Question

Octane (C8H18) is burned with 250 percent theoretical air, which enters the combustion chamber at 25°C. Assume complete combustion and a total pressure of 1.8 atm. determine (a) the air-fuel ratio and (b) the dew-point temperature of the products.

Answer 1

a) The air-fuel ratio is 37.83 kg air/kg fuel

b) The dew-point temperature is 47.02°C

Step-by-step explanation:

a) The combustion equation is equal to:

C₈H₁₈ + 2.5a(O₂ + 3.76N₂) = xCO₂ + yH₂O + 9.4aN₂ + 1.5aO₂

The mass balance for each elements is equal to:

C: 8 = x

H: 18 = 2y, y = 9

O: 5a = 2x + y + 3a, a = 12.5

C₈H₁₈ + 31.25(O₂ + 3.76N₂) = 8CO₂ + 9H₂O + 117.5N₂ + 18.75O₂

The air fuel ratio is equal to:

Taking the molar mass of octane = 114 kg/mol

molar mass of air = 29 kg/mol

`A_(F) =(mass-of-air)/(mass-of-fuel) =(31.25*4.76*29)/(1*114) =37.83kgair/kgfuel`

b) The dew-point temperature is:

`P_(v) =(N_(v)*P_(prod))/(N_(prod) )`

Where Nv = moles of water vapor in products

Nprod = moles in the products

Pprod = total pressure = 1.8 atm = 182.38 kPa

`P_(v)=(9*182.38)/(8+9+117.5+18.75) =10.71kPa`

The saturation temperature at 10.71 kPa is obtained from saturated-water temperature tables and interpolation:

T₁ = 45°C

P₁ = 9.5953 kPa

T₂ = 50°C

P₂ = 12.352 kPa

`(T_(sat)-T_(1) )/(T_(2)-T_(1) ) =(P_(v)-P_(1))/(P_(2)-P_(1)) \\(T_(sat)-45)/(50-45) =(10.71-9.5953)/(12.352-9.5953) \\T_(sat)=47.02`