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Dmitry [639]
2 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]2 years ago
6 0

Answer:

Explanation:

Please check the below file for the attached file

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A spherical tank for storing gas under pressure is 25 m in diameter and is made of steel 15 mm thick. The yield point of the mat
nalin [4]

Answer:

a) (option B) 230 kPa

b) (option A) 100 N/m

Explanation:

Given:

Diameter, d = 25 m

Thicknesses, t = 15 mm

Yield point = 240 MPa

Factor of safety = 2.5

a) To find the maximum internal pressure, let's use the formula:

\sigma l = \frac{\sigma y}{FOS} = \frac{PD}{4t}

\frac{\sigma y}{FOS} = \frac{PD}{4t}

Solving for P, we have:

P = \frac{\sigma y * 4t}{FOS * D}

P = \frac{240 * 4 * 15}{2.5 * 25}

P = 230.4 kPa

≈ 230 kPa

The maximum permissible internal pressure is nearly 230kPa

b) Given:

Thickness, t = 6.35 mm

L = 203 mm

Torque, T = 8 N m

Let's find the mean Area,

mA = (l - t)²

= (203 - 6.5)²

= 38671.22mm²

≈ 0.03867 m² (converted to meters)

To find the average shear flow, let's use the formula:

q = \frac{T}{2* mA}

= \frac{8}{2 * 0.03867}

q = 103.4 N/m approximately 100N/m

The average shear force flow is most nearly 100 N/m

4 0
2 years ago
Read 2 more answers
How does Accenture generate value for clients through Agile and DevOps?
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By permanently locking in stakeholder requirements during a project's planning phase. -through highly detailed process documentation that is updated following every work cycle.
7 0
1 year ago
Read 2 more answers
Write a program with total change amount as an integer input, and output the change using the fewest coins, one coin type per li
pshichka [43]

Answer:

The C code is given below with appropriate comments

Explanation:

#include<stdio.h>

//defining constants

#define DOLLAR 100

#define QUARTER 25

#define DIME 10

#define NICKEL 5

#define PENNIES 1

//converting method

void ExactChange(int userTotal,int coinVals[])

{

//checking dollars

if (userTotal >=100)

{

coinVals[0]=userTotal/DOLLAR;

userTotal=userTotal-(100*coinVals[0]);

}

//checking quarters

if (userTotal >=25)

{

coinVals[1]=userTotal/QUARTER;

userTotal=userTotal-(25*coinVals[1] );

}

//checking dimes

if (userTotal >=10)

{

coinVals[2]=userTotal/DIME;

userTotal=userTotal-(10*coinVals[2]);

}

//checking nickels

if (userTotal >=5)

{

coinVals[3]=userTotal/NICKEL;

userTotal=userTotal-(5*coinVals[3]);

}

//checking pennies

if (userTotal >=1)

{

coinVals[4]=userTotal/PENNIES;

userTotal=userTotal-coinVals[4];

}

}

//main method

int main() {

//defining the variables

int amount;

//asking for input

printf("Enter the amount in cents :");

//reading the input

scanf("%d",&amount);

//validating the input

if(amount<1)

{

//printing the message

printf("No change..!");

}

//when the input is >0

else

{

int coinVals[5]={0,0,0,0,0};

ExactChange(amount,coinVals);

//checking dollars

if (coinVals[0]>0)

{

//printing dollars

printf("%d Dollar",coinVals[0]);

if(coinVals[0]>1) printf("s");

}

//checking quarters

if (coinVals[1]>0)

{

//printing quarters

printf(" %d Quarter",coinVals[1]);

if(coinVals[1]>1) printf("s");

}

//checking dimes

if (coinVals[2]>0)

{

//printing dimes

printf(" %d Dime",coinVals[2]);

if(coinVals[2]>1) printf("s");

}

//checking nickels

if (coinVals[3]>0)

{

//prinitng nickels

printf(" %d Nickel",coinVals[3]);

if(coinVals[3]>1) printf("s");

}

//checking pennies

if (coinVals[4]>0)

{

//printing pennies

printf(" %d Penn",coinVals[4]);

if(coinVals[4]>1) printf("ies");

else printf("y");

}

}

//end of main method

}

6 0
2 years ago
Three return steam lines in a chemical processing plant enter a collection tank operating at steady state at 1 bar. Steam enters
Papessa [141]

Answer:

a) 4 kg/s

b) 99.61 °C

Explanation:

See attached pictures.

5 0
2 years ago
Problem 5) Water is pumped through a 60 m long, 0.3 m diameter pipe from a lower reservoir to a higher reservoir whose surface i
kap26 [50]

Answer:

\epsilon = 0.028*0.3 = 0.0084

Explanation:

\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2

where P_1 = P_2 = 0

V1 AND V2  =0

Z1 =0

h_P = \frac{w_p}{\rho Q}

=\frac{40}{9.8*10^3*0.2} = 20.4 m

20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10

we know thaTV  =\frac{Q}{A}

V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec

20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10

f  = 0.0560

Re =\frac{\rho v D}{\mu}

Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5

fro Re = 7.53*10^5 and f = 0.0560

\frac{\epsilon}{D] = 0.028

\epsilon = 0.028*0.3 = 0.0084

4 0
2 years ago
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