Answer:
Explanation:
= Area of section 1 =
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 =
= Density of fluid =
= Area of section 2 =
Mass flow rate is given by
The mass flow rate through the pipe is
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1
The speed at section 2 is .
The composition of gas in the feed, the percentage conversion and the
theoretical yield are combined to give the product stream composition.
Response:
The composition of gas in the product stream are;
The amount of oxygen used = 30% exceeding the theoretical amount
Number of moles of hydrochloric acid = 4 kmol/h
Percentage conversion = 80%
Required:
The composition of the gas in the product feed.
Solution;
The given reaction is; 4HCl + O₂ 2Cl₂ + 2H₂O
Which gives;
Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h
Which gives;
Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h
Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h
Similarly;
Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h
Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h
The composition of the production stream is therefore;
Learn more about theoretical and actual yield here: