Rankine Joule

V̇ = 2,88 m³/h = 0,0008 m³/s

p = 1,100 bar

A

T = 25,0°C

N_is = 60%

S_gen = 5,10 kW/K

Pe = ?

Log₃ = ṁv (P₂ - P₁)

Liquid

S_gen = ṁ c ln

T_{AO A23} = 4.974 kg/m³

ṁ = v 0,974 kg/s

v =

Riuso T₂

S_gen = ṁ c ln T₂ - ṁ c ln v̇

ln v̇ = 4,926

T_i = 138°C

l₀₁ + l₀₁s = c (T₂ - T₁) = c (T₂ - T₁) = 473 kJ/kg

M_1s =

l - l_A1,15 = 1182 kJ/kg

P_E2 = 8bar

J = ml = 942 kJ

RANKINE IDEALE

P1in = 10kPa - 0,1barP1max = 2000kPa - 20barTSGETA = 1200°CTSET = 20°C

Vc1

Rendu de pompeh1

2 + Lp = h2 - h3Delta sgom = s2 - sgen = 0

Vc2

h2 + Qa = h3D1 + Qm/Tm = sgo = Sso

Vc3

h3 = L + h4s3 - sgen = s4 - sgen = 0

Vc4

h4 = Qe + h1s4 = Delta sgen = Qm/TB + Delta1

Redonde statua. Rosso la ho collochero

T1 TBA AB e vedo T —> 45,80°C = 3 PAT2 E' cambierai sia di T eTB, TBA AB

h2 TBA AB e vedo SLdelta vTBA AB e vedo SLV2 TBA AB e vedo SL

Pag 142 n°2.1

TH = 400°C

TB = 20°C

T3 = 350°C

P_e = 3,5 MPa - p₃ = 35 bar

P_u = 100 kPa - p₄ = 1 bar

Calor mt. sx.

k                        b                        x

• 372        1
• 372        35   4173,13   0
• 623        35   3106&
• 372        1     2403,665    0,28

Stato 1

liquido surriscaldato

T = T_sat con v_ABAB = 99,63°C

h₁ = h₂ = 4173

s = 2   l   = 1,3

Stato 2

T₁ = T₂

p₂ = 35 bar

liquido sottoraffreddato

h = x

h₁ = h₂ = 417,3

P_sat = T₁ = 2 bar

VC1

m_in + Zp = m_in2

m_in1 + s_gen = m_in3 + s_gen = 0

VC2

m_no + Q  2 = m_h + Q₁ = h₃ - h = 2,688,7

m_no + s_gen + q = m_no3

VC3

m_nb + Q = m_nb + t₁

m_nb3 + s_gen = m_nb4 + s_gen = 0

VC4

m_nb + m_nb7 + u_w q m_b + x < h₀ + h₁ = 1,285,1

m_d1 + s_gen + q x m_d1 + Q_b

<  T_B

RANKINE RISURRISCALDAMENTO

P_max = 4 MPa = 40 bar

T_max = 350°C = 623 K

P_int = 2 MPa = 20 bar

P_min = 350°C

P_sev = 160 kPa = 1.6 bar

T_sevf = 120°C = 393.15 K

T_{set B} = 200°C = 293 K

Stato 1

Liquido saturo x = 0

T = T_s = 45.4°C = 318.9 K

h₁ = h₂ + D

h₁ = 192.1 h₄ = 0.165

Stato 2

Liquido saturo umidoP₂ = 40 bar

T₁ = T₂ = 381 K matric

h₂ : reciprocoh₂ = x

h x_L46 50

x₁ x₂x₃ ? /

Stato 3

T = 623 KP = 20 barPhase: ?P_S 7.4ypore/sub 1tl - hlo - o dsophen

IDT satureT = T_s + 654

h₃ = 3.98D₃ = 658

Stato 4

D₃ = D₄ = 5,98

P₃ = 20 bar

D_3VSES medio tva.h_b transvet0emme cossro T. | ? + abdaz S Condensato E.c O puntot

x₁ x₂ x₃ x₄

64, 65 646 hx₁ ? x₄ 246 x₃ 28s

Ininaprivalurex₁ x₂ x₃ x₄250, 235, 123.5

x = 8

Ciclo Joule Con Post Combusitione

β=20

P₁=1 bar

T₂=20°C=293K

Tmax=1200°C=1473K

Mr=0.78

Mc=0.8

TsetA=1773K

TsetB=268K

Mn=11.90 kg/s

Pmin=res

Prmb=626 Pa

T₆=1

P₂=P₁ β

Pmax=max

Pmin=2.645 bar

Tc=1

Lt=L_t+L_h=17,639

Lmaisimis=Cp(T₃-T₅)

Q̇_e=ṁ Cp(T_{nsuperscript_prim}-T₂)↗=865.1

L_n=L_t-L_e=11365

• 0128.040=0.36