Question

Consider the following data of evapotranspiration from an area near New Delhi in the month of November :
1. Slope of the saturation vapour pressure vs temperature curve at the mean air temperature = 1.00 mm of Hg/ ${}^{\circ }$C.
2. Net radiation in mm of evaporable water/day = 2.
3. Parameter including wind velocity and saturation deficit = 2.5 mm/day.
4. Psychrometric constant = 0.5 mm of mercury/ ${}^{\circ }$C.
Using Penman's formula, the potential evapotranspiration (PET) will be

• A. 1.8 mm/day
• B. 2.17 mm/day
• C. 2.24 mm/day
• D. 2.32 mm/day

Answer 1

The correct option is B 2.17 mm/day

$PET=\frac{A{H}_n+E_a\gamma }{A+\gamma }$

where,
$A=1\text{mm of Hg}{/}^{\circ }C$
$H_n=2mm\text{of evaporable water day}$
$E_1=2.5mm/day$
$\gamma =0.5mm\text{of Hg/}$ ${}^{\circ }C$

$\therefore PET=\frac{1\times 2+2.5\times 0.5}{1+0.5}$

$=\frac{2+1.25}{1.5}=\frac{3.25}{1.5}=2.17mm/day$