Enunciado

Turbines can also cavitate when the pressure at point 1 in

Fig. P11.98 drops too low. With NPSH defined by

Eq. (11.20), the empirical criterion given by Wislicenus [4]

for cavitation is

N

ss 5

(r/min)(gal/min)1/2

3NPSH (ft) 43/4 $ 11,000

Use this criterion to compute how high z1 2 z2, the impeller

eye in Fig. P11.98, can be placed for a Francis turbine with

a head of 300 ft, Nsp 5 40, and pa 5 14 lbf/in2 absolute

before cavitation occurs in 608F water.

Passo 1

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

Os dados são os seguintes:

Critério empírico:

N s s = ( r / m i n ) ( g a l / m i n ) 1 2 [ N P S H ( f t ) ] 3 4 ≥ 11000

Velocidade específica, N s p = 40 cabeça, H = 300 f t pressão absoluta, p a = 14 l b f / i n 2

Passo 2

Para água em 60 ∘ F :Temos ρ g ≈ 62.4 l b f / f t 3 e P v ≈ 37 p s f a = 0.25 p s i a . Então, a partir da equação abaixo

N P S H = p a - P v ρ g - Δ z - h f i   = ( 14.0 - 0.25 ) ( 144 ) 62.4 - Δ z - 0   = 31.7 - Δ z

Passo 3

Agora precisamos do NPSH, que encontramos entre os dois critérios de velocidade específica:

N s p = n ( P  in hp  ) 1 / 2 ( 300 f t ) 5 / 4

Com P = η ρ g Q 449 × 300 550

N s s = 11000 = n ( Q  in  g p m ) 1 / 2 ( N P S H ) 3 / 4

Iterando para resolver assumindo η ≈ 90 % (Não podemos encontrar n ou Q , apenas n Q 1 / 2 ). N P S H ≈ 45.0 f t Δ z ≈ 31.7 - 45 = - 13.3 Assim, a altura do olho do impulsor é Δ z = - 13.3 f t

Resposta

Δ z = - 13.3 f t

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