How do I calculate the flow of water from a dam, when given only power and height?

Question

How do I calculate the flow of water (in kg s^-1), from a dam, when I am only given Power = 50 MW at 80% efficient and the water level is 400m above the generator. I am guessing it is something to do with (mgh) but I don't have a mass and I am stuck after that.

Answer 1

What follows is a very rough calculation.

Estimate the pressure 'P' at 400 m depth, in incompressible (assumption) water, for a 'A' = 1 m square column: -

P = A.m.g  + P(atmosphere)

Where g=9.81 m/s² (assuming the acceleration of gravity is constant over the 400 m of height) and 'm' is the mass of the column of water.

At 20 degrees C the density of water is 998.2 kg/m³ (science data table), thus, the column has a mass of:

m = 400 x 1 x 1 x 998.2 kg

Hence: -

m = 399280 kg

Thus the pressure at a depth of 400 m (is very approximately, and ignoring P(atmosphere), which is much smaller 1.01325 x 10^5 Pa or N/m² at 0 m): -

P = 1 x 399280 x 9.81 N/m²

P= 3.9169368 x 10^6 N/m²  (or about thirty atmospheres)

For 50 MW of power at 80 % efficiency the total energy per second is: -

W' = 5.0 x 10^7 x (1/0.8)

Thus: -

W' = 6.25 x 10^7 W (J/s)

During each second this is the passage of 6.25 x 10^7 J of kinetic energy.

Therefore: -

W' =  ÃƒÂ‚½mv² = 6.25 x 10^7 J

Where 'm' is the mass of water passing and 'v' is its’ velocity, which are two unknown variables. However, we may use the impulse equation: -

F.t = m.u - m.v

We know the value of F  =P/A and we know the impulse time as t = 1 second. If we assume that u=0  (I have swapped u and v over here  to maintain notation consistency) after the energy loss to the generator then: -



(3.9169368 x 10^6/1) x 1 = mv

Now: -

The momentum ‘mv’ may be converted to a value for the kinetic energy of the water: -

(mv)²/2m = kinetic energy = 6.25 x 10^7 J

Or

m = (3.9169368 x 10^6)²/(2 x 6.25 x 10^7)

Thus, the mass of water = 1.22739 x 10^5 kg per second or 122 tonnes (or a volume of 122 ish cubic metres) . With a water velocity of about 71 mph and this is consistent with the 85 mph reported for the Hoover dam in the USA.

Needless, to say, this is a very crude approximation.

I hope this is of some help!

Ironically, when I was checking my figures - I found the following Wikipedia entry, ' ... A simple formula for approximating electric power production at a hydroelectric plant is: P = hrk, where P is Power in watts, h is height in meters, r is flow rate in cubic meters per second, and k is a conversion factor of 7500 watts (assuming an efficiency factor of about 76.5 percent and acceleration due to gravity of 9.81 m/s2, and fresh water with a density of 1000 kg per cubic metre. Efficiency is often higher with larger modern turbines and may be lower with very old or small installations due to proportionately higher friction losses). ...'

Assuming k = 0.8 this gives a value of 15625 kg per second. Thus, my approximation seems reasonable!

[Corrected]

Answer 2

No, but you have a mass per second, and mass per second times g times h would be the power.

Singing...

Power to the people,

power to the people, right now...  

Answer 3

u r right, no mass no power, a dam's ability to generate power is ultimately derived from the flow into the backwater

Answer 4

The first thing to recognize is that the Watt is the same thing as a Joule/second. So every second the dam is producing 50MJ of electrical energy. At 80% effeciency that means the dam is consuming 62.5MJ of gravitational energy from the falling water every second.

So using PE = mgh

62.6MJ = mg(400m)

you can figure out how much water is flowing through the dam every second, which is, incidentally, the flow rate.