Stewart Corman
1st February 2007, 08:12
I think this was discussed previously on AWEA:
http://engineering.curiouscatblog.net/2007/01/02/
however, I recently found a forum where someone bebunked the numbers:
""The scenario described is full of easily calculable factors. One
cannot get more energy out than the difference in kinetic and
potential energy provided.
Since the provided scenario is a "stream", I am considering the
difference in velocity (kinetic energy) from the inlet and outlet to
be negligible and thus neglecting it in the calculation.
2000W is 2000 joules/sec. 1 joule is the energy of raising or lowering
0.7376 lbs over 1 ft. So at 100% efficiency, you'd need to move 2212.8
lbs/sec over 8", 276.6 gal/sec. 16,596 gal/min.
Water turbines have had 80%-90% efficiency for over 100 years when
used under ideal circumstances. As such it's notable that while a
small, economical, low draw waterwheel may be something new, it cannot
possibly produce much more power than turbines have in the past. The
70% specified is a somewhat low performer but operating on such low
head may be something new.
So let's take 70% specified in this Wiki. Then we need 23,709 gpm
through the turbine to generate 2KW. That's a pretty powerful stream @
8" of head! This would fill a 50m by 25m by 2m Olympic swimming pool
in 27.8 minutes. I have to note that since the speed of water in a
natural stream is usually limited to a few feet per sec, the width of
the device depicted is perhaps a meter, and the height of the water
channel's cross sectional area must be only a small fraction of the 8"
head then I don't see how such a volume could flow through a device of
the width depicted. I get 89.9 cu meter/min through a 1 meter wide by
2cm high cross section (10% of head) requires 74.8 m/sec flow rate, or
167.3 mph!
Perhaps he meant the device could begin turning at only 8" of head,
but achieved 2KW at a higher head which would require a lower flow rate?"
Sounds like some of the claims for VAWT ??
Stew Corman from sunny Endicott
http://engineering.curiouscatblog.net/2007/01/02/
however, I recently found a forum where someone bebunked the numbers:
""The scenario described is full of easily calculable factors. One
cannot get more energy out than the difference in kinetic and
potential energy provided.
Since the provided scenario is a "stream", I am considering the
difference in velocity (kinetic energy) from the inlet and outlet to
be negligible and thus neglecting it in the calculation.
2000W is 2000 joules/sec. 1 joule is the energy of raising or lowering
0.7376 lbs over 1 ft. So at 100% efficiency, you'd need to move 2212.8
lbs/sec over 8", 276.6 gal/sec. 16,596 gal/min.
Water turbines have had 80%-90% efficiency for over 100 years when
used under ideal circumstances. As such it's notable that while a
small, economical, low draw waterwheel may be something new, it cannot
possibly produce much more power than turbines have in the past. The
70% specified is a somewhat low performer but operating on such low
head may be something new.
So let's take 70% specified in this Wiki. Then we need 23,709 gpm
through the turbine to generate 2KW. That's a pretty powerful stream @
8" of head! This would fill a 50m by 25m by 2m Olympic swimming pool
in 27.8 minutes. I have to note that since the speed of water in a
natural stream is usually limited to a few feet per sec, the width of
the device depicted is perhaps a meter, and the height of the water
channel's cross sectional area must be only a small fraction of the 8"
head then I don't see how such a volume could flow through a device of
the width depicted. I get 89.9 cu meter/min through a 1 meter wide by
2cm high cross section (10% of head) requires 74.8 m/sec flow rate, or
167.3 mph!
Perhaps he meant the device could begin turning at only 8" of head,
but achieved 2KW at a higher head which would require a lower flow rate?"
Sounds like some of the claims for VAWT ??
Stew Corman from sunny Endicott