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loris [4]
3 years ago
13

The homogeneous, reversible, exothermic, liquid phase reaction: A근R Is being carried out in a reactor system consisting of an id

eal PFR followed by an ideal CSTR. The volume of the PFR is 75,000 L, and the volume of the CSTR is 150,000 L. The PFR operates at 150 °C, and the CSTR operates at 125 °C. The volumetric flow rate entering the PFR is 55,000 L/h, the concentration of A in this stream is 6.5 M, and the concentration of R is zero. The reaction is first order in both directions. The rate constant of the forward reaction at 150 °C is 1.28 h and the equilibrium constant based on concertation at 150 °C is 2.5. The rate constant at 125 °C is 0.280 h and the equilibrium constant is 3.51. (a) What is the rate of R leaving the CSTR (mol/h)? (b) The flowrate to the PFR is increased so that the fractional conversion of A is 0.5 in the efluent from the CSTR. All other parameters remain unchanged. What is the new production rate of R?

Engineering
1 answer:
baherus [9]3 years ago
3 0

Answer:

attached below

Explanation:

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Which term describes erosion?
Vera_Pavlovna [14]

Answer:

transports solid materials

7 0
3 years ago
Read 2 more answers
(a)Compute the electrical conductivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in leng
igor_vitrenko [27]

Answer:

a) \sigma = 12.2 (Ω-m)^{-1}

b) Resistance = 121.4 Ω

Explanation:

given data:

diameter is 7.0 mm

length 57 mm

current I = 0.25 A

voltage v = 24 v

distance between the probes is 45 mm

electrical conductivity is given as

\sigma = \frac{I l}{V \pi r^2}

\sigma  = \frac{0.25 \times 45\times 10^{-3}}{24 \pi [\frac{7 \times 10^{-3}}{2}]^2}

\sigma = 12.2(Ω-m)^{-1}[/tex]

b)

Resistance = \frac{l}{\sigma A}

                  = \frac{l}{ \sigma \pi r^2}

= \frac{57  \times 10^{-3}}{12.2 \times \pi [\frac{7 \times 10^{-3}}{2}]^2}

Resistance = 121.4 Ω

8 0
3 years ago
A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p
Stella [2.4K]

Answer:

Cut-off ratio\dfrac{V_3}{V_2}=6.15

Cxpansion ratio\dfrac{V_4}{V_3}=3.25

The exhaust temperatureT_4=1997.5R

Explanation:

Compression ratio CR(r)=20

\dfrac{V_1}{V_2}=20

P_2=P_3=920 psia

T_1=520 R ,T_{max}=T_3,T_3=3200 R

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

\dfrac{T_2}{T_1}=r^{\gamma -1}

\dfrac{T_2}{520}=20^{1.4 -1}

T_2=1723.28R

\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}

\dfrac{V_3}{V_2}=\dfrac{3200}{520}

So cut-off ratio\dfrac{V_3}{V_2}=6.15

\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}

Now putting the values in above equation

\dfrac20=\dfrac{V_4}{V_3}\times 6.15

\dfrac{V_4}{V_3}=3.25

So expansion ratio\dfrac{V_4}{V_3}=3.25.

\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}

\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}

T_4=1997.5R

So the exhaust temperatureT_4=1997.5R

3 0
4 years ago
IN JAVA,
Citrus2011 [14]

Answer:

Explanation:

Code:

import java.io.File;

import java.io.FileWriter;

import java.io.IOException;

import java.util.Scanner;

public class Knapsack {

 

  public static void knapsack(int wk[], int pr[], int W, String ofile) throws IOException

  {

      int i, w;

      int[][] Ksack = new int[wk.length + 1][W + 1];

     

      for (i = 0; i <= wk.length; i++) {

  for (w = 0; w <= W; w++) {

  if (i == 0 || w == 0)

  Ksack[i][w] = 0;

  else if (wk[i - 1] <= w)

  Ksack[i][w] = Math.max(pr[i - 1] + Ksack[i - 1][w - wk[i - 1]], Ksack[i - 1][w]);

  else

  Ksack[i][w] = Ksack[i - 1][w];

