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Natalka [10]
4 years ago
6

At steady state, a cycle develops a power output of 10 kW for heat addition at a rate of 10 kJ per cycle of operation from a sou

rce at 1500 K. Energy is rejected by heat transfer to cooling water at 300 K. Determine the minimum theoretical number of cycles required per minute.
Engineering
1 answer:
aleksandr82 [10.1K]4 years ago
6 0

Answer:

Number of cycle per minute = 75

Explanation:

given data

Power output Wnet = 10 KW

Temperature of hot reservoir = 1500 K

Temperature of cold reservoir = 300 K

solution

we know that Efficiency is express as

n = 1 - To ÷ Tk     ...............1

here  Tk is Temperature of hot reservoir and To is Temperature of cold reservoir

so put here value in equation 1 we get

n = 1- 300  ÷ 1500

n = 0.8

and Rate of heat input will be

Rate of heat input Qin = 10 × No. of cycles per sec

Rate of heat input Qin  = 10 n

and

n = Wnet ÷ 10 n    .......................2

put here value

0.8 = 10 ÷  10 n

solve it we get

n = 1.25

so here Number of cycle per second is = 1.25

and Number of cycle per minute will be = 1.25 × 60

Number of cycle per minute = 75

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how is friction losses in pipes reduced? a. decrease the pipe diameter b. increase the length of the pipes. c. decrease the leng
Citrus2011 [14]

Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.

Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.

Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:

  • <u><em>Decrease the length of the pipes</em></u>: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
  • <u><em>Reduce the surface roughness of the pipes</em></u>:  By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
  • <u><em>Increase the pipe diameter: </em></u>By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.

You can learn more about friction losses at

brainly.com/question/13348561

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3 0
1 year ago
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'))

 outputMonth_num = 12;

}

 

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;

}

 

void Month::outputMonthLetters()

{

 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;

   case 8:

     cout << "Aug" << endl;

     break;

   case 9:  

     cout << "Sep" << endl;

     break;

   case 10:

     cout << "Oct" << endl;

     break;

   case 11:

     cout << "Nov" << endl;

     break;

   case 12:

     cout << "Dec" << endl;

     break;

   default:

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

   }

}

7 0
3 years ago
A current vehicle registration expires at _____ of the first owner listed on the registration form
wariber [46]

Answer:

Midnight on the birthday. Hope this helps!

3 0
3 years ago
public interface Frac { /** @return the denominator of this fraction */ int getDenom(); /** @return the numerator of this fracti
True [87]

Answer:

The Full details of the answer is attached.

7 0
3 years ago
A driver traveling in her 16-foot SUV at the speed limit of 30 mph was arrested for running a red light at 15th and Main, an int
Lynna [10]

Answer:

(a) Yes

(b) 102.8 ft

Explanation:

(a)First let convert mile per hour to feet per second

30 mph = 30 * 5280 / 3600 = 44 ft/s

The time it takes for this driver to decelerate comfortably to 0 speed is

t = v / a = 44 / 10 = 4.4 (s)

given that it also takes 1.5 seconds for the driver reaction, the total time she would need is 5.9 seconds. Therefore, if the yellow light was on for 4 seconds, that's not enough time and the dilemma zone would exist.

(b) At this rate the distance covered by the driver is

s = v_0t + \frac{at^2}{2}

s =44*1.5 + 44(4.4) - \frac{10*4.4^2}{2} = 162.8 (ft)

Since the intersection is only 60 feet wide, the dilemma zone must be

162.8 - 60 = 102.8 ft

4 0
3 years ago
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