Question 3.4: Offshore wave conditions are described by a JONSWAP frequenc......
Offshore wave conditions are described by a JONSWAP frequency spectrum with F = 80 km, U_{10} = 20 m/s and offshore depth h_{0} = 50 m. Find α, \gamma and the peak period of the offshore spectrum, and then determine the values of α and \gamma at an inshore water depth, h_{i}, of 5 m assuming refraction effects can be ignored. Also, determine the reduction factor at the peak frequency.
From Equation 3.23
a=0.076\left\lgroup\frac{gF}{U^{2}_{10} }\right\rgroup ^{-0.22} f_{p}=\frac{3.5g}{U_{10} }\left\lgroup\frac{gF}{U^{2}_{10} }\right\rgroup ^{-0.33} (3.23)
we have f_{p}= (3.5 g/U_{10})(gF/U_{10}²)^{−0.33}=0.141, so T_{p} = 1/f_{p} = 7.1 secs.
In deep water, the wave length of the wave with peak frequency may be calculated from L_{po}= gT_{p}^{2}/2 π = 78.7 m. [As a check, note that h_{o}/L_{po}= 0.635 > 0.5]. We also calculate, from Equation 3.23, α = 0.076(gF/U_{10} ²)^{−0.22} = 0.076(9.81 × 80000/400)^{−0.22} = 0.014. You obtain the same value for α if you use Equation 3.34
α=0.0078\left\lgroup\frac{2\pi U^{2}_{10} }{gL_{p} }\right\rgroup ^{0.49} \gamma =2.47\left\lgroup\frac{2\pi U^{2}_{10} }{gL_{p} }\right\rgroup ^{0.39} (3.34)
instead, which also gives \gamma =2.47\left(2\pi U_{10} ^{2}/gL_{po} \right) ^{0.39} =3.9.
Now, Equation 3.34 is valid for all water depths. Denoting the offshore and inshore values of α and γ by the subscripts ‘o’ and ‘i,’ respectively, we find
α_{0}=0.0078\left\lgroup\frac{2\pi U^{2}_{10} }{gL_{po} } \right\rgroup ^{0.49} α_{i} =0.0078\left\lgroup\frac{2\pi U^{2}_{10} }{gL_{pi} } \right\rgroup ^{0.49}and similarly for \gamma . Taking U_{10} to be the same offshore and inshore and identifying ‘offshore’ with ‘JONSWAP’ and ‘inshore’ with ‘TMA’ gives
α_{TMA}=α_{JONSWAP} \left\lgroup\frac{L_{po} }{L_{pi} } \right\rgroup ^{0.49} \gamma _{TMA}=\gamma _{JONSWAP} \left\lgroup\frac{L_{po} }{L_{pi} } \right\rgroup ^{0.39}.
L_{pi} can be determined using the methods described in Section 2.3.4 to be 46.4 m. [Using a shallow water approximation, L_{pi} ≈ √(gh_{i})T_{p}, gives L_{pi} = 49.7 m]. Hence α_{TMA} = 0.014(78.7/46.4)^{0.49} = 0.018, and \gamma _{TMA}= 4.06. For the peak frequency w_{h}=2\pi f_{p}\sqrt{\left(h/g\right) } =\left(2\pi /7.1\right)\sqrt{\left(5/9.81\right) } =0.6318\lt 1.0. Thus, \Theta =w^{2}_{h} /2=0.20 and the corresponding TMA spectrum is calculated. Figure 3.11 shows the JONSWAP and TMA spectra for this case.
