Propeller (air)flow ... [uni 'project']

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Post Fri Sep 05, 2014 3:29 pm

Does anyone here have a clue about propellers?

They were not covered at all in lectures and we have an extra 'project' that counts for 15% of the final grade in fluid mechanics, that does not seem like much but it's ~30% of the points we need to pass :D

I am unfortunately unable to find really suitable or helpful info online -.- I even read some chapters in the "Pilot's Handbook of Aeronautical Knowledge" ffs and could not find anything that might really help... just the basic drawing

just ignore the bad english, apparently no one bothered to spellcheck before uploading it (we also have a german version that only differs slightly)
Image

i drew this..
Image
Vr is the circumferential speed
W Drag
Q sheer force? (googled that ... german 'Querkraft')
R resulting force
A Lift
S thrust
alpha is the aoa
beta is 'steigungswinkel' .... angle of climb according to dict.cc :D
a) already did that in the drawing... should be right, the german version also asks about the angle at which the flow hits the propeller (phi) should be tan(phi)= V/Vr

b) So what do i take here, do i take the maximum aoa of 16°
Image
or 11° where the lift coefficient is at its maximum?

i first assumed that the aoa will be zero at the tip, took an aoa of 11° for 0,15m and 0° for 0,6m and calculated a straight between those two points since the graph is almost straight between those two points
might be right...might be wrong, dunno what exactly they are asking

The point is, as far as i understood propellers you really want the thrust to be the same across the whole propeller/radius, that's probably what they mean with the assumption of constant axial force, but i only have equations for drag and lift, that would give me the resulting force and from there i could get to the thrust with phi ... but i did not manage to get to any decent result because something seems to be missing

c) now comes the weird part: isn't the twist supposed to allow for a constant aoa? I mean then b) would be pointless or the aoa constant with the blade twist changing, dunno how to calculate the twist...

d) here i guess i'd just have to fiddle with matlab and the second graph to get some correlation
Image
e) i believe i should see the propeller as a series of infinite small wings (airfoils)? find an equation for the thrust depending on the radius r and integrate across the raduis? or something...

f) same for the force Q ... momentum =r*Q? :wtf:

It's really just annoying if you have no info whatsoever in your lecture notes ... maybe someone knows that stuff well and can help :)
RZS10
 
Joined: 7 Dec 2013

Post Fri Sep 05, 2014 4:40 pm

...a Matlab program written by you (and only you)... :mrgreen:


To me the key to solve part b is:

A * cos(phi) - W * sin(phi) = constant
(A= rho/2 * cl * Area * vrel^2 and W = rho/2 * cd * Area * vrel^2)

that leads to: cl * cos(phi) - cd * sin(phi) = constant

The blue triangle gives you: 180° = phi(r) + alpha(r) + beta(r) + 90°

Cl and cd are connected by the two diagrams you posted. Create two tables in Matlab, that allows you to interpolate and define cl(alpha) and cd(cl).
"Torque * 2Pi * f" is nothing without control!
Blanchimont
 
Joined: 9 Nov 2012

Post Fri Sep 05, 2014 4:54 pm

Blanchimont wrote:...a Matlab program written by you (and only you)... :mrgreen:


well as long as i write the matlab program :D

thanks ... i'll try what you suggested :)
RZS10
 
Joined: 7 Dec 2013

Post Fri Sep 05, 2014 7:49 pm

hmm... i still end up lacking something :roll:
Last edited by RZS10 on Fri Sep 05, 2014 10:03 pm, edited 2 times in total.
RZS10
 
Joined: 7 Dec 2013

Post Fri Sep 05, 2014 9:46 pm

Sorry, i think there's a mistake my equation above.

As vrel does change with the radius the constant should be:

( cl * cos(phi) - cd * sin(phi) ) * vrel^2

But before you continue, could you please post the questions in German? Don't want to get lost in tranlation here.
What exactly is the orientation of the axial force? Parallel to the velocity of the free stream, as i assumed, or maybe parallel to the chord of the blade, as can be seen in this link: http://www.aerospaceweb.org/question/ae ... 0194.shtml
"Torque * 2Pi * f" is nothing without control!
Blanchimont
 
Joined: 9 Nov 2012

Post Fri Sep 05, 2014 9:52 pm

yea i already realized that Vrel changes^^

Image

there are slight differences OR the questions aren't really clear... i'd go with that because they even managed to make the exam questions confusing and very unclear (and not on purpose ... they admitted it during the exam -.-)

you get info in the english version you won't get in the german one and the other way round
RZS10
 
Joined: 7 Dec 2013

Post Sat Sep 06, 2014 8:46 am

So, i believe the formula to find the constant axial force resulting from drag and lift is correct, it should be the part of the forces parallel to the free stream velocity. You can now calculate that constant at the spinner, all the angles are available at this location.

