Chapter 3 Propeller Theory

Axial Momentum Theory

1. Propeller is regarded as an "actuator disc" which imparts a
sudden increase in pressure to the fluid passing through it.
2. Assumptions
a. The pressure increasing mechanism is ignored.
b. Acceleration of the fluid and thrust generated by the propeller
are uniformly distributed over the disc
c. flow is frictionless and irrotational.
d. unlimited inflow of fluid to the propeller.
3. Acceleration takes place over some distance forward and some
distance aft of the propeller disc to accommodate contraction
of fluid column passing through the propeller disc.
4. The pressure in the fluid decreases gradually as it approaches
the disc, it is suddenly increased at the disc, and it then
gradually decreases as the fluid leaves the disc.
5. Consider a propeller of area 𝐴𝐴𝑜𝑜 advancing into undisturbed
fluid with a velocity 𝑉𝑉𝐴𝐴 .
6. Conditions far ahead of the propeller are 𝑝𝑝𝑜𝑜 and 𝑉𝑉𝐴𝐴
7. Just ahead of the propeller/disc 𝑝𝑝1 and 𝑉𝑉𝐴𝐴 + 𝑣𝑣1
8. Behind the disc 𝑝𝑝1 ′ and 𝑉𝑉𝐴𝐴 + 𝑣𝑣1
9. Far behind the disc 𝑝𝑝2 and 𝑉𝑉𝐴𝐴 + 𝑣𝑣2
10. Continuity (𝜌𝜌𝜌𝜌𝜌𝜌 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) implies equal velocity just ahead
and just behind the disc
11. Irrotational flow implies pressure far behind the propeller
must be equal to the pressure far ahead, i.e. 𝒑𝒑𝟐𝟐 = 𝒑𝒑𝟎𝟎
12. Mass flow per unit time is given by: 𝒎𝒎̇ = 𝝆𝝆𝑨𝑨𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )
13. This mass is accelerated from 𝑉𝑉𝐴𝐴 to a velocity 𝑉𝑉𝐴𝐴 + 𝑣𝑣2 by the
propeller so thrust T is equal to rate of change of axial
momentum
𝑻𝑻 = 𝒎𝒎(𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟐𝟐 − 𝑽𝑽𝑨𝑨 ) = 𝝆𝝆𝑨𝑨𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )𝒗𝒗𝟐𝟐
14. Power delivered = rate of change of KE
𝟏𝟏
𝑷𝑷𝑫𝑫 = 𝒎𝒎�(𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟐𝟐 )𝟐𝟐 − 𝑽𝑽𝟐𝟐𝑨𝑨 �
𝟐𝟐
 𝟏𝟏
⇒ 𝑷𝑷𝑫𝑫 = 𝝆𝝆𝑨𝑨𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )𝒗𝒗𝟐𝟐 �𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟐𝟐 �
𝟐𝟐
𝟏𝟏
⇒ 𝑷𝑷𝑫𝑫 = 𝑻𝑻 �𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟐𝟐 �
𝟐𝟐
15. This delivered power is also equal to the work done by the
thrust on the fluid per unit, time
𝑷𝑷𝑫𝑫 = 𝑻𝑻(𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )
𝟏𝟏
𝒗𝒗𝟏𝟏 = 𝒗𝒗𝟐𝟐
𝟐𝟐
The same result can be obtained by applying Bernoulli’s Equation
successively at the 4 places (just behind/ just after, far ahead / far
behind)
16. The useful work done by the propeller per unit time is 𝑻𝑻𝑽𝑽𝑨𝑨 .
The efficiency of the propeller is therefore:
𝑻𝑻𝑽𝑽𝑨𝑨 𝑻𝑻𝑽𝑽𝑨𝑨 𝟏𝟏 𝟏𝟏
𝜼𝜼𝒊𝒊 = = = 𝒗𝒗 =
𝑷𝑷𝑫𝑫 𝟏𝟏 𝟏𝟏 𝟏𝟏 + 𝒂𝒂
𝑻𝑻 �𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟐𝟐 � 𝟏𝟏 + 𝑽𝑽
𝟐𝟐 𝑨𝑨
𝒗𝒗𝟏𝟏
17. Where 𝒂𝒂 = is the axial inflow factor
𝑽𝑽𝑨𝑨

