Page 351 - Wind Energy Handbook

PEAK RESPONSE 325
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u ð 1
u 2 2 2
u n S MB (n)dn þ n ó M1
1
t
í ¼ 0 (A5:47)
ó 2 þ ó 2
MB M1
Now
ð ð
R R
2
N
S MB (n) ¼ (C f rU) S u (n) ł (r, r9, n)c(r)c(r9)rr9 dr dr9 (A5:48)
uu
0 0
and
ð
R
1 2
M ¼ C f 2 rU c(r)r dr (A5:49)
0
so

ð ð
R R
N
S u (n) ł (r, r9, n)c(r)c(r9)rr9 dr dr9
M 2 0 0 uu
S MB (n) ¼ 4 ! 2 (A5:50)
U 2 ð R
c(r)r dr
0
Defining
ð ð
R R
N
ł (r, r9, n)c(r)c(r9)rr9 dr dr9
uu
K SMB (n) ¼ 0 0 ! (A5:51)
ð 2
R
c(r)r dr
0
we obtain

M 2
S MB (n) ¼ 4 S u (n)K SMB (n) (A5:52)
U 2

Substituting into Equation (A5.47) gives

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u 2 ð 1
u M 2 2 2
u 4 n S u (n)K SMB (n)dn þ n ó M1
1
t U 2 0
í ¼ (A5:53)
ó 2 þ ó 2
MB M1
Ð 1
Noting from Equation (A5.52) that ó 2 ¼ (4M =U ) S u (n)K SMB (n)dn, the ex-
2
2
MB 0
pression for í becomes