Question

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.

Answer 1

`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`