18. The efficiency 𝜼𝜼𝒊𝒊 is called the "ideal efficiency" because the
only energy loss considered is the kinetic energy lost in the
fluid column behind the propeller, i.e. in the propeller
slipstream
19. Thrust loading coefficient is defined by
𝑻𝑻 𝟐𝟐
𝑪𝑪𝑻𝑻𝑻𝑻 = 𝟏𝟏 𝜼𝜼𝒊𝒊 =
𝝆𝝆𝑨𝑨𝒐𝒐 𝑽𝑽𝟐𝟐𝑨𝑨 𝟏𝟏+�𝟏𝟏+𝑪𝑪𝑻𝑻𝑻𝑻
𝟐𝟐
20. This is an important result, for it shows that the maximum
efficiency of a propeller even under ideal conditions is
limited to a value less than Unity (1)
21. This efficiency decreases as the thrust loading increases.
22. For a given thrust, the larger the propeller the greater its
efficiency, other things being equal.
Textbook Example 3.1
A propeller of 2.0 m diameter produces a thrust of 30.0 kN when
advancing at a speed of 4.0 m/s in sea water. Determine the power
delivered to the propeller, the velocities in the slipstream at the
propeller disc and at a section far astern, the thrust loading
coefficient and the ideal efficiency
Solution:
𝝅𝝅
𝑫𝑫 = 𝟐𝟐. 𝟎𝟎𝒎𝒎, 𝑨𝑨𝒐𝒐 = × 𝑫𝑫𝟐𝟐 = 𝟑𝟑. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝒎𝒎𝟐𝟐 ,
𝟒𝟒
𝒌𝒌𝒌𝒌
𝑻𝑻 = 𝟑𝟑𝟑𝟑. 𝟎𝟎 𝒌𝒌𝑵𝑵, 𝝆𝝆 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝟑𝟑 , 𝑽𝑽𝑨𝑨 = 𝟒𝟒. 𝟎𝟎 𝒎𝒎/𝒔𝒔
𝒎𝒎
𝑻𝑻 = 𝝆𝝆𝑨𝑨𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )𝟐𝟐𝟐𝟐𝟏𝟏
𝟑𝟑𝟑𝟑. 𝟎𝟎 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 × 𝝅𝝅 × (𝟒𝟒. 𝟎𝟎 + 𝒗𝒗𝟏𝟏 ) × 𝟐𝟐𝒗𝒗𝟏𝟏
𝒎𝒎
𝒗𝒗𝟏𝟏 = 𝟎𝟎. 𝟗𝟗𝟒𝟒𝟐𝟐𝟐𝟐 ,
𝒔𝒔
𝒗𝒗𝟏𝟏 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗
𝒂𝒂 = = = 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐,
𝑽𝑽𝑨𝑨 𝟒𝟒. 𝟎𝟎
𝒗𝒗𝟐𝟐 = 𝟏𝟏. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 𝒎𝒎/𝒔𝒔
The slipstream velocity
𝑽𝑽𝑩𝑩 = 𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 = 𝟒𝟒. 𝟎𝟎 + 𝒗𝒗𝟏𝟏 = 𝟒𝟒. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝒎𝒎/𝒔𝒔
 𝑽𝑽𝑪𝑪 = 𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟐𝟐 = 𝟒𝟒. 𝟎𝟎 + 𝒗𝒗𝟐𝟐 = 𝟓𝟓. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 𝒎𝒎/𝒔𝒔
𝟏𝟏 𝟏𝟏
𝜼𝜼𝒊𝒊 = = = 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝟏𝟏 + 𝒂𝒂 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝑻𝑻𝑽𝑽𝑨𝑨 𝟑𝟑𝟑𝟑. 𝟎𝟎 × 𝟒𝟒. 𝟎𝟎
𝑷𝑷𝑫𝑫 = = = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐 𝒌𝒌𝒌𝒌
𝜼𝜼𝐢𝐢 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝑻𝑻 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
𝑪𝑪𝑻𝑻 = = = 𝟏𝟏. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏 𝟏𝟏
𝝆𝝆𝑨𝑨𝒐𝒐 𝑽𝑽𝟐𝟐𝑨𝑨 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 × 𝛑𝛑 × 𝟒𝟒𝟐𝟐
𝟐𝟐 𝟐𝟐
If CT reduces to zero, i.e. T = 0, the ideal efficiency 𝜂𝜂i becomes equal
to 1 (because a = 0).
If, on the other hand, VA tends to zero, 𝜂𝜂i also tends to zero, although
the propeller still produces thrust.
The relation between thrust and delivered power at zero speed of
advance is of interest since this condition represents, the practical
situations of a tug applying a static pull at a bollard or of a ship at a
dock trial.
For an actuator disc propeller, the delivered power is given by:
𝑻𝑻𝑽𝑽𝑨𝑨 𝟏𝟏
𝑷𝑷𝑫𝑫 = = 𝑻𝑻 𝑽𝑽𝑨𝑨 (𝟏𝟏 + �𝟏𝟏 + 𝑪𝑪𝑻𝑻 )
𝜼𝜼𝒊𝒊 𝟐𝟐
𝟐𝟐
𝑺𝑺𝑺𝑺𝒏𝒏𝒄𝒄𝒄𝒄 𝜼𝜼𝒊𝒊 =
𝟏𝟏 + �𝟏𝟏 + 𝑪𝑪𝑻𝑻𝑻𝑻
𝑻𝑻
𝑾𝑾𝑾𝑾𝑾𝑾𝑾𝑾 𝑽𝑽𝑨𝑨 ⟶ 𝟎𝟎, 𝑪𝑪𝑻𝑻 = ⟹ 𝑪𝑪𝑻𝑻 ⟶ ∞
𝟏𝟏
𝝆𝝆𝑨𝑨𝒐𝒐 𝑽𝑽𝟐𝟐𝑨𝑨
𝟐𝟐
𝒔𝒔𝒔𝒔 𝟏𝟏 + �𝟏𝟏 + 𝑪𝑪𝑻𝑻 ⟶ �𝑪𝑪𝑻𝑻
 𝟏𝟏 𝟏𝟏 𝟐𝟐 𝟐𝟐 𝑻𝑻 𝟏𝟏 𝑻𝑻𝟑𝟑
𝑷𝑷𝑫𝑫 = 𝑻𝑻 𝑽𝑽𝑨𝑨 �𝑪𝑪𝑻𝑻 = � 𝑻𝑻 𝑽𝑽𝑨𝑨 =�
𝟐𝟐 𝟒𝟒 𝟏𝟏 𝟐𝟐 𝝆𝝆𝑨𝑨𝒐𝒐
𝝆𝝆𝑨𝑨𝒐𝒐 𝑽𝑽𝟐𝟐𝑨𝑨
𝟐𝟐

