Wave energy

Contents

From the dispersion relation page (D = equation number on that page):

\[ \omega^2 = gk\tanh(kh) \tag{D7} \]

\[ \eta(x, t) = a\cos(\omega t - kx) \tag{D8} \]

\[ u(x, z, t) = a\omega\,\frac{\cosh\big(k(z + h)\big)}{\sinh(kh)}\,\cos(\omega t - kx) \tag{D12} \]

\[ w(x, z, t) = -a\omega\,\frac{\sinh\big(k(z + h)\big)}{\sinh(kh)}\,\sin(\omega t - kx) \tag{D13} \]

\[ p(x, z, t) = -\rho gz + \rho ga\,\frac{\cosh\big(k(z + h)\big)}{\cosh(kh)}\,\cos(\omega t - kx) \tag{D17} \]

All quantities are per unit horizontal area (per unit width of wave crest for the energy flux).

1. Potential energy

PE of the water column, from the bed \(z = -h\) to the surface \(z = \eta\):

\[ \int_{-h}^{\eta} \rho gz\,dz = \rho g\left[\frac{z^2}{2}\right]_{-h}^{\eta} = \frac{1}{2}\rho g\left(\eta^2 - h^2\right) \]

Still water, \(\eta = 0\):

\[ -\frac{1}{2}\rho gh^2 \]

Wave PE = column with the wave \(-\) still-water column:

\[ PE = \frac{1}{2}\rho g\left(\eta^2 - h^2\right) - \left(-\frac{1}{2}\rho gh^2\right) = \frac{1}{2}\rho g\eta^2 \]

Plug (D8):

\[ PE = \frac{1}{2}\rho ga^2\cos^2(\omega t - kx) \]

Average over one period, \(T = \dfrac{2\pi}{\omega}\). \(\;\frac{1}{2}\rho ga^2\) is constant \(\rightarrow\) out of the integral:

\[ \langle PE \rangle = \frac{1}{2}\rho ga^2\cdot\frac{1}{T}\int_0^T \cos^2(\omega t - kx)\,dt \]

\(\cos^2X = \dfrac{1}{2}\big(1 + \cos(2X)\big)\):

\[ \frac{1}{T}\int_0^T \cos^2(\omega t - kx)\,dt = \frac{1}{T}\int_0^T \frac{1}{2}\,dt + \frac{1}{T}\int_0^T \frac{1}{2}\cos\big(2(\omega t - kx)\big)\,dt \]

First integral:

\[ \frac{1}{T}\cdot\frac{T}{2} = \frac{1}{2} \]

Second integral:

\[ \frac{1}{2T}\left[\frac{\sin\big(2(\omega t - kx)\big)}{2\omega}\right]_0^T = \frac{\sin(2\omega T - 2kx) - \sin(-2kx)}{4\omega T} \]

\(\omega T = 2\pi \;\rightarrow\; \sin(4\pi - 2kx) = \sin(-2kx) \;\rightarrow\;\) second integral \(= 0\):

2. Kinetic energy

KE of the water column:

\[ KE = \int_{-h}^{\eta} \frac{1}{2}\rho\left(u^2 + w^2\right)dz \;\approx\; \int_{-h}^{0} \frac{1}{2}\rho\left(u^2 + w^2\right)dz \]

Small amplitudes: \(u^2 + w^2 \propto a^2\) and the strip \(0 \le z \le \eta\) has height \(\propto a\) \(\;\rightarrow\;\) its contribution \(\propto a^3\) is dropped.

Square (D12) and (D13), add:

\[ u^2 + w^2 = \frac{a^2\omega^2}{\sinh^2(kh)}\Big[\cosh^2\big(k(z + h)\big)\cos^2(\omega t - kx) + \sinh^2\big(k(z + h)\big)\sin^2(\omega t - kx)\Big] \]

Average over one period. Only \(\cos^2(\omega t - kx)\) and \(\sin^2(\omega t - kx)\) depend on time \(\rightarrow\) everything else out of the integral. The \(\cos^2\) average is \(\dfrac{1}{2}\), from (1).

