=0.36(7.53)(2.35)

the average daily evaporation from a lake EL= 6.396 mm/day

(ii) Latent heat flux is the flux of heat from the Earth's surface to the atmosphere that is associated with evaporation of water at the surface and subsequent condensation of water vapor in the troposphere. It is an important component of Earth's surface energy budget. Latent heat flux is commonly measured with the Bowen ratio technique, or by eddy covariance.

Latent heat flux Hl= lvEL

Water flux =Evaporation rate EL (mm/day)

Energy flux= Latent heat flux Hl (W/m2)

Here =1.24 kg/m3 , assume lv= 2.5*106 J/Kg and EL= 6.396 mm/day

Now Latent heat flux Hl= lvEL=(1.24)(2.5*106)(6.396)=19.827*106 W/m2.

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