Question 3.3: Calculate the significant wave height and zero upcrossing pe......

SMB method (original version)

\hat{F} = \frac{5000\times 9.81}{100} = 490.5

so

H_{s} =\frac{100\times 0.283}{9.81} \tan h\left\{0.0125\hat{F} ^{0.42} \right\}= 0.5  m

T_{z} =\frac{10\times 7.54}{9.81} \tan h\left\{0.077\hat{F} ^{0.25} \right\}= 2.7  s

SMB method (modified version)

First calculate U_{a} :U_{a} =0.71U_{10}  ^{1.23} =12.06  m/s

\hat{F}=\frac{5000\times 9.81}{12.06^{2} } =337.2

H_{s}=\frac{12.06^{2}\times 0.283 }{9.81}tan h\left\{0.0125\hat{F} ^{0.42} \right\}=0.6  m

T_{z}=\frac{12.06\times 7.54}{9.81}tanh\left\{0.077\hat{F} ^{0.25} \right\}=3.0  s

JONSWAP method (original version)

We have U_{10} = 10 m/s and nondimensional fetch = 490.5. These give

H_{s}=\frac{10.0^{2}\times 0.00178 }{9.81}\hat{F} ^{0.5}=0.4  m

T_{p}=\frac{10.0\times 0.352}{9.81}\hat{F} ^{0.33}=2.8  s

But for a JONSWAP spectrum with an average value of the peak enhancement factorwe have   T_{z} ≈ 0.7775T_{p} = 2.2  s.

JONSWAP method (modified version)

As above, we have U_{a} = 12.06  m/s and nondimensional fetch = 337.2. These give

H_{s}=\frac{12.06^{2}\times 0.0016 }{9.81}\hat{F} ^{0.5}=0.4  m

T_{p}=\frac{12.06\times 0.2857}{9.81}\hat{F} ^{0.33}=2.4  s

But for a JONSWAP spectrum with an average value of the peak enhancement factorwe have  T_{z} ≈ 0.7775T_{p} = 1.9  s.