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Sati [7]
2 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]2 years ago
7 0

Answer: oop

Explanation:oop

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CHALLENGE ACTIVITY 2.8.1: Using constants in expressions. The cost to ship a package is a flat fee of 75 cents plus 25 cents per
mihalych1998 [28]

Answer:

Weight(lb): 10

Flat fee(cents): 75

Cents per pound: 25

Shipping cost(cents): 325

Explanation:

we run this as a jave programming language

import java.util.Scanner;

public class Shipping Calculator {

   public static void main (String [] args) {

       int shipWeightPounds = 10;

       int shipCostCents = 0;

       final int FLAT_FEE_CENTS = 75;

       

        final int CENTS_PER_POUND = 25;

       shipCostCents = FLAT_FEE_CENTS + CENTS_PER_POUND * shipWeightPound

       /* look up the solutioin above */

       System.out.println("Weight(lb): " + shipWeightPounds);

       System.out.println("Flat fee(cents): " + FLAT_FEE_CENTS);

       System.out.println("Cents per pound: " + CENTS_PER_POUND);

       System.out.println("Shipping cost(cents): " + shipCostCents);

   }

}

7 0
2 years ago
1. Add:<br>(i) 5xy, -2xy, -11xy, 8xy<br>(iv) 3a - 2b + c, 5a + 8b -70​
Cerrena [4.2K]

Answer:

(i) 0

(iv) 8a+6b+c-70

Explanation:

Hope this helps you

8 0
1 year ago
The pressure drop across a valve through which air flows is expected to be 10 kPa. If this differential were applied to the two
ozzi

Answer:hit that soulja boy

Explanation:

6 0
2 years ago
Identify the four engineering economy symbols and their values from the following problem statement. Use a question mark with th
Vilka [71]

Answer:

The company will have $311,424 in its investment set-aside account.

Explanation:

To determine the amount of money that the company will have after 7 years with an interest rate of 11% per year, we must calculate the price to start with an increase of 11%, and then repeat the operation until reaching seven years:

Year 0: 150,000

Year 1: 150,000 x 1.11 = 166,500

Year 2: 166,000 x 1.11 = 184,815

Year 3: 184,815 x 1.11 = 205,144.65

Year 4: 205,144.65 x 1.11 = 227,710.56

Year 5: 227,710.56 x 1.11 = 252,758.72

Year 6: 252,758.72 x 1.11 = 280,562.18

Year 7: 280,562.18 x 1.11 = 311,424

3 0
2 years ago
The ingredient weights for making 1 yd (cyd) of concrete by assuming aggregates in SSD state are given below. The volume of air
Pachacha [2.7K]

Answer:

Explanation:

Ans) Given batch weight of each component :

Cement = 700 lb

Water = 315 lb

Coarse aggregate = 1575 lb

Fine aggregate = 1100 lb

Part 1) Amount of water = 328.5 lb

Amount of water is needed to be increased if the aggregates has absorption capacity, To maintain constant water cement ratio, the mixing water is increased because some of the water is absorbed by aggregates.

Amount of water absorbed = 328.5 lb - 315 lb = 13.5 lb

Total amount of aggregates = 1575 + 1100 = 2675 lb

=> % Absorption capacity = 13.5 x 100 / 2675 = 0.5 %

Hence, new amount of Coarse aggregate = (1 - 0.005) x 1575 lb = 1567.125 lb

New amount of fine aggregate = (1 - 0.005) x 1100 = 1094.5 lb

Since, water cement ratio is maintained constant , amount of cement remains unchanged

=> Volume of water = 328.5 / 62.4 = 5.26 ft3

=> Volume of cement = 700 / (3.15 x 62.4) = 3.56 ft3

=> Volume of coarse aggregate = 1567.125 / (2.4 x 62.4) = 10.46 ft3

=> Volume of fine aggregate = 1100 / (2.4 x 62.4) = 7.34 ft3

Volume of air = 2% = 0.02 x 27 = 0.54 ft3

Total concrete volume = 5.26 + 3.56 + 10.46 + 7.34 + 0.54 \approx 27 ft3 = 1 yd3

Hence, calculated amount of each component is correct

Part 2) We know, minus sign indicated that the aggregate will absorb some moisture from concrete, hence mixing water amount needed to be corrected .

=> Amount of water absorbed by coarse aggregate = 0.01 x 1567.125 lb = 15.67 lb

=> Amount of water absorbed by fine aggregate = 0.02 x 1094.50 lb = 21.89 lb

Total amount of water absorbed = 15.67 + 21.89 = 37.56 lb

To maintain same water cement ratio, amount of mixing water is needed to be increased

=> Corrected amount of mixing water = 328.5 lb + 37.56 lb = 366 lb

=> Corrected amount of coarse aggregate = (1 - 0.01) x 1567.125 = 1551.45 lb

=> Corrected amount of fine aggregate = (1 - 0.02) x 1094.5 = 1072.6 lb

Part 3) We know,

Unit weight = Sum of weight of each material / Total volume

=> Sum of weight = 366 + 700 + 1551.45 + 1072.6 = 3690.05 lb

Total volume = 1 yd3 or 27 ft3

=> Expected Unit Weight = 3690.05 lb / 27 ft3 = 136.67 lb/ft3

Also, Concrete Yield = Weight of all components / Unit weight of concrete

=> Yield = 3690.05 / 136.67 = 27 ft3 or 1 yd3

4 0
2 years ago
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