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Default Power absorbed by propeller

Hi,

I have a question that someone might be able to assist with - a
formula for - or an idea. I have a 12 inch x 10 inch 3 blade kaplan
prop for a bow thruster and I am trying to work out in rough handfuls
what power the prop will absorb at say 1500rpm.

So far I have worked out the volume of water displaced per rev/sec/min
- allowing for prop slip and the loss of area at the centre caused by
the boss and also the speed advance of the prop per rev/sec/min. This
gives a mass of water in kg moving at a velocity of metres/sec.

Somehow, I think I should be able to relate this force required to
achieve the movement into the torque and hence power required to turn
the prop, but have not yet achieved the linkage.
Wondered if anyone has any thoughts???

TVMIA

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JAXAshby
 
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If you can relate your calculations to pounds of thrust, figure about
1 hp for every 30 pounds thrust, e.g., 300 pounds of thrust is roughly
equal to 10 hp.


ah, no. thrust times rpm / 60 / 550 = hp. (IIRC) 30# of thrust at 4,500 rpm
is 3x the hp of 30# of thrust at 1,500 rpm.
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Brian Whatcott
 
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On Wed, 12 Jan 2005 10:15:39 -0500, Wayne.B
wrote:


We are talking about thrust into the water, i.e., bollard pull as the
tug boat folks like to call it:

http://www.hydrocompinc.com/knowledg...ollardPull.pdf



It is better not to argue with the lunatic fringe. Full stop.

Brian W


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Evan Gatehouse / Diane Selkirk
 
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wrote in message
ups.com...
Hi,

I have a question that someone might be able to assist with - a
formula for - or an idea. I have a 12 inch x 10 inch 3 blade kaplan
prop for a bow thruster and I am trying to work out in rough handfuls
what power the prop will absorb at say 1500rpm.


Without doing the detailed calculations (because I'm feeling lazy tonight):
10-15 HP. Depends on blade area and tunnel fairing details but the number I
have quoted you is probably close enough for what you're thinking about.


--
Evan Gatehouse
ceilydh **at** 3web **dot** net
---- rewrite my signature to send email


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