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guapka [62]
3 years ago
12

A cold air-standard Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum temperature of t

he air and the rate of heat addition to this cycle when it produces 150 kW of power and the state of the air at the beginning of the compression is 90 kPa and 57 °C. Use constant specific heats at room temperature.
Engineering
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

\dot Q_{in} = 228.659\,kW, T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)

Explanation:

The ideal efficiency of the Diesel cycle is given by this expression:

\eta_{th} = \left\{1 - \frac{1}{r^{k-1}} \cdot \left[\frac{r_{c}^{k}-1}{k\cdot (r_{c}-1)} \right]\right\}\times 100\%

Where r and r_{c} are the compression and cutoff ratios, respectively.

\eta_{th} = \left\{1-\frac{1}{18^{0.4}}\cdot \left[\frac{1.5^{1.4}-1}{1.4\cdot (1.5-1)} \right] \right\}\times 100\%

\eta_{th} = 65.648\,\%

The heat addition to the cycle is:

\dot Q_{in} = \frac{\dot W}{\eta_{th}}

\dot Q_{in} = \frac{150\,kW}{0.656}

\dot Q_{in} = 228.659\,kW

The temperature at state 2 is:

T_{2} = T_{1} \cdot r^{k-1}

T_{2} = (330.15\,K)\cdot 18^{0.4}

T_{2} = 1049.108\,K\,(775.958\,^{\textdegree}C)

And the temperature at state 3 is:

T_{3} = T_{2}\cdot r_{c}

T_{3} = (1.5)\cdot (1049.108\,K)

T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)

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Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Stea
Nataly [62]

Answer:

1) The exergy of destruction is approximately 456.93 kW

2) The reversible power output is approximately 5456.93 kW

Explanation:

1) The given parameters are;

P₁ = 8 MPa

T₁ = 500°C

From which we have;

s₁ = 6.727 kJ/(kg·K)

h₁ = 3399 kJ/kg

P₂ = 2 MPa

T₂ = 350°C

From which we have;

s₂ = 6.958 kJ/(kg·K)

h₂ = 3138 kJ/kg

P₃ = 2 MPa

T₃ = 500°C

From which we have;

s₃ = 7.434 kJ/(kg·K)

h₃ = 3468 kJ/kg

P₄ = 30 KPa

T₄ = 69.09 C (saturation temperature)

From which we have;

h₄ = h_{f4} + x₄×h_{fg} = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg

s₄ =  s_{f4} + x₄×s_{fg} = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)

The exergy of destruction, \dot X_{dest}, is given as follows;

\dot X_{dest} = T₀ × \dot S_{gen} = T₀ × \dot m × (s₄ + s₂ - s₁ - s₃)

\dot X_{dest} = T₀ × \dot W×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)

∴ \dot X_{dest} = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138  - 2554.49) ≈ 456.93 kW

The exergy of destruction ≈ 456.93 kW

2) The reversible power output, \dot W_{rev} = \dot W_{} + \dot X_{dest} ≈ 5000 + 456.93 kW = 5456.93 kW

The reversible power output ≈ 5456.93 kW.

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Answer:

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Explanation:

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For a project in C++ we are supposed toDesign a class named Month. The class should have the following private members:-name: a
mart [117]

Answer:

include <iostream>

using namespace std;

 

class Month

{

public:

 Month (char firstLetter, char secondLetter, char thirdLetter);

 

 Month (int monthNum);

.

 

 Month();

 void outputMonth_num();

 

 

 void outputMonthLetters();

private:

 int month;

};

 

 

int main ()

{

 //

 // Variable declarations

 //

 int monthNum;

 char firstLetter, secondLetter, thirdLetter;    

 char testAgain;              

 

 do {

 

   cout << endl;

   cout << "Testing the default constructor ..." << endl;

   Month defaultMonth;

   defaultMonth.outputMonth_num();

   defaultMonth.outputMonthLetters();

 

   //

   // Construct a month using the constructor with one integer argument

   //

   cout << endl;

   cout << "Testing the constructor with one integer argument..." << endl;

   cout << "Enter a month number: ";

   cin >> monthNum;