𝑻𝑻 𝟏𝟏 𝑻𝑻
⇒ � = √𝟐𝟐
𝑷𝑷𝑫𝑫 𝟐𝟐 𝝆𝝆𝑨𝑨𝒐𝒐

𝟏𝟏 𝑻𝑻𝟑𝟑
⇒ 𝑷𝑷𝑫𝑫 = �
𝟐𝟐 𝝆𝝆𝑨𝑨𝒐𝒐

when 𝑽𝑽𝑨𝑨 ⇢ 𝟎𝟎
Textbook Example 3.2
A propeller of 3.0 m diameter absorbs 700 kW in the static condition
in sea water. What is its thrust?
Solution:
𝝅𝝅
𝑫𝑫 = 𝟑𝟑. 𝟎𝟎𝒎𝒎 𝑨𝑨𝒐𝒐 = × 𝟑𝟑𝟐𝟐 = 𝟕𝟕. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 𝒎𝒎𝟐𝟐 𝑷𝑷𝑫𝑫 = 𝟕𝟕𝟕𝟕𝟕𝟕𝒌𝒌𝒌𝒌
𝟒𝟒
𝝆𝝆 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝒌𝒌𝒌𝒌/𝒎𝒎𝟑𝟑

𝟏𝟏 𝑻𝑻𝟑𝟑
𝑷𝑷𝑫𝑫 = �
𝟐𝟐 𝝆𝝆𝑨𝑨𝒐𝒐

⇒ 𝑻𝑻𝟑𝟑 = 𝟐𝟐 × 𝑷𝑷𝟐𝟐𝑫𝑫 𝝆𝝆𝑨𝑨𝒐𝒐

⇒ 𝑻𝑻𝟑𝟑 = 𝟐𝟐 × (𝟕𝟕𝟕𝟕𝟕𝟕 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏)𝟐𝟐 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 × 𝟕𝟕. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
⇒ 𝑻𝑻𝟑𝟑 = 𝟕𝟕𝟕𝟕𝟎𝟎𝟎𝟎. 𝟑𝟑𝟑𝟑 × 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝐍𝐍 𝟑𝟑 ⇒ 𝑻𝑻 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐𝒌𝒌𝒌𝒌
Momentum Theory Including Rotation
1. Sometimes called the impulse theory
2. Propeller is regarded as imparting axial and angular
acceleration
3. Disc area 𝐴𝐴𝑜𝑜 advancing with 𝑉𝑉𝐴𝐴 and 𝜔𝜔.
4. Superimpose −𝑉𝑉𝐴𝐴 on the whole system so that the propeller is
revolving with an angular velocity 𝜔𝜔 at a fixed position.
5. Axial and angular velocities are 𝑉𝑉𝐴𝐴 + 𝑣𝑣1 𝑎𝑎𝑎𝑎𝑎𝑎 𝜔𝜔1 at the propeller
disc and 𝑉𝑉𝐴𝐴 + 𝑣𝑣2 𝑎𝑎𝑎𝑎𝑎𝑎 𝜔𝜔2 far downstream
6. The mass of fluid through an annular element
𝒅𝒅𝒅𝒅 = 𝝆𝝆𝝆𝝆𝝆𝝆𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )
 𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅[𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟐𝟐 − 𝑽𝑽𝑨𝑨 ] = 𝒅𝒅𝒅𝒅 × 𝒗𝒗𝟐𝟐 = 𝝆𝝆𝝆𝝆𝝆𝝆𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )𝒗𝒗𝟐𝟐
𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅𝒓𝒓𝟐𝟐 [ 𝝎𝝎𝟐𝟐 − 𝟎𝟎] = 𝝆𝝆𝝆𝝆𝝆𝝆𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )𝝎𝝎𝟐𝟐 𝒓𝒓𝟐𝟐
The work done by the element thrust is equal to the increase in the
axial kinetic energy of the fluid flowing through the annular
element/unit time, this is given by (change in KE = WD)
𝟏𝟏
𝒅𝒅𝒅𝒅(𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 ) = 𝒅𝒅𝒅𝒅�(𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟐𝟐 )𝟐𝟐 − 𝑽𝑽𝑨𝑨 𝟐𝟐 �
𝟐𝟐
𝟏𝟏
𝝆𝝆𝝆𝝆𝝆𝝆𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )𝒗𝒗𝟐𝟐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 ) = 𝝆𝝆𝝆𝝆𝝆𝝆𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )𝒗𝒗𝟐𝟐 (𝟐𝟐𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟐𝟐 )
𝟐𝟐
𝟏𝟏
⇒ 𝒗𝒗𝟏𝟏 = 𝒗𝒗𝟐𝟐
𝟐𝟐
The work done per/unit time by the element torque is similarly equal
to the increase in the rotational kinetic energy of the fluid per/time :
 𝟏𝟏
𝒅𝒅𝒅𝒅𝝎𝝎𝟏𝟏 = 𝒅𝒅𝒅𝒅𝒓𝒓𝟐𝟐 � 𝝎𝝎𝟐𝟐 𝟐𝟐 − 𝟎𝟎�
𝟐𝟐
𝟏𝟏
𝒅𝒅𝒅𝒅𝝎𝝎𝟏𝟏 = 𝝆𝝆𝝆𝝆𝝆𝝆𝒐𝒐 (𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 )𝝎𝝎𝟐𝟐 𝒓𝒓𝟐𝟐 𝝎𝝎𝟐𝟐
𝟐𝟐
𝟏𝟏
𝒅𝒅𝒅𝒅𝝎𝝎𝟏𝟏 = 𝒅𝒅𝒅𝒅𝝎𝝎𝟐𝟐
𝟐𝟐
𝟏𝟏
𝝎𝝎𝟏𝟏 = 𝝎𝝎𝟐𝟐
𝟐𝟐

