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How Hydro power Plants Work
Hydropower plants use this same principle of
bottlenecking the water in order to harness the water’s
energy. By using some simple mechanics they can convert the
energy that’s flowing into electricity. This is done by
water flowing through a dam which turns a turbine, which
then turns a generator, producing electricity.
The Basic Components
of a Conventional Hydropower Plant:
http://www.micro-hydro-power.com/Hydro-Power.htm
Dam - Most
hydropower plants rely on a dam that holds back water,
creating a large reservoir. Often, this reservoir is used as
a recreational lake, such as Lake Roosevelt at the Grand
Coulee Dam in Washington State.
Intake - Gates on
the dam open and gravity pulls the water through the
penstock, a pipeline that leads to the turbine. Water builds
up pressure as it flows through this pipe.
Turbine - The
water strikes and turns the large blades of a turbine, which
is attached to a generator above it by way of a shaft. The
most common type of turbine for hydropower plants is the
Francis Turbine, which looks like a big disc with curved
blades. A turbine can weigh as much as 172 tons and turn at
a rate of 90 revolutions per minute (rpm), according to the
Foundation for Water & Energy Education (FWEE).
Generators - As
the turbine blades turn, so do a series of magnets inside
the generator. Giant magnets rotate past copper coils,
producing alternating current (AC) by moving electrons.
(You'll learn more about how the generator works later.)
Transformer - The
transformer inside the powerhouse takes the AC and converts
it to higher-voltage current.
Power lines - Out
of every power plant come four wires: the three phases of
power being produced simultaneously plus a neutral or ground
common to all three. (Read How Power Distribution Grids Work
to learn more about power line transmission.)
Outflow - Used
water is carried through pipelines, called tailraces, and
re-enters the river downstream.
As streams and rivers
flow down mountain sides they feed into large water in the
reservoirs. The water reservoir is now considered to be the
potential energy for creating the hyrdro power.
When the gates open, the water than begins to flow through
the penstock where the energy is no longer still but it is
in motion, the water is now in the form of kinetic
energy. To determine the amount of
electricity that is generated, there are several factors
that need to be accounted for. The two main factors are the
volume of water flow
and the amount of hydraulic
head.
“The head refers to the distance between the water surface
and the turbines. As the head and flow increase, so does the
electricity generated. The head is usually dependent upon
the amount of water in the reservoir” (Bonsor).
Power = Gravity*Water
Flow Rate*Head Height
We are familiar with
gravity of course,but not flow rate and head height. The
hydraulic head is the distance that water flows from the
higher elevation where it is stored, to the hydro-turbine
generator. Hydraulic head is usually divided into three
categories: low, medium, and high head.
Elevations above
around 100 meters are usually considered high head, while
low head system elevations range around 10 meters in height.
A high head hydro turbine system requires less volume of
water flow to operate because the momentum collected from
gravity through the longer distance of falling water makes
up for the loss of volume.
The high head systems
also require a smaller turbine because of the lower volume
of water flow needed.
A low head hydro
turbine is usually used in a flowing river with little
elevation change, or in moving ocean tides.
A low head system
with a high volume of water usually requires a much larger
turbine generator to efficiently convert the water energy
into electricity.
The Hydroelectric Dam Problem
Wolfe, Joe. "Work, Energy, and
Power." Http://www.animations.physics.unsw.edu.au/jw/work.htm#dam. Science UNSW,
2011. Web.
The water level in a hydroelectric dam is 100 m above the height at which water comes out of the pipes. Assuming that the turbines and generators are 100% efficient, and neglecting viscosity and turbulence, calculate the flow of water required to produce 10 MW of power. The output pipes have a cross section of 5 m2. This problem has the work-energy theorem, uses power, and requires a bit of thought. Let's do it.
Let's consider what
is happening in steady state for this system. Over a time
dt, some water of mass dm exits the lower pipe at speed v.
This water is delivered to the top of the dam at negligible
speed. So the nett effect is to take dm of stationary water
at height h and deliver it at the bottom of the dam at
height zero and speed v. Looks straightforward. Let's go.
Let the flow be
dm/dt. The work done by the water, dW, is minus the energy
increase of the water, so
dW = − dE = − dK − dU
= − (½dm.v2 − 0) −
(0 − dm.gh) = dm(gh − ½v2)
The power delivered
is just P = dW/dt. so
P = (gh − ½v2)dm/dt
Of course the flow
dm/dt depends on v. Let's see how: In time dt, the water
flows a distance vdt along the pipe. The cross section of
the pipe is A, so the volume of water that has passed a
given point is dV = A(vdt). Using the definition of density,
ρ = dm/dV, we have
dm/dt = ρdV/dt = ρA.(vdt)/dt
= ρAv.
Substituting in the equation above
gives us
P = ρAv(gh −
½v2) or
½v3 − ghv + P/ρA = 0.
However you look at
it, it's a cubic equation, which sounds like a messy
solution. However, let's think of what the terms mean. The
first one came from the kinetic energy term. The second is
the work done by gravity. The third is the work done on the
turbines. Now, if I had designed this dam, I'd have wanted
to convert as much gravitational potential energy as
possible into work done on the turbines, so I'd make the
pipes wide enough so that the kinetic energy lost by the
water outflow would be negligible. Let's see if my guess is
correct.
If the first term is
negligible, then we simply have hgv = P/ρA. So v = P/ρghA =
2 m.s−1. So the first term would be 4 m3.s−3, the second
would be − 2000 m3.s−3 and the third would be 2000 m3.s−3.
So yes, the guess was correct and, to the precision required
of this problem, the answer is v = 2 m.s−1.
Following Problem was
performed by Professor Joe Wolfe, UNSW