Dave if we have 800,000 liters of water falling 250 meters through a pipe 20 cm in diameter hitting a turbine how much electricity would we generate? And what is the optimum diameter of the pipe?

I realized after rereading this that I might have seemed a bit lazy here, but I've gotten into the habbit of asking questions of my students when I've already an idea of the answer just to keep their interest up!

So here goes!

To get an idea of how much power in kilowatt-hours can be derived from hydropower we need to know the height, flow and efficiency of the system.

Flow is the tricky one as anyone who has studied fluid dynamics in pipes will attest!

To get an estimate on how long it takes for something to fall from a given height to the ground we can use;

Time = SquareRoot ((Height/(0.5*Gravity))

= SquareRoot ((250meters/(0.5*9.

)

7.142 seconds

Now obviously we need to discount air resistance/drag and the friction from the pipe's

walls!

If the pipes inner walls were coated with a hydrophobic material a lot of this friction could be offset.

A single droplet of water naturally evolves into the most aerodynamic shape possible as it moves through the air, it is very difficult to estimate how aerodynamic water falling down a pipe would be, however.....

let's estimate a drag coefficient of Cd=0.18

How fast is the water moving at the bottom of the pipe?

Well let's see how far the water has fallen at time 6.142 seconds, 1 second before exiting.

Height = 0.5*Gravity*(time Squared)

= 184.848 meters

So the water's velocity

**if we just averaged the acceleration rate after 6.142 seconds** at the end of the pipe would be (250-184.848)= 65.15 meters per second.

Now just to be clear let's look at Time 7.142 seconds and Time 8.142 seconds

@7.142 seconds = 250 meters

@8.142 seconds = 324.83 meters

So the water's velocity

**if we just averaged the acceleration rate after 7.142 seconds** the water's velocity after falling 324.83 meters would be (324.83-250)=

**74.83 meters per second**.

Now we have a velocity between 65.15 meters per second and 74.83 meters per second to base our estimates on so let's take an average of

**70 meters per second**.

Okay now these figures don't take into account air resistance, air pressure, humidity and temperature!

Mass doesn't come into the equation if we don't think about an object falling through air, but here we have a column of water 250 meters high with a diameter of 20 cm.

Cylinder pi * radius2 * height

Volume = (pi * radius squared)*height = (3.1416 * (diameter/2)squared)* height

= (3.1416 * 0.10Squared)* 250 meters

= 7.8540 cubic meters or 7,854,000 cubic centimeters

As 1 liter equals 1000 cubic centimeters this gives us 7,854 liters

As 1 liter of water has a mass of 1 kilogram this gives us a mass of 7,854 kg

An object that is falling through the atmosphere is subjected to two external forces. The first force is the gravitational force, expressed as the weight of the object, and the second force is the aerodynamic drag of the object. The weight equation defines the weight W to be equal to the mass m of the object times the gravitational acceleration g:

W = m * g

W = 7854kg * 9.8

Weight = 76,969.2

The drag equation tells us that drag D is equal to a drag coefficient Cd times one half the air density (estimated at 1.2kg/cu m) r times the velocity V squared times a reference area A (pi * 0.10 meters Squared) on which the drag coefficient is based:

D = Cd * 0.5 * r(kg/cu m) * V Squared * A

Drag = 0.18 * 0.5 * 1.2 * 70^2 * 0.0314 cubic meters

Drag = 16.625308

The motion of any moving object can be described by Newton's second law of motion, force F equals mass m times acceleration a:

F = m * a

We can do a little algebra and solve for the acceleration of the object in terms of the net external force and the mass of the object:

a = F / m

Weight and drag are forces which are vector quantities. The net external force is then equal to the difference of the weight and the drag forces:

F = W - D

The acceleration of the object then becomes:

a = (W - D) / m

a = (76,969.2 - 16.625) / 7854

a = 9.79

The drag force depends on the square of the velocity. So as the body accelerates it’s velocity and the drag increase. It quickly reaches a point where the drag is exactly equal to the weight. When drag is equal to weight, there is no net external force on the object, and the acceleration becomes zero.

So in our case our estimates are going to be fairly accurate even if we take air resistance into account.

Height = 0.5*Acceleration*(time Squared)

= 0.5*9.79*(6.142^2)

= 184.66 meters as opposed to 184.848 meters

Taking into account pipe friction, Temperature and Humidity factors let's lower our estimated velocity for our water from 70 meters per second to

**62 meters per second**Now 1 meter of a 20cm diameter pipe holds 31.4 liters of water.

and 62 meters of a 20cm diameter pipe holds 1946.8 liters.

Therefore we can estimate that at the bottom of the pipe we have a flow of

**1946.8 liters per second**If we have a reservoir of 800,000 liters then dividing by 1946.8 we find that

**after 411 seconds or 6 minutes and 51 seconds - all the water has gone!**The only formula I've found that estimates power from falling water is from an American website (

http://www.wvic.com/hydro-works.htm) and it's in imperial.

Power = (Height of Dam) x (River Flow) x (Efficiency) / 11.8

Power = The electric power in kilowatts per Hour (one kilowatt equals 1,000 watts).

Height of Dam = The distance the water falls measured in feet.

River Flow = The amount of water flowing in the river measured in cubic feet per second.

Efficiency = How well the turbine and generator convert the power of falling water into electric power. For older, poorly maintained hydroplants this might be 60% (0.60) while for newer, well operated plants this might be as high as 90% (0.90).

11.8 Converts units of feet and seconds into kilowatts.

Converting and inputting our variables

Height = 250 meters = 820.2 feet

Flow = 1946.8 liters per second = 68.75 cubic feet per second

Efficiency = 90%

Power = ( 820.2 * 68.75 * 90% )/11.8

= 4300.8 kilowatts per hour or 71.68 kilowatts per minute

BUT our system only runs for 6 minutes and 51 seconds!

So our system (As it Stands) should generate about

**491 kilowatts a day**, which would provide enough electricity for 24 homes all year round!

I say as it stands as next week I'd like to address the efficiency issues, but for a quick taster the amount of energy it takes to evaporate water is not a constant for all conditions.

Most of the figures quoted would make your average leaf curl up and die!

Blaine