calculating maximum Hydropower Turbine output with a given water tank

[yon1000]

Hello,

I'm calculating feasibility of using grey water in high rise buildings to generate hydroelectricity.
let's assume a given water volume of 100 cubic meters a day and a water head of 100 meter.
what should be the diameter of the pipe and thus what should be the flow rate in order to maximise the total power generated through the turbine?
lets assume emptying the tank will be done once ( or more ) a day.

 

[LittleInch]

Nice idea, won't work.

The only way it could is that the drain pipe is full of water, but then the lower floors would get flooded!!



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[dhengr]

Yon1000:
Your idea involves a very expensive turbine and generator for what little electricity it would likely generate. It also involves a lot of expensive mech. equip. and plumbing/piping and tankage. The 100m head suggests that the storage tank is at the 28 fl., so do you pump everything below that elevation back up to the 28 fl? If you are really talking about real gray water, that means sinks without garbage disposals (maybe not kitchen sinks with grease, food waste and all), showers and bath tubs and the like; not real sanitary sewerage, which should go directly to the city sanitary sewer. If you can get the tenants to be accepting of the gray water look in the toilet bowls, a better use might to use that gray water for toilets (water closets) and the like or watering the yard and local golf course. Put a smaller tank every 5 or 10 fls. to catch the gray water from the above 5/10 fls. and then to service to the next 5/10 fls. below. That would be a double use of the water and a volume savings in potable water and sewerage charges, without the expensive turbine and generator. But, I suspect, still a pretty long payback period.

 

[TugboatEng]

Allowing the water to fall in to the turbine would create an impulse effect just like a water wheel without backing up water into drains above. There may be noise issues and the low terminal velocity of water in the pipe will severely limit efficiency.

 

[LittleInch]

Water falling down a vertical pipe is a complex thing and often ends with annular flow, small drops and no real impulse effect.

Sounds like a good idea but just doesn't stand up to any scrutiny I'm afraid.

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[TugboatEng]

I guess your could run the pipe on a spiral around the outside of the building and have several smaller units at each level where best velocity is.

 

[GregLocock]

Just the simple number is 100*10*1000*100 J, or about $4 worth of electricity. Pumped storage systems have to be enormous.

Cheers

Greg Locock


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[TheTick]

Gaming velocity and flow paths and pipe size yields zero net gain.

The total amount the energy in the system can be calculated by the potential energy of the water, which is simply the mass of the water times the height times gravity

PE = mgh

There is no more energy available than that. That's all there is. Greg did the math (thank you). Don't forget to pay the entropy tax.

 

[TheTick]

addendum:
That's 27.5 kWh per day
$3.58 (USD) per day at my local rate of 13 cents per kWh (residential retail)

 

[TugboatEng]

I'm only inventing preposterous solutions to demonstrate how preposterous the problem is.

 

[yon1000]

Ok I understand it's a ridiculous idea and not economical at all.
But let's assume I wanted to make plans for a system like that even If it's not economical , how can I calculate the needed pipe size in order to maximise the total energy generated in a day.
and also given X floors in a building and a given 1.6 cubic meters of water per floor in a day. how should I calculate the number of tanks in a building and in what floor I should put them in order to maximize total energy generated .
I've seen one Indian article regarding this proposed system and it suggested using just one tank in the middle so half of the water from the floors of the building is utilized and the other half are not utillized at all.
It looks like a problem of some linear programming, since going higher would allow higher head of water but less water to be utillized and vice versa.
I've been using the general equation of hydropower turbine: P = η * ρ * g * h * Q where defining the Q in order to maximise total energy generated is the problem for me

thank you very much for your help.

 

[LittleInch]

yon,

You can make it any size you want as it varies by the POWER of the turbine you plan to install.

The amount of ENERGY per day is fixed by the volume of the water collected in one day. POWER is then that energy divided by the time scale you want to extract that energy.

OR just pick a pipe size, 50,80 or 100mm is my best guess and then calculate a flowrate based on say 1m/sec velocity of a full pipe.

You want to minimise frictional losses in your pipe to be able to use the majority of the potential energy at the pipe end on the ground floor in the turbine and 1m/sec won't loose you that much pressure.

Then calculate how long it will take to empty your tank based on one days capacity in the tank. I wouldn't go for anything more than 60 minutes my self, but look at the numbers.

To optimise the tank volume to something reasonable as this is taking up floor space you could rent out, you might then need to run your turbine for up to say 10 times a day, by then dividing your total volume of water by the number of times per day.

Just be sure to come back and tell us what you find.



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[PingSlayer]

LittleInch:
What formulas should one use in order to calculate each step you've suggested?

 

[rb1957]

how do you define "grey water" ? How would you separate out objects in the flow ? How do you ensure that you're sending only water to the turbines ?) How would you deal with these separated objects (they'll build up over time)?



another day in paradise, or is paradise one day closer ?

 

[SwinnyGG]

Grey water has a pretty clear definition- it's any waste water from a stream that does not include fecal matter.

So it's likely to have food waste, trash, etc in it.

 

[TheTick]

Headloss (friction loss) is inversely related to pipe diameter and directly proportional to speed². The most efficient system uses slowest flow through larges pipes.

I picture largest feed pipes possible, but using water at the slowest rate possible (i.e. 100m³/day).

 

[TheTick]

@TugboatEng: noted

 

[LittleInch]

turbine power you have the eqauation - you said all you were missing was flowrate.

So find out the ID of different sized pipes in metres, find the area (Pi x R^2). times this by the velocity (1). Multiply by 3600 (60 x 60) and then you get m3/hr.

It's all pretty basic at this point....

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[LittleInch]

I agree that largest diameter is best, but once you get below 1m/second over the sort of distance the OP is taking about (50m or so) it becomes a very low pressure drop compared to the static pressure.

Basically aim for a pressure drop to be less than 5% of the static pressure and you're in a good place.

Not sure what penstocks are designed on but I think it's more than 1m/second. But then they usually are 36" or bigger x 3 or 4. Or not very long /(e.g. bottom of a dam)

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[TheTick]

Gotta say, in in spite of the economic impracticality, this is a fun idea to explore.