Question

Two streams flow into a 500.0-gallon tank. The first stream is 10.0 wt % ethanol and 90.0 wt% hexane (mixture density is rho1 0.680 g/cm3) and the second is 90.0 wt% ethanol, 10.0 wt% hexane (mixture density rho2 0.780 g/cm3). After the tank has been filled, which takes 22.0 minutes, an analysis of its contents determines that it is 40.0 wt% ethanol and 60.0 wt% hexane. Estimate the density of the final mixture and the mass and volumetric flow rates of the two feed streams.

Answer 1

Answer:

Density of final mixture=0.7175$\frac{g}{cm^3}$.  Mass/Volumetric flow rate stream 1= 39793.6$\frac{g}{min}$/ 58.52 litre/min. Mass/Volumetric flow rate stream 2= 21925.8$\frac{g}{min}$/ 28.11 litre/min

Explanation:

The first thing to do is the density's determination. For this you can establish a system of equation. For stream 1, you know that 10%wt of ethanol + 90%wt hexane give a density of 0.68 $\frac{g}{cm^3}$. Calling "X" to the ethanol's density and "Y" Hexane`s one:

$0.1*X+0.9*Y= 0.68(Eq. a)$

The same applies to stream 2,so:

$0.9*X+0.1*Y= 0.78(Eq. b)$

Solving: X=0.7925 $g/cm^3$ ; Y= 0.6675$g/cm^3$.

Then, since the final concentration is 40%wt ethanol and 60%wt hexane, the density of the final mixture is:

$\rho_{f}=0.4*0.7925+0.6*0.6675\Rightarrow \rho_{f}=0.7175 \frac{g}{cm^3}$

Then, you know the volume of the tank (500gallons=1892500cm^3), the density (0.7175 g/cm^3) and mixture (40%wt ethanol;60%hexane). From here you can get the mass of each gas in the tank.

Mass of ethanol=0.4*1892500* 0.7175 g=543147.5 g (grams)

Mass of hexane=0.6*1892500* 0.7175 g=814721.25 g (grams)

These masses come from both stream so it is necessary to know the net volume from each stream. For this, the following equation are needed.

$0.1*X*V_1+0.9*X*V_2=543147.5 g$  (Mass of ethanol)

$0.9*Y*V_1+0.1*Y*V_2=814721.25 g$ (Mass of hexane)

Solving, the volume from stream 1 ($V_1$) is 1287528.15 cm^3 and the volume from stream 2 ($V_2$) is 618452.00 cm^3.

Then, the time for complete the tank is 22 min, so the volume rate is:

For stream 1: $Q_1=V_1/(22min*1000cm^3/litre)=58.52 litre/min$

For stream 2: $Q_2=V_2/(22min*1000cm^3/litre)=28.11 litre/min$

Finally, multiplying the volume rate by the density of each current the mass flow is:

$\dot{m_1}=0.68\frac{g}{cm^3} *58.52 litre/min*1000cm^3/litre=39793.6 \frac{g}{min} \\\dot{m_2}=0.78\frac{g}{cm^3} *28.11 litre/min*1000cm^3/litre=21925.8 \frac{g}{min}$