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4 years ago

Describe the importance of ferrite and austenite stabilizing elements in steels

Engineering

1 answer:

podryga [215]4 years ago

3 0

Answer:

The importance of ferrite and austenite stabilizing elements in steels .

Explanation:

Alloying -

The process which improves the properties of the steel by changing the chemical composition of the steel via adding some elements .

The properties can be improved by - Stabilizing Austenite and Stabilizing Ferrite .

Stabilizing austenite -

The process by which temperature is increased , in which Austenite exists .

Elements with the same crystal structure as of the austenite ( FCC ) raises its A4 value i.e. the temperature of the formation of austenite from its liquid phase and reduces the value of A3 .

Hence, the elements are -

Cobalt , Nickel , Manganese , Copper.

The examples of the Austenitic steels are -

Hadfield Steel ( 13% Mn , 1.2% Cr , 1% C ) and Austenitic Stainless steel.

Stabilizing ferrite –

The process by which temperature is decreased , in which austenite exists .

Elements with the same crystal structure as of the ferrite (BCC - Cubic body centered ) lowers its A4 value i.e. the temperature of the formation of austenite from its liquid phase and increases the value of A3 .These elements have lower solubility of carbon in austenite, that lead to increase in the amount of carbides in the steel.

Hence, the elements are -

Aluminium , Silicon , Tungsten , Chromium , Molybdenum , Vanadium

The examples of the Ferritic steels are -

F-Cr alloys , transformer sheets steel ( 3% Si ).

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Steam enters an adiabatic turbine at 8 MPa and 500C with a mass flow rate of 3

Vinil7 [7]

Answer:

a)temperature=69.1C

b)3054Kw

Explanation:

Hello!

To solve this problem follow the steps below, the complete procedure is in the attached image

1. draw a complete outline of the problem

2. to find the temperature at the turbine exit use termodinamic tables to find the saturation temperature at 30kPa

note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

3. Using thermodynamic tables find the enthalpy and entropy at the turbine inlet, then find the ideal enthalpy using the entropy of state 1 and the outlet pressure = 30kPa

4. The efficiency of the turbine is defined as the ratio between the real power and the ideal power, with this we find the real enthalpy.

Note: Remember that for a turbine with a single input and output, the power is calculated as the product of the mass flow and the difference in enthalpies.

5. Find the real power of the turbine

3 0

3 years ago

Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...

leva [86]

Answer: Pi= 4 - 4/3 + 4/5 - 4/7 + 4/9 ...

Explanation:

Is the same as the example,

If Π/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 ...

Then

(Π/4 )*4= 4*(1 - 1/3 + 1/5 - 1/7 + 1/9 ...)

Π =4 - 4/3 + 4/5 - 4/7 + 4/9 ...

The way to write this is

Sum(from n=0 to n=inf) of (-1)^n 4/(2n+1)

(photo)

6 0

4 years ago

The distribution of actual weights of 8‑ounce wedges of cheddar cheese produced at a dairy is Normal, with mean 8.1 ounces and s

s344n2d4d5 [400]

Answer:

sampling distribution

Explanation:

Sampling distribution is distribution of multiple samples' satistics of a population.

3 0

4 years ago

Joe Bruin has a big lawn in front of his house that is 30 meters wide and 20 meters long. Josephine makes him go out and mow the

zysi [14]

<u>Explanation:</u>

5 Horsepower for 30 mins,

(5)(745.7) = 3.7285 KW power delivered

General Efficiency of IC engine = 20%

Power required = $\frac{3 \cdot 7285}{0 \cdot 2}=18 \cdot 6425 kw$

Energy required per week,

=P × Time = 18.64 × 60 × 30 = 33.5565 MJ

Lawn area = (30) (20) = 600 $m^{2}$

let sunlight hours be 8 hours

Hence, solar power input on lawn,

=5.62×3600 = 20232 kJ/ $m^{2}$ /day

energy input in lawn = (600) (20232) (7)

= 84974.4 mJ/week

Chemical efficiency by photosynthesis = 4%

Chemical content in grass = (84974.4) (0.04)

= 3398.97 mJ

Mass of the clippers $=(30)(20)(1 \cdot 096)^{2}(667)$

$=478632 \cdot 33$ pounds

Removing water content,

dried grass clippings $$=95726.46$ pound$

= 11533.25 gallons

Trash cans repaired

$=\frac{11533}{50} =230.66\\=231 cans$

By burning the gas, total energy input = 3398.97 MJ × 0.2

= 679.794 MJ

Efficiency of steeling engine = 20%

Energy output by engine = 679.794 ×0.2

= 135.96 mJ

Energy required by mover = 33.5565 mJ

Hence, Energy (output) ⇒ energy required

5 0

3 years ago

For this question you must write a java class called Rectangle and a client class called RectangleClient. The partial Rectangle

Alex Ar [27]

Answer:

Java program is given below. You can get .class after you execute java programs, You can attach those files along with .java classes given , Those .class files are generated ones.

Explanation:

//Rectangle.java class

public class Rectangle {

private int x;

private int y;

private int width;

private int height;

// constructs a new Rectangle with the given x,y, width, and height

public Rectangle(int x, int y, int w, int h)

{

this.x=x;

this.y=y;

this.width=w;

this.height=h;

}

// returns the fields' values

public int getX()

{

return x;

}

public int getY()

{

return y;

}

public int getWidth()

{

return width;

}

public int getHeight()

{

return height;

}

// returns a string such as “Coordinate is (5,12) and dimension is 4x8” where 4 is width and 8 is height. public String toString()

public String toString()

{

String str="";

//add x coordidate , y-coordinate , width, height and area to str and return

str+="Coordinate is ("+x+","+y+")";

str+=" and dimension is : "+width+"x"+height;

str+=" Area is "+(width*height);

return str;

}

public void changeSize(int w,int h)

{

width=w;

height=h;

}

======================

//main.java

class Main {

public static void main(String[] args) {

//System.out.println("Hello world!");

//create an object of class Rectangle

Rectangle rect=new Rectangle(5,12,4,8);

//print info of rect using toString method

System.out.println(rect.toString());

//chamge width and height

rect.changeSize(3,10);

//print info of rect using toString method

System.out.println(rect.toString());

}

==========================================================================================

//Output

Coordinate is (5,12) and dimension is : 4x8 Area is 32

Coordinate is (5,12) and dimension is : 3x10 Area is 30

========================================================================================

6 0

3 years ago