The power of the dam is
P=ρghQ,where Q - flow of water in m3/s.
Assume that the battery has V=12 V,q=50 A⋅h. Its power will be
P=tqV.Equate:
ρghQ=3600 sqV,the flow of water will be
Q=3600ρghqV=3600⋅1000⋅9.8⋅0.550⋅12=3600 s0.12 m3, the total volume for 1 hour is
v=Q⋅3600=0.12 m3.
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