The total power expended by the element must be equal to the increase
in the total kinetic energy (axial and rotational) per unit time, or the
work done by the element thrust and torque on the fluid passing
through the element ,per unit time:
 𝒅𝒅𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅(𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 ) + 𝒅𝒅𝒅𝒅𝝎𝝎𝟏𝟏
𝒅𝒅𝒅𝒅(𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 ) = 𝒅𝒅𝒅𝒅(𝝎𝝎 − 𝝎𝝎𝟏𝟏 )
𝝎𝝎𝟏𝟏
𝑑𝑑𝑑𝑑𝑉𝑉𝐴𝐴 (𝝎𝝎 − 𝝎𝝎𝟏𝟏 )𝑉𝑉𝐴𝐴 𝟏𝟏 − 𝟏𝟏 − 𝒂𝒂/
𝜂𝜂 = = = 𝝎𝝎 =
𝑑𝑑𝑑𝑑𝑑𝑑 ( )
𝑽𝑽𝑨𝑨 + 𝒗𝒗𝟏𝟏 𝜔𝜔 𝟏𝟏 + 𝒗𝒗 𝟏𝟏 𝟏𝟏 + 𝒂𝒂
𝑽𝑽𝑨𝑨
Substituting the known expressions for Ao etc,
𝒅𝒅𝑨𝑨𝒐𝒐 = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐, 𝒗𝒗𝟏𝟏 = 𝒂𝒂𝑽𝑽𝑨𝑨 , 𝒗𝒗𝟐𝟐 = 𝟐𝟐𝟐𝟐𝑽𝑽𝑨𝑨
𝝎𝝎𝟏𝟏 = 𝒂𝒂/ 𝝎𝝎, 𝝎𝝎𝟐𝟐 = 𝟐𝟐𝟐𝟐/ 𝝎𝝎
𝒅𝒅𝒅𝒅 = 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝑽𝑽𝟐𝟐 𝑨𝑨 𝒂𝒂[𝟏𝟏 + 𝒂𝒂]
𝒅𝒅𝒅𝒅 = 𝟒𝟒𝟒𝟒𝟒𝟒𝒓𝒓𝟑𝟑 𝒅𝒅𝒅𝒅𝑽𝑽𝑨𝑨 𝝎𝝎𝒂𝒂/ [𝟏𝟏 + 𝒂𝒂]
𝒅𝒅𝒅𝒅𝑽𝑽𝑨𝑨 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝑽𝑽𝟐𝟐 𝑨𝑨 𝒂𝒂[𝟏𝟏 + 𝒂𝒂]𝑽𝑽𝑨𝑨 𝒂𝒂 𝑽𝑽𝟐𝟐 𝑨𝑨
𝜼𝜼 = = = / 𝟐𝟐 𝟐𝟐
𝒅𝒅𝒅𝒅𝒅𝒅 𝟒𝟒𝟒𝟒𝟒𝟒𝒓𝒓 𝒅𝒅𝒅𝒅𝑽𝑽𝑨𝑨 𝝎𝝎𝒂𝒂 [𝟏𝟏 + 𝒂𝒂]𝝎𝝎 𝒂𝒂 𝝎𝝎 𝒓𝒓
𝟑𝟑 /
 𝒂𝒂 𝑽𝑽𝟐𝟐 𝑨𝑨 𝟏𝟏 − 𝒂𝒂/
𝜼𝜼 = / 𝟐𝟐 𝟐𝟐 =
𝒂𝒂 𝝎𝝎 𝒓𝒓 𝟏𝟏 + 𝒂𝒂
𝒂𝒂/ �𝟏𝟏 − 𝒂𝒂/ �𝝎𝝎𝟐𝟐 𝒓𝒓𝟐𝟐 = 𝒂𝒂(𝟏𝟏 + 𝒂𝒂)𝑽𝑽𝟐𝟐 𝑨𝑨
This is an important result which is useful in solving problems
Example 3.3
A propeller (D = 4 m, N = 180 rpm) is advancing into sea 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 and rotational
