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Ann [662]
3 years ago
7

A flow field is characterized by the stream function ψ= 3x2y−y3. Demonstrate thatthe flow field represents a two-dimensional inc

ompressible flow. Show that the magnitude ofthe velocity at a point depends only on the distance from the origin of the coordinates.
Plot the strim ψ = 2.

Engineering
1 answer:
boyakko [2]3 years ago
8 0

Answer:

δu/δx+δu/δy = 6x-6x =0

9r^2

Explanation:

The flow is obviously two-dimensional, since the stream function depends only on the x and y coordinate. We can find the x and y velocity components by using the following relations:  

u =δψ/δy = 3x^2-3y^2

v =-δψ/δx = -6xy

Now, since:  

δu/δx+δu/δy = 6x-6x =0

we conclude that this flow satisfies the continuity equation for a 2D incompressible flow. Therefore, the flow is indeed a two-dimensional incompressible one.  

The magnitude of velocity is given by:  

|V| = u^2+v^2

    =(3x^2-3y^2)^2+(-6xy)^2

    =9x^4+18x^2y^2+9y^2

    =(3x^2+3y^2)^2

    =9r^2

where r is the distance from the origin of the coordinates, and we have used that r^2 = x^2 + y^2.  

The streamline ψ = 2 is given by the following equation:  

3x^2y — y^3 = 2,

which is most easily plotted by solving it for x:  

x =±√2-y^3/y

Plot of the streamline is given in the graph below.  

Explanation for the plot: the two x(y) functions (with minus and plus signs) given in the equation above were plotted as functions of y, after which the graph was rotated to obtain a standard coordinate diagram. The "+" and "-" parts are given in different colors, but keep in mind that these are actually "parts" of the same streamline.  

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Answer:

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B) W' = 17113.87 KW

Explanation:

We are given;

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Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

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Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

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W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

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Answer:

Below is the required code:

Explanation:

Using for loop

#include <iostream>

using namespace std;

int main()

{

    //Initial crud size

    int init = 0;

    int newCrud;

    int next=0;

    //Number of days to simulate

    int no_days;

    int day;

    cout << "Enter initial amount of green crud: ";

    cin >> newCrud;

    cout << "Enter number of days to simulate: ";

    cin >> no_days;

    for (day = 10; day<=no_days; day++)

    {

         if (day % 10 == 0)

         {

             next = newCrud + init;

         }

             newCrud = init;

             init = next;

    }

    if (no_days < 5)

    cout << "\nCrud reproduce only after 5 days minimum.Hence the current amount is "

    << newCrud << " pounds.";

    else

    cout << "On day " << no_days << " you have " << init

    << " pounds of green crud." << endl;

    cout << "\nWould you like to continue? (y or n): ";

    cin >> ans;

         return 0;

}

Output:

         Enter initial amount of green crud: 5

         Enter number of days to simulate: 220

    On day 220 you have 10485760 pounds of green crud.

Using while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans='y';

    while (ans == 'Y' || ans == 'y')

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud:

         ";

         cin >> newCrud;

         cout << "Enter number of days to simulate:

         ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         cout << "On day " << no_days << " you have "

         << init

         << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or

         n): ";

         cin >> ans;

    }

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

Using do while loop

Program:

#include <iostream>

using namespace std;

int main()

{

    char ans;

    do

    {

         //Initial crud size

         int init = 0;

         int newCrud;

         int next=0;

         //Number of days to simulate

         int no_days;

         int day;

         cout << "Enter initial amount of green crud: ";

         cin >> newCrud;

         cout << "Enter number of days to simulate: ";

         cin >> no_days;

         for (day = 10; day<=no_days; day++)

         {

             if (day % 10 == 0)

             {

                  next = newCrud + init;

             }

                  newCrud = init;

                  init = next;

         }

         if (no_days < 5)

         cout << "\nCrud reproduce only after 5 days

         minimum.Hence the current amount is "

         << newCrud << " pounds.";

         else

         

         cout << "On day " << no_days << " you have " <<

         init << " pounds of green crud." << endl;

         cout << "\nWould you like to continue? (y or n):

         ";

         cin >> ans;

    } while (ans == 'Y' || ans == 'y');

    return 0;

}

Output:

Enter initial amount of green crud: 5

Enter number of days to simulate: 220

On day 220 you have 10485760 pounds of green crud.

Would you like to continue? (y or n): y

Enter initial amount of green crud: 5

Enter number of days to simulate: 225

On day 225 you have 10485760 pounds of green crud.

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3 years ago
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