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A person with a remote montain cabin plansto

install her own hydroelectric plant. A nearby stream is 3.00mwide
and 0.500m deep. Water floes at 1.20m/s over the brink of
awaterfall 5.00m high. The manufacturer promises only
25.0%efficiency in converting the PE of the water-Earth system
intoelectric energy. Find the power she can generate.

User AndrewP
by
8.5k points

1 Answer

6 votes

Answer:


P=22050\ W

Step-by-step explanation:

Given:

  • width of the stream,
    w=3\ m
  • depth of the stream,
    d=0.5\ m
  • height of the waterfall,
    h=5\ m
  • velocity of flow,
    v=1.2\ m.s^(-1)
  • efficiency promised by the manufacturer,
    \eta=25\%

Now the volume flow rate of the water through the fall:


\dot V=v* w* d


\dot V=1.2* 3* 0.5


\dot V=1.8\ m^3.s^(-1)

Now we know the density of water is 1000 kilogram per cubic meter.

So, the mass flow rate:


\dot m=\dot V* 1000


\dot m=1.8* 1000


\dot m=1800\ kg.s^(-1)

Hence potential energy delivered to the system per second:


\dot {PE}=\dot m* g* h


\dot {PE}=1800* 9.8* 5


\dot {PE}=88200\ J.s^(-1)

Now the power generated according to efficiency:


P=\dot{PE}* \eta


P=88200* 0.25


P=22050\ W

User Lakeia
by
7.7k points