inflow factors, and the efficiency of the element.
Solution:
𝑟𝑟𝑟𝑟𝑟𝑟 𝑚𝑚
𝐷𝐷 = 4.0𝑚𝑚 𝑁𝑁 = 180 𝑟𝑟𝑟𝑟𝑟𝑟 = 3 , 𝑉𝑉𝐴𝐴 = 6.0
𝑠𝑠 𝑠𝑠
𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘
𝑟𝑟 = 0.7 × 2.0 = 1.4𝑚𝑚, = 200 , 𝜔𝜔 = 2𝜋𝜋 𝑛𝑛 = 6𝜋𝜋 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠𝑠𝑠𝑠𝑠
𝑑𝑑𝑑𝑑 𝑚𝑚
4𝜋𝜋 × 1025 × 1.4 × 6.02 𝑎𝑎(1 + 𝑎𝑎) = 200 × 1000
𝑎𝑎 = 0.2470
𝑎𝑎′ (1 − 𝑎𝑎′) 𝜔𝜔2 𝑟𝑟 2 = 𝑎𝑎 (1 + 𝑎𝑎) 𝑉𝑉𝐴𝐴2
𝑎𝑎′ �1 − 𝑎𝑎′ �(6𝜋𝜋)2 × 1.42 = 0.2470(1 + 0.2470) × 6.02
𝑎𝑎′ = 0.01619
𝑑𝑑𝑑𝑑
= 4𝜋𝜋 𝜌𝜌𝑟𝑟 3 𝑉𝑉𝐴𝐴 𝜔𝜔𝑎𝑎′ (1 + 𝑎𝑎) =
𝑑𝑑𝑑𝑑
4𝜋𝜋 × 1025 × 1.43 × 6.0 × 6𝜋𝜋 × 0.01619 × 1.2470
= 80.696 𝑘𝑘𝑘𝑘 𝑚𝑚
1 − 𝑎𝑎′ 1 − 0.01619
𝜂𝜂 = = = 0.7889
(1 + 𝑎𝑎) 1 + 0.2470

𝑑𝑑𝑑𝑑
𝑉𝑉𝐴𝐴 200 × 6.0
= 𝑑𝑑𝑑𝑑 = = 0.7889
𝑑𝑑𝑑𝑑
𝜔𝜔 80.696 × 6𝜋𝜋
𝑑𝑑𝑑𝑑
Blade element theory – W. Froude (1878)
1. Thrust generating mechanism, and thrust dependence on
shape of the propeller blades is studied.
2. A propeller blade is regarded as a series of blade elements
that produce hydrodynamic force.
3. The axial component of this hydrodynamic force is the
element thrust.
4. Moment about the propeller axis of the tangential component
is the element torque.
5. The integration of the element thrust and torque for all the
blades gives the total thrust and torque of the propeller.
 • Chord (width) ‘c’
• span (length) ‘s’
• angle of attack 𝛼𝛼
• incident flow of velocity
• density 𝜌𝜌.

