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FenTrans, MecFlu, TransCal e TransMassa - Frank White - Mecânica dos Fluidos- Ed: 8º - Capítulo 8.Problemas - Ex. 54

The original Flettner rotor ship was approximately 100 ft

long, displaced 800 tons, and had a wetted area of 3500 ft2. As

sketched in Fig. P8.54, it had two rotors 50 ft high and 9 ft in

diameter rotating at 750 r/min, which is far outside the range

of Fig. 8.15. The measured lift and drag coeffi cients for each

rotor were about 10 and 4, respectively. If the ship is moored

and subjected to a crosswind of 25 ft/s, as in Fig. P8.54, what

will the wind force parallel and normal to the ship centerline

be? Estimate the power required to drive the rotors.

Passo 1

Saaaalve, tranquilo? Vamos lá resolver mais uma questão juntos?! 😊

Os dados são os seguintes:

Comprimento do navio do rotor do flettner, L = 100 f t

Diâmetro e altura do rotor são, D = 9 f t e h = 50 f t Coeficiente de arrasto da água, C D ,   barco   = 0.006 Área molhada A wetted  = 3500 f t 2 O rotor gira a N = 750 r / m i n

Velocidade do vento cruzado, U ∞ = 25 f t / s

Coeficiente de elevação, C L = 10 Coeficiente de arrasto, C D = 4

Passo 2

Força de elevação

F L = C L ρ 2 U ∞ 2 D L F L = 10 0.00238 2 ( 25 ) 2 ( 9 f t ) ( 50 f t ) ( 2  rotor e s  ) F L ≈ 6700  lbf( paralelo  

Força de arrasto

F D = C D ρ 2 U ∞ 2 D L F D = 4 0.00238 2 ( 25 ) 2 ( 9 f t ) ( 50 f t ) ( 2  rotor e s  ) F D ≈ 2700 l b f (  normal  )

Passo 3

Suponha que o cilindro simule uma placa plana de comprimento 2 π R = 2 π ( 4.5 ) = 28.3 f t . Tensão de cisalhamento

τ w   = C f ρ 2 U 2 τ w   ≈ 0.027 R e L 1 / 2 ρ 2 ( Ω R ) 2

Onde

Ω R = 750 2 π 2 ( 4.5 ) Ω R = 353 f t s R e L = 0.00238 ( 353 ) ( 28.3 ) 3.71 E - 7 R e L ≈ 6.42 E 7

Portanto, τ w ≈ 0.308 l b f f t 2

Passo 4

torque

T = τ w ( π D L ) R T = 0.308 π ( 9 ) ( 50 ) ( 4.5 ) T ≈ 1958 f t - l b f

Potência total

P = T Ω P = ( 1958 ) 750 2 π 60 ( 2  rotors  ) / 550 h p f t - l b f / s P ≈ 560 h p

Resposta

F L ≈ 6700  lbf( paralelo   )

F D ≈ 2700 l b f (  normal  )

P ≈ 560 h p

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