Butane gas C4 H10 is burned completely with 40 percent exces

Butane gas (C4 H10) is burned completely with 40 percent excess air during a steady-flow combustion process. The fuel and air enter the combustion chamber at 25 degree Celsius and the products leave at 1400 K. Determine

a. The balanced reaction equation

b. The air-fuel ratio

c. The heat transfer for this process.

Solution

  Step1 :

          Balanced equation is :

       C₄H₁₀ + (1/) O₂ (O₂ + 3.76 N₂) a CO2 + b H2O + c N2 + d O2

STEP2:

First we need to find the stoichiometric air ( = 1, d = 0 )

C balance: 4 = a, H balance: 10 = 2b => b = 5

O balance: 2O2 = 2a + b = 8 + 5 = 13 => O2 = 6.5

Now we can do the actual air: (1/) = 2 => O2 = 2 × 6.5 = 13

                    N balance: c = 3.76 O2 = 48.88, O balance: d = 13 - 6.5 = 6.5

Step3:

Energy Eq.: q = HR - HP = HoR - HoP - HP

HoR = -126 200 + 0 + 0 = -126 200 kJ/kmol fuel—stanard values at 298 k

HoP = 4 (-393 522) + 5(-241 826) + 0 + 0 = -2 783 218 kJ/kmol fuel

The rest of the values are from standard temperature table at 298 K

hCO2 = 33397, hN2 = 21463, hO2 = 22703, hH2O = 26000 kJ/kmol

HP = 4 × 33 397 + 5 × 26 000 + 48.88 × 21 463 + 6.5 × 22 703

= 1 460 269 kJ/kmol fuel

From the energy equation we get

q = -126 200 –(-2 783 218) - 1 460 269 = 1 196 749 kJ/kmol butane