Payment to beneficiaries who were named by<br> the insured person

g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress

zvonat [6]

Answer:

a) $\mathbf{\sigma _ 1 = 4800 psi}$

$\mathbf{ \sigma _2 = 0}$

b) $\mathbf{\sigma _ 1 = 6000 psi}$

$\mathbf{ \sigma _2 = 3000 psi}$

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = 0$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 4800 psi}$

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

$\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}$

$\sigma_{long} = \sigma _2 = \frac{Pd}{4t}$

$\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}$

$\mathbf{\sigma _ 1 = 6000 psi}$

$\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}$

$\mathbf{ \sigma _2 = 3000 psi}$

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0

3 years ago

Given a mass-spring-damper system. The impulse response of strength 1 can be obtained from a unit step response by: ______

Alina [70]

Answer:

Multiplying impulse response by t ( option D )

Explanation:

We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .

When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .

5 0

3 years ago

Casein, a dairy product used in making cheese, contains 25% moisture when wet. A dairy sells this product for $40/100 kg. If req

Nataly_w [17]

Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.

<h3>What is the mass of casein in wet casein?</h3>

The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg

Mass of water 250 kg

The mass of casein is constant while the moisture content can be changed.

At 12% moisture content;

750 kg = 88%%

100 % = 100 ×750/88 = 852.27 kg

Therefore, the mass of dried casein produced os 852.3 kg.

Learn more about mass at: brainly.com/question/24658038

#SPJ1

3 0

2 years ago

A pipeline (NPS = 14 in; schedule = 80) has a length of 200 m. Water (15℃) is flowing at 0.16 m3/s. What is the pipe head loss f

dangina [55]

Answer:

Head loss is 1.64

Explanation:

Given data:

Length (L) = 200 m

Discharge (Q) = 0.16 m3/s

According to table of nominal pipe size , for schedule 80 , NPS 14, pipe has diameter (D)= 12.5 in or 31.8 cm 0.318 m

We know, $head\ loss = \frac{f L V^2}{( 2 g D)}$

where, f = Darcy friction factor

V = flow velocity

g = acceleration due to gravity

We know, flow rate Q = A x V

solving for V

$V = \frac{Q}{A}$

$= \frac{0.16}{\frac{\pi}{4} (0.318)^2} = 2.015 m/s$

obtained Darcy friction factor

calculate Reynold number (Re) ,

$Re = \frac{\rho V D}{\mu}$

where, $\rho$ = density of water

$\mu$ = Dynamic viscosity of water at 15 degree C = 0.001 Ns/m2

so reynold number is

$Re = \frac{1000\times 2.015\times 0.318}{0.001}$

= 6.4 x 10^5

For Schedule 80 PVC pipes , roughness (e) is 0.0015 mm

Relative roughness (e/D) = 0.0015 / 318 = 0.00005

from Moody diagram, for Re = 640000 and e/D = 0.00005 , Darcy friction factor , f = 0.0126

Therefore head loss is

$HL = \frac{0.0126 (200)(2.015)^2}{( 2 \times 9.81 \times 0.318)}$

HL = 1.64 m

7 0

3 years ago

Your program will be a line editor. A line editor is an editor where all operations are performed by entering commands at the co

soldier1979 [14.2K]

Answer:

Java program given below

Explanation:

import java.util.*;

import java.io.*;

public class Lineeditor

{

private static Node head;

class Node

{

int data;

Node next;

public Node()

{data = 0; next = null;}

public Node(int x, Node n)

{data = x; next =n;}

}

public void Displaylist(Node q)

{if (q != null)

{

System.out.println(q.data);

Displaylist(q.next);

}

public void Buildlist()

{Node q = new Node(0,null);

head = q;

String oneLine;

try{BufferedReader indata = new

BufferedReader(new InputStreamReader(System.in)); // read data from terminals

System.out.println("Please enter a command or a line of text: ");

oneLine = indata.readLine(); // always need the following two lines to read data

head.data = Integer.parseInt(oneLine);

for (int i=1; i<=head.data; i++)

{System.out.println("Please enter another command or a new line of text:");

oneLine = indata.readLine();

int num = Integer.parseInt(oneLine);

Node p = new Node(num,null);

q.next = p;

q = p;}

}catch(Exception e)

{ System.out.println("Error --" + e.toString());}

}

public static void main(String[] args)

{Lineeditor mylist = new Lineeditor();

mylist.Buildlist();

mylist.Displaylist(head);

}

7 0

3 years ago

Payment to beneficiaries who were named by the insured person