Question

In a combustion furnace, 2094 standard ft3 per hour

Answer 1

Explanation:

The chemical reaction is as follows.

            $CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$

It is given that 2094 $ft^{3}/hr$. And, it is known that 1 $m^{3}/s$ = 127133 $ft^{3}/hr$

Hence, convert 2094 $ft^{3}/hr$ into $m^{3}/s$ as follows.

                $\frac{2094 ft^{3}/hr}{127133 ft^{3}/hr} \times 1 m^{3}/s$

                  = $0.0165 m^{3}/s$

As ideal gas equation is PV = nRT. So, calculate the number of moles as follows.

                   n = $\frac{PV}{RT}$

                      = $\frac{1 atm \times 0.0165 m^{3}}{0.0821 \times 298 K}$

                       = 0.673 mol/sec

According to the stoichiometry of the given reaction, 1 mol of methane reacts with 2 mol of oxygen.

So, 1 mol $CH_{4}$ = 2 mol $O_{2}[\tex]Hence, [tex]O_{2}[\tex] required theoretically = [tex]2 \times 0.673 mol/s$ = 1.346 mol/s

Hence, air required theoretically = $\frac{1.346}{0.21}$ = 6.4095 mol/s.

Since, 6% of excess air is being supplied. Therefore, total air supplied will be calculated as follows.

                Total air supplied = $6.4095 mol/s [1 + \frac{6}{100}]$

                                              = 6.794 mol/s

Now, calculate the volume using ideal gas law equation as follows.

                            PV = nRT

           $1 atm \times V = 6.794 mol/s \times 8.21 \times 10^{-5} Latm/K mol \6 times 298 K$

                           V = 0.166229 $m^{3}/s$

Converting calculated volume into $ft^{3}/hr$ as follows.

                  1 $m^{3}/s$ = 127133 $ft^{3}/hr$

So,        0.166229 $m^{3}/s$ = $0.166229 m^{3}/s \times 127133 \frac{ft^{3}/hr}{1 m^{3}/s}$  

                                        = 21133.191 $ft^{3}/hr$

Thus, we can conclude that 21133.191 $ft^{3}/hr$ of air are drawn from outside  per hour by the fan that supplies the air.