6. Wing develops hydrodynamic force

7. Normal & parallel components (to V) are 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝐿𝐿 and 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝐷𝐷.
𝐿𝐿 𝐷𝐷
𝐶𝐶𝐿𝐿 = , 𝐶𝐶𝐷𝐷 =
1 1
𝜌𝜌𝜌𝜌𝑉𝑉 2 𝜌𝜌𝜌𝜌𝑉𝑉 2
2 2
 where 𝐴𝐴 = 𝑠𝑠 × 𝑐𝑐 is the area of the wing plan form.
8. 𝐶𝐶𝐿𝐿 and 𝐶𝐶𝐷𝐷 are determined experimentally in a wind tunnel.
𝑃𝑃
9. Propeller with Z blades, diameter D and pitch ratio
𝐷𝐷

advances with a velocity 𝑉𝑉𝐴𝐴 rotating at ′𝑛𝑛′ rps.

10. Blade element between 𝒓𝒓 and 𝒓𝒓 + 𝒅𝒅𝒅𝒅 when expanded will
have an incident flow whose axial and tangential velocity
components are 𝑽𝑽𝑨𝑨 and 𝟐𝟐𝟐𝟐𝒏𝒏𝒏𝒏 respectively, giving a resultant
velocity 𝑽𝑽𝑹𝑹 at an angle of attack 𝜷𝜷.
 𝑽𝑽𝑨𝑨
𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 =
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
11. The blade element will then produce a lift dL and a drag dD
𝟏𝟏
𝒅𝒅𝒅𝒅 = 𝑪𝑪𝑳𝑳 𝝆𝝆𝝆𝝆𝝆𝝆𝝆𝝆𝑽𝑽𝟐𝟐𝑹𝑹
𝟐𝟐
𝟏𝟏
𝒅𝒅𝒅𝒅 = 𝑪𝑪𝑫𝑫 𝝆𝝆𝝆𝝆𝝆𝝆𝝆𝝆𝑽𝑽𝟐𝟐𝑹𝑹
𝟐𝟐
12. dT and dQ per blade are given by
𝒅𝒅𝒅𝒅
𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 − 𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄(𝟏𝟏 − 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕)
𝒅𝒅𝒅𝒅
𝟏𝟏 𝒅𝒅𝒅𝒅
𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 + 𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅 = 𝒅𝒅𝒅𝒅 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄(𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 + )
𝒓𝒓 𝒅𝒅𝒅𝒅
 𝑽𝑽𝑨𝑨 𝒅𝒅𝒅𝒅
𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠 𝑡𝑡ℎ𝑎𝑎𝑎𝑎 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 = , Let, 𝒕𝒕𝒕𝒕𝒕𝒕 𝜸𝜸 =
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 𝒅𝒅𝒅𝒅

For Z Baldes,
1
𝑑𝑑𝑑𝑑 = 𝑍𝑍 𝐶𝐶𝐿𝐿 𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌 𝑉𝑉𝑅𝑅2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (1 − 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
2
1
𝑑𝑑𝑑𝑑 = 𝑟𝑟𝑟𝑟 𝐶𝐶𝐿𝐿 𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌 𝑉𝑉𝑅𝑅2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
2
13. The efficiency of the blade element is:
𝒅𝒅𝒅𝒅𝑽𝑽𝑨𝑨 𝑽𝑽𝑨𝑨 𝟏𝟏 − 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
𝜼𝜼 = = � �
𝒅𝒅𝒅𝒅𝟐𝟐𝝅𝝅𝝅𝝅𝒓𝒓 𝟐𝟐𝝅𝝅𝝅𝝅𝝅𝝅 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 + 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
𝑽𝑽𝑨𝑨 𝟏𝟏 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕
= × =
(
𝟐𝟐𝝅𝝅𝝅𝝅𝝅𝝅 𝒕𝒕𝒕𝒕𝒕𝒕 𝜷𝜷 + 𝜸𝜸) 𝒕𝒕𝒕𝒕𝒕𝒕(𝜷𝜷 + 𝜸𝜸)
14. If the propeller works in ideal conditions drag = 0, and
𝑡𝑡𝑡𝑡𝑡𝑡 𝛾𝛾 = 0, 𝜂𝜂 = 1.
15. Momentum theory states that if a propeller produces a thrust
greater than zero, its efficiency even in ideal conditions must
be 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'.
With induced velocities

