Wilbur Hubbard wrote:
> "R.C. Payne" <rcp27@[EMAIL PROTECTED]
> wrote in message
> news:frul71$guv$1@[EMAIL PROTECTED]
>>Wilbur Hubbard wrote:
>>>"Mees de Roo" <mees.deroo.laatditweg@[EMAIL PROTECTED]
> wrote
in
>>>message news:47e1a8a3$0$24416$5fc3050@[EMAIL PROTECTED]
>>>>Sir,
>>>>
>>>>To everyone that knows anything about mechanics, you are obviously
right.
>>>>Although I like your argumentation, you are however wasting your time.
>>>>When you read carefully the part about a boat being "blown back", you
>>>>would have realized that this same argument would aplly to any boat
>>>>sailing to windward, and not only to a boat sailing directly in the
wind.
>>>>Since I cannot believe that anyone posting in uk.rec.sailing does
think
>>>>that a boat cannot sail higher that halfwind, I cannot believe that
such
>>>>a person would really think that, given sufficient artifacts (like a
keel
>>>>or a propellor) to avoid too much drag, a boat cannot sail directly in
>>>>the wind (I am not talking about efficiency in doing such a thing).
After
>>>>all, if you compare the keel with a fixed airplane wing, and the
>>>>propellor with a helicopterrotor, it is clear that your propellor
would
>>>>act as a (less efficient) keel. Hence I am sorry to say that the
warning
>>>>about a troll, given out by other members of this newsgroup, seems to
be
>>>>correct; he probably will be arguing that helicopters cannot fly
straigt
>>>>upwards, next.
>>>>
>>>>Mees de Roo
>>>>
>>>>
>>>Poppycock! Draw yourself some force vector diagrams and you will see
how
>>>foolish your argument is. No sailboat can said directly into the wind (
0
>>>degrees ) without all the vector arrows facing directly aft ( 180 ).
When
>>>all the forces are facing aft the vessel will proceed aft. It's that
>>>simple. With a windmill system all the vector arrows will be facing aft
>>>but one that will be facing forward. That will be the arrow indicating
>>>the force of the water propeller. However that forward arrow will
always
>>>be shorter than the combined aft arrows so there is a net aft force
>>>acting upon the system. Thus the system proceeds backwards. It's going
>>>backwards slower than without the water propeller but it's going
>>>backwards nonetheless.
>>I've drawn a force diagram. Using an air speed of 11 knots upstream and
9
>>knots down stream of the turbine, of diameter 2 m. Air density is 1.2
>>kg/m^3. One knot is about 0.5 m/s. Newton tells me that force is rate
of
>>change of momentum. Mass flow rate is about 10 knots ( = 5 m/s) x 1.2
>>kg/m^3 * pi * 1^2 = 18.8 kg/s. Change in momentum per kg of air will be
>>11 kn - 9 kn = 2 kn = 1 m/s. This means that the force on the disc will
>>be 18.8 N (the number I had previously rounded to 19).
>>
>>Take the First Law of Thermodynamics. This tells me that the work I can
>>get out will be the change in kinetic energy of the flow. 1 kg of air
>>enters at 11 kn = 5.5 m/s. The air leaves with 9 kn = 4.5 m/s. Its
>>kinetic energy when entering will be 1/2 * 1kg * 5.5^2 = 15.125 J.
>>Leaving, it will have 1/2 * 1kg * 4.5^2 = 10.125. This means that each
>>kilogram of air passing the turbine will liberate 5 J of energy to the
>>turbine. There are 18.8 kg per second, so the power of the turbine is
94
>>W.
>>
>>If we put 90 W into our propeller (so I'm losing some, because the
>>connection is not perfectly efficient). the propeller is 20 cm
diameter,
>>and the water enters at 2 kn (= 1 m/s). The density of water is 1000
>>kg/m^3. So the mass flow rate is 1 * 1000 * pi * 0.1^2 = 31.4 kg/s.
>>
>>1 kg of water entering will come in with a kinetic energy of 1/2 * 1 *
1^2
>>= 0.5 J. At 31.4 kg/s, the incoming water brings 15.7 J/s. If we add
our
>>90 W on to that, we have 105.7 J/s leaving. This corresponds to 3.366
>>J/kg leaving (with our mass flow rate of 31.4 kg/s). Putting this into
>>our kinetic energy equation, we have V = SQRT(3.366/(1/2 * 1)) = 2.59
m/s,
>>or 5.2 knots. So our propeller will be ejecting the water at 5.2 knots.
