Civil Engineering - Water Resources Engineering - Discussion
Discussion Forum : Water Resources Engineering - Section 3 (Q.No. 49)
49.
In estimating the rate of evaporation from the reservoir surface, a pan 1.5 metres in diameter, was filled upto 8.0 cm. During a specified period of time, the rainfall recorded was 5 cm. 3 cm of water was removed from the pan to keep the depth of water. At the end of the time, the depth was 9 cm. If the pan coefficient is 0.6, the evaporation loss is
Discussion:
5 comments Page 1 of 1.
Bidyut Barman said:
8 years ago
Initial depth of water = 8 cm.
Rainfall= 5 cm,
Water removed= 3 cm,
Net addition water in pan= Rainfall-Water removed =5-3 = 2 cm,
Depth of the end time = Initial depth+Net addition water=8+2=10 cm,
Actual depth end time =Depth of end time - Initial depth of water =10-9=1cm,
Evaporation losses from the reservoir= Panchanan Coefficient x Initial depth of water =0.6x1=0.6 cm=6mm.
Rainfall= 5 cm,
Water removed= 3 cm,
Net addition water in pan= Rainfall-Water removed =5-3 = 2 cm,
Depth of the end time = Initial depth+Net addition water=8+2=10 cm,
Actual depth end time =Depth of end time - Initial depth of water =10-9=1cm,
Evaporation losses from the reservoir= Panchanan Coefficient x Initial depth of water =0.6x1=0.6 cm=6mm.
(3)
Rdx said:
4 years ago
Water evaporated = 8 + 5 - 3 - 9 = 1cm.
Evaporation loss = 0.6 x 1 = 0.6cm = 6mm.
Evaporation loss = 0.6 x 1 = 0.6cm = 6mm.
(2)
Latha said:
8 years ago
Pan evaporation=8-5-3-9 = 1cm.
Evaporation loss = 0.6 x 1 = 0.6cm = 6mm.
Evaporation loss = 0.6 x 1 = 0.6cm = 6mm.
Wokneh said:
3 years ago
Thanks, everyone for explaining the answer.
Shahul said:
7 years ago
Thanks for the explanation.
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