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sergiy2304 [10]
3 years ago
7

(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and pr

oeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each? (c) In bullet format compare and contrast the expected mechanical behavior of hypoeutectoid and hypereutectoid steels in terms of: (i) Yield strength (ii) Ductility (iii) Hardness (iv) Tensile strength (d) If you want to choose an alloy to make a knife or ax blade would you recommend a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your recommendation in 1-3 bullet points. (e) If you wanted a steel that was easy to machine to make a die to press powders or stamp a softer metal, would you choose a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your choice in 1-3 bullets.
Engineering
2 answers:
Gnoma [55]3 years ago
7 0

Answer:

See explanation below

Explanation:

Hypo-eutectoid steel has less than 0,8% of C in its composition.

It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.

Ferrite has a higher tensile strength than cementite but cementite is harder.

Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:

Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel

Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel

Hyper-eutectoid steel is harder than Hyper-eutectoid steel

Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.

When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because

1. It is harder

2. It has low cost

3. It is lighter

When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because

1. It is ductile

2. It has high tensile strength

3. It is durable

Romashka-Z-Leto [24]3 years ago
4 0

Answer:

(a)

Steels having carbon within 0.02% – 0.8% which consist of ferrite and pearlite are known as hypoeutectoid steel.

Steels having greater than 0.8% carbon but less than 2.0% are known as hypereutectoid steel.

(b)

The proeutectoid ferrite formed at a range of temperatures from austenite in the austenite+ferrite region above 726°C. The eutectoid ferrite formed during the eutectoid transformation as it cools below 726°C. It is a part of the pearlite microconstiutents . Note that both hypereutectoid and hypoeutectoid steels have proeutectoid phases, while in eutectoid steel, no proeutectoid phase is present.

Proeutectoid signifies is a phase that forms (on cooling) before the eutectoid austenite decomposes. It has a parallel with primary solids in that it is the first phase to crystallize out of the austenite phase. If the steel is hypoeutectoid it will produce proeutectoid ferrite and if it is hypereutectoid it will produce proeutectoid cementite.  The carbon concentration for both ferrites is 0.022 wt% C.

(c)

(i) Yield strength: The hypoeutectoid steel have good yield strength and hypereutectoid steels have little higher yield strengh.

(ii) Ductility: The hypoeutectoid steel is more ductile and the ductility has decreased by a factor of three for the eutectoid alloy. In hypereutectoid alloys the additional, brittle cementite on the pearlite grain boundaries further decreases the ductility of the alloy. The proeutectoid cementite restricts plastic deformation to the ferrite lamellae in the pearlite.

(iii) Hardness:  hypoeutectoid steels are softer and hypereutectoid steel contains low strength cementite at grain boundary region which makes it harder than hypoeutectoids.

(iv) Tensile strength: Grain boundary regions of hypereutectoid steel are high energy regions prone to cracking because of cementite in the grain boundaries, its tensile strength decreases drastically even though pearlite is present. Hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains, so grain boundaries being the high energy state region, it has a higher tensile strength.

(d)

I would recommend hypereutectoid steel alloy to make a knife or ax blade

1- Hardness is required at the surface of the blades.

2- Ductility is not needed for such application.

3- Due to constant impact, the material will not easily yield to stress.

(e)

I would choose a hypoeutectoid steel alloy to make a steel that was easy to machine.

1- hypoeutectoid steel alloys have high machinability, hence better productivity

2- It will be used on softer metals, hence its fitness for the application

3- Certain amount of ductility is required which hypoeutectoid steel alloys possess.

Explanation:

See all together above

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7 0
3 years ago
Find the thickness of the material that will allow heat transfer of 6706.8 *10^6 kcal during the 5 months through the rectangle
Vinvika [58]

Answer:

The thickness of the material is 6.23 cm

Explanation:

Given;

quantity of heat, Q = 6706.8 *10⁶ kcal  

duration of the heat transfer, t = 5 months

thermal conductivity of copper, k = 385 W/mk

outside temperature of the heater, T₁ = 30° C

inside  temperature of the heater, T₂ = 50° C

dimension of the rectangular heater = 450 cm by 384 cm

1 kcal = 1.163000 Watt-hour

6706.8 *10⁶ kcal  = 7800008400 watt-hour

I month = 730 hours

5 months = 3650 hours

Rate of heat transfer, P = \frac{7800008400 \ Watt-Hour}{3650 \ Hours}  = 2136988.6 \ W

Rate of heat transfer, P = \frac{K*A *\delta T}{L}

where;

P is the rate of heat transfer (W)

k si the thermal conductivity (W/mk)

ΔT is change in temperature (K)

A is area of the heater (m²)

L is thickness of the heater (m)

P = \frac{KA(T_2-T_1)}{L} \\\\L =  \frac{KA(T_2-T_1)}{P}\\\\L =  \frac{385(4.5*3.84)(50-30)}{2136988.6}\\\\L = 0.0623 \ m

L = 6.23 cm

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8 0
3 years ago
Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is
GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

   public static void main(String[] args) {

       final int MAXCHEESE = 10;

       String[] names = new String[MAXCHEESE];

       double[] prices = new double[MAXCHEESE];

       double[] amounts = new double[MAXCHEESE];

       // Three Special Cheeses

       names[0] = "Humboldt Fog";

       prices[0] = 25.00;

       names[1] = "Red Hawk";

       prices[1] = 40.50;

       names[2] = "Teleme";

       prices[2] = 17.25;

       System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

       System.out.println(names[0] + ": $" + prices[0] + " per pound");

       System.out.println(names[1] + ": $" + prices[1] + " per pound");

       System.out.println(names[2] + ": $" + prices[2] + " per pound");

       Random ranGen = new Random(100);

       // error at initialising i

       // i should be from 0 to MAXCHEESE value

       for (int i = 0; i < MAXCHEESE; i++) {

           names[i] = "Cheese Type " + (char) ('A' + i);

           prices[i] = ranGen.nextInt(1000) / 100.0;

           amounts[i] = 0;

           System.out.println(names[i] + ": $" + prices[i] + " per pound");

       }        

   }

}

7 0
3 years ago
An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,
blsea [12.9K]

Answer:

<em>the % recovery of aluminum product is 80.5%</em>

<em>the % purity of the aluminum product is 54.7%</em>

<em></em>

Explanation:

feed rate to separator = 2500 kg/hr

in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the  machine

of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg  is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

% recovery of aluminium = 140kg/174kg x 100% = <em>80.5%</em>

% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream

% purity of the aluminium product = 140kg/256kg

x 100% = <em>54.7%</em>

8 0
3 years ago
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