The \(\sin^2\) average, same steps with \(\sin^2X = \dfrac{1}{2}\big(1 - \cos(2X)\big)\):

\[ \big\langle u^2 + w^2 \big\rangle = \frac{a^2\omega^2}{2\sinh^2(kh)}\Big[\cosh^2\big(k(z + h)\big) + \sinh^2\big(k(z + h)\big)\Big] \]

Combine \(\cosh^2 + \sinh^2\), with \(X = k(z + h)\):

\[ \cosh^2X = \frac{e^{2X} + 2 + e^{-2X}}{4}, \qquad \sinh^2X = \frac{e^{2X} - 2 + e^{-2X}}{4} \]

\[ \cosh^2X + \sinh^2X = \frac{e^{2X} + e^{-2X}}{2} = \cosh(2X) \]

\[ \big\langle u^2 + w^2 \big\rangle = \frac{a^2\omega^2}{2\sinh^2(kh)}\cosh\big(2k(z + h)\big) \]

Integrate over the depth:

\[ \int_{-h}^{0}\cosh\big(2k(z + h)\big)\,dz = \left[\frac{\sinh\big(2k(z + h)\big)}{2k}\right]_{-h}^{0} = \frac{\sinh(2kh) - \sinh(0)}{2k} = \frac{\sinh(2kh)}{2k} \]

\[ \langle KE \rangle = \frac{1}{2}\rho\cdot\frac{a^2\omega^2}{2\sinh^2(kh)}\cdot\frac{\sinh(2kh)}{2k} = \frac{\rho a^2\omega^2\sinh(2kh)}{8k\sinh^2(kh)} \]

\(\sinh(2X) = 2\sinh X\cosh X\):

\[ 2\sinh X\cosh X = 2\cdot\frac{e^{X} - e^{-X}}{2}\cdot\frac{e^{X} + e^{-X}}{2} = \frac{e^{2X} - e^{-2X}}{2} = \sinh(2X) \]

\[ \langle KE \rangle = \frac{\rho a^2\omega^2\cdot 2\sinh(kh)\cosh(kh)}{8k\sinh^2(kh)} = \frac{\rho a^2\omega^2\cosh(kh)}{4k\sinh(kh)} = \frac{\rho a^2\omega^2}{4k\tanh(kh)} \]

Dispersion relation (D7), \(\omega^2 = gk\tanh(kh)\):

3. Total wave energy

(2) + (4):

Half potential, half kinetic.

4. Energy flux

Rate of work done by the pressure on a vertical plane \(x = \text{const}\):

\[ P = \int_{-h}^{\eta} p\,u\,dz \;\approx\; \int_{-h}^{0} p\,u\,dz \]

(Same small-amplitude argument as for \(KE\).)

The fluid also carries energy across the plane: \(\frac{1}{2}\rho\left(u^2 + w^2\right)u\) is a product of three small quantities \(\rightarrow\) dropped; \(\rho gz\,u\) averages to zero, since \(\langle u \rangle = 0\) at fixed \(z\) (same steps as the hydrostatic part below). Only the pressure work remains.

Split \(p\) from (D17) into hydrostatic and dynamic parts:

\[ p\,u = \underbrace{(-\rho gz)\,u}_{\text{hydrostatic}} + \underbrace{\rho ga\,\frac{\cosh\big(k(z + h)\big)}{\cosh(kh)}\cos(\omega t - kx)\,u}_{\text{dynamic}} \]

Hydrostatic part. Plug (D12), average over one period; everything except \(\cos(\omega t - kx)\) is constant in time \(\rightarrow\) out of the integral:

\[ \langle -\rho gz\,u \rangle = -\rho gz\,a\omega\,\frac{\cosh\big(k(z + h)\big)}{\sinh(kh)}\cdot\frac{1}{T}\int_0^T \cos(\omega t - kx)\,dt \]

\[ \frac{1}{T}\int_0^T \cos(\omega t - kx)\,dt = \frac{1}{T}\left[\frac{\sin(\omega t - kx)}{\omega}\right]_0^T = \frac{\sin(\omega T - kx) - \sin(-kx)}{\omega T} \]

