Should i show my face?

An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va

Ksju [112]

Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

$\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}$

State 2:

$\left.\begin{array}{l}P_{2}=5 \text { bor } \\S_{2}=S_{1}\end{array}\right\} \quad h_{2}=262.78 \mathrm{kJ} / \mathrm{kg}$

State 3:

$\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}$

State 4:

Throttling process $h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}$

(a)

Magnitude of compressor power input

$\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}$

$w_{c}=4 \cdot 013 \mathrm{kw}$

(b)

Refrigerator capacity

$Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}$

$Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega$

$\ Q_{in} =6 \cdot 583 \text { tons }$

(c)

Cop:

$\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}$

$\beta=5 \cdot 758$

3 0

3 years ago

Question text

lisabon 2012 [21]

Answer:

That's a really nice question sadly I don't know the answer I'm replying to you cuz I'm tryna get points so... Sorry

3 0

3 years ago

This assignment is designed to test your understanding of graphical user interface components and Event Driven programming in Ja

lys-0071 [83]

Answer:

Java program is given below

Explanation:

JavaPadGUI.java

package javapad.gui;

import java.awt.BorderLayout;

import java.awt.Container;

import java.awt.Dimension;

import java.awt.event.ActionEvent;

import java.awt.event.ActionListener;

import java.awt.event.WindowAdapter;

import java.awt.event.WindowEvent;

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

import javax.swing.JButton;

import javax.swing.JFrame;

import javax.swing.JLabel;

import javax.swing.JOptionPane;

import javax.swing.JPanel;

import javax.swing.JScrollPane;

import javax.swing.JTextArea;

public class JavaPadGUI extends JFrame implements ActionListener{

//constants

private static final String TITLE = "Macrosoft JavaPad XP";

private static final String SLOGAN = "Macrosoft: Resistance is futile";

//button names

private static final String NEW = "New";

private static final String LOAD = "Load";

private static final String SAVE = "Save";

private static final String QUIT = "Quit";

private static final String FILENAME = "hardcode.txt";

//messages

private static final String QUIT_MSG = "Quitting, Save?";

private static final String LOAD_ERR = "Could not access file " + FILENAME;

//fields

private JTextArea txtContent;

private JButton butNew, butLoad, butSave, butQuit;

public JavaPadGUI()

{

super(TITLE);

createUI();

setSize(new Dimension(400, 300));

setDefaultCloseOperation(EXIT_ON_CLOSE);

}

private void createUI()

{

Container c = getContentPane();

c.setLayout(new BorderLayout(30,30));

//create the buttons panel

JPanel butPanel = new JPanel();

butPanel.add(butNew = new JButton(NEW));

butPanel.add(butLoad = new JButton(LOAD));

butPanel.add(butSave = new JButton(SAVE));

butPanel.add(butQuit = new JButton(QUIT));

//set command names for the buttons, these will be when button is clicked

butNew.setActionCommand(NEW);

butLoad.setActionCommand(LOAD);

butSave.setActionCommand(SAVE);

butQuit.setActionCommand(QUIT);

JPanel sloganPanel = new JPanel();

sloganPanel.add(new JLabel(SLOGAN));

//create textarea with scrollbar

txtContent = new JTextArea(15, 25);

txtContent.setLineWrap(true);

JScrollPane scroll = new JScrollPane(txtContent);

//now add all components

c.add(butPanel, BorderLayout.NORTH);

c.add(scroll, BorderLayout.CENTER);

c.add(sloganPanel, BorderLayout.SOUTH);

//set the actionlistner to buttons to handle button click event

butNew.addActionListener(this);

butLoad.addActionListener(this);

butSave.addActionListener(this);

butQuit.addActionListener(this);

}

public void actionPerformed(ActionEvent e)

{

String cmd = e.getActionCommand();

if(cmd.equals(NEW))

txtContent.setText("");

else if (cmd.equals(LOAD))

load();

else if(cmd.equals(SAVE))

save();

else if (cmd.equals(QUIT))

quit();

}

private void quit()

{

int option = JOptionPane.showConfirmDialog(this, QUIT_MSG);

if(option == JOptionPane.YES_OPTION)

{

save();

}

System.exit(0);

}

private void save()

{

try {

PrintWriter w = new PrintWriter(new File(FILENAME));

w.write(txtContent.getText());

w.close();

} catch (FileNotFoundException e) {

JOptionPane.showMessageDialog(this,"I/O Error", LOAD_ERR, JOptionPane.ERROR_MESSAGE);

}

private void load()

{

try {

Scanner s = new Scanner(new File(FILENAME));

String str = "";

while(s.hasNextLine())

str += s.nextLine();

txtContent.setText(str);

s.close();

} catch (FileNotFoundException e) {

JOptionPane.showMessageDialog(this, LOAD_ERR, "I/O Error",JOptionPane.ERROR_MESSAGE);

}

RunJavaPad.java

package javapad;

import javapad.gui.JavaPadGUI;

public class RunJavaPad {

public static void main(String[] args) {

JavaPadGUI gui = new JavaPadGUI();

gui.setVisible(true);

}

7 0

3 years ago

A sample of cast iron with .35 wt%C is slow cooled from 1500C to room temperature. What is the fraction of proeutectoid Cementit

Marizza181 [45]

Answer:

So the fraction of proeutectoid cementite is 44.3%

Explanation:

Given that

cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.

We know that

Fraction of proeutectoid cementite phase gievn as

${w_p}'=\dfrac{{C_o}'-0.022}{0.74}$

Now by putting the values

${w_p}'=\dfrac{0.35-0.022}{0.74}$

${w_p}'=0.443$

So the fraction of proeutectoid cementite is 44.3%

6 0

3 years ago

When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner

ipn [44]

Answer:

$\dot W = 1.667\, kW$

Explanation:

A well-sealed house means that there is no mass interaction between air indoors and outdoors. Hence, cooling process is isochoric. The heat removed by the air conditioner is:

$\dot Q_{L} = \frac{m_{air}}{\Delta t} \cdot c_{v, air} \cdot (T_{o}-T_{f})\\\dot Q_{L} = \frac{800\, kg}{(30\, min)\cdot (\frac{60\, s}{1 \, min} )}\cdot (0.7 \frac{kJ}{kg \cdot K}) \cdot (15\, K)\\\dot Q_{L} = 4.667\, kW$

The power drawn by the air conditioner is:

$\dot W = \frac{\dot Q_{L}}{COP_{R}} \\\dot W = \frac {4.667\, kW}{2.8}\\\dot W = 1.667\, kW$

3 0

3 years ago