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sp2606 [1]
3 years ago
15

A simple ideal Rankine cycle which uses water as the working fluid operates its condenser at 40°C and its boiler at 250°C. Calcu

late the work produced by the turbine, the heat supplied in the boiler, and the thermal efficiency of this cycle when the steam enters the turbine without any superheating

Engineering
1 answer:
BlackZzzverrR [31]3 years ago
8 0

The thermal efficiency is 37.5%.

<u>Explanation:</u>

The T-s diagram with steam was attached below.

The table of saturated water temperature is referred and the interpret value is h_{1} and V_{1} at the temperature of 40° C.

h_{1} = h_{f} = 167.53 KJ/Kg

V_{1} = V_{f} = 0.001008 m^{3}/ Kg

For P_{1} and P_{2} the interpret value at the table of saturated water temperature is

P_{1}=P_{\text {Sat at } 40} = 7.385 K Pa

P_{2}=P_{\text {Sat at } 300} = 8.588 K Pa

The specific work input is expressed:

w_{\mathrm{p,in}}=v_{1}\left(P_{2}-P_{1}\right)

= 0.001008 m^{3}/ Kg(8.588 K Pa - 7.385 K Pa)

w_{p,in} = 8.65 KJ/Kg

Enthalpy steam is expressed in state 2.

h_{2} = h_{1} + w_{p,in}

h_{2} = 167.53 KJ/Kg + 8.65 KJ/Kg

h_{2} = 176.18 KJ/Kg

The table of saturated water temperature is referred and the interpret value is s_{3} and h_{3} at the temperature 300° C.

s_{3} = s_{g} = 5.7059 KJ/Kg.K

h_{3} = h_{g} = 2749.6 KJ/Kg

For S_{f} and S_{fg} the interpret value at the table of saturated water temperature is

S_{f} = 0.5724 KJ/Kg.K

S_{fg} = 7.6832 KJ/Kg.K

At last the process of heat rejection, the steam quality is expressed.

x_{4}=\frac{s_{4}-s_{f}}{s_{fg}}

x_{4}=\frac{5.7059-0.5724}{7.6832}

x_{4} = 0.6681

The table of saturated water temperature is referred and the interpret value is h_{fg} at the temperature of 40° C as 2406 KJ/Kg.

Enthalpy steam is expressed in state 4.

h_{4} = h_{f} + x_{4}h_{fg}

h_{4} = 167.53 KJ/Kg + 0.6681(2406 KJ/Kg)

h_{4} = 1775.1 KJ/Kg

For the Rankine cycle, the specific heat input is expressed.

q_{in} = h_{3} - h_{2}

q_{in} = 2749.6 - 176.18

q_{in} = 2573.4 KJ/Kg

For the Rankine cycle, the specific heat output is expressed.

q_{out} = h_{4} - h_{1}

q_{out} = 1775.1 - 167.53

q_{out} = 1607.57 KJ/Kg

The specific net work output is expressed.

w_{T,out} = h_{3} - h_{4}

w_{T,out} = 2749.6 -  1775.1

w_{T,out} = 974.5 KJ/Kg

The thermal efficiency is expressed.

\eta_{\mathrm{th}}=1-\frac{q_{out}}{q_{\mathrm{in}}}

\eta_{\mathrm{th}}=1-\frac{1,607.6 \mathrm{kJ} / \mathrm{kg}}{2,573.4 \mathrm{kJ} / \mathrm{kg}}

=1 - 0.6246

= 0.3753

Hence, the thermal efficiency is 37.5%

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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.

7 0
3 years ago
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