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Sati [7]
3 years ago
7

A 100 kmol/h stream that is 97 mole% carbon tetrachloride (CCL) and 3% carbon disulfide (CS2) is to be recovered from the bottom

of a distillation column. The feed to the column is 16 mole% CS2 and 84% CC14, and 2% of the CC14 entering the column is contained in the overhead stream leaving the top of the column. (a) Draw and label a flowchart of the process and do the degree-of-freedom analysis. (b) Calculate the mass and mole fractions of CCl4 in the overhead stream, and determine the molar flow rates of CCl4 and CS2 in the overhead and feed streams. (c) Suppose the overhead stream is analyzed and the mole fraction of CS2 is found to be significantly lower than the value calculated in Part (b). List as many reasons as you can for the discrepancy, including possible violations of assumptions made in Part (b).
Engineering
1 answer:
Ugo [173]3 years ago
7 0

Answer: oop

Explanation:oop

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Moral creativity and innovation are based on original discoveries,  whereas immoral innovation is based on unscrupulous actions.

<h3>What is innovation?</h3>

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2 years ago
Is a compass a analog or a digital sensor?
ollegr [7]
A compass is a analog sensor
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2 years ago
.........................................
hoa [83]

Answer:

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3 0
3 years ago
A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

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To put out a class D metal fire, you must smother the fire and eliminate the oxygen element in the fire.

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Smothering in this context involves adding a solution like carbon dioxide (CO2) into the fire, this results in a reduction of oxygen in the atmosphere surrounding the class D fire.

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2 years ago
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