FIND: Wind setup at the new time level.
SOLUTION: The wind stress coefficient given by Equation 3-59 is
k = K + K, 1 - -f
= 1.1 X W6 + 2.5 X 10-‘ 1----- —
\
W /
y
50.5
= 2.27 X 10"6
From Equation 3-91
4KAt | Q”+ï |
G = 1 + ---------------------------------------------(D/+k+D?+3/2)" (bi+K+bi+3/2)
G =
4 (0.003) (0.2) 10.0707 I
=
[(0.0162)2 + (0.0132)2] (26.3 + 21.0)
The two terms needed in the évaluation of
given by
q”*1 (Equation 3-89) are
Y (bi+J6 + b;+3/j) (kW2 cos e)"++/
0 2
=
(26.3 + 21.0) (2.27 X10'6) (50.5)2 (1) = 0.0274
and
(Af+^ +Ai+y2)” (S^ S«>3/2)”
79,000 (0.2) (0.426 + 0.278) (3.32 ~ 3.76) = _
(2) (10) (5,280)
The volume flow rate from Equation 3-89 is
Q"+7_____ 1_ [0 070 + 0.0274-0.0463) = 0.0513 mi? /hr.
1.0129
3-139
SOLUTION: The wind stress coefficient given by Equation 3-59 is
k = K + K, 1 - -f
= 1.1 X W6 + 2.5 X 10-‘ 1----- —
\
W /
y
50.5
= 2.27 X 10"6
From Equation 3-91
4KAt | Q”+ï |
G = 1 + ---------------------------------------------(D/+k+D?+3/2)" (bi+K+bi+3/2)
G =
4 (0.003) (0.2) 10.0707 I
=
[(0.0162)2 + (0.0132)2] (26.3 + 21.0)
The two terms needed in the évaluation of
given by
q”*1 (Equation 3-89) are
Y (bi+J6 + b;+3/j) (kW2 cos e)"++/
0 2
=
(26.3 + 21.0) (2.27 X10'6) (50.5)2 (1) = 0.0274
and
(Af+^ +Ai+y2)” (S^ S«>3/2)”
79,000 (0.2) (0.426 + 0.278) (3.32 ~ 3.76) = _
(2) (10) (5,280)
The volume flow rate from Equation 3-89 is
Q"+7_____ 1_ [0 070 + 0.0274-0.0463) = 0.0513 mi? /hr.
1.0129
3-139