Page 119 - Fundamentals of Ocean Renewable Energy Generating Electricity From The Sea

110 Fundamentals of Ocean Renewable Energy

potential function can satisfy the governing equations and provide us with the
hydrodynamic flow field. This is referred to as potential flow theory. Through
some mathematical manipulation, you can easily show that the curl of a gradient
of any scalar function is always zero. In other words,
∇× ∇Φ = 0 and ∇× u = 0 ⇒ u =∇Φ (5.5)
where Φ is called the potential function, which as mentioned is a scalar function.
In potential flow theory, the flow velocity is the gradient of this potential
function. The momentum equation is already satisfied because d ω = d0 = 0.
dt dt
For the continuity equation, we have
2
∇· u = 0 ⇒∇ · u =∇ · (∇Φ) = 0 ⇒∇ Φ = 0 (5.6)
This is called the Laplace equation, and for a 2D wave field can be expanded as
follows:

2
2
∂ Φ ∂ Φ
2
∇ Φ = + = 0 (5.7)
∂x 2 ∂z 2
In linear wave theory, it is assumed that the flow is inviscid, incompressible,
and irrotational; therefore, a potential function, which satisfies the periodic
boundary conditions, will be the solution.
Consider a sinusoidal progressive wave (Fig. 5.1). The basic parameters such
as wave length (L), wave height (H), still water depth (d), and water surface
elevation (η) are presented in this figure. For a progressive wave, it can be shown
that the potential function is given by
H g cosh k(d + z)
Φ = sin(kx − σt) (5.8)
2 σ cosh kd
where k is the wave number and σ is the wave angular frequency. Although we
skipped the derivation of this equation, let us just show that the above solution
satisfies the Laplace equation. Taking the derivatives of the potential function
leads to

2
∂ Φ 2 H g cosh k(d + z)
=−k sin(kx − σt) (5.9)
∂x 2 2 σ cosh kd
2
∂ Φ 2 H g cosh k(d + z)
= k sin(kx − σt) (5.10)
∂z 2 2 σ cosh kd
2
2
∂ Φ ∂ Φ
⇒ + = 0 (5.11)
∂x 2 ∂z 2
Also, the potential function (Eq. 5.8) is periodic. This is because Φ(x, z, t) =
2π
Φ(x, z, t + T) = Φ(x, z, t + ), which is demonstrated as follows
σ

2π
sin kx − σ t + = sin (kx − σt − 2π) = sin(kx − σt) (5.12)
σ