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Dmitry [639]
4 years ago
10

a. (24 points) Describe the microstructure present in a 10110 steel after each step in each of the following heat treatments (no

te that i-iv are all individual heat treatments on different specimens, not a sequence of heat treatments on a single specimen; each will have a separate answer): i) Heat to 900 ºC, quench to 400 ºC and hold for 1000 sec. and then quench to 25 ºC. ii) Heat to 900 ºC, and then quench to 25 ºC. iii) Heat to 900 ºC, quench to 700 ºC and hold for 1 sec., quench to 600 ºC and hold for 1 sec, quench to 400 ºC and hold for 10 sec., quench to 300 ºC and hold for 100 sec, and then quench to 25 ºC. iv) Heat to 900 ºC, quench to 675 ºC and hold for 1 sec, quench to 300 ºC and hold for 1000 sec, and then quench to 25 ºC. b. (10 points) Describe the difference in final microstructures for 1080 and 1020 steels when cooled to room temperature at cooling rates of: i) 4 ºC/s ii) 25 ºC/s iii) 53 ºC/s iv) 120 ºC/s c. (6 points) Why is there a cementite start line for the 10110 steel but none for the 1050 steel?

Engineering
1 answer:
Mrac [35]4 years ago
6 0

Answer:

Explanation:

Please check the below file for the attached file

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Please add comments to your program to explain it. Thank you!
Step2247 [10]

Answer / Explanation:

(1) We should first understand that the input filename are passed in as the first command arguments at command line, respectively.

To do this, we import the data file:

So we have,

import java.io.*;

/**

*  Makes a copy of a file.  The original file and the name of the

*  copy must be given as command-line arguments.  In addition, the

*  first command-line argument can be "-f"; if present, the program

*  will overwrite an existing file; if not, the program will report

*  an error and end if the output file already exists.  The number

*  of bytes that are copied is reported.

*/

public class CopyFile {

  public static void main(String[] args) {

     String sourceName;   // Name of the source file,  

                          //    as specified on the command line.

     String copyName;     // Name of the copy,  

                          //    as specified on the command line.

     InputStream source;  // Stream for reading from the source file.

     OutputStream copy;   // Stream for writing the copy.

     boolean force;  // This is set to true if the "-f" option

                     //    is specified on the command line.

     int byteCount;  // Number of bytes copied from the source file.

     

     /* Get file names from the command line and check for the  

        presence of the -f option.

(2)   If the command line is not one

        of the two possible legal forms, print an error message and  

        end this program. */

   

     if (args.length == 3 && args[0].equalsIgnoreCase("-f")) {

        sourceName = args[1];

        copyName = args[2];

        force = true;

     }

     else if (args.length == 2) {

        sourceName = args[0];

        copyName = args[1];

        force = false;

     }

     else {

        System.out.println(

                "Usage:  java CopyFile <source-file> <copy-name>");

        System.out.println(

                "    or  java CopyFile -f <source-file> <copy-name>");

        return;

     }

     

     /* Create the input stream.  If an error occurs, end the program. */

     

     try {

        source = new FileInputStream(sourceName);

     }

     catch (FileNotFoundException e) {

        System.out.println("Can't find file \"" + sourceName + "\".");

        return;

     }    

     /* If the output file already exists and the -f option was not

        specified, print an error message and end the program. */

   

     File file = new File(copyName);

     if (file.exists() && force == false) {

         System.out.println(

              "Output file exists.  Use the -f option to replace it.");

         return;  

    }      

     /* Create the output stream.  If an error occurs, end the program. */

     try {

        copy = new FileOutputStream(copyName);

     }

     catch (IOException e) {

        System.out.println("Can't open output file \"" + copyName + "\".");

        return;

     }

     

   (3)   /* Copy one byte at a time from the input stream to the output

        stream, ending when the read() method returns -1 (which is  

        the signal that the end of the stream has been reached).  If any  

        error occurs, print an error message.  Also print a message if  

        the file has been copied successfully.  */  

     byteCount = 0;

