Thanks Max,

A nice clear explanation.

So - on the water getting warm thing...

Lets say we have 1 litre of water in a tank 10 meters above the second tank.

I litre of water weighs (has a mass of) 1kg.

Potential energy is mgh = 1 x 9.8 x 10 = 980 Joules.

Lets assume that all the lost energy went into frictional losses in the water 
(assume perfect insulation and 0 thermal mass for the pipe).

It takes 4186 Joules to heat 1kg of water by 1 degree Celsius (are your 
gallon-foot-pound-Fahrenheit heads spinning yet?)

So 980 Joules gives 980/4186 degrees temperature rise = 0.23 degrees Celsius.

This is independent of the amount of water, 1 drop or 1 Billion Gallons, more 
water gives more energy but takes more energy to heat it.

The only variables here are the height difference and the temperature rise.

So for every 10m (33ft) of drop, the water temperature will rise 0.23 degrees 
Celsius (0.58 degrees Fahrenheit).

You are right - we wouldn't notice the temperature difference unless we 
measured it carefully.

Following your best case that we could extract 2/3 of this energy for 
electrical power then each litre of water with a head of 10m could produce 980 
Joules. 

So a flow of 1 litre per second with a 10 meter head could theoretically 
produce nearly 1000 Watts.

I wonder how close to that ideal any micro-hydro project has ever come?

Dominic Read
Georgia, USA

  ----- Original Message ----- 
  From: Max Enfield 
  To: [email protected] 
  Sent: Wednesday, February 02, 2005 1:00 AM
  Subject: Re: [microhydro] Confused about potential energy.



  ReadFamily wrote:
  > 
  > I was thinking about the pumped water storage idea and then started to 
think about the energy in the system.
  > 
  > If there is a full tank at the top and an empty tank at the bottom then the 
potential energy in the system is mgh (mass of water x gravitational constant x 
height difference).
  > 
  > So if you let the water flow from top tank to bottom tank through a simple 
pipe then the potential energy is gone. (assuming bottom tank is at height 0).
  > 
  > Now put the water back into the top tank and this time put a turbine in the 
pipe. The water flows from the top tank to the bottom tank and power is 
generated by the turbine and extracted from the system. But the difference in 
potential energy in the system is the same. All the water from the top is now 
in the bottom. Where did the energy that was extracted from the system by the 
turbine come from?
  > 
  > I'm sure that the answer is somewhere in the statement in the first case 
where the water flows through a simple pipe. The potential energy in the system 
has changed - so where did the energy go? I dont know.
  > 
  > Dominic Read
  > Georgia, USA. 

  Yes the difference in potential energy is the same.  

  In the first case where the water flows unimpeded down the pipe most of the
  potential energy is lost as friction in the pipe (eventually to be dissipated 
as
  heat) with some being used to provide the kinetic energy for the water to 
move.

  In the second case, the turbine impedes the flow of water and so less energy 
is
  lost as friction or needed for kinetic energy.  However the water must flow 
for
  the turbine to work and so some of the potential energy is still consumed as
  friction or converted to kinetic energy.  In the best case the turbine is able
  to extract up to two thirds of the potential energy.

  I have expressed the view several times to this group that hydraulics is
  intuitively obvious only to a gifted few - so don't be put off if at first you
  find the ideas somewhat alien.  The two thirds limit to the power that can be
  extracted can be proved rigorously, based on the fact that both the frictional
  loss and the kinetic energy are proportional to the square of the water
  velocity.

  Regards,

  Max Enfield
  Planetary Power





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