Esercitazioni Macchine e Sistemi Energetici

Esercitazione NSE

10-10-18

1)

Piano meridiano

S₁ = π/4 (D_w b)² = π D_w b

S₂ = π (4 D_w b) = π D_w b

Macchina Assiale:

Diametro medio

D_h = (D₀ - D_b) / 2

D₂ = D_w + b

D_b = D_w - b

Macchina Radiale:

S₁ = π D_w b₁

S₂ = π D₂ b₂

• D_w = 200 m
• α₂ = α₃ = cost.
• b = 50 mm = 0.05 m
• p = 1 bar
• V = c/s m³/s
• n = 3000 giri/min.
• Δβ = 20°
• L/D = ?
• ε = V₂ z - V₁ yt = 0.35 V/kg

* Caso ideale componente assiale costante.

* Macchina assiale, fluido incomprimibile.

* V₁ = 20

V_R = V₂ cos β

V₁₀ = V₂₀ U/S₁ = 15,9 m/s

S₁ = π D_w b = 0,031 m²

U = ω D_w = 34 m/s

ω = 2 π n / 60

W₁ = W_z W₂ W_t = 33,2 m/s

β₃ = tg⁻¹ (W₂ / U ) = 63,0°

Ven>3.2.2.MONO → W ₁₃ m/s

W_z0 = V_z0 = W_zz = 20

β₂ = β₁ + Δ β = 43°

W_zz = W_z0 + tg β₂ (W_{z2 - Wzz)}

V_{z1 = Vi3 - U2 U zz = 14,4}

Macchina Fluido Incomprimbile. Macchina Radiale

Q = 1500 giri/min

V₁ = 1000 m³/h

D₂ = 200 mm

D₃ = 300 mm

β₂ = 30°

β₃ = 45°

N₀ = V̇ = 160 m³/s

S.it. [Unknown Words]

U₂ = [Unknown Words] 2πm D₂ = 23,6 m/s

2 60 2

|W| = √(V₁U₂² + U₂) = 16,4 m/s

U₁ = 14,6 m/s - U₁

V₂ = πbD₂ V₂ - V₂U₂ U₂ = 1,9 m/s

σ₂ σ₂

V₂ = ω D₂ = 2πm D₂ = 23,6 m/s

2 60 2

V₂ = √(V_2x² + V_2y²) = 23,8 m/s

V₂ = tg^-1 V_2x W₂ 58°

V₂

ℓ₀ (H₀) - U₂ V_2x + V_2x + V₄ = 550 J/kg

ℓ = V₂² - V₁² + V₂² U₀² - U₁² W₂² - W₃²

2 2 2 2

= 553 J/kg

Esercitazione MSE

ΔH=70 m

N=2500 giri/min

HF=0.033 (V²/gD) (l/D)^1.10

ΔHL=60 m

V=3.5 m³/h

ε/c_d=7

Q_c=0.99

γ₁=0.99

Φ:

• V̅
• U̅₃
• U̅₂ V̅₂
• N̅₂

N₂/N₁=ND₂/ND₂

N/N₁

U

V=3.5 m³/h=9.72 cm/s

HP=74 m

H_F=HP+c_dV²/2g=42.6

λ/4g=(λ+A)(V²/2g)

D/2g=cc=230

V=V̅/SN=4πD

N

V̅=3.84 m/s

DN=πD=gD=150g 1cm²

ε=5.16 μ/D=cc=125

D=cc=64.6

V̅=V̅=74.44 m/s

z=VP

Re=2Re=309g/cm²

λ=εc=cz06

HP=110+c_dV²/γ=167.8

9

9IILLIONE=H/D

D=HP/c₂c=1.34

A=B

• H=C₃6+V²

H=c=cb3(V²8)+c3(l/N-64)+110

A NR MINIMUM CON B

Φ=NB-V₃/NB-NA V_B V.350 g/c

NB=NA V_B

• HP=74w-(ele)=gHP=721g W/g
• η_j=Rlw/e g

Rs(r-Rw)/lnv

cc22 g1/g

HQ(g050g

• Pee마다=η_e=qld/j
• เนร₂=Q_rk W

V̅=26 m³/h

η₅=g

Esercitazione MSE

1) Pompa Centrifuga

• N = 2000 giri/min
• β2 = 52°
• b2 = b2 = 30 mm
• D2 = 40 cm
• D1 = 40 mm
• Dg = 250 mm

Q = 500 l/s, 0.5 m³/s

η = e_s i

P = 3 bar

Vz = Vo = Vg = 0

Vacrido? Pin? Pnum?

U2 = ωD2 = 2πN D2 / 60 = 14.66 m/s

Q = 2

√ = W2r(Do2 -> W_z = 3.49 m/s

W2 = W_2r = Vz = 0.16 m/s

W2t = W2scwβ2 = 4.65 m/s

U2 = W_z - U2 = V2 < Vz < U2_t = 19.6 ± U/s

R = Vz U2 = 2.0 cc U2 ( Vz

b) P_u?

P_u = (V_IT - V_ZT) * 13.95 Q / ξ_g

P_u = V_u = 10.8 MW

c) η_TS ? η_TT ?

η_s = ε Δh = ε c^pgT

h2,0s = Richiamo da Moller

η_S =^ε⁄_Δh = ε^chT_{0 - T2}g

hT2,0s = h2,0s - V²⁄₂

η_TT = η_TS

d) χ ?

χ =^Δh_rot⁄_{Δh^rot+Δh^stat} =^{h₁ - h₂}⁄_{h0 - h2} = Q⁴

χ_cinematico =^{U2 - W2⁄₂⁄_{U2 - V02} =W₂[ - ](U2 - V0₂)⁄₂gg}

χ_C.LVG^MAIOR =^Δh_rot⁄_{ε_GT} =^{h₁ - h₃}⁄_{h0 - h2} = C₄₆

χ_{LAVOREO EXTAND} =^{U₂ - V_u2} =^{[_U1 + ](U2 - V0₂)g}[ C₄₆ ]g^g

2) STADIO A _REAZIONE

Kp = V_u^Q

A_u^-_bor

TO = 350°C

P_0bor

l = 3000 g_mr⁴

_min⁄^nv

_TR = SIME_TR^ICO_N^{E_{l P2}}

K_u

C