Esercitazioni di Meccanica dei Fluidi I

H = z + ⅟g y ^{V^2} - JLi - EZi

n_o = 0.98

g = Q ⅟Z_v

Z₂ = Z₁ + f_a + V₂²/2g

Z₂ = Z₃ + f_a + V₂²/2g → Z₁ qV

Q = 4.16 m³/s

V_m = 1.36 m/s

νreal νΔ

RA = 2.9 / L_D

V_o = 5.42 → E₂/D₂

n mon CI TROVARI

JEpO

Z_u = Z_u + V₃³/y + w(V_u/V_z)^3/2/y^3/2 + L_σZ_u + q₅ V²/2gy

qui

CONTROLLA N_r NEL ROSTRIMENTO

RAPPORTO DI VANNI

5C

h = 0.5

5C

h = 0.3

z₁ = 0.00085

z₃ = 0.2

Caso CO COMPL.

CO SCLEMO CIR LA VAZIONE DEL FLUSSO

a⊔

OCIO!

FVAL SELOME CO RCONTROTA

PEROA

z_c = 2

TuMO SUP / TUBO INF

1) z_m = z_u + z_t + z_i + 0.5 V₁²

2) z_r = z_v + z_i + z_t + (V₁ - V₃)² + 3z_L

SUPPONDO PARO Q = 1

IN PROEX

Q = 1

Q₁ = 1

Q₂ = 0.364

L = 0.926

A′ = A₁ • C_c

Formulas and Calculations

• Z_S = Z_B + V²/2g + L_B + 0.5 V_L²/g

• Z_M = Z_S + V²/2g + 3L - ΔH_p

Q = Q₁ + Q₂ + Q₃

Z₁ = Z_C + V₁²/2g + J₂L₂

Z_S = Z_B + V₂/_p + V₁²/g

Q = A₃v sqrt(2g(Z₃-Z_S))

Z_B = Z_A + ^V₃2⁄_2g + L₃J₃ + q_s ^V₃2⁄_2g + L q_s ^V₄2⁄_2g

λ₃ = ^0.001⁄_0.01 = 0.01

λ₃ = 0.037869

λ₄ = ^0.001⁄_0.2 = 0.005

Hp → Mf λ₄ = 0.03034

(Z_B - Z_A) = ^Q2⁄_{2g A3²} + λ₃L₃ ^Q2⁄_{2g D A3²} + 0.5 ^Q2⁄_{2g A3²} + L₄λ₄ ^Q2⁄_{2g D A4²} + 0.5 ^Q2⁄_{2g A4²}

2g (Z_B - Z_A) → Q → Q = 0.04002 m³⁄_s

VENERE = V₃ = 5.0954 m⁄_s V₄ = 9.2732 m⁄_s

A₃ = (D₃)² ⁄ 4 = 0.007854

A₄ = 0.032416

R_B3 = 5.09 × 10⁵

R_e4 = 2.51 × 10⁵

Z_q = ∑ Z_B + 0.5 ^V₂2⁄_2g + L J₅ + Z_q J₃ + ln(^{V₁ - V_k2}⁄_2g) + ^V2⁄_2g - ΔHp

CONSIDERo IL MOMA MISTRO:

Z_a + ^p_a⁄_jo + ^V₂2⁄_jg = Z_b + ^p_a⁄_jo + ^Vo2⁄_jg + L₅J₅

Z_a - H_B = δ → S = ^{V_s2 - V_o2}⁄_2g + L₅J₅ S = λ₅ V_L ²⁄_{2g D}

δ = μ - γ Δ = 0.504

Z_m = 25 .......... L₁ = 13 ........ D = 0.15 m ....... L₂ = 8 m ....... L_e = 1 m ....... ε = 0.0075 m

Z_B = 10 .......... Z_U = 5 m ........... p_atm = 5 × 10⁴ Pa ........... γ = 0.3

?

η = Q &LowBar;L W₃₀₈₁ = ^0.10 &sub{m³}

Z_m = Z_B + ^P_g⁄_γ + ^V₁2⁄_2g + L₁n + 0.5 ^V₂2⁄_g

p_ci = ^P_B⁄_γ

L_S = ^{25 × 103}⁄ = ^V₁⁄_2g ln ^0.1⁄

+ ε⁄&sub>