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jok3333 [9.3K]
3 years ago
15

Ethane (component A - C2H6) and hydrogen (component B) are fed to a differential reactor where they react on the catalyst to for

m methane (component C - CH4) at 300K. The amount of catalyst is 10 grams. The reaction is irreversible.The total molar feed rate into the reactor is 1.70 mol/h. Total pressure is one atm. The mole fraction of ethane at the inlet is 0.5. There are no inerts in the feed.
Required:
a. Write the stoichiometric equation for this reaction.
b. Find the concentrations of ethane and hydrogen at the inlet.
Engineering
1 answer:
Fofino [41]3 years ago
4 0
HELP ILL GIVE MOST BRAINLY AND 50 POINTS
HURRY PLEASE component c it is a compound so it will break
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The slope distance and zenith angle between points A and B were measured with a total station instrument as 17685.20 ft and 93°
dimaraw [331]

True

Explanation:

three point bending is better than tensile for evaluating the strength of ceramics. it is got a positive benefit to tensile for evaluating the strength of ceramics.

4 0
3 years ago
Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at
sergeinik [125]

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

h_{f} = 173.358 \\h_{fg} = 2402.522

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

h_{2a} =  489.752\\h_{2b} =  313.2

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a}  - h_{4a}) - (h_{2a}  - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26  - 2241.448938 ) - (489.752  - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3a} -  h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4a} -  h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.06485

Part b) @ 4 MPa

mass flow

\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b}  - h_{4b}) - (h_{2b}  - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14  - 2405.54119 ) - (313.12  - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}

Heat transfer rate through boiler

Q_{in}  = mass flow * (h_{3b} -  h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W

Heat transfer rate through condenser

Q_{out}  = mass flow * (h_{4b} -  h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W

Thermal Efficiency

n = \frac{W_{net}  }{Q_{in} } = \frac{100*10^3}{1542011.787}  \\\\n = 0.038275

6 0
3 years ago
Determine the change in the enthalpy of helium, in kJ/kg, as it undergoes a change of state from 100 kPa and 20 oC to 600 kPa an
MatroZZZ [7]

Answer:

The change in enthalpy of helium is 4073.86kJ/kg

Explanation:

∆H = Cp(T2 - T1)

Cp = 3.5R = 3.5×8.314 = 29.099kJ/kgmolK ÷ 2 (1kgmol of helium = 2kg of helium) = 14.5495kJ/kgK, T2 = 300°C = 300+273K = 573K, T1 = 20°C = 20+273K = 293K

∆H = 14.5495kJ/kgK(573K - 293K) = 14.5495kJ/kgK × 280K = 4073.86kJ/kg

7 0
4 years ago
Milestone 1: Write code which asks for a length input until it gets an integer 10 or greater, then creates 2 arrays of this leng
jonny [76]

Answer:

import java.util.Scanner;

import java.lang.Math;

class Main {

  public static void main(String[] args) {

      int length = 0;

      boolean lengthCheck = true;

      Scanner scan = new Scanner(System.in);

      while (lengthCheck == true)

      {

          System.out.println("Enter an array length (must be 10 or greater):");

          length = scan.nextInt();

          if (length >= 10)

          {

              lengthCheck = false;

          }

      }

      int[] firstArray = new int[length];

      int[] secondArray = new int[length];

      System.out.print("\nFirst Array: ");

      for (int i = 0; i < length; i++)

      {

          firstArray[i] = (int) (Math.random() * 100) + 1;

          System.out.print(firstArray[i] + " ");

      }

      System.out.print("\n\nSecond Array: ");

      for (int i = 0; i < length; i++)

      {

          secondArray[i] = (int) (Math.random() * 100) + 1;

          System.out.print(secondArray[i] + " ");

      }

      System.out.println("\n");

     

      boolean[] isAdded = new boolean[100];

      int[] merge = new int[(firstArray.length + secondArray.length)];

     

      int j=0;

      for (int i = 0; i < length; i++)

      {

          if(!isAdded[firstArray[i] - 1]) {

              merge[j] = firstArray[i];

              j++;

              isAdded[firstArray[i] - 1] = true;

          }

         

          if(!isAdded[secondArray[i] - 1]) {

              merge[j] = secondArray[i];

              j++;

              isAdded[secondArray[i] - 1] = true;

          }

         

      }

     

      System.out.print("Merged Array: ");

     

      for (int i = 0; i < 2*length && merge[i] != 0; i++)

      {

          System.out.print(merge[i] + " ");

      }

      System.out.println("\n");

     

  }

}

3 0
3 years ago
9.19 Generate Bode magnitude and phase plots (straight-line approximations) for the following voltage transfer functions. (a) H(
Norma-Jean [14]

Answer:

attached below

Explanation:

8 0
3 years ago
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