Complete Question
Air is bubbled through a drum of liquid hexane at a rate of 0.200 kmol / min. the gas stream leaving the drum contains 10.0% mole's hexane vapor. air may be considered insoluble in liquid hexane. use an integral balance to estirmate the time reguired to vaporize in min to vaporize 10.0 m^3 of the liquid.if the total moles of the products 0.222 kmol / min and the specific gravity of liquid hexane is 0.659 and the accumulation term equals -76.45 kmol ch4
Answer:
The value is [tex]t = 3851.7 \ min [/tex]
Explanation:
From the question we are told that
The rate at which the air is bubbled is [tex]R = 0.200 \ kmol/min[/tex]
The rate at which the gas stream is leaving the drum is [tex]r = 0.222 \ kmol/min[/tex]
The percentage of hexane contained in the gas leaving the drum [tex]10\%[/tex]
The volume of liquid hexane to be vaporized is [tex]V = 10 \ m^3[/tex]
The specific gravity of liquid hexane is[tex]SG_{H} = 0.659[/tex]
Generally the density of the liquid is mathematically evaluated as
[tex]\rho_h = \rho_w * SG_{H}[/tex]
Here [tex]\rho_w[/tex] is the density of water with value
[tex]\rho_w = 1000 \ kg/m^3 [/tex]
So
[tex]\rho_h = 1000* 0.659[/tex]
[tex]\rho_h =659 [/tex]
The mass of the liquid is mathematically represented as
[tex]m = \rho_h * V[/tex]
[tex]m = 659 * 10V[/tex]
[tex]m = 6590 \ kg [/tex]
The number of moles of liquid hexane is mathematically represented as
[tex]n = \frac{m}{ M}[/tex]
Here M is the molar mass of hexane with value [tex]M = 0.086 \ kg[/tex]
So
[tex]n = \frac{6590 }{ 0.086}[/tex]
[tex]n = 76628 mols [/tex]
[tex]n = 76.628 \ k mols [/tex]
The rate at which hexane liquid is vaporized is mathematically represented as
[tex]z = 10\% * r[/tex]
[tex]z = 0.1 * 0.222[/tex]
[tex]z = 0.0222 \ kmol/min [/tex]
Generally the time required to vaporize the the liquid(i.e n) is mathematically represented as
[tex]t = \frac{n}{z}[/tex]
=> [tex]t = \frac{76.628}{0.0222}[/tex]
=> [tex]t = 3851.7 \ min [/tex]