>>
>>The water flow speed will change from 1 m/s to 2.59 m/s, so it will
>>increase by 1.59 m/s. For 31.4 kg/s of water increasing by 1.59 m/s,
this
>>gives us a force of 50.0 N.
>>
>>So, let's look at these force arrows you mention.
>>
>>The force acting to push the boat backwards will be 18.8 N.
>
> Wrong, my dear chap. What about the other force pu****ng on the structure
and
> hull? That 18.8 is only the force pu****ng on the propeller
The other force is part of the 31.2 N force. As the shape of the boat
has not been specified, this can not be computed, but it has been
allowed for.
>>The force acting to push the boat forwards will be 50.0 N.
>
> Simply not so. You have failed to take into account that you have to
> subtract your 18.8
> plus some more for the total force pu****ng against the entire system.
From
> this 50. You are assuming something is keeping the system at rest as if
it
> were attached to the ground or something.
Uh no. I'm considering the three forces on the boat: the force due to
momentum change on the turbine (18.8 N), the force due to momentum
change on the propeller (50.0 N) and the force due to other windage and
water frictional forces (which can be up to 31.2 N for the boat to
accelerate upwind).
>>Provided the drag on the hull and the wind force on other parts of the
>>upperworks of the boat are less than 31.3 N, the boat will have a net
>>force *forwards*.
>
> Sorry, but the wind force on the total boat and structure s greater than
> that. Do the math since you're so good at it.
So what would a reasonable value for this to be? I would suggest that
30 N is not an unreasonable value for this.
>>This assumption assumes that the power transmission from wind turbine to
>>propeller is 96% efficient, so it does not require a perfect power
>>transmission.
>
> 96 percent efficiency is a dream, my good man. Try about 70% and you
will be
> closer to the mark. And, try not to forget that some of the force is
used to
> spin the propeller. It doesn't go round using no force at all to turn
it.
The force requried to keep the propeller spinning is part of the loss.
70% efficiency gives a propeller thrust of 40.1 N. Still more than
double the pressure force on the turbine of 18.8 N. The propeller is
already spinning in this scenario, perhaps from the boat having
previously been sailing on a different point of sailing. Newton's first
law requires no force to make it spin once it has started.
>>There are no forces from the ground here. There are no external sources
>>of energy.
>
> And there needs to be or else the thing will just go backwardards albeit
a
> bit slower had the contraption not been installed.
The only way it might possibly go backwards is if the air drag on the
upperworks of the body is greater than the difference in thrust (31.2 N,
or if you prefer 21.3 N with the more conservative power train
efficiency). If the air resistance is any less than this value, the
vessel will make way to windward.
>>The reason this works, if you look at the equations, is that energy goes
>>as the square of the speed, while force goes linearly with speed.
Provided
>>the magnitude of the wind speed is large (10 knots) and the water speed
is
>>small (2 knots), by changing the wind speed and the water speed by
similar
>>ammounts (1 and 1.6 knots) you get similar forces out of them. But
>>because the energy liberated goes as the *square* of the speed, you get
>>lots more power out of the fast flow (the air) than is put into the slow
>>flow (the water).
>
> The reason it doesn't work is you failed to account for all the forces
> pu****ng the system backwards and you added a mythological forward force
as
> if the thing were firmly attached to the ground and the ground was
resisting
> its backwards movement with equal force to those pu****ng it backwards.
All the forces I can think of on the boat are:
1: the force on the turbine (pointing astern)
2: the force on the propeller (pointing ahead)
3: the drag force on the hull due to water (pointing astern if making
headway)
4: the drag force on the upperworks from windage (pointing astern).
force 1 has a value of 18.8 N, and I have accounted for it.
force 2 has a value of either 50.0 N (or 40.1 N if you assume an
iniefficient power train)
if force 4 is less than 31.2 N (or 21.3 N if you would rather), then the
boat will definitely be able to make way to windward.
if force 3 + force 4 is less than 31.2 N (or 21.3 N if you would
rather), then the boat will sail to windward at greater than 2 knots.
The are also
5: the force of gravity
6: the archimedian buoyancy force
7: any hydrodynamic or aerodynamic lift forces
which all act in the vertical plane, and are ignored.
There are no other forces on the boat. No force from the ground, no
force from God, no force from any other external factor.
Robin
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