"Liam" archimedes screw wind turbine design

[Oldgreybeard]

Sorry Gus but it is more BS. 1500kW per year at 5m/s? Even if the largest one quoted (1.5m ) extracted 100% of the energy from the wind 24 hours a day, 365 days of the year at 5m/s it would only get to just under 1200kWh in the year. They are quoting that for the smaller "domestic" one which if it extracted 100% would generate under 0.3kWh per annum.
On a good site they may just do 50% of that figure but sat on top of a roof in turbulent air ??????? No chance.

Do the math, no more chance of working than the mad idea of VAWT around lamp posts by the road.

If you see any "new" design for a wind generating device do a quick calculation to check their numbers.

Energy in the wind, i.e. the maximum that can be generated is given by the formula. P =1/2*p*A*V³

Where P is power in Watts
p is density of air. 1.23kg/m³
A is swept area of the blades or screw or whatever
V is wind speed. (cubed)

And to add to this the Betz Limit has to be taken into account, so the reality is even worse. It is physically impossible to extract ALL the energy from the wind (or water flow), as if you did then it would stop dead, there would be no where for the wind (or water) to go. Betz worked out that this meant there was a practical upper limit to extracting potential energy from a moving fluid, no more than 59.3% of the theoretical maximum from the equation above can be extracted by any machine: https://en.wikipedia.org/wiki/Betz%27s_law

[Joeboy]

Sorry Gus but it is more BS. 1500kW per year at 5m/s? Even if the largest one quoted (1.5m ) extracted 100% of the energy from the wind 24 hours a day, 365 days of the year at 5m/s it would only get to just under 1200kWh in the year. They are quoting that for the smaller "domestic" one which if it extracted 100% would generate under 0.3kWh per annum.
On a good site they may just do 50% of that figure but sat on top of a roof in turbulent air ??????? No chance.

Do the math, no more chance of working than the mad idea of VAWT around lamp posts by the road.

If you see any "new" design for a wind generating device do a quick calculation to check their numbers.

Energy in the wind, i.e. the maximum that can be generated is given by the formula. P =1/2*p*A*V³

Where P is power in Watts
p is density of air. 1.23kg/m³
A is swept area of the blades or screw or whatever
V is wind speed. (cubed)

And to add to this the Betz Limit has to be taken into account, so the reality is even worse. It is physically impossible to extract ALL the energy from the wind (or water flow), as if you did then it would stop dead, there would be no where for the wind (or water) to go. Betz worked out that this meant there was a practical upper limit to extracting potential energy from a moving fluid, no more than 59.3% of the theoretical maximum from the equation above can be extracted by any machine: https://en.wikipedia.org/wiki/Betz%27s_law

Now THAT is interesting and also groovy. Thanks!

[Countrypaul]

Sorry Gus but it is more BS. 1500kW per year at 5m/s? Even if the largest one quoted (1.5m ) extracted 100% of the energy from the wind 24 hours a day, 365 days of the year at 5m/s it would only get to just under 1200kWh in the year. They are quoting that for the smaller "domestic" one which if it extracted 100% would generate under 0.3kWh per annum.
On a good site they may just do 50% of that figure but sat on top of a roof in turbulent air ??????? No chance.

Do the math, no more chance of working than the mad idea of VAWT around lamp posts by the road.

If you see any "new" design for a wind generating device do a quick calculation to check their numbers.

Energy in the wind, i.e. the maximum that can be generated is given by the formula. P =1/2*p*A*V³

Where P is power in Watts
p is density of air. 1.23kg/m³
A is swept area of the blades or screw or whatever
V is wind speed. (cubed)

I presume area is in m2 (square metres) and wind speed is in m/s (metres per second).

[Oldgreybeard]

Sorry Gus but it is more BS. 1500kW per year at 5m/s? Even if the largest one quoted (1.5m ) extracted 100% of the energy from the wind 24 hours a day, 365 days of the year at 5m/s it would only get to just under 1200kWh in the year. They are quoting that for the smaller "domestic" one which if it extracted 100% would generate under 0.3kWh per annum.
On a good site they may just do 50% of that figure but sat on top of a roof in turbulent air ??????? No chance.

Do the math, no more chance of working than the mad idea of VAWT around lamp posts by the road.

If you see any "new" design for a wind generating device do a quick calculation to check their numbers.

