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Leya [2.2K]
3 years ago
9

Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A

ir enters the heat exchanger at 95 kPa and 20°C at a rate of 0.6 m3/s. The combustion gases (cp = 1.10 kJ/kg·°C) enter at 160°C at a rate of 0.95 kg/s and leave at 95°C. Determine the rate of heat transfer to the air and its outlet temperature.
Engineering
1 answer:
uysha [10]3 years ago
7 0

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C   ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

T is the exit temperature of the air.

Heat transfer

Q=[m Cp ΔT] for gas

Q=0.95 x 1.1 x ( 160 -95 )

Q=67.95 W

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Answer:

(b) 56%

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here we have given T₁ (Higher temperature)= 600+273=873

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Efficiency of power cycle is given by =1-\frac{T2}{T1}

=1-\frac{383}{873}

=1-0.43871

=.56

=56%

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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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Explanation:

JavaPadGUI.java

package javapad.gui;

import java.awt.BorderLayout;

import java.awt.Container;

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import java.awt.event.ActionEvent;

import java.awt.event.ActionListener;

import java.awt.event.WindowAdapter;

import java.awt.event.WindowEvent;

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import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

import javax.swing.JButton;

import javax.swing.JFrame;

import javax.swing.JLabel;

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import javax.swing.JScrollPane;

import javax.swing.JTextArea;

public class JavaPadGUI extends JFrame implements ActionListener{

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

 

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  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";

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

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

 

 

  //fields

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  private JButton butNew, butLoad, butSave, butQuit;

 

  public JavaPadGUI()

  {

      super(TITLE);

      createUI();

      setSize(new Dimension(400, 300));

      setDefaultCloseOperation(EXIT_ON_CLOSE);

         

  }

 

  private void createUI()

  {

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

     

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

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      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);

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      butSave.setActionCommand(SAVE);

      butQuit.setActionCommand(QUIT);

     

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  {

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  {

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          String str = "";

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}

RunJavaPad.java

package javapad;

import javapad.gui.JavaPadGUI;

public class RunJavaPad {

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7 0
3 years ago
1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b
Rainbow [258]

Answer:

1.  b. False

2. b. False

3.  b. False

4.  b. False

5. a. True

6. a. True

7.  b. False

8.  b. False

9. a. True

Explanation:

1. The fatigue properties of a material  are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6.  The bending fatigue could be handled with specific load requirements  for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the  maximum stress, as in torsional fatigue, which can be performed on  axial-type specially designed machines also, using the proper fixtures if  the maximum twist required is small, in which linear motion is changed to rotational motion.

7.  A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0
3 years ago
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