Hi,
I have a big tank and 1/3rd of the tank is filled with water. Now I have a huge funnel shaped pipe immersed inside the
water. There is valve at the end of the funnel which is closed. There is a turbine inside the funnel below the water level
of the tank. The funnel is also filled with water.
Now if the valve is opened a constant flow of water is released into the tank from the funnel and this makes the turbine
rotate. Suppose equal amount of water is added to the funnel so that the water level in the funnel does not change, how
am I supposed to calculate the input energy of the moving water on the turbine?
If the placement of turbine is shifted will the input energy change?
I have added a picture for clarity, please explain
Do you really want to know?
Quote
Calculating the amount of available power A hydropower resource can be evaluated by its available power (https://en.wikipedia.org/wiki/Power_%28physics%29). Power is a function of the hydraulic head (https://en.wikipedia.org/wiki/Head_%28hydraulic%29) and rate of fluid flow (https://en.wikipedia.org/wiki/Rate_of_fluid_flow). The head is the energy per unit weight (or unit mass) of water. The static head is proportional to the difference in height through which the water falls. Dynamic head is related to the velocity of moving water. Each unit of water can do an amount of work equal to its weight times the head.
The power available from falling water can be calculated from the flow rate and density of water, the height of fall, and the local acceleration due to gravity. In SI units, the power is:
(https://upload.wikimedia.org/math/9/4/c/94cfcc1dc82a060f518967d9a7dddd5b.png)
(There is an equation image here, check the WIKI reference)
where
P is power in watts
η is the dimensionless efficiency of the turbine
ρ is the density of water in kilograms per cubic metre
Q is the flow in cubic metres per second
g is the acceleration due to gravity
h is the height difference between inlet and outlet
To illustrate, power is calculated for a turbine that is 85% efficient, with water at 62.25 pounds/cubic foot (998 kg/cubic metre) and a flow rate of 2800 cubic-feet/second (79.3 cubic-meters/second), gravity of 9.80 metres per second squared and with a net head of 480 ft (146.3 m).
From WIKI:
https://en.wikipedia.org/wiki/Hydropower#Calculating_the_amount_of_available_power (https://en.wikipedia.org/wiki/Hydropower#Calculating_the_amount_of_available_power)
You will also find lots of useful formulae and calculators on this site:
http://www.engineeringtoolbox.com/hydraulic-pumps-d_1628.html (http://www.engineeringtoolbox.com/hydraulic-pumps-d_1628.html)
@Tinsel Koala
Thank you
but I want to know if there will be any difference in input energy if I allow the water to fall freely from
above without the funnel and if I have a funnel and place the turbine inside.
Sorry im a little ignorant on these things
Nothing wrong with ignorance: it is how we know we don't know something! Unrecognized and uncorrected willfull ignorance is a different matter though, and not your problem.
I'm not an expert in fluid flow either but there are some things I do know. There will be some friction from the walls of the tube leading down from the funnel, and some power will be lost here. But the tube will also channel the water to the turbine better than just an open stream of water would. So maybe that's an equal tradeoff there.
But if you are thinking that the funnel shape, up above the tube, will be beneficial by concentrating or channelling the water pressure into the tube, that's wrong, the pressure in the tube depends only on the height (aka head) of water above that point, not the shape of the reservoir. Shallow flat reservoirs vs. tall skinny ones may each have some advantages or disadvantages in certain designs, like the Heron's fountain, but the funnel shape itself doesn't gain you anything. There may be some flow rate advantage to be gained by shaping the channel at the turbine itself, like a venturi in air, but the energy available will still be the same, even though the power level might change.
ETA: The total energy available in Joules is just mgh, where m is the mass of water in kilograms, g is the acceleration due to gravity (about 10 meters per second per second), and h is the height of the center of mass of the water above the turbine in meters. There are more or less efficient ways of getting this energy into the turbine: an open flow, a tight pipe, bucket by bucket, etc. so the energy that the turbine actually produces will be less than the mgh of the water.
Thank you very much Tinsel Koala :)