Energy in the wind, i.e. the maximum that can be generated is given by the formula. P =1/2*p*A*V³

Where P is power in Watts
p is density of air. 1.23kg/m³
A is swept area of the blades or screw or whatever
V is wind speed. (cubed)

I presume area is in m2 (square metres) and wind speed is in m/s (metres per second).

Yes, if you're metric. When I started work we were still using the imperial version of the drag equation, where area was in ft², drag was in lbs, density was in slugs/ft³ and velocity was in ft/s . . .

[Countrypaul]

Sorry Gus but it is more BS. 1500kW per year at 5m/s? Even if the largest one quoted (1.5m ) extracted 100% of the energy from the wind 24 hours a day, 365 days of the year at 5m/s it would only get to just under 1200kWh in the year. They are quoting that for the smaller "domestic" one which if it extracted 100% would generate under 0.3kWh per annum.
On a good site they may just do 50% of that figure but sat on top of a roof in turbulent air ??????? No chance.

Do the math, no more chance of working than the mad idea of VAWT around lamp posts by the road.

If you see any "new" design for a wind generating device do a quick calculation to check their numbers.

Energy in the wind, i.e. the maximum that can be generated is given by the formula. P =1/2*p*A*V³

Where P is power in Watts
p is density of air. 1.23kg/m³
A is swept area of the blades or screw or whatever
V is wind speed. (cubed)

I presume area is in m2 (square metres) and wind speed is in m/s (metres per second).

Yes, if you're metric. When I started work we were still using the imperial version of the drag equation, where area was in ft², drag was in lbs, density was in slugs/ft³ and velocity was in ft/s . . .

I was always "taught" to put the units in the equation as well as the numbers. that way if the units don't match the expected you know something is wrong. So, just to check

kg/m3 * m2 * (m/s)3 = kg*m2*s2

that way also help show when you overlook a mili or kilo, or the like.

So could the designers have over estimated by a 1000 ???

[smegal]

I presume area is in m2 (square metres) and wind speed is in m/s (metres per second).

Yes, if you're metric. When I started work we were still using the imperial version of the drag equation, where area was in ft², drag was in lbs, density was in slugs/ft³ and velocity was in ft/s . . .

I was always "taught" to put the units in the equation as well as the numbers. that way if the units don't match the expected you know something is wrong. So, just to check

kg/m3 * m2 * (m/s)3 = kg*m2*s2

that way also help show when you overlook a mili or kilo, or the like.

So could the designers have over estimated by a 1000 ???

This should be less of an issue with SI units to be fair.

[Mr Gus]

Damn your good, rational common sense & maths gentlemen,
I saw this just before hitting the hay this morning, was ever hopeful but you live in hope despite all the signage & experience, a litany of BS in wt design pre-dating but not exclusive to swindlesave.

Caught by the "if it works with water" (not but how much does it lose) ..as I put my head down, ..because I ceased to think! (damn it)

[Oldgreybeard]

AS a rough guide as to judging whether any design is any good, just do a quick and dirty calculation. Say a new wind turbine design is 100m in diameter. It will have a swept area of 7,854m²

Say the wind speed is 10m/s and the air density at that altitude and temperature is 1.225kg/m³.

That wind flow has a potential power, P (in watts) over an area of 7,854m² given by P = 0.5 . ρ . A . v³ = 0.5 . 1.225kg/m² . 7854m/s . 10³ = 4,810,575W, or about 4.8MW.

Now apply Betzs' Law, which states, in effect, that a 100% efficient machine can only extract 59.3% of the potential power in a fluid stream, and we get a maximum available power of 4,810,575W . 0.593 = 2,852,671W

So a 100m diameter, 100% efficient wind turbine could theoretically extract about 2.85MW from a 10m/s wind. In reality, the very best wind turbines are pretty efficient now, they can run at about 85% or so (after allowing for the Betz Limit). That means that a 10m/s wind that has the potential to deliver about 4.8MW can, in reality only deliver less than half that as usable power.

The bottom line is that if you see any advertisement for a wind (or water - same rules apply, just change the density) turbine that seems to suggest figures better than these then it is almost certainly BS.

[nowty]

And as a sanity check of OGB calculations, Ripple WT1 has a 100m diameter rotor and has a max output of 2.5 MW.

[Oldgreybeard]

And as a sanity check of OGB calculations, Ripple WT1 has a 100m diameter rotor and has a max output of 2.5 MW.

Thanks, that's a relief, shows I didn't screw up the arithmetic!