DOMAINS OF VALIDITY FOR WAVE THEORIES
79
From Figure 3.10, we see that this is an intermediate depth wave and that Stokes
second-order theory is appropriate.
2. Compute the wave length A, the wave number fc, and the circular frequency u. For this intermediate depth wave, use the dispersion relation, équation
(3.17), to compute the wave length. Using k - 2tt/A from équation (3.2) and
c = X/T from équation (3.7), the dispersion relation can be expressed implicitly
in ter ms of A and the given wave parameters, or
îgX
f27rd\
/32.2.
/2tt(50)\
A —
— tanh —— = 6< / ——A tanh —7—y 2tf
\ A J
y 2tt
\ A /
This équation, when solved for A by trial and correction, leads to the following
results:
A = 174.6 ft;
k = 2tt/X = 0.0360 ft"1 ;
u = kX/T = 1.048 rad/sec
3. Use the appropriate expressions from Table 3.2, together the given values:
H = 3 ft, and d = 50 ft. With the values of A, fc, and u just computed, the
surface élévation 77 is thus:
77 = 1.5 cos (0.036x - 1.048t) + 0.0502 cos (0.072a; - 2.096t)
These results for the surface élévation show that, due to the second term on the
right, there is approximately a 3 percent correction to the linear theory due to
second order effects. The results for the pressure p at the depth z = 30 — 50 =
— 20 ft in water of mass density p = 1.94 slug/ft3 are found by substituting the
foregoing numerical results into the full expression for p in Table 3.2. In the
numerical results below, the four terms on the right side of the équation, from
left to right, represent, respectively: the hydrostatic pressure of linear theory,
or —pgz = —1.94(32.2)(—20) = 1249.36 lb/ft2; the dynamic pressure of linear
theory; the second order dynamic correction to the linear theory; and the second
order hydrostatic correction to linear theory:
p = 1249.36 + 49.52 cos (kx - ut) + 0.07cos 2(kx - ut) - 0.47 lb/ft2
Again we observe that the second order corrections to linear theory theory are
very small, less than 1 percent in this example problem.
Example Problem 3.3. Détermine the celerity c for a wave with a period of
15 sec and a wave height of 2 ft, propagating in water 20 ft deep.
1. Deduce the appropriate wave theory by computing the following parameters:
d/T2 = 20/152 = 0.0889
H/T2 = 2/225 = 0.00889
79
From Figure 3.10, we see that this is an intermediate depth wave and that Stokes
second-order theory is appropriate.
2. Compute the wave length A, the wave number fc, and the circular frequency u. For this intermediate depth wave, use the dispersion relation, équation
(3.17), to compute the wave length. Using k - 2tt/A from équation (3.2) and
c = X/T from équation (3.7), the dispersion relation can be expressed implicitly
in ter ms of A and the given wave parameters, or
îgX
f27rd\
/32.2.
/2tt(50)\
A —
— tanh —— = 6< / ——A tanh —7—y 2tf
\ A J
y 2tt
\ A /
This équation, when solved for A by trial and correction, leads to the following
results:
A = 174.6 ft;
k = 2tt/X = 0.0360 ft"1 ;
u = kX/T = 1.048 rad/sec
3. Use the appropriate expressions from Table 3.2, together the given values:
H = 3 ft, and d = 50 ft. With the values of A, fc, and u just computed, the
surface élévation 77 is thus:
77 = 1.5 cos (0.036x - 1.048t) + 0.0502 cos (0.072a; - 2.096t)
These results for the surface élévation show that, due to the second term on the
right, there is approximately a 3 percent correction to the linear theory due to
second order effects. The results for the pressure p at the depth z = 30 — 50 =
— 20 ft in water of mass density p = 1.94 slug/ft3 are found by substituting the
foregoing numerical results into the full expression for p in Table 3.2. In the
numerical results below, the four terms on the right side of the équation, from
left to right, represent, respectively: the hydrostatic pressure of linear theory,
or —pgz = —1.94(32.2)(—20) = 1249.36 lb/ft2; the dynamic pressure of linear
theory; the second order dynamic correction to the linear theory; and the second
order hydrostatic correction to linear theory:
p = 1249.36 + 49.52 cos (kx - ut) + 0.07cos 2(kx - ut) - 0.47 lb/ft2
Again we observe that the second order corrections to linear theory theory are
very small, less than 1 percent in this example problem.
Example Problem 3.3. Détermine the celerity c for a wave with a period of
15 sec and a wave height of 2 ft, propagating in water 20 ft deep.
1. Deduce the appropriate wave theory by computing the following parameters:
d/T2 = 20/152 = 0.0889
H/T2 = 2/225 = 0.00889