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notsponge [240]
3 years ago
6

An equimolar liquid mixture of benzene an toluene is separated into two product streams by distillation. Inside the column a liq

uid stream flows downward and a vapor stream rises. At each point in the column some of the liquid vaporizes and some of the vapor condenses. The vapor leaving the top of the column, which contains 97mol% benzene, is completely condensed and split into two equal fractions: one is taken off as the overhead product stream, and the other (the reflux) is recycled to the top of the column. The overhead product stream contains 89.2% of the benzene fed to the column. The liquid leaving the bottom of the column is fed to a partial reboiler in which 45% of it is vaporized. The vapor generated in the reboiler (the boilup) is recycled to become the rising vapor stream in the column, and the residual reboiler liquid is taken off as the bottom product stream. The compositions of the streams leaving the reboiler are governed by the relation: (Yb/(1-Yb))/(Xb/(1-Xb))=2.25 where Yb and Xb are the mole fractions of benzene in the vapor and liquid streams, respectively.Draw and completely label the flowchart and do a degree of freedom analysis for each of the four systems (overall process, column, condenser, and reboiler), then write the equations used to solve the molar amounts of the overhead and bottoms products, the mole fraction of benzene in the bottoms product, and the percent recovery of toluene in the bottoms product.
Engineering
1 answer:
mamaluj [8]3 years ago
7 0

do your abc's                                   yes i see                                                                                        

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marissa [1.9K]

Answer:

See the pictures attached

Explanation:

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PLEASE HELP QUICK!!
ivolga24 [154]

R01= 14.1 Ω

R02=  0.03525Ω

<h3>Calculations and Parameters</h3>

Given:

K= E2/E1 = 120/2400

= 0.5

R1= 0.1 Ω, X1= 0.22Ω

R2= 0.035Ω, X2= 0.012Ω

The equivalence resistance as referred to both primary and secondary,

R01= R1 + R2

= R1 + R2/K2

= 0.1 + (0.035/9(0.05)^2)

= 14.1 Ω

R02= R2 + R1

=R2 + K^2.R1

= 0.035 + (0.05)^2 * 0.1

= 0.03525Ω

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2 years ago
If you touch a downed power line, covered or bare, what's the likely outcome?
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3 years ago
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An automobile tire with a volume of 0.8 m3 is inflated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if t
saveliy_v [14]

Answer:

2.83 kg

Explanation:

Given:

Volume, V = 0.8 m³

gage pressure, P = 200 kPa

Absolute pressure = gage pressure + Atmospheric pressure

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Temperature, T = 23° C = 23 + 273 = 296 K

Now,

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PV = mRT

Where,

m is the mass

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thus,

301 × 10³ × 0.8 = m × 287 × 296

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m = 2.83 kg

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3 years ago
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