Esercizi svolti di fondamenti dell'ingegneria di processo

ESERCIZIO 1

ω_CH4=0.60

ω_H2=0.20

ω_CO=0.20

ω_H2O=0.20

T [°C]: T [°F]: 32 = 260 °C

P [atm]= 22.96 atm

Q = 283.17

1) &ṁu;_CH4 [kg/h]

&ṁu;_CH4 = ω_CH4 · &ṁu;_mix

= ω_CH4 · PM_mix · &ṅu;_mix

&ṁu;_mix = β · ^P V̇_m/_RT

PM_mix = 7.023

[...]= 0.538 kg/h

ESERCIZIO 2

PM_A=192 kg/kmol

V̇_mix= 10.5 L/min

w_A= 0.612

ρ̇_mix=102.0 kg/m³

1) &ṅu;^v [kmol/min]

ṁAPMA ρ_m^·= ω_A

168 PLB O 0.6557

ESERCIZIO 3

m_mix = 100 lb/h

x_CH4 = 0.02

x_H2 = 0.38

x_CO = 0.50

x_CO2 = 0.10

1) V̇_mix [ft³/s]

V̇_mix =

1) ṅ_HZ [kmol/h]

X_HZ = X_H₂ + X_mix = X_H₂

Pvmix = 0.38 ⋅ 19.73 atm ⋅ 5569.6 J/mol

R ⋅ T 0.0821 ⋅ 583.2

= 88.66 kmol/h ⋅ 1 kmol

10³ mol

2.466 kmol/s

2) W_HZ

W_HZ = ṁ_HZ = Pvmix

ṁ_mix ṁ_mix 100.26 ⋅ 0.6537 kg ⋅ 1

ṁ_mix ⋅ 3500 s

4.95 x 10^-6

0.039

Esercizio 4

X_n-ex = 0.6 X_n-ep = 0.6

m_mix = 3600 kg/h

1) C_i ρ_i ρ_mix

La sol ideale => c_i = x_i ⋅

ρ_iρ_mix ⋅ V_i ⋅ = ρ_i ⋅ x_i ⋅ V

V_mix ⋅ V_i

ρ_mix = ∑ρ_imix = 0.674 kg/dm³

2) C_i C_mix

c_i = ṅ_i; ṁ_i = ρ_i ⋅ m_i

V_mix m_i

C_n-ex = 3.058⋅10^-3 kmol/dm³

ṁ_mix kg

C_n-ep = 0.096⋅10^-3 kmol/dm³

C_mix = ∑ki C_i = 7.55 kmol/m³

3) W_i x_i

w_i = ṁ_i = ρ_i

ṁ_mix ⋅ ṁ_mix = ρ_i

n_mix | c_i ⋅ ṁ_mix

x_i = ṅ_i = Σc_imix

W_n-ex = 0.3914

W_n-ep = 0.6089

X_n-ex = 0.6275

X_n-ep = 0.5796

4) ṅ_mix V_mix

ṅ_mix = ṁ_mix ṁ_mix = 36.09 kmol/h

ρ_mix ∑ρ_i/P_i

V_mix = ṁ_mix = 5.065 m³/h

ρ_mix

ṅmix e BILANCIO SUL MIX

In ṅin + ṅin = ṅin

H2O ṅH2O + ṅH2O = ṅH2O

CO ṅCO + ṅCO = ṅCO

CH3OH

MIX ṅ̂emixȳm = ṅ̂mixȳm

BILANCI SULl'INTERO PROCESSO

| 1 | 2 | 3

CH4 in ṅe ̂1mix ȳCH4̂1 = n2mixȳCH4₂

H2 n2mixȳH2₂ = n3mixȳH23

CO n2mixȳCO₂ - ṅrCO = n3mixȳCO₃

CH3OH ṅ3mixȳCH3OH₃ = ṅ̂CH3OH

MLX ṅemix - ṅRDet promoter D̆ mix̂ + ṅR̂mix - ṅmix = Ø

rR = n3mix ṅ̂R = ṅ3mix ȳch₄ȳcó - ṅ3mixȳcó

2 ṅ̂R = ṅ2mixȳcool - n̆émieKn3mixȳcó

ṅmoinix = n̆émieȳCH3OH3 = n3mixȳC3

composta al picciolo é la stessa della corrente 3

n3mix = n1mix - ṅmix = 32.8 kmol/h

In

H2 n2mixȳH2₂ - n3mixȳH23 - ṅe̊3mixȳH22

CO y2CH4 = 0.0069

y3H2 = 0.7983

yCO = 0.1708

svelto la fraz ondasse del CHiLs imegens

ṅmixȳCH4n + ṅ̂mixȳCH4 - ṅ̂mixȳCH4

= n1mixȳCH4n - n̆emieṡremainu̇ hinxumu̇rîumeiȳ̆himeu̇

ṅmixȳch4 = n̆emieṡremainu̇ŸlCH4-Chi̇hliŭɨrik

_______

ṅA ˧mix = 28.11 kmol/h

SESSIO8

ṅ̂mix=650 kmol/h

y0= 0.005

O2 + + 4H2 = CH3OH + H2O

∆H_m.i = 0

∆H_C2H4,HR2 = (Cp ∙ dT) = 58.29751 kJ/mol

∆H_C6H6,HR2 = (Cp ∙ dT) = 63.57335 kJ/mol

∆H_C8H12,HR2 = (Cp ∙ dT) = 1.65685 kJ/mol

P = 2200 ???

METODO HOF

∆H_C2H4 = -1638.51 kJ/mol

∆H_C6H6 = 20.07095 kJ/mol

∆H_C8H12 = 31.27695 kJ/mol

∆H_H2 = 1.65685 kJ/mol

Q = 322.668 kJ/mol

Esercizio III

25°C, 5 atm

V_mix = 3,1 m³/h

y_CO = 0.33

y_H2 = 0.67

CO + 2H₂ → CH₃OH

T = 127°C P = 5 atm

1) V_CO

ṅ_mix = 𝑬_mix

BILANCIO REATTORE

CO ṅ_mix)y_CO = ṅ_CO

H₂

CH₃OH

MIX = (ṅ_mix = 0)

RELAZIONI STECHIOMETRICHE

ṅ_CH3OH = r_CO

BILANCIO ENERGIA / PORTATA TERMICA

= 1.05 ṅ_mix

P.F. 25°C 1 atm (s) per tutte le spec.

2) Q̇

METODO: HOF    RIF: 25°C, 1 atm (c)    STP

Q̇ = ṅ _v^w ∑ y_i^wΔĤ_i^u + ṅ_v^u∑ y_i^uΔĤ_i^c - ṅ_v^e∑ y_i^eΔĤ_i^c

ΔĤ_m,1c^m = ΔĤ_v.beu,1 + ∫₂₅^80,10 C_p dt + PM/ρ

Q̇ (0,2155 - 1 atm) = 52.635 kJ/mol

ΔĤ_m,1c^m =

ΔĤ_beu,1 = 46.704 kJ/mol

ΔĤ_beu,2 = 9.683 kJ/mol

Q̇ = 0.111 kJ/s    (Quoto = 0.55 kJ/s)

Essendo un'assenza di C.C., potevo scegliere arbitrariamente lo stato di riferimento

Esercizio 16

Reattore Water-Gas Shift: reattore continuo isoterma

• ṅ_CO2 = 100 mol/h
• ṅ_H2O = 150 mol/h
• ṅ_CO2 = 45 mol/h
• @ 350°C   1 atm

X_co = 0.80    Q = 1.2 kW   t.c. 1s01

1) TUT, utilizzo solo primo termine (a) del cs

Bilancio Reattore

• ṅ_c - ṅ_ce = ṅ_co = Ø
• ṅ_H2O - ṅ_H2 = ṅ_H2 = Ø
• ṅ_N2 - ṅ₂ = Ø
• ṅ_O2 - ṅ₂ = Ø

Relazioni Stechiometriche & Conversione

• ṅ_co-ṅ_co = ṅ_co-ṅ_co2
• X_co = 80 mol/h
• ṅ_co = 80 mol/h
• ṅ_CO2 = 20 mol/h = 5.55 • 10^-2 mol/s
• ṅ_H2O - ṅ_H2 = 0.0364 mol/s
• ṅ_H2 = 80 mol/h = 6.672 mol/s

2)

BLU MISCELATORE

\(\dot{n}_{in}^{\alpha} + \dot{n}_{in}^{\beta} + \dot{n}_{in}^{\gamma} =\varnothing\)

\(\dot{n}_{in} = \dot{n}_{in}^{\alpha} + \dot{n}_{in}^{\beta} + \dot{n}_{in}^{\gamma}\)

\(y_{in} = \frac{\dot{n}_{mix} y_{in} + \dot{n}_{mix} y_{in}}{\dot{n}_{mix}}\)

\(\dot{n}_N\theta{2} = 0.15 = \frac{\dot{n}_{N2}}{\dot{n}_{mix}} \Rightarrow \dot{n}_{N2} = \dot{n}_{N2} = \frac{108.333 \text{ mol/s}}{0.15}\)

BLU MISCELATORE

\(\dot{n}_N\theta{2} = \dot{n}_N\theta{2}\)

\(\dot{n}_{mix} = \dot{n}_{N2} = \dot{n}_{mix} y_{N2} = 386.9 \text{ mol/s}\)

\(y_{N2} = \frac{\dot{n}_{N2}}{\dot{n}_{mix}} = 0.236\)

\(\dot{n}_{mix} = \dot{n}_{mix} + \dot{n}_{B}^{\alpha} = 956.58 \text{ mol/s}\)

\(y_{O} = \frac{\dot{n}_{in}}{\dot{n}_{mix}} = \frac{\dot{n}_{mix} + \dot{n}_{mix} \beta_{in}}{\dot{n}_{mix}} = \frac{\dot{n}_{mix} y_{in}^{\gamma}}{\dot{n}_{mix}} = \frac{\dot{n}_{mix}^{\delta}}{\dot{n}_{mix}}\)

= 0.075

8)

RI = 5.5.5

A) \(\Delta H^o\)

\(\Delta H^o_m = \sum_{1}^{l=1} \Delta H_{cH_m}^{l=1} \sum_{1}^{l=1} \Delta H_{f}^{o} \frac{\Delta H_{o}^{m}=2(-66.29 \text{ KJ/moe})}{\omega}\)

ESERCIZIO 23

\(\dot{n}_{mix} = 200 \text{ mol/s}\)

LIQUINO

\(y_{\gamma =ep} = 0.50\)

\(y_{\gamma -ex} = 0.60\)

T=100°C, P = 7atm

1\) \(\dot{n}_{y}^{\gamma}\)

\(\dot{n}_{mix} y_{\gamma}^{ep} = \dot{n}_{mix}^{V} y_{\gamma -ex} = \dot{n}_{mix}^{V} y_{\gamma -ex} = \dot{n}_{mix}^{ep} y_{\gamma -ex} + \dot{n}_{mix}^{ep} y_{\gamma -ep} = \dot{n}_{v}^{ep} y_{\gamma -ep} = \dot{n}_{mix}^{\gamma}\)

\(\dot{n}_{mix} = 256.64-1.55 \dot{n}_{mix}^{v}\)

\(\dot{n}_{mix} = 163.98 - 0.60 \dot{n}_{mix} = 101 \dot{n}_{mix}\)