\(\omega T = 2\pi \;\rightarrow\; \sin(2\pi - kx) = \sin(-kx) \;\rightarrow\;\) the integral is \(0\):

\[ \langle -\rho gz\,u \rangle = 0 \]

Dynamic part. Plug (D12):

\[ p_{dyn}\,u = \rho ga\cdot a\omega\,\frac{\cosh^2\big(k(z + h)\big)}{\cosh(kh)\sinh(kh)}\,\cos^2(\omega t - kx) \]

Average over one period, with (1):

\[ \langle p_{dyn}\,u \rangle = \frac{\rho ga^2\omega\cosh^2\big(k(z + h)\big)}{\cosh(kh)\sinh(kh)}\cdot\frac{1}{2} = \frac{\rho ga^2\omega\cosh^2\big(k(z + h)\big)}{2\cosh(kh)\sinh(kh)} \]

Integrate over the depth, with \(\cosh^2X = \dfrac{1}{2}\big(1 + \cosh(2X)\big)\):

\[ \int_{-h}^{0}\cosh^2\big(k(z + h)\big)\,dz = \left[\frac{z}{2} + \frac{\sinh\big(2k(z + h)\big)}{4k}\right]_{-h}^{0} = \frac{h}{2} + \frac{\sinh(2kh)}{4k} \]

\[ \langle P \rangle = \frac{\rho ga^2\omega}{2\cosh(kh)\sinh(kh)}\left[\frac{h}{2} + \frac{\sinh(2kh)}{4k}\right] \]

\(2\cosh(kh)\sinh(kh) = \sinh(2kh)\):

\[ \langle P \rangle = \frac{\rho ga^2\omega}{\sinh(2kh)}\left[\frac{h}{2} + \frac{\sinh(2kh)}{4k}\right] = \rho ga^2\omega\left[\frac{h}{2\sinh(2kh)} + \frac{1}{4k}\right] \]

Take out \(\dfrac{\rho ga^2\omega}{4k}\):

\[ \langle P \rangle = \frac{\rho ga^2\omega}{4k}\left[1 + \frac{2kh}{\sinh(2kh)}\right] \]

With \(E = \frac{1}{2}\rho ga^2\) from (5) and \(c = \dfrac{\omega}{k}\):

5. Group velocity

\[ c_g = \frac{d\omega}{dk} \]

Differentiate \(\omega^2 = gk\tanh(kh)\) with respect to \(k\), using \(\dfrac{d}{dk}\tanh(kh) = h\,\text{sech}^2(kh)\):

\[ 2\omega\frac{d\omega}{dk} = g\tanh(kh) + gkh\,\text{sech}^2(kh) \]

\[ \frac{d\omega}{dk} = \frac{g\tanh(kh)}{2\omega}\left[1 + \frac{kh\,\text{sech}^2(kh)}{\tanh(kh)}\right] \]

First factor, \(g\tanh(kh) = \dfrac{\omega^2}{k}\):

\[ \frac{g\tanh(kh)}{2\omega} = \frac{\omega^2}{2k\omega} = \frac{\omega}{2k} = \frac{c}{2} \]

Second factor, \(\sinh(2kh) = 2\sinh(kh)\cosh(kh)\):

\[ \frac{kh\,\text{sech}^2(kh)}{\tanh(kh)} = \frac{kh}{\cosh^2(kh)}\cdot\frac{\cosh(kh)}{\sinh(kh)} = \frac{kh}{\sinh(kh)\cosh(kh)} = \frac{2kh}{\sinh(2kh)} \]

Compare (6) and (7):

\(\rightarrow\) wave energy travels at the group velocity.

Deep water: \(\dfrac{2kh}{\sinh(2kh)} \rightarrow 0 \;\rightarrow\; c_g = \dfrac{1}{2}c\)

Shallow water: \(\sinh(2kh) \rightarrow 2kh \;\rightarrow\; c_g \rightarrow c\)