Question

3. The temperature and relative humidity average for the month of March have the following values: Monthly average mean air temperature at 4m-7.2 °C Monthly average maximum, lake temperature-4.4°C Monthly average maximum pan temperature 5.5 oC Monthly average relative humidity-70% During the month, the evaporation measured from the pan was 1.3 cm. Determine the monthly evaporation from the reservoir by a. The simple formula, assuming that K-0.68 b. The revised formula, assuming that K'-1.5 4. Solve #3 using the aerodynamic method for evaporation at 4 m. The wind speed is 2.4 m/s at 4 m level and zo-0.02 cm. Assume the mean temperature near the surface of the lake to be 85% of the Monthly average maximum lake temperature. Solve #3 using the energy balance method. The average net radiation is 22 win. Disregard the soil heat index 5. 6. Solve #3 using the combination method.

Answer 1

Ans 3 a) Given,

evaporation from pan (E_p) = 1.3 cm

K = 0.68

Hence, monthly evaporation(E) using simple method = E_p x K

E = 1.3 x 0.68

E = 0.884 cm

Ans3 b) Using revised formula,

saturation vapor pressure(E_sp) at lake temperature 4.4 deg C is,

E_sl = 0.834 kPa (can be found from standard table)

Saturation vapor pressure at maximum pan temperature at 5.5 deg C ,

E_sp = 0.899 kPa (from standard table)

saturation vapor pressure at mean air temperature at 7.2 deg C

E_sm = 1.015 kPa (from standard table)

Relative humidity = 70%

Vapor pressure of air(E_z) = (70/100) x 1.015

= 0.7105 kPa

Revised formula can be given by,

E_L = [K' (E_sl - E_z) / (E_sp - E_z ) ] x E_p

= [1.5(0.834 - 0.7105) / (0.899 - 0.7105)] x 1.3

= 1.277 cm

Ans 4) Evaporation using aerodynamics method,

Mean lake temperature = 4.4 x 0.85

= 3.74 deg C

Saturation vapor pressure at 3.74 deg C is,

E_s = 0.799 kPa

Air density at 7.2 deg C($\rho$_a) is 1.260 kg/m³

standard atmospheric temperature (P)= 101.325 kPa

Bulk evaporation coefficient, C = 0.0014

Mass transfer coefficient, M = 0.622 $\rho$_aC/($\rho$_w P)

Putting values,

M = 0.622 (1.26 x 0.0014)/(1000x101.325)

= 1.08 x 10^-8 kPa^-1

Velocity of wind at 4m height ,U2= U1 x [ln(Z₂/Z₀) / ln (Z₁/ Z₀)]

Putting values,

U2 = 2.4 x [ln(400/0.02)/ ln(400/0.02)]

= 2.4 m/s

Now, E_a = M x (E_s - E_z) x U2

= 1.08 x 10^-8 x (0.799 - 0.7105) x 2.4

= 2.3 x 10^-9 m/s

or 2.3 x 10^-6 mm/s

or 0.198 mm /day

Therefore, using aerodynamics method ,daily evaporation rate is 0.198 mm/day

So, monthly evaporation rate = 31 x 0.198

= 6.138 mm or 0.6138 cm

Ans 5) Evaporation rate using energy balance method,

Given average radiation(Rn) = 22W/m²

E_r = Rn - G / ($\rho$w $\lambda$ (1 + B)

where, G = ground heat flux (zero because ground flux neglected)

$\lambda$ = latent heat of vaporization ( 2484 kJ/kg at 7.2 deg celsius)

B = Bowen ratio (zero because sensible heat flux neglected)  

Putting values,

E_r = 22 - 0 / (1000 x 2484000)

= 8.85 x 10^-9 m/s

or 8.85 x 10^-6 mm/s

or 0.765 mm/day

So, monthly evaporation rate = 0.765 x 31

= 23.715 mm

or 2.37 cm

Ans 6) Using combination method,

We know,

E = $\Delta$ E_r/ ($\Delta$ + $\gamma$ ) + $\gamma$ E_a / ($\Delta$ + $\gamma$ )

where, E_r = evaporation by energy balance method

E_a = evaporation by aerodynamic method

  $\gamma$ = Psychometric constant ( 0.0668 kPa / C)

$\Delta$ = gradient saturated vapor pressure (0.153 kPa/C)

Putting values,

E = 0.153(23.715)/ (0.153 + 0.0668) + 0.0668(6.138)/(0.153 + 0.0668)

= 0.264 mm /day

For a month, evaporation rate = 0.264 x 31

= 8.184 mm or 0.818 cm