proftvkb blade element theory
TRANSCRIPT
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6. The wing develops a hydrodynamic force whose componentsnormal and parallel to, V are the lift L and the drag D.
Consider a wing of chord (width) cand span (length) s at an angle of
attack to an incident flow ofvelocity V in a fluid of density .
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=
12 2 , = 122
where = istheareaofthewingplanform.
7.
These coefficients depend upon the shape of the wingsection, the aspect ratio s/c and the angle of attack, and areoften determined experimentally in a wind tunnel.
8. These experimental values may then be used in the bladeelement theory, which may thus be said to rest on observedfact.
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9.
Consider a propeller with Z blades, diameter D and pitchratio advancing into undisturbed water with a velocity while rotating at rps.
10. The blade element between the radii rand r + dr whenexpanded will have an incident flow whose axial andtangential velocity components are and 2 respectively,giving a resultant velocity at an angle of attack .
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11.
The blade element will then produce a lift dL and a drag dD = 122
= 122
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12. If the thrust and torque produced by the elements between rand r + dr for all the Z blades are dT and dQ, the1 = = (1
)
1 = + = ( )
, Putting tan =
= 12 (1 )
= 12
( )13.
The efficiency of the blade element is then:
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= 2=
2
1
= 21
( + ) ( + )
14. If the propeller works in ideal conditions = 0, = 1.15.
Momentum theory states that if a propeller produces a thrustgreater than zero, its efficiency even in ideal conditions mustbe less than 1.
16.
The
primary
reason
for
this
discrepancy
lies
in
the
neglect
of
the
induced
velocities,
i.e.
the
inflow
factors
a,
a'.
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17. With induced velocities
= 12 (1 )
= 12
( ) = 2=
2
1 =
+
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and
/
/ =1
1 +
+
18. Expression for efficiency consists of three factors:o 1/(1 + ), which is associated with the axial induced
velocityo
(1 a'), which reflects the loss due to the rotation of theslipstream
o
+which indicates the effect of blade element drag.
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19. If there is no drag and 0, the expression for efficiency,becomes identical to the expression obtained from themomentum theory.
20.
Necessary to know , , for blade elements atdifferent radii so that and can be determined and
integrated with respect to the radius r.21. , may be obtained from experimental data, and
with the help of the momentum theory.22.
Unfortunately, this procedure does not yield realistic resultsbecause it neglects a number of factors. Example 3
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Example
A propeller of diameter 4.0 m rotating at 180 rpm is advancing intosea water at a speed of 6.0 m/s. The element of the propeller at 0.7R
produces a thrust of 200 kN/m. Determine the torque, the axial androtational inflow factors, and the efficiency of the element.Solution:
= 4.0 = 180 =3.0s = 6.0 = 0.7 = 0.7 2.0 = 1.4
= 200
= 2 = 6 / ,
4 1025 1.4 6.02(1 + ) = 200 1000
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= 0.2470 (1 ) 2 2 = (1 + ) 2
1 62 1.4 2 = 0.2470(1 + 0.2470) 6.0 2
= 0.01619= 4
3 1 + =
4 1025 1.43
6.0 6 0.01619 1.2470= 80.696
=
1
1 + =
1 0.01619
1 +0.2470= 0.7889=
= 2006.080.6966= 0.7889
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Example4
A four bladed propeller of 3.0 m diameter and 1.0 constant pitchratio has a speed of advance of 4.0 m per sec when running at 120
rpm. The blade section at 0.7R has a chord of 0.5 m, a nolift angle of2 degrees, a liftdrag ratio of 30 and a lift coefficient that increases atthe rate of 6.0 per radian for small angles of attack. Determine the
thrust, torque and efficiency of the blade element at 0.7R (a)neglecting the induced velocities and (b) given that the axial androtational inflow factors are 0.2000 and 0.0225 respectively.Solution:
= 4 = 3.0 = 1.0 V A 4.0, = 120 = 2.0/
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== 0.7 = 0.5 = 2
= 30,
= 6.0 ,
= 1025 /3(a) Neglecting induced velocities:
==
1.0 0.7= 0.4547 = 24.4526
= 2= 4.02 2.0 0.7 1.5= 0.3032 = 16.8648
130 = 0.03333, 1.9091
= = 7.5878
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= 0+ = 6.0
2 + 7.5878180 = 1.0040
2 = 2 + 22= 4.0 2 + (2 2.0 1.05) 2
= 190.0998 2/2=
12
(1 )
= 12 2 ( + )Substituting the numerical values calculated:
= 185.333 = 66.148 / = tantan + = 0.8918