112
CHAPTER 4. HYDRODYNAMIC MODELS
dEb
2 (
HT-~N T Sa \Lsinh(l^)
dx
(4.56)
after making use of the linear dispersion relation (a2 = gk tanh kh) and
replacing the wavenumber with k — 2ir/L.
The mean energy loss over the same flume length, dx, due to the two
side walls was formulated by Keulegan as
dEw
dt
de .
— dz
dt
dx
(4-57)
Introducing the expression for de/dt from Eqn. 4.52, integrating, and using
the linear dispersion relation gives
dEw
[irv 2
(4.58)
Summing the energy loss due to both the bottom and side walls gives
the total energy loss
dE
[tuJ 2
-dï = -p\l-T9a
2irB
+ L sinh (4^) dx
(4.59)
Neglecting the contribution of atmospheric pressure on the free surface,
Keulegan formulated the average rate of work being done on the incremental
fluid volume by the fluid to the negative side of the volume at position, x,
as
dWx
1
dt ~ T
•o
C dz\ dt
(4.60)
— h
where
B
C
flume width
wave speed
and u0 is the instantaneous horizontal velocity given by
u0 = uosin(at — kx)
(4-61)
Substituting for uo and integrating with respect to time gives
dWx _ pBC [° _2 ,
dt ~ 1 j_h U°
(4.62)
CHAPTER 4. HYDRODYNAMIC MODELS
dEb
2 (
HT-~N T Sa \Lsinh(l^)
dx
(4.56)
after making use of the linear dispersion relation (a2 = gk tanh kh) and
replacing the wavenumber with k — 2ir/L.
The mean energy loss over the same flume length, dx, due to the two
side walls was formulated by Keulegan as
dEw
dt
de .
— dz
dt
dx
(4-57)
Introducing the expression for de/dt from Eqn. 4.52, integrating, and using
the linear dispersion relation gives
dEw
[irv 2
(4.58)
Summing the energy loss due to both the bottom and side walls gives
the total energy loss
dE
[tuJ 2
-dï = -p\l-T9a
2irB
+ L sinh (4^) dx
(4.59)
Neglecting the contribution of atmospheric pressure on the free surface,
Keulegan formulated the average rate of work being done on the incremental
fluid volume by the fluid to the negative side of the volume at position, x,
as
dWx
1
dt ~ T
•o
C dz\ dt
(4.60)
— h
where
B
C
flume width
wave speed
and u0 is the instantaneous horizontal velocity given by
u0 = uosin(at — kx)
(4-61)
Substituting for uo and integrating with respect to time gives
dWx _ pBC [° _2 ,
dt ~ 1 j_h U°
(4.62)
