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SSSSS [86.1K]
2 years ago
12

Two aerial photographs were taken 30 seconds apart over one east-bound lane of l-80 near Grand Island, NE. The following results

were recorded.
Position from start of road section Vehicle 2000 2300 1700 1200 600 2940 3200 2400 2100 1730 1000 2 3 4
Plot the trajectories of the vehicles on graph paper and compute the average flow (vph), density (veh/mi) and space mean speed (mph) over the 3000 ft length of the lane.
Engineering
1 answer:
NikAS [45]2 years ago
8 0

Answer:

the average flow (vph) = 222.69 veh/hr.  

The average velocity = 111348 ft/hr.

density = 2 × 10⁻³ veh/ft.

Explanation:

The first thing to do in this particular question is to determine the average velocity.

The average velocity = [ ( 2940 - 2000/30) + ( 3200 - 2300/30) + ( 2400 - 1700/30) + ( 2100 - 1200) + ( 1730 - 600/30) + ( 1000 - 0/30).

The average velocity = [31.33 + 30 + 23.33 + 30 + 37.66 + 33.66]/ 6 = 30.99 ft/sec.

Thus, 30.99 ft/sec × 60 × 60 = 111348 ft/hr.

The next thing to do is to determine the density. therefore, the density = 6/ 3000 = 2 × 10⁻³ veh/ft.  

The average flow (vph) =  111348 ft/hr × 2 × 10⁻³ veh/ft.  = 222.69 veh/ hr.

Also, the space mean speed (mph) over the 3000 ft length of the lane = 6/ [ 1/31.33 + 1/30 + 1/23.33 + 1/30 + 1/37.66 + 1/33.66] = 6/ 0.1977 = 30.34 ft/sec.

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Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu
Sonja [21]

Answer:

a

The rate of work developed is \frac{\r W}{\r m}= 300kJ/kg

b

The rate of entropy produced within the turbine is   \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

Explanation:

     From  the question we are told

          The rate at which heat is transferred is \frac{\r Q}{\r m } = -  30KJ/kg

the negative sign because the heat is transferred from the turbine

          The specific heat capacity of air is c_p = 1.1KJ/kg \cdot K

          The inlet temperature is  T_1 = 970K

          The outlet temperature is T_2 = 670K

           The pressure at the inlet of the turbine is p_1 = 400 kPa

          The pressure at the exist of the turbine is p_2 = 100kPa

           The temperature at outer surface is T_s = 315K

         The individual gas constant of air  R with a constant value R = 0.287kJ/kg \cdot K

The general equation for the turbine operating at steady state is \

               \r Q - \r W + \r m (h_1 - h_2) = 0

h is the enthalpy of the turbine and it is mathematically represented as          

        h = c_p T

The above equation becomes

             \r Q - \r W + \r m c_p(T_1 - T_2) = 0

              \frac{\r W}{\r m}  = \frac{\r Q}{\r m} + c_p (T_1 -T_2)

Where \r Q is the heat transfer from the turbine

           \r W is the work output from the turbine

            \r m is the mass flow rate of air

             \frac{\r W}{\r m} is the rate of work developed

Substituting values

              \frac{\r W}{\r m} =  (-30)+1.1(970-670)

                   \frac{\r W}{\r m}= 300kJ/kg

The general balance  equation for an entropy rate is represented mathematically as

                       \frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma  = 0

          =>          \frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)

    generally (s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]

substituting for (s_1 -s_2)

                      \frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} +  c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]

                      Where \frac{\sigma}{\r m} is the rate of entropy produced within the turbine

 substituting values

                \frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]

                    \frac{\sigma}{\r m}=  0.0861kJ/kg \cdot K

           

 

                   

   

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3 years ago
Which system of linear inequalities is represented by the graph? y > x – 2 and y x + 1 y x + 1 y > x – 2 and y < x + 1
KatRina [158]

Answer:

The graph representing the linear inequalities is attached below.

Explanation:

The inequalities given are :

y>x-2   and y<x+1

For tables for values of x and y and get coordinates to plot for both equation.

In the first equation;

y>x-2

y=x-2

y-x = -2

The table will be :

x    y

-2  -4

-1    -3

0     -2

1      -1

2      0

The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)

Use a dotted line and shade the part right hand side of the line.

