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uysha [10]
2 years ago
8

What is the t max for a carbon steel heater tube?

Engineering
1 answer:
boyakko [2]2 years ago
4 0

Overheating of Carbon Steel Heater Tubes. ... document and find the creep analysis in the appendix-J. Maximum temperature of the CS permitted by the API-581 is 410 Deg C after which you need to
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A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 30.00 a
IrinaVladis [17]

Answer:

The original length of the specimen l_{o} = 104.7 mm

Explanation:

Original diameter d_{o} = 30 mm

Final diameter d_{1} = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length l_{1} = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × 10^{3} M pa

Shear modulus G = 25.4 G pa = 25.4 × 10^{3} M pa

We know that the relation between the shear modulus & elastic modulus is given by

G = \frac{E}{2(1 + \mu)}

25.5 = \frac{65.5}{2 (1 + \mu)}

\mu = 0.28

This is the value of possion's ratio.

We know that the possion's ratio is given by

\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }

{\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}

{\frac{change \ in \ length}{l_{o} } = 0.00476

\frac{l_{1} - l_{o}  }{l_{o}  } = 0.00476

\frac{l_{1} }{l_{o} } = 1.00476

Final length l_{o} = 105.2 m

Original length

l_{o} = \frac{105.2}{1.00476}

l_{o} = 104.7 mm

This is the original length of the specimen.

5 0
3 years ago
2. BCD uses 6 bits to represent a symbol. a) True b) False​
Goshia [24]

Answer:

true because BCD used 6 bits to represent a symbol .

Explanation:

mark me brainlist

4 0
2 years ago
Write a program that uses a function called Output_Array_Info. Output_Array_Info Properties: Input Parameters: 1. A pointer to a
Artyom0805 [142]

Answer:

C++ code explained below

Explanation:

Please note the below program has been tested on ubuntu 16.04 system and compiled using g++ compiler. This code will also work on other IDE's

-----------------------------------------------------------------------------------------------------------------------------------

Program:

-----------------------------------------------------------------------------------------------------------------------------------

//header files

#include<iostream>

//namespace

using namespace std;

//function defintion

void Output_Array_Info(int *array_ptr, int size)

{

//display all array elements

cout<<"Array elements are: "<<endl;

for(int i =0; i<size; i++)

{

cout<<*(array_ptr+i)<<endl;

}

//display address of each element

cout<<endl<<"memory address of each array elemnt is: "<<endl;

for(int i =0; i<size; i++)

{

cout<<array_ptr+i<<endl;

}

}

//start of main function

int main()

{

//pointer variables

int *pointer;

//an array

int numbers[] = { 5, 7, 9, 10, 12};

//pointer pointing to array

pointer = numbers;

//calculate the size of the array

int size = sizeof(numbers)/sizeof(int);

//call to function

Output_Array_Info(numbers, size);

return 0;

}

//end of the main program

8 0
3 years ago
An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e
Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in 2.25 * 10^4 BTU. It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.  

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - 1 BTU = 778.17  lbf*ft

2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:  

Ep = m*g*(h_2 - h_1)\\ W = m*g  

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.  

h_2 is the final elevation and h_1 is the initial elevation.

Replacing W in the Ep equation

Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\

Finally, the next step is to replace the variables of the problem.  

h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft

The elevation at the high point of the road is 12186.5 in ft.  

3 0
3 years ago
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
Nina [5.8K]

dutile is the correct answer

6 0
2 years ago
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