Question

Carburetted water gas is produced in the same way as blue water gas except that it is done in the presence of cracked oil vapors in a carburettor. A typical gas analysis shows 4.7% CO2, 7.8% C2H4, 0.3% O2, 36.5% H2, 35.5% CO, 8.6% CH4, 6.6% N2. If this gas is saturated with water at 20C and 742 torrs and burned in 10.434 m3 air at 30C, 101 kPa and 60% RH per m3 fuel, calculate: (a) % excess O2, (b) orsat analysis of stack gas if 85% od C burns to CO2; all H2 burns to H2O, (c) m3 stack gas/ m3 fuel gas, (d) Pwater in stack gas

Answer 1

Answer:

think so option a is correct hope this HELPS you

Answer 2

Given:

For air:

P = 101×10³ Pa

R = 8.314 Pam³/mol

V = 10.434 m³

T = 30°C = 303K

For fuel:

P = 742 torr = 98925.2 Pa

V = 1 m³

T = 20°C = 293K

To find:

(a) % excess O₂

(b) orsat analysis of stack gas if 85% of C burns to CO2; all H₂ burns to H₂O

(c) m³ stack gas/ m³ fuel gas

(d) P_water in stack gas

Solution:

PV = nRT

For air:

$n_a_i_r = \frac{101*10^{3} * 10.434 }{8.314*303}$

n_air = 418.33 mol

n_O₂ = 0.21 × 418.33 = 87.9 mol

n_N₂ = 0.79 × 418.33 = 418.33 mol

For fuel:

$n= \frac{98925.2*1}{8.314*293}$

n = 40.61 mol

Pv = vapor pressure of water at 30°C = 4239.651 Pa

∴ Fuel composition

CO₂ = 4.7% = 1.91 mol

C₂H₄ = 7.8% = 3.16 mol

O₂ = 0.3% = 0.122 mol

H₂ = 36.5% = 14.8 mol

CO = 35.5% = 14.4 mol

CH₄ = 8.6% = 3.5 mol

N₂ = 6.6% = 2.7 mol

C₂H₂ + 3O₂ → 2CO₂ + 2H₂O

H₂ + 1/2 O₂ → H₂O

CO + 1/2 O₂ → CO₂

CH₄ + 2O₂ → CO₂ + 2H₂O

C + O₂ → CO₂

C + 1/2 O₂ → CO

H₂ + 1/2 O₂ → H₂O

a) Theoretical O₂ required

= O₂ required for complete combustion

= O₂ for C₂H₄ + O₂ for H₂ + O₂ for CO + O₂ for CH₄ - O₂ available

= 9.48 + 7.4 + 7.2 + 7 - 0.122

= 30.958 mol

O₂ supply by air = 87.9 mol

%excess = (O₂ supply by air - theoretical O₂) / (theoretical O₂) × 100

%excess = 1.84 × 100 = 184%

b) if 85% of C burned to CO₂

∴ 15% of C burnt to CO

we have total C = 1.91 + (3 × 3.16) + 14.4 + 3.5 = 26.13 mol

∴ CO₂ produced = 0.85 × 26.13 = 22.21 mol

∴ total CO₂ in stack gas = 22.21 + 1.91 = 24.12 mol

CO produced = 0.15 × 26.13 = 3.92 mol

total CO in stack gas = 3.92 mol

H mole is fuel = 4 × 3.16 + 2 × 14.8 + 4×3.5 = 56.24 mol

∴ mole of H₂ = 56.24 / 2 = 28.12 mol

∴ H₂O produced = 28.12 mole

O₂ consume = O₂ for CO₂ + O₂ for CO + O₂ for H₂O

O₂ consume = 22.21 + (3.92 / 2) + (28.12 / 2) = 38.23 mol

∴ O₂ in stack gas = O₂ supplied - O₂ consume + O₂ available in fuel

O₂ in stack gas = 87.9 - 38.23 + 0.122 = 49.8 mol

Stack gas composition

CO₂ = 24.12 mol

CO = 3.92 mol

O₂ = 49.8 mol

N₂ = 2.7 mol

H₂O = 28.12 mol

For orsat analysis H₂O is not taken

So, total moles of stack gas after removing water = 24.12 + 3.92 + 49.8 + 2.7

                                                                                  = 80.54 mol

Total mol of stack gas taking H₂O = 80.54 + 28.12 = 108.66 mol

% CO₂ = 24.12 / 80.54 = 30%

% CO = 3.92 / 80.54 = 4.8%

% O₂ = 49.8 / 80.54 = 61.8%

% N₂ = 2.7 / 80.54 = 3.35%

(c) $\frac{V_s m^{3}\ of \ stack \ gas}{V_f m^{3}\ of \ fuel \ gas} = \frac{mol \ of \ stack \ gas}{mol \ of \ fuel} *\frac{T_S}{T_F} * \frac{P_F}{P_S}$

(760 torr = 1 atm =  101325 Pa)

$\frac{V_s m^{3}\ of \ stack \ gas}{V_f m^{3}\ of \ fuel \ gas} = \frac{108.66}{40.61}*\frac{673}{293}*\frac{98925.2}{101325}$  = 6

$P_w_a_t_e_r = xP_T = \frac{28.12}{108.66}*101325 = 26236.2Pa$

P_water = 26236.2Pa