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

Question 7 of 10: Preliminary Propeller Selection for a Product Tanker

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

Notes on this paper

National Exams — May 2019 — 16-Nav-A2 Hydrodynamics of Ships I: Resistance and Propulsion. Three-hour, closed-book exam; a data sheet, a Wageningen B4-55 propeller chart and a Burrill cavitation chart are supplied. Format: Questions 1–8 are compulsory (attempt all eight), then one of Questions 9 or 10. All ten are solved below for completeness.

Reference texts: Larsson & Raven, Ship Resistance and Flow (SNAME) — model-scale resistance testing, Froude/Reynolds scaling, boundary-layer estimates and the ITTC 1978 performance-prediction method; Lewis (ed.), Principles of Naval Architecture, Vol. II — Resistance, Propulsion and Vibration (SNAME) — propeller geometry, wave-pattern interference, open-water B-series design and wake-induced blade loading; Carlton, Marine Propellers and Propulsion (Butterworth-Heinemann) — Wageningen B-series charts, the Burrill back-cavitation criterion and open-water model testing.

Question 7: Preliminary Propeller Selection for a Product Tanker (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. $V_S=15.25$ kn $=7.845$ m/s; $D=4.0$ m; ship resistance $R=305.6$ kN at design speed; thrust deduction $t=0.124$; speed of advance $V_A=5.90$ m/s; shaft speed $n=147.5$ rpm $=2.458$ rps; sea water at 15 °C ($\rho=1025\ \text{kg/m}^3$). Propeller is a 4-bladed Wageningen B4-55 series ($Z=4$, $A_E/A_O=0.55$ by definition of the series; developed area $=$ expanded area, as instructed).

Quantity	Value
Design speed $V_S$	7.845 m/s (15.25 kn)
Diameter $D$	4.0 m
Resistance $R$	305.6 kN
Thrust deduction $t$	0.124
Speed of advance $V_A$	5.90 m/s
Shaft speed $n$	2.458 rps

Find. (i) $P/D$ ratio, corresponding torque and delivered power at the design condition; (ii) open-water efficiency $\eta_o$ and hydrodynamic propulsive efficiency $\eta_D$.

Approach. Convert resistance to required thrust via the thrust deduction fraction, form the advance coefficient $J$ from the fixed $n$ and $D$, compute the required thrust coefficient $K_T$, then enter the B4-55 open-water chart at that $(J,K_T)$ point to read off $P/D$ and $10K_Q$. Delivered power and $\eta_o$ follow from the chart point directly; $\eta_D=\eta_H\eta_B$ needs the wake fraction recovered from $V_A$ and $V_S$.

Required thrust and advance coefficient. $$T=\frac{R}{1-t}=\frac{305.6}{1-0.124}=348.9\ \text{kN}.$$ $$J=\frac{V_A}{nD}=\frac{5.90}{(2.458)(4.0)}=\boxed{0.600}.$$
Required thrust coefficient. $$K_{T,req}=\frac{T}{\rho n^2D^4}=\frac{348{,}900}{(1025)(2.458)^2(4.0)^4}=\boxed{0.220}.$$
Read the B4-55 chart at $J=0.600$, $K_T=0.220$. The $P/D\approx0.88$ curve passes through $K_T=0.220$ at $J=0.600$; at the same point $10K_Q\approx0.31$.
Torque and delivered power. $$Q=K_Q\rho n^2D^5=(0.031)(1025)(2.458)^2(4.0)^5=197\ \text{kN}\cdot\text{m}.$$ $$P_D=2\pi nQ=2\pi(2.458)(197{,}000)=\boxed{3.04\ \text{MW}}.$$
(ii) Open-water efficiency, directly from the chart point. $$\eta_o=\frac{K_TJ}{2\pi K_Q}=\frac{(0.220)(0.600)}{2\pi(0.031)}=\boxed{0.678}.$$
Hydrodynamic propulsive efficiency. Wake fraction from $V_A=V_S(1-w)$: $w=1-V_A/V_S=1-5.90/7.845=0.248$. Hull efficiency $\eta_H=(1-t)/(1-w)=0.876/0.752=1.165$. Taking relative-rotative efficiency $\eta_R\approx1.0$ (not separately given), $$\eta_D=\eta_H\eta_B\approx\eta_H\eta_o=(1.165)(0.678)=\boxed{0.789}.$$ Check: $P_E=RV_S=(305{,}600)(7.845)=2.397\ \text{MW}$, so $\eta_D=P_E/P_D=2.397/3.04=0.789$ — matches.

Figure 4 — Non-dimensional design point ($J=0.600$, required $K_T=0.220$) located on the supplied B4-55 open-water chart (page 10 of the exam); the $P/D\approx0.88$ curve passes through this point, giving $10K_Q\approx0.31$ at the same $J$.

Quantity	Result
(i) Advance coefficient $J$	0.600
(i) Selected pitch ratio $P/D$	≈ 0.88
(i) Delivered torque $Q$	≈ 197 kN·m
(i) Delivered power $P_D$	≈ 3.04 MW
(ii) Open-water efficiency $\eta_o$	≈ 0.68
(ii) Hydrodynamic propulsive efficiency $\eta_D$	≈ 0.79

Check

Part (i) requires reading $K_T$ and $10K_Q$ off the supplied Wageningen B4-55 open-water chart at $J=0.600$. The printed chart's curve-crossing density is too fine to trace to better than about $\pm0.01$–$0.02$ in $K_T$/$K_Q$ by eye — the same tolerance a candidate reading the printed chart in the exam room would have. The values used ($P/D\approx0.88$, $10K_Q\approx0.31$) give $\eta_D=0.789$ from the chart-based power balance, matching $\eta_D=P_E/P_D=0.789$ from the given resistance and speed.

Topic. Fixed-pitch propeller selection from a required thrust using the Wageningen B-series open-water chart.

Key relations. $T=R/(1-t)$; $J=V_A/(nD)$; $K_T=T/(\rho n^2D^4)$; $K_Q=Q/(\rho n^2D^5)$; $\eta_o=K_TJ/(2\pi K_Q)$; $\eta_H=(1-t)/(1-w)$; $\eta_D=\eta_H\eta_B$.

Why this works. With $n$ and $D$ both fixed by the machinery and hull-clearance constraints, $J$ depends only on the wake-adjusted speed of advance — it is independent of pitch. The required $K_T$ is likewise fixed by the thrust needed to overcome resistance. Plotting that $(J,K_T)$ point against the chart's family of $P/D$ curves is exactly how the pitch that delivers the required thrust at the given shaft speed and diameter is selected.

Common pitfall. Using ship speed $V_S$ instead of speed of advance $V_A$ in the advance coefficient — $J$ must use the wake-reduced inflow speed the propeller actually sees, not the ship's speed over the water.

Source. Lewis (ed.), PNA Vol. II, Ch. 2–3 (propeller series and $K_T$/$K_Q$ design methods); Carlton, Marine Propellers and Propulsion, Ch. 5–6.

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