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25-Nav-A3 Hydrodynamics of Ships (II)_ Ship Motion · December 2019

Question 4 of 6: Irregular Seaway — Wave Energy, Significant Heave, Significant Vertical Acceleration

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

Notes on this paper

National Exams — December 2019 — 16-Nav-A3 Hydrodynamics of Ships (II): Ship Motion. Three-hour, closed-book exam; one two-sided 8.5″×11″ formula sheet and an approved calculator are permitted. Format: Questions 1–5 are compulsory; Question 6 offers a choice of (a) or (b) — both are solved below for completeness.

Reference texts: Bhattacharyya, Dynamics of Marine Vehicles (Wiley) — wave kinematics/pressure, roll response and magnification-factor theory, irregular-seaway spectral analysis; Lewis (ed.), Principles of Naval Architecture, Vol. III — Motions in Waves and Controllability (SNAME) — Froude-Krylov theory, added mass, hydroelasticity, linear maneuvering derivatives; Lloyd, Seakeeping: Ship Behaviour in Rough Weather — roll magnification factor and encounter-frequency spectra; Lewandowski, The Dynamics of Marine Craft — the prime-system nondimensional maneuvering equations used in Question 6(b).

In Question 2, sensor 2's dynamic pressure amplitude is 25,600 N/m², the value used throughout (it gives a water depth of about 10.0 m).

Question 4: Irregular Seaway — Wave Energy, Significant Heave, Significant Vertical Acceleration (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. $H_{1/3}=8\ \text{m}=800$ cm; ITTC spectrum $S(\omega)=(A/\omega^5)e^{-B/\omega^4}$ (cm²-s) with $A=8.1\times10^{-3}g^2$, $g=981$ cm/s², $B=3.11\times10^4/H_{1/3}^2$; heave RAO tabulated at $\omega=0.3$ to $0.8$ rad/s (step 0.1): $0.90,0.90,0.92,0.95,0.98,0.75$. Seawater $\rho=1025$ kg/m³.

$\omega$ (rad/s)0.30.40.50.60.70.8
Heave RAO (m/m)0.900.900.920.950.980.75

Find. (a) wave energy per unit sea-surface area $E$; (b) significant heave amplitude $\zeta_{a,1/3}$; (c) significant vertical acceleration at the CG.

Approach. The spectrum's zeroth moment $m_0$ (its integral over $\omega$) is the mean-square wave elevation, from which the wave energy follows directly; combining $S(\omega)$ with the heave RAO point-by-point gives the heave response spectrum, whose zeroth moment gives the significant heave amplitude, and whose moment weighted by $\omega^4$ gives the significant vertical acceleration.

  1. Part (a) — wave energy. The ITTC spectrum integrates in closed form: $\int_0^\infty S(\omega)\,d\omega=A/(4B)$. With $B=3.11\times10^4/800^2=0.04859\ \text{s}^{-4}$ and $A=8.1\times10^{-3}(981)^2=7795\ \text{cm}^2/\text{s}^4$, $$m_0=\frac{A}{4B}=\frac{7795}{4(0.04859)}=40{,}104\ \text{cm}^2=4.010\ \text{m}^2$$ (check: $4\sqrt{m_0}=4\sqrt{4.010}=8.01$ m $\approx H_{1/3}$, confirming the spectrum is correctly normalized). The wave energy per unit surface area is $$E=\rho g\,m_0=(1025)(9.81)(4.010)=\boxed{40{,}300\ \text{J/m}^2}\ (40.3\ \text{kJ/m}^2).$$
  2. Part (b) — heave response spectrum and its zeroth moment. $S_\zeta(\omega)=[\text{RAO}(\omega)]^2\,S(\omega)$ at each tabulated point:
    $\omega$0.30.40.50.60.70.8
    $S(\omega)$ (m²·s)0.79611.4111.466.8903.7882.113
    $S_\zeta(\omega)=\text{RAO}^2 S$0.6459.2399.7036.2183.6381.188
    Integrating $S_\zeta(\omega)$ over the tabulated range by the trapezoidal rule ($\Delta\omega=0.1$): $$m_{0,\zeta}=\int S_\zeta\,d\omega \approx 2.971\ \text{m}^2.$$ The significant (single) amplitude of a narrow-band Gaussian process is $2\sqrt{m_0}$, so $$\zeta_{a,1/3}=2\sqrt{m_{0,\zeta}}=2\sqrt{2.971}=\boxed{3.45\ \text{m}}.$$
  3. Part (c) — significant vertical acceleration. The acceleration spectrum of a harmonic response is $\omega^4$ times its displacement spectrum, so $$m_{4,\zeta}=\int \omega^4 S_\zeta(\omega)\,d\omega \approx 0.2768\ \text{m}^2/\text{s}^4$$ (same trapezoidal rule, now weighting each tabulated $S_\zeta$ point by $\omega^4$), giving $$a_{a,1/3}=2\sqrt{m_{4,\zeta}}=2\sqrt{0.2768}=\boxed{1.05\ \text{m/s}^2}\ (0.107\,g).$$
Q4 ITTC spectrum S(w), H_1/3 = 8 m0.300.400.500.600.700.800.02.85.58.311.013.8wave frequency w (rad/s)S(w) (m^2 s)0.8011.4111.466.893.792.11
Figure 4a — the ITTC spectrum $S(\omega)$ for $H_{1/3}=8$ m, peaking near $\omega\approx0.44$ rad/s.
Heave RAO at CG (given)0.300.400.500.600.700.800.00.20.50.70.91.1wave frequency w (rad/s)heave RAO (m/m)0.900.900.920.950.980.75
Figure 4b — the given heave RAO at the ship's CG.
Check: the integrals in (b) and (c) are evaluated only over the six tabulated points ($\omega=0.3$ to $0.8$ rad/s), as that is the only range for which the RAO is given; the spectrum itself has some energy outside this band (mainly below $\omega=0.3$), so these are the best estimates obtainable from the data supplied, consistent with how the question is posed.
QuantityResult
(a) Wave energy $E$40.3 kJ/m²
(b) Significant heave amplitude3.45 m
(c) Significant vertical acceleration1.05 m/s² (0.107 g)