8. The operation of a TXV is controlled by the

A strain gage is mounted at an angle of 30° with respect to the longitudinal axis of the cylindrical pressure. The pressure vess

GuDViN [60]

Answer:

1790 μrad.

Explanation:

Young's modulus, E is given as 10000 ksi,

μ is given as 0.33,

Inside diameter, d = 54 in,

Thickness, t = 1 in,

Pressure, p = 794 psi = 0.794 ksi

To determine shear strain, longitudinal strain and circumferential strain will be evaluated,

Longitudinal strain, eL = (pd/4tE)(1 - 2μ)

eL = (0.794 x 54)(1 - 0.66)/(4 x 1 x 10000)

eL = 3.64 x 10-⁴ radians

Circumferential strain , eH = (pd/4tE)(2-μ)

eH = (0.794 x 54)(2 - 0.33)/(4 x 1 x 10000)

eH = 1.79 x 10-³ radians

The maximum shear strain is 1790 μrad.

4 0

3 years ago

2.4: Add a method called setValue(), and the description of setValue is: public int setValue(long searchKey) In this method, the

Yanka [14]

Answer:

Below is java code that must be used for the given question:

// highArray.java

// demonstrates array class with high-level interface

// to run this program: C>java HighArrayApp

////////////////////////////////////////////////////////////////

class HighArray

{

private long[] a; // ref to array a

private int nElems; // number of data items

//-----------------------------------------------------------

public HighArray(int max) // constructor

{

a = new long[max]; // create the array

nElems = 0; // no items yet

}

//-----------------------------------------------------------

public setValue find(long searchKey)

{ // find specified value

int j;

for(j=0; j<nElems; j++) // for each element,

if(a[j] == searchKey) // found item?

break; // exit loop before end

if(j == nElems) // gone to end?

return false; // yes, can't find it

else

return true; // no, found it

} // end find()

//-----------------------------------------------------------

public void insert(long value) // put element into array

{

a[nElems] = value; // insert it

nElems++; // increment size

}

//-----------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element,

System.out.print(a[j] + " "); // display it

System.out.println("");

}

//-----------------------------------------------------------

} // end class HighArray

////////////////////////////////////////////////////////////////

class HighArrayApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

HighArray arr; // reference to array

arr = new HighArray(maxSize); // create the array

arr.insert(77); // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);

arr.display(); // display items

int searchKey = 35; // search for item

if( arr.find(searchKey) )

System.out.println("Found " + searchKey);

else

System.out.println("Can't find " + searchKey);

} // end main()

} // end class HighArrayApp

Explanation:

6 0

2 years ago

Air within a piston cylinder assembly executes a Carnot refrigeration cycle between hot and cold reservoirs at TH=600 K and TC=3

Nataly [62]

Answer:

See explaination

Explanation:

for a reverse carnot cycle T-S diagram is a rectangle which i have shown

net work for a complete cycle must be equal to net heat interaction.

Kindly check attachment for the step by step solution of the given problem.

5 0

3 years ago

Reduce the force F ij = + (2 5 ) kN to point A(2m,3m) that acts on point B( 3m,5m) - .

Alexxx [7]

Given :

Force, $\vec{F}= (2\hat{i} + 5\hat{j})\ kN$ .

Force is acting at point A( 2 m, 3 m ) and B( 3 m, 5 m )

To Find :

The work done by force F .

Solution :

Displacement vector between point A and B is :

$\vec{d} = (3-2)\hat{i} + (5-3)\hat{j}\\\\\vec{d} = \hat{i} + 2\hat{j}$

Now, we know work done is given by :

$W = \vec{F}.\vec{d}\\\\W= (2\hat{i} + 5\hat{j}).(\hat{i}+\hat{2j})\\\\W = (2\times 1) +( 5\times 2) \ kJ\\\\W = 12 \ kJ$

W = 12000 J

Therefore, work done by force is 12000 J .

6 0

2 years ago

A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well

ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

= 14.9 * 20.1 = 299.5 m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

= 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = $\frac{Q}{4\pi T} * W (u)$

where : Q = 2725.5

T = 299.5

W(u) = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0

3 years ago