Esercitazioni Fisica Tecnica

Esercitazione Fisica Tecnica

m_NZ = 50 g = 50·10^-3 kg

p₀ = 8 bar = 8·10⁵ Pa

T₀ = 20°C = 293 K

P = 50 W

t₀ = 5 mm = 480 s

P₂ = 8 bar → 150 bar

VEDO GAS COME IDEALE

cp = cv + R

H₂O

cp = 7/2 R

O₂

cp = 7/2 R

B

cp = 3/2 R

CO₂

cp = 5/2 R

CO

cp = 5/2 R

ΔU = ?

L_D = ?

S_IRR = ?

ΔU = L − Q ← L_ND – L_G

Q = CU (T₂ – T₁)

ΔU = CP (T₂ – T₁)

n CU (T₂ – T₁) = L_ND – L_G

n CU (T₂ – T₁) = L_ND – P (V₂ – V₁)

n CU (T₂ – T₁) = L_ND = HR (T₂ – T₁)

T₂ = T₁ + L_ND/M (CU + R)/CP = 293 K, 24000

P = L_WC/C (L_DC = 24000 J)

L_D = ∫_V1^V₂ PdV = P (V₂ – V₁)

P_N = RT – V = RT/p

V = RT/m_NZ /p

NZ = CP + 4/2 R

Uso cal/paso

ΔU = n (CU (T₂ – T₁)) = 42,3 3/Kg c.p (754,8 – 293) K 50·10^-3

ΔU = 14144 J

L_D = P (V₂ – V₁) = HR (T₂ – T₁) = 50·10^-3 Kg 8314/28 ΔT = 6854 J

Bilancio entropico: S₁ + ΔS_G + S_IRRE = S₂ →...

S_IRR = M ΔS = M (CP ln (T₂/T₁) – R ln (P₂/P₁))

S_IRR = M (CP ln (T₂/T₁)) = 49 J/K

2)

mO₂ = 0,025 kg

P_O2 = 10 BAR

T_i = 30°C = 303K

m_He = 0,334 kg

P_He = 10 BAR

T_i = 150°C = 423K

c_v = c_vmix = 4,85

P = 100 W

Cu₀₂ = (

Cu₁ = 9/5

Cu_HE = Cu₂) =

Cu = = 8,314

Cu_HE = C

C_v = 8,9

v _P T_i =

CHe _{RFPT2 =}

S_IRR = ΔS₁ + ΔS₂

AU_TOT – L + Q^es

ΔU₁ + ΔU₂ = P_e

ΔU₁ = G₂1 =

mO₂C₀₂ (TF - Ti₁)

–

AU₂ = L - G₂1 =

mHeC_vHe (TF - T_2i)

T_F =

mO₂CuO₂(TF - Ti₁)

+ m_HEC_vHE (TF - T_2i)

S_IRR =

(mO₂CuO₂TF + mHeCuHePT₂Ti₁)

PT_i +

mO₂CuO₂ +

mHeCuHeT_2i = 330,5K

G₂1 =

ΔU₁ =

mO₂CuO₂ΔT_c2

G₂1 = 0,025

K

64,9 5

83 =

330,5 - 303K

K = 365J

3)

P ₁ = 1 BAR

T₁ = 25°C = 298K

P ₂ = 1,25 m

ΔU = Cm²2

Cu_L = ΔU

= g_el in

C^x v

P^NFT_KS

3/5

= (Cu^2559,461C^u)

M

P₁ = RT

ln P₁ =

ln (

R

v_i =

Vn

m

Tⁿ = 9:7

9:7

(

=^P_oFR

non ho la massa ad un po^P_eF

SPECIFCHE

C

PUNTO DI

Δu(v)^{e = PB26}

C^{R ONDRESUB231C_me2}

R

u = CuPO

MrPot

n - x

Cu - U₂

V₂ = CU = C

R =

CuCuCu = 46,003

+ 33/

S_IRR

R

1O

2=

P₂of

Cu1_NOTPFNON

P^m

v 816

P²

p

m

ln (

V ₁/V ₂

P₁

ln (

ln(P₂/P)

V ₁/V ₂

ren

P ² (ln

en(

En

V₁/P _f

= rm

V ₂/ V ^1/en

(ln (

ut

ESERCITAZIONE FISICA TECNICA

1° TC= 18°C - COST

Ġ= 3 KW

ṁ = 3 kg/s

The 10°C TZF= 15°C

SIRR?

PORTATA COSTANTE

Δsc = dsC

G- Δsc + Q * = 1,96 KW

20°

V_i = 1500 m³ | ṁ˙G=1,196 m³/s

T_i = 20°C - 293 K

P_e = 1,2 MW

P₂ = 12 MW

T₁₅ ? T₁₅ = ?

Ẏ_B = Ẏ_V T₁₂ = 136,0 KW

β₁ = P₁ -> P_x - β₂ = P₂

β_TOT =_... =2β

P_B L₁, L₂

Esercitazione Fisica Tecnica

1) P₁ = 10⁵ N/m²

T₁ = 27 °C = 300 K

T_max = (373.3 - 147) °C = 1400 K

β = 16 → P₂ = β P₁ = 1.6 10⁶ N/m²

1 → 2 Compressions

Livello 1 → T₂ = T₁ (P₁/P₂)^γ-1/γ = 602 K

3 → 4 Espansione

P₄ = P₁ = 10⁵ N/m²

T₃ = T₄ (P₄/P₃)^γ-1/γ = 770 K

Δh = CP(T₂ - T₁) - 364.6 kJ/kg

P_out = Δh(4-3) CP(T₄ - T₃) - 936.4 kJ/kg

Q_k = Δh CP(T₃ - T₂) - 1043.1 kJ/kg

η = 0.5147 = 54.4% oppure η = 1 - ηc = chi coeff.

Considero Adesso un Ciclo Reale

Compressions Reale:

• η_c = 0.685
• P_out = 300 MW
• η otto = → 1/0 P_e n_c / λ

Espansion Reale:

• η_i = → Espansion LID (codice)

2°C q out = q in?

P in? cop?

TDC

Q̇ out=25 KW

Tc=-10°C

Th=61°C

Δt=5°C

1. T1=Tc. ΔSe=-4°C
2. P1=Psat(-4°C)=0,2524 MPa
3. h1=hsat(-4°C)=244,4 kJ/kg
4. s1=s sat(-4°C)=0,9423 kJ/kg K

3. T3=56°C
4. P3=1,8278 MPa
5. h3=h3(56°C)=130,93 kJ/kg
6. s2=s sat(56°C)=0,6622 kJ/kg K

4. h4=h3=130,93 kJ/kg
5. T4=T1=-4°C
6. h es(-4°C)
7. h4=hs=1
8. h4=hs=28

COP PDC=4,6

Ricicloco con:

T c=9°C Th=35°C

T i=T c−D T=−1°C

P i=P sat(−1°C)=c.337654 MPa

h 1=249,33 kJ/kg

q out=h 2−h 3=151,92 kJ/kg

q out=h 1−h 4h in=35,97 kJ/kg

P 3=0,253 MPa

P out=1.161 MPa

q in=h 1−h 4=113,97 kJ/kg

e in=h 2−h 4 = d h v+q in=3,42 kJ/kg

Ricicloco con:

T c=9°C T f=40°C

1. T i=T c-D T=-1°C
2. h i=273,4
3. P 2=1.61 MPa
4. h 2s=0.6622
5. T f=T c-P 3=0.253 MPa

P out=1,61 MPa

6. mi=Q̇ out=0,0991 kg/s