Question

A liquid mixture of density 0.85 g/mL flowing at a rate of 700 lbm/h is formed from mixing a stream of benzene S = 0.879 and another of n-hexane S = 0.659. Determine the volumetric flow rates of all three streams as well as the mass flow rate of benzene and n-hexane.

Answer 1

Volumetric flow:

Mixture = 373.55 l/h; Benzene = 310.05 l/h; n-hexane = 63.5 l/h

Mass flow

Mixture = 317.52 kg/h; Benzene = 275.66 kg/h; n-hexane = 41.85 kg/h

Step-by-step explanation:

First, we can express the mixture mass flow as

`M=700(lbm)/(h)*(453.6g)/(1lbm) = 317,520g/h=317,52kg/h`

The volumetric flow of the mixture is

`V=M/\rho = (317520g/h)/(0.85g/ml) =373,553ml/h=373.55l/h`

The mass balance can be written as

`M_1+M_2=M\\`

And the volumetric flow as

`V_1+V_2=V\\\\\rho_1*M_1+\rho_2*M_2=\rho*M\\\\\rho_1(M-M_2)+\rho_2*M_2=\rho*M\\\\(\rho_2-\rho_1)*M_2=(\rho-\rho_1)*M\\\\M_2=((\rho-\rho_1))/((\rho_2-\rho_1)) *M`

If fluid 2 is n-hexane, we have

`M_2=((\rho-\rho_1))/((\rho_2-\rho_1)) *M\\\\M_2=((0.85-0.879))/((0.659-0.879)) *317.52kg/h\\\\M_2=0.1318*317.52kg/h=41.85kg/h`

The mass flow of n-hexane is 41.85 kg/h.

Its volumetric flow is 63.5 litre/h

`V=M/\rho=41.85(kg)/(h)*(1)/(0.659g/ml) *(1000g)/(1kg)\\\\V = 63505 ml/h=63.5 l/h`

The mass flow of benzene can be deducted by difference:

`M_1= M-M_2= 317.52-41.85=275.66 kg/h\\\\V_1=V-V_2= 373.55 - 63.5=310.05 l/h`