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chubhunter [2.5K]
3 years ago
11

Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value

. The tube surface is initially scored; however, over time the effects of scoring diminish and the boiling eventually exhibits behavior similar to that associated with a polished surface. Determine the tube surface temperature immediately after installation and after prolonged service. Assume nucleate boiling at outer surface of tube.
Engineering
1 answer:
masya89 [10]3 years ago
8 0

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} }  )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

q_{n} =uK(\frac{g(\rho_{L}-\rho _{v})     }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr  } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

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

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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;

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  //button names

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  private static final String LOAD = "Load";

  private static final String SAVE = "Save";

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

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  private JTextArea txtContent;

  private JButton butNew, butLoad, butSave, butQuit;

 

  public JavaPadGUI()

  {

      super(TITLE);

      createUI();

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      setDefaultCloseOperation(EXIT_ON_CLOSE);

         

  }

 

  private void createUI()

  {

      Container c = getContentPane();

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

     

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      JPanel butPanel = new JPanel();

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  }

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  {

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package javapad;

import javapad.gui.JavaPadGUI;

public class RunJavaPad {

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3 years ago
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The heat liberated by the LP gas is:

Q_{LP} = \frac{3599.435\,kJ}{0.16}

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