For the other locations between r>0,15m you have to find the combination of cd and cl that solves the equation

( cl * cos(phi) - cd * sin(phi) ) * vrel^2 = constant@spinner

or comes close to it. Let Matlab vary the angle of attack from -8 to +16, and get cl and cd from the tables you define.
"Torque * 2Pi * f" is nothing without control!
Blanchimont
 
Joined: 9 Nov 2012

Post Sat Sep 06, 2014 10:54 pm

so... i ended up with this:
a)
Relative Velocity as function of r
Image

angle of airflow
Image

b)
aoa as function of r
Image

i did as you suggested, calculated
( cl * cos(phi) - cd * sin(phi) ) * vrel^2 = constant@spinner
then used two loops, one going through phi and vrel and therefore getting (constant/vrel^2) for each radius, the second one going through all pairs of cl&cd and calculating ( cl * cos(phi) - cd * sin(phi) ) ...then i took the absolute value of the difference between those results and checked if its smaller than a deviation i allowed (because ==0 would have given no results ofc)

c)
installation angle (should be the angle between Vrel and V)
Image

d)
cd as function of r
Image

dunno what that "ditch" is doing there but whatever :D

still have no clue about e) and f)
RZS10
 
Joined: 7 Dec 2013

Post Sun Sep 07, 2014 8:41 am

Shouldn't this graph start at the point r=0,15m and aoa=16°?

Image
"Torque * 2Pi * f" is nothing without control!
Blanchimont
 
Joined: 9 Nov 2012

Post Sun Sep 07, 2014 9:54 am

it should...normally...yes... but there are no values for cd for a cl above ~0,96 ... i just wrote that i'm limited there by the second diagram and will just start with 8° and assume a linear relationship between cl and aoa...

a friend of mine talked to our profs assistant a few days ago, hoping to get some info from him but the only thing he told her was to start with the highest lift at r=0,15, not the max. aoa that is in the diagram, she asked some more and he confirmed that the assumption of linearity is reasonable ...

i now changed it to cl 1,05 and cd 0,044

managed to got rid of the kink in cd...just had to read the values more precisly :D

what confuses me: "the drag coefficient PER UNIIT LENGHT as a function of radius"
Last edited by RZS10 on Sun Sep 07, 2014 12:56 pm, edited 2 times in total.
RZS10
 
Joined: 7 Dec 2013

Post Sun Sep 07, 2014 12:04 pm

Looks good to me so far!

Next step: During today's race, think about how

d Area = d r * l

and my signature can help you solving parts e and f.
"Torque * 2Pi * f" is nothing without control!
Blanchimont
 
Joined: 9 Nov 2012

Post Sun Sep 07, 2014 3:09 pm

Well i thought that there's the constant thrust at each infinitesimal piece of the propeller or airfoil ... so maybe i have to integrate the thrust across the radius or something
RZS10
 
Joined: 7 Dec 2013

Post Sun Sep 07, 2014 4:06 pm

I can't get to any solution #-o
It's probably very simple .....

I mean... torque M would be the force Q * r/2 and power is in your signature ^^ so P=M*omega or 2pi f
i would still need that goddamn thrust and overall Q :D
RZS10
 
Joined: 7 Dec 2013

Post Sun Sep 07, 2014 5:11 pm

Blanchimont i neeeeed you :D

have to hand that sh** in tomorrow at 9 am

i just did some random stuff and got some results :D better to have something than nothing :D

R = sqrt (A²+W²) and Q = sqrt(S² - R²)

M = Q * r


got 16534 N for the resulting axial force
torque is 4362 Nm
and power is 1644300 W :mrgreen:
RZS10
 
Joined: 7 Dec 2013

Post Sun Sep 07, 2014 6:29 pm

RZS10 wrote:Well i thought that there's the constant thrust at each infinitesimal piece of the propeller or airfoil ... so maybe i have to integrate the thrust across the radius or something


This is the way to go, but your 1,6 MW engine seems to be too powerful :lol:

Did you consider that only the part of the drag and lift force which is perpendicular to the shaft (Q in your sketch) causes the torque?
"Torque * 2Pi * f" is nothing without control!
Blanchimont
 
Joined: 9 Nov 2012

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