25-Nav-B3 Finite Element Analysis for Ship Structures · May 2016

Question 1 of 7: Propulsion

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

Question 1: Propulsion (20 marks: a–5, b–15)

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.

Part (a). At a blade section 0.7R from the shaft axis, two velocity components combine into the resultant inflow the blade section actually "sees." The rotational component is the local peripheral speed, $2\pi n r$ (perpendicular to the shaft axis, in the plane of rotation), where $r=0.7R$ and $n$ is the rate of rotation. The translational component is the speed of advance $V_a$ (parallel to the shaft axis), which already accounts for the wake fraction reducing the ship's speed as seen by the propeller. Vector-summing these gives the resultant velocity $V_R=\sqrt{V_a^2+(2\pi n r)^2}$, inclined to the plane of rotation by the hydrodynamic advance angle $\beta=\tan^{-1}\!\big(V_a/2\pi n r\big)$.

The pitch angle $\theta$ is defined purely by the blade geometry: for a helix of pitch $P$ at radius $r$, $\theta=\tan^{-1}\!\big(P/2\pi r\big)$. For the blade to work efficiently, $\theta$ must sit a few degrees above $\beta$ (the inflow angle) — that difference is the angle of attack $\alpha$ that produces useful lift/thrust; too large an $\alpha$ stalls the section (cavitation risk), too small produces negligible thrust.

Part (b).

Given.

Find. Whether the propeller, turning at the given rate, delivers enough thrust (and stays within the installed power) for the ship to reach its 10 m/s design speed.

Approach. Compute the advance coefficient the propeller actually sees at the design speed, confirm it matches the manufacturer's open-water point, then compare the thrust it delivers there against the thrust the hull requires, with the delivered power as a cross-check against the installed power.

1. Speed of advance and advance coefficient. The wake fraction reduces the ship speed seen by the propeller: $V_a=V_s(1-w)=10(1-0.25)=7.5\ \text{m/s}$. The advance coefficient is then $J=\dfrac{V_a}{ND}=\dfrac{7.5}{4.5\times3.5}=0.476$, matching the manufacturer's tabulated peak-efficiency point $J=0.475$ — the propeller is indeed being run at its most efficient open-water point for this speed/rpm combination.
2. Thrust required by the hull. Thrust deduction means the propeller must supply more thrust than the bare-hull resistance: $T_{req}=\dfrac{R_T}{1-t}=\dfrac{200}{1-0.20}=\boxed{250\ \text{kN}}$.
3. Thrust the propeller delivers at this $J$. $T_{avail}=K_t\,\rho\,N^2D^4=0.075\times1025\times4.5^2\times3.5^4=233{,}605\ \text{N}=\boxed{233.6\ \text{kN}}$.
4. Compare. $T_{avail}=233.6\ \text{kN} < T_{req}=250\ \text{kN}$ — a shortfall of about 6.6%. At $N=4.5$ rps the propeller cannot generate enough thrust to hold the ship at 10 m/s; the ship's actual self-propulsion speed will settle slightly below the design speed (a lower $V_s$ reduces $R_T$ and raises the propeller's thrust as $J$ falls, until the two curves cross).
5. Cross-check with delivered power. Torque at this point: $Q=K_q\,\rho\,N^2D^5=0.012\times1025\times4.5^2\times3.5^5=130{,}819\ \text{N}\!\cdot\!\text{m}$. Delivered (shaft) power: $P_D=2\pi NQ=2\pi(4.5)(130{,}819)=\boxed{3699\ \text{kW}}$, which is 1.4% below the 3750 kW installed — the engine has essentially no spare power margin left to drive the propeller faster and recover the thrust shortfall.