Esercitazione 1-2 Idrodinamica

ESERCITAZIONE 1

ESERCIZIO 1

ρ_Olio = 816.1.92 N/m³

ρ_Glicerina = 1226.2.5 N/m³

ρ_Aria = 11.84 N/m³

In pressioni relative calcolare P_A

P_B = 0 -> P_atm = 0

P_C = ρ_Glicerina∆_BC -> ρ_Glicerina (9.0 m - 1.5 m)

P_C = 91 968.75 Pa

P_D = P_C - ρ_Glicerina (∆_DE) = P_C - ρ_Glicerina (3.6 - 1.5) m

P_D = 66 217.5 Pa

P_A = P_D - ρ_Olio (∆_AC) -> ρ_Olio (7.5 - 3.6 m)

P_A = 34 286.012 Pa

ESERCITAZIONE 1

ESERCIZIO 1

ρ_olio= 8161.92 N/m³

ρ_glicerina= 12262.5 N/m³

ρ_aria = 11.84 N/m³

In pressioni relative calcolare P_A

l.d. riferimento

P_B = 0 ⇒ Patm = 0

P_C = ρ_glicerina ⋅ ∆bc ⇒ glicerina (9.0 m - 1.5 m)

P_C = 91 968.75 Pa

P_D = P_C - ρ_glicerina (∆DC) = P_C - ρ_glicerina (3.6 - 1.5) m

P_D = 66217.5 Pa

P_A = P_D - ρ_olio (∆ac) ⇒ ρ_ol11o (7.5 - 3.6 m)

P_A = 34286.012 Pa

Andamenti Pressione

m

30

7.5

3.5

48.5

p(ξ)

ξ(m)

h_atm

h_g.c.

h

h = ξ + ^P⁄_γ

Esercizio 2

Δh₁ = 0.5 m

Δh_m = 10 cm

ρ_acqua = 997.10³ Kg/m³

ρ_m = 1398.25 Kg/m³

P_A = γ_acqua (Δh₁ + d + Δh_m)

P_C = P_A - γ_m Δh_m - γ_acqua d =

P_C = γ_acqua (Δh₁ + d / Δh_m) - γ_m Δh_m - γ_acqua (d)

P_C = -74118.6 Pa

Esercizio 3

• a = 2.5 m/s²
• L = 0.4 m
• V = 0.56 m³
• H = 1 m
• rho = 722 ± 2 m

V₀ = L h₀ 1 → h₀ = V₁=0.8 m

a) H_max = h₀ + ^a/_g (^L/₂ - y') = 0.89 m (y' = 0)

b) Equazione

P = -rho gy' + rho g (h₀ + ^aL/_2g - z'){ y' = 0z' = 0

P_max = rho g (H_max) = 8722 Pa

c) H_min = h₀ + ^a/_g(^L/₂ - y') → y' = L = 0.8 + ^a/_g(-^L/₂)= 0.71 m

d) P_min = -rho gy' + rho g (h₀ + ^aL/_2g-z'){ y' = Lz' = 0

→ P_min = -gal + rho g (H_max) = 6373 Pa

es 4 (stesso carrello es 3)

H_MAX = h_o + ^a/_g [^L/₂ - y']

1 m = (0.8 + ^a/_g [^L/₂]) m

0.2 m = ^aL/_2g -> a = ^{0.2 · 2g}/_L = 5.1 m/s²

es 5 (stesso carrello es 3)

a = 2.5 m/s²L = 0.4 mH = 1 m

H_MAX = h_o + ^a/_g (^L/₂) y'^o

y' = 0

h_o = H_MAX - ^a/_g (^L/₂) -> h_o = 0.92 m

V_MAX = (0.92 · L · 1) m³ = 0.64 m³

R = 0.8 m

ω = 7.8 s^-1

V = 0.3 m³

H = 1 m

Disegno il contenitore con il fluido fermo.

V_m = πR²h_o

-> h_o = V / πR² = 0.6 m

1. H_max = h_o + ω²R² / 4g = 0.8 m
2. H_min = h_o - ω²R² / 4g = 0.4 m

P_max = ϱg(h_o + β ω²R²/4) = 4815 Pa

P_min = ϱg(h₀ - β ω²R²/4) = 3898 Pa

I risultati possono essere un po’ diversi

ma volendo evitare l’errore e ammettendo

la professoressa potrebbe aver usato

valore diverso di g

7) Stesso recipiente Es 6

a)

• R = 0.4 m; ω = 78 s^-1; H = 1 m
• H_max = h_o + ^ω2R2/_4g → h_o = H_max - ^ω2R2/_4g = 0.8 m
• V_max = πR²h_o = 0.4 m³

b)

• R = 0.4 m; V = 0.3 m³; H = 1 m
• H_max = h_o + ^ω2R2/_4g → ω = 2√g(H_max-H_o)/R
• = 9.9 s^-1

c)

• ω = 75 s^-1; V = 0.3 m³
• V = πR²h_o → h_o = ^V/_πR²
• H_min = h_o - ^ω2R2/_4g
• 0 = ^V/_πR² - ^ω2R2/_4g → ^V/_πR² = ^ω2R2/_4g
• → R² = 2√gV/ω√π
• → R = √0.2m² = 0.53 m

Es 2

Lato acqua

F₁ = r₀(a - ℓ)ℓ + ¹/₂ r₀ℓ²

F₁ = r(a - ℓ)ℓ + ¹/₂ rℓ² l

M_F1 = v [r₀(a - ℓ)ℓ(^ℓ/₂) + ¹/₂ r₀ℓ³(¹/₃)]

Lato dio

F₂ = r₀(a + b - ℓ)ℓ + ¹/₂ r₀ℓ²

F₂ = (r₀(a + b - ℓ)ℓ + ¹/₂ r₀ℓ²) l

M_F2 = -_vr[r₀(a + b - ℓ)(ℓ)(^ℓ/₂) + ¹/₂ r₀ℓ³(¹/₃)]

M_x = M_{x 0}

2 [ γ (a-e) e²/2 + 1/3 x e³ ] k_e = [ x_o (a-e) e²/2 - x_o b e²/2 - 1/3 xo e ] k_e

b: =

=> b: 0.089 m

a

b

α

LATO CAS

P_CAS = γ_m (-Δh₁) = -γ_m Δh_c

F = - γ_m Δh_c b/sin θ

γ_CAS = (-γ_m Δh_c b/sin θ) (b/2 sin θ) =

=> M_CAS = - k (^{γ_m Δh_c b2}/_{2 sin θ})

Asse Liquido

F_a = ^γab/_sinθ + ^λb2/_2sinθ

M_ϕ = ^γab/_sinϕ ( ^b/_2sinω + ^λb2/_2sinϕ) b ²/3sinθ

=> M_ϕ = -κ ^γab2/_2sin²θ + ^λb3/_6sinθ

=> Allora M = -M_gas - M_ϕ

h_s M = κ [ ^{γmΛb b2}/_2sin²θ + ^γab2/_2sin²θ + ^λb3/_6sinθ ]

M^! = κ [ 4011.2 ] Nm

ES 4

DATI

• Q
• b
• a
• P_h
• θ

F = (P_h - γb) * ^b/_sinθ + ¹/₂ * ^γb2/_sinθ

= ^P_hb/_sinθ - ^γb2/_sinθ + ¹/₂ * ^γb2/_sinθ

M_F = [(P_h - γb) * ^b2/_2sin²θ + ⁴/₆ * ^γb3/_sinθ] * [^-1/_k]

M = -M_F = ¹/_k [-23888] Nm

Pass: P_h - γb - γa = 18133 Pa