25-Nav-A2 Hydrodynamics of Ships (I)_ Resistance and Propulsion · Undated paper

Question 10 of 10: Wake-Induced Variation in Blade Angle of Attack and Lift

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 10: Wake-Induced Variation in Blade Angle of Attack and Lift (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Twin-screw shaft/strut/propeller arrangement; four points A, B, C, D at the $0.7R$ radius fraction, with A at top-dead-centre directly behind the supporting strut, C at bottom-dead-centre, B and D at the 3- and 9-o'clock positions. Measured local axial (advance) velocities: $V_{a,A}=0.2V_A$, $V_{a,B}=0.8V_A$, $V_{a,C}=1.2V_A$, $V_{a,D}=0.8V_A$. Tangential (rotational) speed at $0.7R$, $U=2\pi n(0.7R)$, is constant regardless of position (it depends only on shaft speed, not on the wake field).

Find. How the axial-velocity variation changes (i) the local flow angle and angle of attack, and (ii) the resulting incremental lift/thrust, as the blade sweeps through one revolution.

Approach. At each angular position the blade section sees a resultant velocity that is the vector sum of the (constant) tangential speed $U$ and the (position-dependent) axial inflow $V_a$; a lower $V_a$ steepens the resultant velocity relative to the blade's fixed pitch line, raising the angle of attack, which (below stall) raises the lift coefficient and hence the local thrust — so the strut-shadow wake pattern is mapped, point by point, through to a cyclic (once-per-revolution) thrust and torque load on each blade.

1. Set up the flow angle at each point. The local hydrodynamic flow angle is $$\beta=\tan^{-1}\!\left(\frac{V_a}{U}\right),$$ with $U$ fixed by shaft speed alone. Since the blade's geometric pitch angle $\varphi$ is fixed by blade geometry, the angle of attack is $\alpha=\varphi-\beta$: a smaller $V_a$ gives a smaller $\beta$ and therefore a larger $\alpha$, and vice versa.
2. Apply this at each point. At A (strut wake, $V_a=0.2V_A$, most reduced) $\beta$ is smallest, so $\alpha$ is largest — the blade section is momentarily working at its highest angle of attack and highest local lift coefficient of the whole revolution. At C ($V_a=1.2V_A$, above the free-stream mean, clear of the strut) $\beta$ is largest, so $\alpha$ is smallest — lowest lift of the revolution. At B and D ($V_a=0.8V_A$, moderately reduced, symmetric either side of the strut) $\alpha$ takes an intermediate value.
3. Translate to lift and thrust. Below stall, lift coefficient increases monotonically with $\alpha$, and (from Question 6) the elemental thrust $dT=L\cos\varphi-D\sin\varphi$ scales with $L$. So as a blade sweeps A→B→C→D→A once per revolution, its local lift and thrust rise sharply passing behind the strut (A), fall to a minimum opposite it (C), and pass through intermediate values at B and D — a once-per-revolution (blade-passing-frequency, and for a $Z$-bladed propeller, $Z\times$ shaft-frequency overall) cyclic load, rather than the steady load an axisymmetric (uniform) wake would produce.
4. Consequence. This cyclic loading is the classic strut-wake-induced unsteady propeller force: it drives blade-rate vibration transmitted through the shaft and bearings into the hull, and because point A also carries the highest angle of attack, it is the point in the disk most prone to (unsteady) sheet-cavitation inception — a standard design concern on twin-screw, strut-supported shafting that blade skew is often used to mitigate.