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Olin [163]
3 years ago
11

12. A structural component is fabricated from an alloy that has a plane strain fracture toughness of It has been determined that

this component fails at a stress of 300 MPa when the maximum length of a surface crack is 0.95 mm. What is the maximum allowable surface crack length (in mm) without fracture for this same component exposed to a stress of 300 MPa and made from another alloy with a plane-strain fracture toughness of 35 MPa m (31.9 ksi in. ). 40 MPa m (36.4 ksi in. ). 45 MPa m. 57.5 MPa m

Engineering
1 answer:
Artyom0805 [142]3 years ago
3 0

Answer:

Q1 = 4.233 mm

Explanation:

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Rsidential Solar Solution:a. A type of photovoltaic solar panel manufactured in China receives a rating of 250W. The rating proc
aksik [14]

Answer:

a) \eta = 13.455\%, b) E_{day} = 812.716\,kJ, c) C_{month. total} = 19.505\, USD, d) t = 40.588\,years

Explanation:

a) The area of the solar panel is:

A = (20\,ft^{2})\cdot (\frac{0.3048\,m}{1\,ft} )^{2}

A = 1.858\,m^{2}

The energy potential is determined herein:

\dot E_{o} = (1000\,\frac{W}{m^{2}} )\cdot (1.858\,m^{2})

\dot E_{o} = 1858\,W

The efficiency of the solar panel is:

\eta = \frac{\dot E}{\dot E_{o}}\times 100\%

\eta = \frac{250\,W}{1858\,W}\times 100\%

\eta = 13.455\%

b) The energy generated by the solar panel is presented below:

E_{day} = (0.135)\cdot (150\,\frac{W}{m^{2}} )\cdot (20\,ft^{2})\cdot \left(\frac{0.3048\,m}{1\,ft} \right)^{2}\cdot (6\,h)\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{1\,kJ}{1000\,J} )

E_{day} = 812.716\,kJ

c) The energy generated per month and per panel is:

E_{month} = 30\cdot E_{day}

E_{month} = 30 \cdot (812.716\,kJ)\cdot \left(\frac{1\,kWh}{3600\,kJ}  \right)

E_{month} = 6.773\,kWh

Monthly energy savings due to the use of 18 panels are:

C_{month, total} = 18\cdot E_{month}\cdot c

C_{month, total} = 18\cdot (6.773\,kWh)\cdot (\frac{0.16\,USD}{1\,kWh} )

C_{month. total} = 19.505\, USD

d) The payback of the solar energy system is:

t = \frac{9500\,USD}{12\cdot (19.505\,USD)}

t = 40.588\,years

6 0
3 years ago
A specimen has circular cross-sectional area with initial diameter d = 26 mm and a gauge length Lo = 200 mm. A force P = 127 kN
steposvetlana [31]
The ahahte ekiebecwkwlwtfe einshgeh dinzgdwbsine ensue d
4 0
3 years ago
A plane wall of length L, constant thermal conductivity k and no thermal energy generation undergoes one-dimensional, steady-sta
KIM [24]

Answer:

Temperature distribution is T(x)=\dfrac{q}{k}(L-x)+T

Heat flux=q

Heat rate=q A  

Explanation:

We know that for no heat generation and at steady state

\dfrac{\partial^2 T}{\partial x^2}=0

\dfrac{\partial T}{\partial x}=a

T(x)=ax+b

a and are the constant.

Given that heat flux=q

We know that heat flux given as

q=-K\dfrac{dT}{dx}

From above we can say that

a= -\dfrac{q}{K}

Alos given that when x= L temperature is T(L)=T

T= -\dfrac{q}{K}L+b

b=T+\dfrac{q}{K}L

So temperature T(x)

T(x)=-\dfrac{q}{K}x+T+\dfrac{q}{K}L

T(x)=\dfrac{q}{k}(L-x)+T

So temperature distribution is T(x)=\dfrac{q}{k}(L-x)+T

Heat flux=q

Heat rate=q A         (where A is the cross sectional area of wall)

   

6 0
3 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
Batteries don’t run forever. Why do you think batteries run down?
Marina86 [1]

Because there is no such thing as immortal batteries.

When a battery is connected to a circuit, the charge moves through the circuit, and a chemical reaction occurs inside that separate the charges. The strength of this reaction diminishes over time and the battery eventually dies.

3 0
3 years ago
Read 2 more answers
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