1
𝑑𝑑𝑑𝑑 = 𝑍𝑍 𝐶𝐶𝐿𝐿 𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌 𝑉𝑉𝑅𝑅2 𝑐𝑐𝑐𝑐𝑐𝑐𝛽𝛽𝑖𝑖 (1 − 𝑡𝑡𝑡𝑡𝑡𝑡𝛽𝛽𝑖𝑖 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
2
1
𝑑𝑑𝑑𝑑 = 𝑟𝑟𝑟𝑟 𝐶𝐶𝐿𝐿 𝜌𝜌𝜌𝜌𝜌𝜌𝜌𝜌 𝑉𝑉𝑅𝑅2 𝑐𝑐𝑐𝑐𝑐𝑐𝛽𝛽𝑖𝑖 (𝑡𝑡𝑡𝑡𝑡𝑡𝛽𝛽𝑖𝑖 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
2
𝑑𝑑𝑑𝑑𝑉𝑉𝐴𝐴 𝑉𝑉𝐴𝐴 1 − 𝑡𝑡𝑡𝑡𝑡𝑡𝛽𝛽𝑖𝑖 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
𝜂𝜂 = = � �=
𝑑𝑑𝑑𝑑2𝜋𝜋𝜋𝜋 2𝜋𝜋𝜋𝜋𝜋𝜋 𝑡𝑡𝑡𝑡𝑡𝑡𝛽𝛽𝑖𝑖 + 𝑡𝑡𝑎𝑎𝑎𝑎𝑎𝑎 𝑡𝑡𝑡𝑡𝑡𝑡(𝛽𝛽𝑖𝑖 + 𝛾𝛾)
 𝑉𝑉𝐴𝐴 𝑉𝑉𝐴𝐴 (1+𝑎𝑎) (1+𝑎𝑎)
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = and 𝑡𝑡𝑡𝑡𝑡𝑡𝛽𝛽𝑖𝑖 = = 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
2𝜋𝜋𝜋𝜋𝜋𝜋 2𝜋𝜋𝜋𝜋𝜋𝜋�1−𝑎𝑎/ � �1−𝑎𝑎/ �

�1 − 𝑎𝑎′ � 𝑡𝑡𝑡𝑡𝑡𝑡𝛽𝛽𝑖𝑖
𝜂𝜂 =
(1 + 𝑎𝑎) 𝑡𝑡𝑡𝑡𝑡𝑡(𝛽𝛽𝑖𝑖 + 𝛾𝛾)
17. Expression for efficiency consists of three factors:
• 1/(1 + 𝑎𝑎), which is associated with the axial induced
velocity
• (1 -a'), which reflects the loss due to the rotation of the
slipstream
𝑡𝑡𝑡𝑡𝑡𝑡𝛽𝛽𝑖𝑖
• indicates the effect of blade element drag.
𝑡𝑡𝑡𝑡𝑡𝑡(𝛽𝛽𝑖𝑖 +𝛾𝛾 )
18. If there is no drag and 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 0, the expression for efficiency,
becomes identical to the expression obtained from the
momentum theory.
19. Necessary to know 𝐶𝐶𝐿𝐿 , 𝐶𝐶𝐷𝐷 , 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎′ for blade elements at
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
different radii so that and can be determined and
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

integrated with respect to the radius r.

