Hydro Power
Power Calculations
This section is still under
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1.0 Fundamentals
To allow a thorough understanding of the issues involved in specifying
and sizing the hardware required for a successful hydro power installation
the following chapter takes you through the basic laws of physics which
have to be obeyed if we are to generate any electricity from our installation.
Hydro Power is essentially based upon the fundamental laws of
physics which state that a 'body' may contain energy (amongst other means)
by virtue of its velocity (through 'space') and/or by its relative height.
These are termed Kinetic and Potential Energy respectively. Now the
first law of thermodynamics states that 'Energy cannot be created of destroyed,
but it's form may be changed'. So knowing this we can already see
that we may convert the 'Kinetic Energy' of fast moving water in a stream,
or the 'Potential Energy' of a small pond high up a hillside to some other
type of energy such as electricity !
So how much energy does the water contain ??
The potential Energy of a body (water in our case) is proportional
to its relative height and its mass, and may be expressed by the following
equation :
P.E. = m x g x h Joules (J)
Where :
m = mass (kg)
g = acceleration due to gravity (9.81m/s2)
h = relative height (m)
The Kinetic Energy of a body may be expressed by the following equation
:
K.E. = ½m x v2 Joules (J)
Where :
m = mass (kg)
v = velocity (m/s)
We can work out h (See chapter X for ways of determining this) but
how do we know mass ? Well we can work out the volume flowing in
a period of time (see chapter Y for ways of determining this) and given
that:
Where :
r = Density of Water (kg/litre)
Vol = Volume (litre)
(Note that the actual volume units aren't importand, but that they
MUST be consistent with those used for density. If Specific Gravity
is used, then volume needs to be in litres.)
So the previous equations can now be written as :
P.E. = Vol x r x
g x h Joules (J)
K.E. = ½ x (Vol x r)
x
v2 Joules (J)
However this is simply the energy in a volume of water, given that
we've determined what volume flows in a given period of time (in our case,
one second) these equations now become :
P.E. = (Vol / sec) x r
x
g x h Joules/sec (J/s)
K.E. = ½((Vol / sec) x r)
x
v2 Joules/sec (J/s)
Now, as 1 Joule per second = 1 Watt, the above two equations actually
give us power in Watts, which many of us are much more familiar with.
So now that we understand the differences between potential and
kinetic energy, and how to calculate their values, for a given system we
finally know how many Watts of power, i.e. electricity, we will generate,
right ? Well no, not quite.....!
Yes we know exactly how much energy is in the water at a point
in time & space, and this as a very first check will indicate how much
energy is available from the water, but in reality not all of this is available
for conversion, e.g. in a turbine water needs some velocity on exit otherwise
it wouldn't get out of the turbine ! Also for a turbine fed from
a pipeline such as a pelton wheel, there will be losses in the pipeline
giving a lower pressure (than that provided by static head alone) at nozzle
the turbine. So even on a very well designed installation using a
'high efficiency' turbine, transmission & generator, we may still lose
some 50% of the initial energy by the time we have generated volts &
amps. In some installations, such as flow of river undershot water
wheels, overall efficiency may be as low as 5% or 10%, ultimately it all
depends upon the technology being used.
To provide an accurate estimate of how much electricity we will
actually generate requires several other factors to be taken into consideration.
These will be detailed in the next section and as you will see, depending
upon the hardware that is actually used (some of which will be dictated
by site conditions) different factors need to be considered.
2.0 Extracting the Energy (NOTE:
This and following sections not yet complete.)
2.1.1 Water Wheels - Overshot/Breastshot
2.1.2 Water Wheels - Undershot
2.2.1 Turbine - Impulse Devices
2.2.3 Turbine - Reaction Devices
2.2.3.1 Run of River Water Current Turbine - i.e. Submerged
Propeller
This category covers devices which typically float on a tethered
pontoon of some sort.
We can expand the Kinetc Energy equation further to account
for the swept area of the propeller blades: Assuming that the turbine
is perpendicular to the direction of flow of the river, (in both planes),
the volume of water passing through the blades of the turbine, of swept
area A (m2), per second, is dependent upon the velocity
of the river, v m/s. In other words:
Vol / sec (litres/sec) = A x v (m3/sec)
x 1,000 (litres/m3)
which gives us ;
Vol / sec = A x v x 1000 ltr/sec
Incorporating this into the Kinetic Equation ;
K.E. = ½((Vol / sec) x r)
x
v2 Watts (W)
Gives us ;
K.E. = ½((A x v x 1,000 ) x r)
x
v2 Watts (W)
Simplyfying further, to get ;
K.E. = ½ x A x 1,000 x r
x
v3 Watts (W)
Now, in a run of river device, not all of this energy is available
for extraction, as the water cannot simply stop when it leaves the turbine,
it needs to posses some velocity (and hence energy) to move away from the
turbine exit. We also have inefficiencies in the turbine itself,
denoted by its coefficient of performance, Cp.
(It can be shown theoretically that the maximum coefficient of performance
for an unshrouded reaction device in a run of river configuration, is 0.59.)
The above equation, assumes that the turbine is perpendicular
to the current flow. This may or may not be the case depending upon
turbine design, so to allow for this we need to introduce a further factor:
Swept Area, A (m2) = pi x d2/4
Area presented to the flow of water, when no longer perpendicular to it
Area perpendicular to the flow, Ap = pi x
d2/4
x Cos phi
Where :
d = diameter of rotor (m)
phi = angle axis of turbine is to the flow of water, i.e. the
angle between the turbine shaft and surface of the water.
2.2.3.2 Kaplan and Propeller Turbine
The difference between these two is simply that the 'Kaplan' has blades
which pivot to alter their angle of incidence to the flowing water.
They both differ from the above device (Sect. 2.2.3.1) in that they
are always shrouded, and are fitted in civil works between water of differing
levels, e.g. tidal barage, albeit of small diffential head, (typically
1-2M).
3.0 Energy Losses
3.1 Prior to reaching the turbine or water wheel
3.2 In the device
3.3 Due to Conversion
To be continued.....
Contact: Mike Munro BSc (Hons.) for further information.
© Mike Munro
This page last updated 2nd
November 2006
