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Fundamentos de Engenharia Aeronáutica - Volume único

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165

2

1

0

3

2

min

⎟ ⎟ ⎠

=

D

P

C

K

S

W

v r

ρ

(4.29b)

Desse modo tem-se que:

4

1

0

2

1

2

min

⎟ ⎟ ⎠

=

D

T

C

K

S

W

v r

ρ

(4.30)

e

4

1

0

2

1

3

2

min

⎟ ⎟ ⎠

=

D

P

C

K

S

W

v r

ρ

(4.31)

Relacionando-se as Equações (4.30) e (4.31), pode-se escrever que:

4

1

0

2

1

4

1

0

2

1

3

2

2

min

min

⎟ ⎟ ⎠

=

D

T

D

P

C

K

S

W

v

C

K

S

W

v

r

r

ρ

ρ

(4.32)

4

1

0

2

1

4

1

0

2

1

2

3

2

min

min

=

D

D

T

P

C

K

S

W

C

K

S

W

v

v

r

r

ρ

ρ

(4.32a)

4

1

0

2

1

4

1

0

4

1

2

1

2

3

1

2

min

min

=

D

D

T

P

C

K

S

W

C

K

S

W

v

v

r

r

ρ

ρ

(4.32b)

4

1

3

1

min

min

⎟ ⎠ ⎞

= r

r

T

P

v

v (4.32c)

Portanto:

min

min 76

,

0 r

r

T

P

v

v

= (4.33)

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