 

   Month testMonth1(monthNum);

   testMonth1.outputMonth_num();

   testMonth1.outputMonthLetters();

 

   //

   // Construct a month using the constructor with three letters as arguments

   //

   cout << endl;

   cout << "Testing the constructor with 3 letters as arguments ..." << endl;

   cout << "Enter the first three letters of a month (lowercase): ";

   cin >> firstLetter >> secondLetter >> thirdLetter;

   cout << endl;

 

   Month testMonth2(firstLetter, secondLetter, thirdLetter);

   testMonth2.outputMonth_num();

   testMonth2.outputMonthLetters();

 

   //

   // See if user wants to try another month

   //

   cout << endl;

   cout << "Do you want to test again? (y or n) ";

   cin >> testAgain;

 }

 while (testAgain == 'y' || testAgain == 'Y');

 

 return 0;

}

 

 

Month::Month(char firstLetter, char secondLetter, char thirdLetter)

{

if ((firstLetter == 'j')&&(secondLetter == 'a')&&(thirdLetter == 'n'))

  outputMonth_num = 1;

if ((firstLetter == 'f')&&(secondLetter == 'e')&&(thirdLetter == 'b'))

  outputMonth_num = 2;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'r'))

  outputMonth_num = 3;

if ((firstLetter = 'a')&&(secondLetter == 'p')&&(thirdLetter == 'r'))

  outputMonth_num = 4;

if ((firstLetter == 'm')&&(secondLetter == 'a')&&(thirdLetter == 'y'))

  outputMonth_num = 5;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(thirdLetter == 'n'))

  outputMonth_num = 6;

if ((firstLetter == 'j')&&(secondLetter == 'u')&&(.thirdLetter == 'l'))

  outputMonth_num = 7;

if ((firstLetter == 'a')&&(secondLetter == 'u')&&(thirdLetter == 'g'))

  outputMonth_num = 8;

if ((firstLetter == 's')&&(secondLetter == 'e')&&(thirdLetter == 'p'))

  outputMonth_num = 9;

if ((firstLetter == 'o')&&(secondLetter == 'c')&&(thirdLetter == 't'))

  outputMonth_num = 10;

if ((firstLetter == 'n')&&(secondLetter == 'o')&&(thirdLetter == 'v'))

 outputMonth_num = 11;

if ((firstLetter == 'd')&&(secondLetter == 'e')&&(thirdLetter == 'c'))

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}

 

Month::inputMonthByNumber

{

if (Month_num > 12 && Month_num < 1)

cout << "Invalid number for Month, please choose 1-12)\n";

}

 

void Month::outputMonth_num()

{

 if (month >= 1 && month <= 12)

   cout ><< "Month: " << month << endl;

 else

   cout << "Error - The month is not a valid!" << endl;

}

 

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{

 switch (month)

   {

   case 1:

     cout << "Jan" << endl;

     break;

   case 2:

     cout << "Feb" << endl;

     break;

   case 3:

     cout << "Mar" << endl;

     break;

   case 4:

     cout << "Apr" << endl;

     break;

   case 5:

     cout << "May" << endl;

     break;

   case 6:

     cout << "Jun" << endl;

     break;

   case 7:

     cout << "Jul" << endl;

     break;

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     cout << "Aug" << endl;

     break;

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     cout << "Sep" << endl;

     break;

   case 10:

     cout << "Oct" << endl;

     break;

   case 11:

     cout << "Nov" << endl;

     break;

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     break;

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   }

}

7 0
3 years ago
Suppose a large amount of power is required. Which engine would you choose between Otto and Diesel? Why?
Firdavs [7]

Answer:

Otto engine

Explanation:

As we know that

Power = Torque x speed

So we can say that when speed of engine then power of engine also will increases.

The speed of Otto engine is more as compare to Diesel engine so the power of Otto engine is more.But on the other hand torque of Diesel engine is more as compare to Otto engine but the speed is low so the product of speed and torque is more for Otto engine .It means that when requires large amount of power then Otto engine should be use.

6 0
4 years ago
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