Water Depth, Wave Height and Wavelength from Two Pressure Sensors: 25-Nav-A3 Hydrodynamics of Ships (II)_ Ship Motion, December 2019, Q2

Question 2: Water Depth, Wave Height and Wavelength from Two Pressure Sensors (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. Wave period $T=8$ s; Sensor 1 rests on the seabed; Sensor 2 is $h=7.62$ m above the seabed (figure). Dynamic pressure amplitudes $p_1=20{,}700$ N/m² (Sensor 1) and $p_2=25{,}600$ N/m² (Sensor 2). Assume linear (Airy) wave theory, water density $\rho=1000$ kg/m³, $g=9.81$ m/s².

Figure 2 — Sensor 1 on the seabed ($z=-d$); Sensor 2 at height $h=7.62$ m above the seabed ($z=-d+h$); $d$ = total water depth.

Find. Water depth $d$, wave height $H$, and wavelength $\lambda$.

Approach. Linear wave theory gives the dynamic pressure amplitude at elevation $z$ (measured from SWL, positive up) as $p(z)=\rho g\eta_a\cosh[k(z+d)]/\cosh(kd)$; the RATIO of the two sensor readings eliminates $\eta_a$ and $d$, leaving one equation for $k$ alone (since both sensors share the same $k$), after which the dispersion relation with the given period fixes $d$, and either sensor reading then fixes $\eta_a$.

Eliminate $\eta_a,d$ using the pressure ratio. At the seabed ($z=-d$): $p_1=\rho g\eta_a/\cosh(kd)$. At Sensor 2 ($z=-d+h$): $p_2=\rho g\eta_a\cosh(kh)/\cosh(kd)$. Dividing, $$\frac{p_2}{p_1}=\cosh(kh) \;\;\Rightarrow\;\; kh=\cosh^{-1}\!\left(\frac{p_2}{p_1}\right)=\cosh^{-1}\!\left(\frac{25{,}600}{20{,}700}\right)=\cosh^{-1}(1.2367)=0.6752.$$
Wave number and wavelength. With $h=7.62$ m, $$k=\frac{0.6752}{7.62}=0.08860\ \text{m}^{-1}, \qquad \lambda=\frac{2\pi}{k}=\boxed{70.9\ \text{m}}.$$
Water depth from the dispersion relation. The wave period gives $\omega=2\pi/T=2\pi/8=0.7854$ rad/s. The dispersion relation $\omega^2=gk\tanh(kd)$ solves for $kd$: $$\tanh(kd)=\frac{\omega^2}{gk}=\frac{0.7854^2}{(9.81)(0.08860)}=0.7096 \;\;\Rightarrow\;\; kd=\tanh^{-1}(0.7096)=0.8865,$$ $$d=\frac{0.8865}{k}=\frac{0.8865}{0.08860}=\boxed{10.0\ \text{m}}.$$ (Check: $d/\lambda=10.0/70.9=0.141$, an intermediate-depth wave — consistent with $\tanh(kd)=0.71$ being well short of the deep-water limit of 1.)
Wave amplitude and height from either sensor. Using Sensor 1, $\cosh(kd)=\cosh(0.8865)=1.4194$: $$\eta_a=\frac{p_1\cosh(kd)}{\rho g}=\frac{(20{,}700)(1.4194)}{(1000)(9.81)}=2.995\ \text{m}, \qquad H=2\eta_a=\boxed{5.99\ \text{m}}.$$

Quantity	Result
Water depth $d$	10.0 m
Wavelength $\lambda$	70.9 m
Wave height $H$	5.99 m

Concept & Theory

Topic. Sub-surface pressure attenuation with depth (linear wave theory) used inversely, as a wave/depth-sensing method.

Key relations. $p(z)=\rho g\eta_a\cosh[k(z+d)]/\cosh(kd)$; dispersion $\omega^2=gk\tanh(kd)$; $c=\lambda/T=\omega/k$.

Why this works. The hyperbolic-cosine depth-attenuation factor depends only on the elevation difference between two gauges (not on the wave amplitude), so a single ratio of two simultaneous pressure readings isolates $k$; everything else (depth, then amplitude) follows in sequence from quantities already known or just found.

Common pitfall. Forgetting that the pressure amplitude decays with depth measured from the SWL, not from the seabed — the correct argument of $\cosh$ at Sensor 2 is $k(z+d)=kh$ (height above the bed), which is why the ratio collapses to $\cosh(kh)$ rather than a difference of two unknown $\cosh$ terms.

Source. Bhattacharyya, Dynamics of Marine Vehicles, Ch. 2 (linear wave theory, sub-surface pressure).

Question 2 of 6: Water Depth, Wave Height and Wavelength from Two Pressure Sensors

Question 2: Water Depth, Wave Height and Wavelength from Two Pressure Sensors (10 marks)