diagrammi di papmel

[Nettuno]

Hello everyone.. .

I am optimizing the number of turns of a naval propeller using the papmel diagram and precisely the "macchinista" to find the kdq coefficient.
Input data are the power of the engine, the speed of leftover, the density of the sea water and the diameter of the propeller to optimize. I have all the data to calculate kdq. the problem is that it comes too small and it seems not to fall back into the diagram.
from the data I have, it turns out kdq=0.545, but on the chart it begins to be diagramd by 1.0. then I came a doubt: will be that the value found by the formula must be multiplied by a factor of 10 because maybe so is kdq diagram?
would you explain to me "the stump "?

thanks to all

 

[Drag]

If you don't already have it, try to take a look at this pdf, chapter 2.1 explains the exercise similar to yours.

greetings
http://dinmats.altervista.org/downlo.../relazioni.pdf

 

[Nettuno]

Yes dragon, I've seen this pdf...in fact I started right from here, but it doesn't explain my doubt because he already has the number of turns from which j can get. I also consulted another pdf, which would be the book of the University of Trieste where everything is explained on papmel, but it does not explicitly say that you have to multiply kdq for 10. the doubt came to me only as the book shows a tabellina in which the diagram is indicated (the one of the approach of the engineer) as 10kq - j.
for more clarity in the next message I send you the link with the page number.. .

 

[Nettuno]

dinmats.altervista.org/download/propulsione_navale/book_07.pdf

The text is this. The page with the tabellina I told you is 195.
papmel diagrams start from p 190.

is therefore wrong to multiply kdq by 10?

a greeting and soon.. .

 

[Drag]

the formula you use is:
kdq=d*va*rad(ro*va)/p)
pay attention to the units of measurement, you have to convert them to the technical system,

 

[Nettuno]

Excuse me, dragon.

not to be insistent, but I cannot find myself with value even if I had already applied the units of measurement of the technical system.

Could you please perform a second calculation of the formula you wrote in the last message, with the data I have available?

speed of scrap............................. 5.136 m/s
diameter.................. 0.72 m
sea water density.................... 1026 kg/m3
power..................... 330 hp


formula

kdq = d * v left * square root [(densità * V avanzo)/potenza]I'd be very grateful. .

I just hope to mistake a conversion factor because I can't get into the diagram with the value I got.

See you soon.

 

[Drag]

I think you did:

== sync, corrected by elderman == @elder_man

so doing you have turned the cv into watts, as I told you the formula is referred to the technical system, where the power measurement unit is hp leaving the conversion to cv, consider them equal and get:

== sync, corrected by elderman ==

greetings

 

[Nettuno]

ah ok...so the density remains equal ( 1026 ) and it is only the power left in horses! I had divided the density for the acceleration of gravity and the power I had divided it for 75 converting the horses in kgf*m/s.. .
That's why I couldn't find myself... :biggrin:

Thank you so much dragon, now I can enter the diagram with 14.77...

:finger: