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

Question 8 of 10: Cavitation Check by Burrill's Method

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

Question 8: Cavitation Check by Burrill's Method (10 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. Propeller from Question 7: $D=4.0$ m, $n=2.458$ rps, $P/D=0.88$, $Z=4$, $A_E/A_O=0.55$, $T=348.9$ kN, $V_A=5.90$ m/s. Hub depth $h_0=3.6$ m; $p_v=11$ kPa; $p_{atm}=101$ kPa; $\rho=1025\ \text{kg/m}^3$ (sea water, 15 °C).

Find. (i) the approximate extent of back (suction-side) cavitation; (ii) the expanded area ratio needed to hold cavitation to 5%.

Approach. Compute the local cavitation number $\sigma_{0.7R}$ and thrust-loading coefficient $\tau_c$ at the $0.7R$ blade section from the data-sheet formulae, converting the expanded blade area $A_E$ to the projected area $A_P$ that Burrill's method uses, then locate the point on the Burrill diagram against the family of back-cavitation-percentage contours. Because $\sigma_{0.7R}$ does not depend on blade area, only $\tau_c$ moves when the area is changed for part (ii).

1. Static pressure at the hub and local cavitation number. $$p_0=p_{atm}+\rho gh_0=101{,}000+(1025)(9.806)(3.6)=137{,}184\ \text{Pa}.$$ $$V_R^2=V_A^2+(0.7\pi nD)^2=(5.90)^2+(0.7\pi\cdot2.458\cdot4.0)^2=502.5\ \text{m}^2/\text{s}^2\ \Rightarrow\ V_R=22.42\ \text{m/s}.$$ $$q_{0.7R}=\tfrac12\rho V_R^2=\tfrac12(1025)(502.5)=257{,}500\ \text{Pa}.$$ $$\sigma_{0.7R}=\frac{p_0-p_v}{q_{0.7R}}=\frac{137{,}184-11{,}000}{257{,}500}=\boxed{0.490}.$$
2. Projected blade area and thrust-loading coefficient. $$A_O=\frac{\pi D^2}{4}=12.57\ \text{m}^2,\qquad A_E=0.55A_O=6.91\ \text{m}^2.$$ $$A_P=A_E(1.067-0.229\,P/D)=6.91(1.067-0.229\times0.88)=5.98\ \text{m}^2.$$ $$\tau_c=\frac{T}{A_Pq_{0.7R}}=\frac{348{,}900}{(5.98)(257{,}500)}=\boxed{0.226}.$$
3. Locate the point on the Burrill chart. At $\sigma_{0.7R}=0.49$, $\tau_c=0.226$ falls just above the "suggested upper limit for merchant ship propellers" curve, essentially on the $\boxed{5\%\text{ back-cavitation}}$ contour — a fully loaded but acceptable merchant-ship design point, not a heavily cavitating one.
4. (ii) Re-size the blade area for exactly 5% cavitation. Since $\sigma_{0.7R}$ is independent of blade area (it depends only on depth, pressure and $V_R$), only $\tau_c$ needs to move down to the 5% contour value at the same $\sigma_{0.7R}=0.49$, read as $\tau_{c,5\%}\approx0.20$. Because $\tau_c\propto1/A_P$: $$A_{P,new}=\frac{T}{\tau_{c,5\%}\,q_{0.7R}}=\frac{348{,}900}{(0.20)(257{,}500)}=6.77\ \text{m}^2.$$ $$A_{E,new}=\frac{A_{P,new}}{1.067-0.229\,P/D}=\frac{6.77}{0.865}=7.83\ \text{m}^2\ \Rightarrow\ \left(\frac{A_E}{A_O}\right)_{new}=\frac{7.83}{12.57}=\boxed{0.62}.$$