     

     try {

        while (true) {

           int data = source.read();

           if (data < 0)

              break;

           copy.write(data);

           byteCount++;

        }

        source.close();

        copy.close();

        System.out.println("Successfully copied " + byteCount + " bytes.");

     }

     catch (Exception e) {

        System.out.println("Error occurred while copying.  "

                                  + byteCount + " bytes copied.");

        System.out.println("Error: " + e);

     }    

  }  // end main()  

} // end class CopyFile

8 0
4 years ago
A soil has the following Green-Ampt parameters Effective porosity 0.400 Initial volumetric moisture content-15% Hydraulic Conduc
Rudik [331]

Answer:

The graphs are attached below

Explanation:

Ans) We know,

F(t) = K(t - to) + zΔ∅ ln[(1 + F(t)/zΔ∅]

where K = hydraulic conductivity =0.1 cm/hr

z = capillary suction =20cm

 Δ∅ = ∅ (1 - Se)

 ∅ = effective porosity , Se = initial moisture content

 Δ∅ = 0.40(1 - 0.15)

Δ∅ = 0.34

Now, F(t) = 0.1(t - to) + 20(0.34) ln[ 1 + F(t)/6.8]

F(t) = 0.1(t - to) + 6.8 ln [1+ 0.147 F(t)]

Also, infiltration rate (f),

f = K [(zΔ∅ + F)/F]

f = 0.40 [6.8 + F]/F

Condition of ponding,

Ponding time tp = K zΔ∅ i(i-k)

where, i = rainfall rate (1cm/hr)

tp = 0.40(6.8) / 0.60

tp= 4.53 hours

Now, cumulative infiltraton at ponding Fp = i tp

Fp = 1 x 4.53 or 4.53 cm

For infiltration at time less then ponding time , infiltration rate = rainfall rate

For t = 0.25 hr , f = 1cm/hr ; F = 0.25 x 1 = 0.25 cm

For t = 0.50 hr , f = 1cm/hr ; F = 0.50 cm

For t = 0.75 hr , f = 1cm/hr ; F = 0.75 cm

For t = 1 hr , f = 1cm/hr ; F = 1cm/hr

For t > tp ,

Equivalent time origin(to),

to = tp - 1/K [ Fp - z Δ∅ ln (1+ Fp/zΔ∅)

to = 4.53 - 1/0.40[ 4.53 - 6.8 ln(1 + 0.66)

to = 4.53 - 2.5 ( 4.53 - 3.44)

to = 1.82 hr

Hence,

F = 0.10( t - 1.82) + 6.8 ln[ 1+ 0.147F(t)]

Solving above equation for t by assuming F, and further solving equation for infiltration rate

Ans (a) Following is required curve :

O.S 1-5 2 time Chr)ホー Lt 2 time Chr)

Ans (c) Following is required table :

Ans (d) For time step t = 0.25 hrs

At t < tp ; i = f = 1 cm/hr

F = i x t

F = 1 x 0.25

F = 0.25 cm  

 

5 0
4 years ago
Manufacturing employees who perform assembly line work are referred to as
mamaluj [8]

Answer:

C. assembly line workers.

Explanation:

8 0
3 years ago
Read 2 more answers
If 5000 N of thrust is acting to the left, and 4300 N of drag is acting to the right, what is the magnitude and direction of the
Kipish [7]

Answer:

700 N acting to the left.

8 0
3 years ago
A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit
alexdok [17]

Answer:

5.31\frac{kg}{m^3}

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to \frac{m}{M}, where m is the mass and M the molar mass of the gas, and the density is \frac{m}{V}.

For air M=28.66*10^{-3}\frac{kg}{mol} and \frac{5}{9}R=K

So, 598.59 R*\frac{5}{9}=332.55K

pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}

7 0
4 years ago
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