Question

A standard, ground-based evaporation pan, corresponding to Indian Standards, is installed at the banks of a surface reservoir. The water surface area on a particular day is 100 hectares. The recorded evaporation loss from the pan, on a certain day, is nearly 4.0 cm. What is the anticipated evaporation loss from the reservoir for that day?

• A. $\left(1.8to2\right)\times {10}^4{m}^3\text{per day}$
• B. $\left(2.5to2.75\right)\times {10}^4{m}^3\text{per day}$
• C. $\left(3.0to3.25\right)\times {10}^4{m}^3\text{per day}$
• D. $\left(3.8to4.05\right)\times {10}^4{m}^3\text{per day}$

Answer 1

The correct option is C $\left(3.0to3.25\right)\times {10}^4{m}^3\text{per day}$

Evaporation from reservoir
$=C_P\times \text{pan evaporation}$

where $C_P=\text{pan coefficient}=0.8\text{(Average value) for ISI standard pan}$
$\text{for I.S Pan,}{C}_p=0.65to1.1$

$\therefore \text{Depth of evaporation}$
$=0.8\times 4\times {10}^{-2}\text{m per day}$

The total volume of evaporation
$=\text{Depth of evaporation}\times \text{water surface area}$
$=0.8\times 4\times {10}^{-2}\times 100\times {10}^4{m}^3\text{per day}$
$=3.2\times {10}^4{m}^3\text{per day}$