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vekshin1
3 years ago
10

The Energy Losses Associated with Valves and Fittings: a)- are generally associated with a K factor b)- are generally associated

with an external surface finish. c)- both answers 1 and 2 d)- all of the above.
Engineering
1 answer:
madam [21]3 years ago
8 0

Answer:

a)Are generally associated with factor.

Explanation:

We know that losses are two types

1.Major loss  :Due to friction of pipe surface

2.Minor loss  :Due to change in the direction of flow

As we know that when any hindrance is produced during the flow of fluid then it leads to generate the energy losses.If flow is along uniform diameter pipe then there will not be any loss but if any valve and fitting placed is the path of fluid flow due to this direction of fluid flow changes and  it produce losses in the energy.

Lot' of experimental data tell us that loss in the energy due to valve and fitting are generally associated with K factor.These losses are given as

Losses=K\dfrac{V^2}{2g}

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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 aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass
slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

                              = \frac{14 *12}{1.136} = 147.8

buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}

P_{cr} = 13.62 kips

3 0
3 years ago
A list is sorted in ascending order if it is empty or each item except the last one is less than or equal to its successor. HERE
Free_Kalibri [48]

Using the knowledge of computational language in python it is possible to write a code that writes a list and defines the arrange.

<h3>Writing code in python:</h3>

<em>def isSorted(lyst):</em>

<em>if len(lyst) >= 0 and len(lyst) < 2:</em>

<em>return True</em>

<em>else:</em>

<em>for i in range(len(lyst)-1):</em>

<em>if lyst[i] > lyst[i+1]:</em>

<em>return False</em>

<em>return True</em>

<em>def main():</em>

<em>lyst = []</em>

<em>print(isSorted(lyst))</em>

<em>lyst = [1]</em>

<em>print(isSorted(lyst))</em>

<em>lyst = list(range(10))</em>

<em>print(isSorted(lyst))</em>

<em>lyst[9] = 3</em>

<em>print(isSorted(lyst))</em>

<em>main()</em>

See more about python at brainly.com/question/18502436

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7 0
2 years ago
A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603                                         225

.203                                         368

.162                                          442

answer : 191 mPa

Explanation:

<u>Determine the stress present at the time of fracture for the original plate</u>

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603  bending stress = 225

at 0.203  bending stress = 368

at 0.162  bending stress = 442

<u>applying equation 1   determine the value of n for several combinations</u>

 ( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

 ( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) =  ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x =  0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

     = 225 * 0.84747

     <em>=  191 mPa </em>

3 0
3 years ago
Which type of vehicle is used routinely at construction and mining sites?
satela [25.4K]

We are required to state the type of vehicle which is routinely used at the construction site.

The vehicle which is routinely used at the construction site is called hydraulic vehicle.

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  • Types of fluids used in hydraulic vehicles includes: Water fluid, petroleum fluid, synthetic fluid.

Read more:

brainly.com/question/20536356

5 0
3 years ago
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