Question 5 of 9: Preliminary Propeller Design for a Coastal Tanker

Question 5: Preliminary Propeller Design for a Coastal Tanker (15 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. $V_S=12$ kn; MCR $=790$ kW, derating $=0.90$, service allowance $=0.15$; shaft efficiency $\eta_S=0.98$, transmission efficiency $\eta_M=0.97$; shaft speed $n=360$ rpm $=6.0$ rps; max. diameter $D=1.7$ m; shaft depth $h_0=2.40$ m; calm-water resistance with appendages $R=55.0$ kN at 12 kn; wake fraction $w=0.24$; sea water at 15 °C ($\rho=1025\ \text{kg/m}^3$). Propeller is a 4-bladed Wageningen B4-55 series ($Z=4$, $A_E/A_O=0.55$).

Find. (i) pitch ratio $P/D$; (ii) thrust delivered; (iii) open-water efficiency $\eta_o$; (iv) propulsive efficiency $\eta_p$.

Approach. Work down the power train from the nameplate rating to the power actually delivered to the propeller; combine that with the fixed shaft speed and diameter to get the required torque coefficient; enter the B4-55 open-water chart at the resulting advance coefficient $J$ to select the pitch ratio whose $10K_Q$ curve matches, then read $K_T$ and $\eta_o$ from the same operating point. Because delivered power $P_D$ and effective power $P_E=RV_S$ are both known independently, the overall propulsive efficiency follows directly without needing the thrust-deduction fraction.

1. Power actually available for the calm-water design condition. The 790 kW MCR already includes a 15% service allowance (reserve for fouling/weather) above the calm-water design point, and the engine is run continuously at 90% of MCR: $$P_{B}=\frac{\text{MCR}\times\text{derating}}{1+\text{service allowance}}=\frac{790\times0.90}{1.15}=\boxed{618.3\ \text{kW}}.$$
2. Power delivered to the propeller. Passing through the shaft and transmission (gearbox) losses, $$P_D=P_B\,\eta_S\,\eta_M=(618.3)(0.98)(0.97)=\boxed{587.7\ \text{kW}}.$$
3. Speed of advance and advance coefficient. $$V_A=V_S(1-w)=(6.173)(0.76)=4.691\ \text{m/s},\qquad J=\frac{V_A}{nD}=\frac{4.691}{(6.0)(1.7)}=\boxed{0.460}.$$
4. Required torque and torque coefficient. From $P_D=2\pi nQ$, $$Q=\frac{P_D}{2\pi n}=\frac{587{,}700}{2\pi(6.0)}=15{,}590\ \text{N}\cdot\text{m},\qquad K_{Q,\text{req}}=\frac{Q}{\rho n^2D^5}=\frac{15{,}590}{(1025)(6.0)^2(1.7)^5}=0.0298,$$ so $10K_{Q,\text{req}}=0.298$.
5. Select $P/D$ from the chart, read $K_T$ and $\eta_o$. On the B4-55 chart at $J=0.46$, the $10K_Q$ curve for $\boxed{P/D\approx0.90}$ passes through $10K_Q=0.298$ (part i); at the same point, $K_T\approx0.224$ and $\eta_o\approx0.55$. (Chart values are read to the precision the supplied sheet allows; a full-size printed chart would tighten these to about $\pm0.01$.)
6. Thrust delivered. $$T=K_T\,\rho\,n^2D^4=(0.224)(1025)(6.0)^2(1.7)^4=\boxed{68.9\ \text{kN}}.$$ As a consistency check, $R=T(1-t)$ gives an implied thrust-deduction fraction $t=1-R/T=1-55.0/68.9\approx0.20$ — a thoroughly typical value for a single-screw merchant hull, supporting the chart reading. That is part (ii).
7. Open-water efficiency. Read directly from the chart at the same point (part iii): $$\eta_o=\frac{K_TJ}{2\pi K_Q}\approx\boxed{0.55}.$$
8. Propulsive efficiency. With $P_E=RV_S$ and $P_D$ from Step 2, both already known independently of the chart, $$P_E=RV_S=(55{,}000)(6.173)=339.5\ \text{kW},\qquad \eta_p=\eta_D=\frac{P_E}{P_D}=\frac{339.5}{587.7}=\boxed{0.578}.$$ That is part (iv).