20. 𝐶𝐶𝐿𝐿 , 𝐶𝐶𝐷𝐷 may be obtained from experimental data, and 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎′
with the help of the momentum theory.
21. Unfortunately, this procedure does not yield realistic results
because it neglects a number of factors.
Textbook Example 3.4
A four bladed propeller of 3.0 m diameter and 1.0 constant pitch
ratio 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 no-lift angle of
2 degrees, a lift-drag ratio of 30 and a lift coefficient that increases at
the 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 and
rotational inflow factors are 0.2000 and 0.0225 respectively.
Solution:
𝑃𝑃 𝑚𝑚
𝑍𝑍 = 4, 𝐷𝐷 = 3.0𝑚𝑚, = 1.0, VA = 4.0 , 𝑛𝑛 = 120 𝑟𝑟𝑟𝑟𝑟𝑟 = 2.0/𝑠𝑠
𝐷𝐷 𝑠𝑠
𝑜𝑜 𝐶𝐶𝐿𝐿 𝜕𝜕𝐶𝐶𝐿𝐿
𝑥𝑥 = 0.7, 𝑐𝑐 = 0.5𝑚𝑚 𝛼𝛼𝑜𝑜 = 2 , = 30, = 6.0 / 𝑟𝑟𝑟𝑟𝑟𝑟,
𝐶𝐶𝐷𝐷 𝜕𝜕𝜕𝜕
𝜌𝜌 = 1025 𝑘𝑘𝑘𝑘/𝑚𝑚3
(a) Neglecting induced velocities:
𝑃𝑃
𝐷𝐷 1.0
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = = = 0.4547 𝜑𝜑 = 24.4526°
𝜋𝜋𝜋𝜋 𝜋𝜋 × 0.7
𝑉𝑉𝐴𝐴 4.0
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = = = 0.3032 𝛽𝛽 = 16.8648°
2𝜋𝜋𝜋𝜋𝜋𝜋 2𝜋𝜋 × 2.0 × (0.7 × 1.5)
𝐶𝐶𝐷𝐷 1
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = = = 0.03333, 𝛾𝛾 = 1.9091° , 𝛼𝛼 = 𝜑𝜑 − 𝛽𝛽 = 7.5878°
𝐶𝐶𝐿𝐿 30
 𝜕𝜕𝐶𝐶𝐿𝐿 2 + 7.5878
𝐶𝐶𝐿𝐿 = (𝛼𝛼0 + 𝛼𝛼 ) = 6.0 = 1.0040
𝜕𝜕𝜕𝜕 180
𝜋𝜋
𝑉𝑉𝑅𝑅2 = 𝑉𝑉𝐴𝐴2 + (2𝜋𝜋𝑛𝑛𝑛𝑛)2 = 4.02 + (2𝜋𝜋 × 2.0 × 1.05)2 =
= 190.1 𝑚𝑚2 /𝑠𝑠 2
𝑑𝑑𝑑𝑑 1
= 𝑍𝑍 𝐶𝐶𝐿𝐿 𝜌𝜌𝜌𝜌 𝑉𝑉𝑅𝑅2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (1 − 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
𝑑𝑑𝑑𝑑 2
𝑑𝑑𝑑𝑑 1
= 𝑟𝑟 𝑍𝑍 𝐶𝐶𝐿𝐿 𝜌𝜌𝜌𝜌 𝑉𝑉𝑅𝑅2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (𝑡𝑡𝑡𝑡𝑡𝑡 𝛽𝛽 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
𝑑𝑑𝑑𝑑 2
Substituting the numerical values, we get
𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑
= 185.333 = 66.148 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚
𝑑𝑑𝑑𝑑 𝑚𝑚 𝑑𝑑𝑑𝑑
tanβ
𝜂𝜂 = = 0.8918
tan(β + γ)
(b) 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 𝑎𝑎 = 0.2, 𝑎𝑎́ = 0.0225
𝑉𝑉𝑅𝑅2 = [(1 + 𝑎𝑎)𝑉𝑉𝐴𝐴 ]2 + [(1 − 𝑎𝑎́ )2𝜋𝜋𝜋𝜋𝜋𝜋]2
⇒ 𝑉𝑉𝑅𝑅2 = [(1 + 0.2)4.0]2 + [(1 − 0.0225)2𝜋𝜋 × 2 × 1.05]2
= 189.3935 𝑚𝑚2 /𝑠𝑠 2
𝑉𝑉𝐴𝐴 (1 + 𝑎𝑎) (1 + 0.2)4.0
tan(𝛽𝛽𝑖𝑖 ) = = = 0.3722
(1 − 𝑎𝑎́ )2𝜋𝜋𝜋𝜋𝜋𝜋 (1 − 0.0225)2𝜋𝜋 × 2 × 1.05
𝛽𝛽𝑖𝑖 = 20.4131𝑜𝑜
𝛼𝛼 = 𝜑𝜑 − 𝛽𝛽𝑖𝑖 = 24.4526 − 20.4131 = 4.0395
𝜕𝜕𝐶𝐶𝐿𝐿 2 + 4.0395
𝐶𝐶𝐿𝐿 = (𝛼𝛼𝑜𝑜 + 𝛼𝛼 ) = 6.0 × = 0.6325
𝜕𝜕𝜕𝜕 180
𝜋𝜋
𝑑𝑑𝑑𝑑 1
= 𝑍𝑍 𝐶𝐶𝐿𝐿 𝜌𝜌𝜌𝜌 𝑉𝑉𝑅𝑅2 𝑐𝑐𝑐𝑐𝑐𝑐𝛽𝛽𝑖𝑖 (1 − 𝑡𝑡𝑡𝑡𝑡𝑡𝛽𝛽𝑖𝑖 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
𝑑𝑑𝑑𝑑 2
 𝑑𝑑𝑑𝑑 1
= 𝑟𝑟 𝑍𝑍 𝐶𝐶𝐿𝐿 𝜌𝜌𝜌𝜌 𝑉𝑉𝑅𝑅2 𝑐𝑐𝑐𝑐𝑐𝑐𝛽𝛽𝑖𝑖 (𝑡𝑡𝑡𝑡𝑡𝑡 𝛽𝛽𝑖𝑖 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡)
𝑑𝑑𝑑𝑑 2
𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘
= 113.64
𝑑𝑑𝑑𝑑 𝑚𝑚
𝑑𝑑𝑑𝑑 𝑘𝑘𝑘𝑘. 𝑚𝑚
= 48.991
𝑑𝑑𝑑𝑑 𝑚𝑚
(1 − 𝑎𝑎́ ) tanβi 1 − 0.0225 0.3722
𝜂𝜂 = = × = 0.7383
(1 + 𝑎𝑎) tan(βi + γ) 1 + 0.2 0.4104