Question

A furnace is fired with fuel oil. The Orsat analysis of flue gases by volume is as given below: 10.6% CO2, 6% O2, 83.4% N2. Calculate: (a) % excess air; (b) The C:H ratio in the fuel oil, assuming that fuel does not contain nitrogen.​

Answer 1

Answer:

the C:H ratio in the fuel oil,assuming that fuel does not contain nitrogen

Answer 2

Given,

Orsat analysis of fuel gases by volume is given as

$CO_{2} =10.6$%

$O_{2} =6$%

$N_{2}=83.4$%

To find,

a) % of excess air.

b) the C:H ratio in fuel oil.

Solution,

Let us start by balancing the individual elements,

Balancing $N_{2}$

$N_{2}$ in fuel gas = $83.4 moles$

$N_{2}$ from air = $83.4moles$

Balancing $O_{2}$

$O_{2}$ from air = $83.4$ × $\frac{21}{79}$

⇒$O_{2_{air} } = 22.16 moles$

$O_{2}$ in fuel = $10.6$ × $\frac{1}{1+6}$

⇒$O_{2_{fuel} }= 1.52moles$

$O_{2}$ unaccounted = $O_{2_{air} }-O_{2_{fuel}}$

⇒$O_{2}$ unaccounted = $22.16-1.52$

⇒$O_{2}$ unaccounted = $20.64moles$

$O_{2}$ theoretical = $CO_{2}$ in gas - unaccounted $O_{2}$

⇒$O_{2_{th}}=10.6-1.52$

⇒$O_{2_{th}}=9.08moles$

Balancing $H_{2}$

Moles of $H_{2}$ in the fuel = unaccounted $O_{2}$ × $2$

⇒$H_{2_{fuel}}=20.64$ × $2$

⇒$H_{2_{fuel}}=41.28 moles$

Balancing $C$

Moles of C in fuel = $10.6moles$

Now,

% of excess air is given by,

% excess = $\frac{O_{2_{air}}-unaccounted O_{2}}{unaccountedO_{2}} .100$

⇒% excess = $\frac{22.16-20.64}{20.64} .100$

⇒% excess = $\frac{1.52}{20.64} .100$

⇒% excess = $0.0736$ × $100$

⇒% excess = $7.36$

And,

C to H ratio,

In $moles$ = $\frac{moles of C}{moels of H_{2}}$

⇒In $moles= \frac{10.6}{41.28}$

⇒ In $moles = 0.25$

In weight = $\frac{12.moles of C}{2.moles of H_{2}}$

⇒In $weight = \frac{(12)(10.6)}{(41.28)(2)}$

⇒In $weight = \frac{127.2}{82.56}$

⇒In $weight = 1.54$

Therefore,

a) % of excess air is 7.36%

b) The C:H ratio is 1.54