  }

  }

     

      int maxProfit = Ksack[wk.length][W];

      int tempProfit = maxProfit;

      int count = 0;

      w = W;

      int[] projectIncluded = new int[1000];

      for (i = wk.length; i > 0 && tempProfit > 0; i--) {

         

      if (tempProfit == Ksack[i - 1][w])

      continue;    

      else {

          projectIncluded[count++] = i-1;

      tempProfit = tempProfit - pr[i - 1];

      w = w - wk[i - 1];

      }

     

      FileWriter f =new FileWriter("C:\\Users\\gshubhita\\Desktop\\"+ ofile);

      f.write("Number of projects available: "+ wk.length+ "\r\n");

      f.write("Available employee work weeks: "+ W + "\r\n");

      f.write("Number of projects chosen: "+ count + "\r\n");

      f.write("Total profit: "+ maxProfit + "\r\n");

     

  for (int j = 0; j < count; j++)

  f.write("\nProject"+ projectIncluded[j] +" " +wk[projectIncluded[j]]+ " "+ pr[projectIncluded[j]] + "\r\n");

  f.close();

      }    

  }

 

  public static void main(String[] args) throws Exception

  {

      Scanner sc = new Scanner(System.in);

      System.out.print("Enter the number of available employee work weeks: ");

      int avbWeeks = sc.nextInt();

      System.out.print("Enter the name of input file: ");

  String inputFile = sc.next();

      System.out.print("Enter the name of output file: ");

      String outputFile = sc.next();

      System.out.print("Number of projects = ");

      int projects = sc.nextInt();

      int[] workWeeks = new int[projects];

      int[] profit = new int[projects];

     

      File file = new File("C:\\Users\\gshubhita\\Desktop\\" + inputFile);

  Scanner fl = new Scanner(file);

 

  int count = 0;

  while (fl.hasNextLine()){

  String line = fl.nextLine();

  String[] x = line.split(" ");

  workWeeks[count] = Integer.parseInt(x[1]);

  profit[count] = Integer.parseInt(x[2]);

  count++;

  }

 

  knapsack(workWeeks, profit, avbWeeks, outputFile);

  }

}

Console Output:

Enter the number of available employee work weeks: 10

Enter the name of input file: input.txt

Enter the name of output file: output.txt

Number of projects = 4

Output.txt:

Number of projects available: 4

Available employee work weeks: 10

Number of projects chosen: 2

Total profit: 46

Project2 4 16

Project0 6 30

8 0
3 years ago
A full journal bearing has a shaft diameter of 3.000 in with a unilateral tolerance of -0.0004 in. The l/d ratio is unity. The b
sashaice [31]

Answer:

i) 154°F

ii) 0.0114 mm

iii) 0.075 btu/s

iv) 0.080 in^3/s

Explanation:

<u>i)Determine the average film Temperature </u>

( from Viscosity-temperature chart in US customary units for SAE10 )

at Temp =  154°F

absolute viscosity = 4.25 rev

and ΔT = 2 ( 154 - operating temp ) = 28°F

where : operating temp = 140°F as given in question

also from the chart applying Raimondi and Boyd boundary conditions

ΔT = 29°F  hence we can pick 154°F as the average film temperature

<u>ii) Calculate the minimum film thickness</u>

Cmin = Bore diameter - Journal shaft diameter / 2

         = 3.003 - 3 / 2 = 0.0015 in

Given that : h₀ / Cmin = 0.76

there h₀ = 0.0015 * 0.76 = 0.0114 mm

<u>iii)Determine the heat loss rate </u>

Also known as power loss ratio ( H ) = ( 2π*w*f*r*N ) / ( 778 *12 )

( 2π * 0.0132 * 675 * 1.5 * 10 ) / ( 9336 )

Heat loss rate = 0.075 btu/s

<u>iv)Calculate lubricant side-flow rate for minimum clearance assembly </u>

Side flow rate = 0.315 * Total volume flow rate

                       = 0.315 * ( 3.8 * 1.5 * 0.0015 * 10 * 3 )

                      = 0.080 in^3/s.

6 0
3 years ago
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