Do the same for the second inequality equation and plot then shade the part satisfying the inequality.

The graph attached shows results.

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3 years ago
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Answer:eyesight

Explanation:

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3 years ago
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Write a GUI-based program that plays a guess-the-number game in which the roles of the computer and the user are the reverse of
AURORKA [14]

Answer:

import javax.swing.*;

import java.awt.*;

import java.util.Random;

import java.awt.event.*;

public class Guess extends JFrame

{

   private static final long serialVersionUID = 1L;

   private JButton newGame;

   private JButton enter;

   private JButton exit;

   private JTextField guess;

   private JLabel initialTextLabel;

   private JLabel enterLabel;

   private JLabel userMessageLabel;

   private int randNum;

   private int userInput;

   private int maxtries = 0;

   public Guess()

   {

       super("Guessing Game");

       newGame = new JButton("New Game");

       exit = new JButton("Exit Game");

       enter = new JButton("Enter");

       guess = new JTextField(4);

       initialTextLabel = new JLabel("I'm thinking of a number between 1 and 100. Guess it!");

       enterLabel = new JLabel("Enter your guess.");

       userMessageLabel = new JLabel("");

       randNum = new Random().nextInt(100) + 1;

       setLayout(new FlowLayout());

       add(initialTextLabel);

       add(enterLabel);

       add(guess);

       add(newGame);

       add(enter);

       add(exit);

       add(userMessageLabel);

   

       setSize(500, 300);

       addWindowListener(new WindowAdapter()

       {

           public void windowClosing(WindowEvent e)

           {

               System.exit(0);

           }

       });

       newGameButtonHandler nghandler = new newGameButtonHandler();

       newGame.addActionListener(nghandler);

       ExitButtonHandler exithandler = new ExitButtonHandler();

       exit.addActionListener(exithandler);

       enterButtonHandler enterhandler = new enterButtonHandler();

       enter.addActionListener(enterhandler);

   }

   class newGameButtonHandler implements ActionListener

   {

       public void actionPerformed(ActionEvent e)

       {

           setBackground(Color.ORANGE);

           guess.setEnabled(true);

           guess.setText("");

           enter.setEnabled(true);

           maxtries = 0;

           userMessageLabel.setText("");

           randNum = new Random().nextInt(100) + 1;

       }

   }

   class ExitButtonHandler implements ActionListener

   {

       public void actionPerformed(ActionEvent e)

       {

           System.exit(0);

       }

   }

   class enterButtonHandler implements ActionListener

   {

       public void actionPerformed(ActionEvent e)

       {

           userInput = Integer.parseInt(guess.getText());

           checkGuess(randNum);

      if(userInput > 100 )

          {

                               userMessageLabel.setText("invalid entry");

          }

       }

   }

   public void checkGuess(int randomNumber)

   {

       maxtries++;

     if(maxtries==10){

           userMessageLabel.setText("You Lose!!");

           guess.setEnabled(false);

           enter.setEnabled(false);

         

       }else if (userInput == randomNumber)

           {

               userMessageLabel.setText("Correct !");

           }

       else if (userInput > randomNumber)

           {

               userMessageLabel.setText("Too high");

           }

       else if (userInput < randomNumber)

           {

               userMessageLabel.setText("Too Low");

           }

   }

   public static void main(String[] args)

   {

       Guess game = new Guess();

       game.setVisible(true);

   }

}

8 0
3 years ago
Water leaves a penstock (the flow path through a hydroelectric dam) at a velocity of 100 ft/s. How deep is the water behind the
Marysya12 [62]

Answer:

155fts

Explanation:

We apply the bernoulli's equation to get the depth of water.

We have the following information

P1 = pressure at top water surface = 0

V1 = velocity at too water surface = 0

X1 = height of water surface = h

Hf = friction loss = 0

P2 = pressure at exit = 0

V2 = velocity at exit if penstock = 100ft/s

X2 = height of penstock = 0

g = acceleration due to gravity = 32.2ft/s²

Applying these values to the equation

0 + 0 + h = 0 + v2²/2g +0 + 0

= h = 100²/2x32.2

= 10000/64.4

= 155.28ft

= 155

8 0
2 years ago
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