"R.C. Payne" <rcp27@[EMAIL PROTECTED]
> wrote in message
news:frupus$sj2$1@[EMAIL PROTECTED]
> Wilbur Hubbard wrote:
>> "R.C. Payne" <rcp27@[EMAIL PROTECTED]
> wrote in message
>> news:frul71$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
Good try but you've conveniently forgotten at least one very large force.
That being the waves hitting the hull from directly ahead of it. If you've
done any sailing at all you will know how a close-hauled yacht can
sometimes
be brought almost to a complete standstill due to the action of the waves
striking it. And from straight ahead, even more so than close-hauled,
there
will be a hobby horse effect going on where the bows slam into the waves
in
a rhythmic manner trying to part them and things quickly grind to a halt.
It
takes a helluva lot of energy to part waves. The effect of going straight
into wind and waves is to experience a shorter apparent wavelength. Sorry,
but even with your optimistic figures there simply isn't enough energy
available to sail directly into the wind.
Wilbur Hubbard
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