Sistemi Energetici - Esercizi svolti

Tema di Esame 28/01/2018

Esercizio 1

⁶ ₅

³ ₄

NB in rosso sono indicati i risultati riportati nelle soluzioni del tema d'esame: importante capire il procedimento e ottenere risultati ragionevolmente vicini a quelli "ufficiali".

T₁ = 15°C = 288,15 K

T_2s = T₁ β^γ-1/_γ = 288,15 ⋅ ^4,1/_1.4 = 524,87 K = 248,82°C

Φ_e = ^Φ_es/_Φ1 = ^4,1/Φ₁ - 1 = 0,2857 → β^Φ_e-1

y_adc = ^{Cpa (T_2s - T₁)}/_{Cpc (T2 - T1)} → T₂ = T₁ + ¹/_yadc (T_2s - T₁) =

= 15 + ¹/_0,814 (248,82 - 15) = 293,03°C

Hp. no perdite di carico → P_i = P₃ = P₂ e P₅ = P₆ = P₄

→ anche la turbina ha rapporto delle pressioni pari α β

! T₃ incognita → Ricavo da procedura iterativa:

• ^T(o)₄/_T^(o)1 bilancio energetico in cdc:

m_eCpe (T₃ - T_ref) + ṁ PCI = ṁ Cpg (T_u - T_ref) ; T_ref = 25°C

→ L Cpe (T₃ - T_ref) + PCI = (L+D) Cpe (T_u - T_ref)

T^(o)₃ = T_ref + ^{(L+D) Cpg (T(o)_u - T_ref) - PCI}/_{2 Cpe}

Formulas

T₅₅ = T₄ (1/β)^{(γ_t-1/γ_t)} ; φ_z = γ_t-1/γ_t = 1.33 - 1/1.33 = 0.2681

→ψ_ad,t = 1-(1/β)^(φ_M/φ_t) = 0.2481

1-(1/β)^(0.2681) = 0.905

1-(1/4.3)^(0.2481)

ψ_ad,t = φ_z (T₄-T₅)

T₅ = T₄ - ψ_ad,t (T₄-T₅₅)

• ε = T₃-T₂/T₅-T₂ → T₃ = T₂ + ε (T₅ - T₂) and compare with T₃⁽⁰⁾
• if |T₃⁽¹⁾ - T₃⁽⁰⁾| < 10°C, OK
• HP: T₄⁽⁰⁾ = 1000°C = 1323.15 K → T₃⁽⁰⁾ = 25 + (60 + 1).1.18(100-25) - 48388/60.4.005 = 506.10°C
• T₅₅ = 1323.15 (8)^(0.2681) = 819.68 K = 546.53°C
• T₅ = 1000 - 0.905 (1000 - 546.53) = 599.11°C
• T₃⁽¹⁾ = 293.03 + 0.60 (599.11 - 293.03) = 476.08°C → |T₃⁽¹⁾ - T₃⁽⁰⁾| > 10°C → NO

With T₄⁽⁰⁾ = 1050°C = 1323.15 K → T₃⁽⁰⁾ = 466.42°C

• T₅₅ = 789.83°C ; T₅ = 567.35°C
• T₃⁽¹⁾ = 457.62°C → |T₃⁽¹⁾ - T₃⁽⁰⁾| > 10°C → NO (skip hot steam)

With T₄⁽⁰⁾ = 1060°C = 1333.15 K → T₃⁽⁰⁾ = 658.35°C

• T₅₅ = 795.80 K = 522.65°C ; T₅ = 573.70°C
• T₃⁽¹⁾ = 461.43°C → |T₃⁽¹⁾ - T₃⁽⁰⁾| < 10°C → OK

→ T₃ = 461.43°C ; T₅ = 573.70°C

From balance CdC I have T₄ = 25 + (60 + 1).1.18 (461.43 - 25) + 48388/60.4.005 = 1062.58°C

punto L

h_L = h_E - Y_c1 (h_E-h_LS)

= 3324,57 - 0,80 (3324,57 - 2944,58) = 2999,58 kJ/kg

W_ele + Y_c1 m_hp (h_c1 - h_L) = 296,6 · 41,104 (3324,57 - 2999,58) = 3441,1 kW

punto R

p_R = p₂ - 1 bar = 100 kPa ; T_R = 90°C

T_R = T_Sat (100 kPa) La liq sottoraffa

h_R = h_ls (100 kPa) - Cpw(T_Sat - T_R) = 417,44 - 4,18(99,6 - 90) = 377,25 kJ/kg

̇Q_v = ̇m_lp (h_L-h_R)

̇m_v =

26000

h_L-h_R

26000

2999,58 - 377,25 = 9,915 kg/s

punto M

Miscellamento:

̇m_HP - ̇m_LF

H_L

M

H_E

̇m_LP

̇m_HP - ̇m_LF + ̇m_P

H_M

̇m_HP - ̇m_LF(h_L + ̇m_lph_E) = (̇m_HP - ̇m_LF + ̇m_lph_M)

̇m_HP - ̇m_LF(41,104 - 9,915) - 2999,58 + 2197,2943,22 = 2380,19 kJ/kg

41,104 - 9,915 + 2197

p_M = p_L = 1000 kPa ; intercplo fra 250 e 300°C (vap, surrisc.)

h₂₅₀ = 2943,22 kJ/kg ; Δ 250 = 6,9266 kJ/kg K (già trovata [prima])

h (1000 kPa, 250°C) = h₂₅₀ = 3054,70 kJ/kg 1s (1000 kPa 300°C) = 31247 kJ/kg K

h_M - h₂₅₀ = h₂₅₀ + Δ

h₃₀₀ = h₂₅₀ h₃₀₀ - h₂₅₀

&#916; m = &#916; 250

3054,70 - 2943,22

Δ 7,1247 - 6,9266) = 6,9961 kJ/kg K

Bilancio all'albero gen. gen.

η_mẆ_T1 = Ẇ_c → η_mṁ_fgc_pg(T₃-T₄) = ṁ_ac_pa(T₂-T₁)

T_4s = T₃ - ṁ_ac_pa(T₂-T₁) = 1105,11 - 120⋅1.005 (542,20 - 15) = 684,38 °C

ṁ_mc_pg 0,99⋅122⋅1,48 = 957,53 K

γ_T1 = c_pg(T₃-T₄) → T_4s = T₃ - 1 (T₃-T₄) =

c_pg(T₃-T_4s) γ

= 1105,11 - 1 (1105,11 - 684,38) = 675,89 °C = 889,04 K

0,86

T_4s = T₃ 1/ β_T1 → β_T1 = T₃ = 1379,26^1,33

T_4s (1379,26^1,33-1) (5,853)

889,04

P_t4 = P₃ = 22,7677 = 3,8899 bar

β_T1 5,853

T₅ = T₄ - ΔT_cls = 684,38 - 8 = 676,38 °C = 949,53 K

P₅ = P₄ = 3,8899 bar (Hp. no perdite di carico)

P₆ = P_z + ΔP_gf = 0,9997 + 100 ⋅ 9,81 ⋅ 10^-5 = 1,0095 bar

β_t2 = P₅ = 3,8899 = 3,853 ; φ_g = γ - 1^1,33-1 = 0,2481

P₆ 1,0095 γ 1,33

η_od;t2 = 1 - 1 (1/3,853)^0,284-0,88

(φ+1)&sup>-γ&sup>1

1 - 0,2481 = 0,897

1 - 1/ (3,853)

T_6s = T₅ 1/3,853 = 949,531/