Steam enters a turbine from a 2 inch diameter pipe, at 600 psia, 930 F, with a velocity of 620 ft/s. It leaves the turbine at 12 psia with a quality of 1.0, through an outlet duct 1 ft in diameter. Calculate the turbine power output

Answer: the thermal conductivity of the second material is 125.9 W/m.k

Explanation:

Given that;

The two rods could be approximated as a fins of infinite length.

TA = 75°C,    θA = (TA - T∞) = 75 - 25 = 50°C

TB = 70°C     θB = (TB - T∞) = 70 - 25 = 45°C

Tb = 100°C    θb = (Tb - T∞) = (100 - 25) = 75°C

T∞ = 25°C

KA = 200 W/m · K,   KB = ?

Now

The temperature distribution for the infinite fins are given by

θ/θb = e^(-mx)

θA/θb= e^-√(hp/A.kA) x 1  --------------1

θB/θb = e^-√(hp/A.kB) x 1---------------2

next we  take the natural logof both sides,  

ln(θA/θb) = -√(hp/A.kA) x 1 ------------3

In(θB/θb) = -√(hp/A.kB) x 1 ------------4

now we divide 3 by 4

[ ln(θA/θb) /in(θB/θb)] = √(KB/KA)

we substitute

 [ In(50/75) /In(45/75)] = √(KB/200)

In(0.6666) / In(0.6) = √KB / √200

-0.4055/-0.5108 = √KB / √200

0.7938 = √KB / 14.14

√KB = 11.22

KB = 125.9 W/m.k

So the thermal conductivity of the second material is 125.9 W/m.k

A. The heat transfer rate from natural gas is 2105.26 MW

B. The heat transfer rate to river is 1305.26 MW

Efficiency formula

Efficiency = (power output / power input) × 100

A. How to determine the heat transfer from natural gas

• Efficiency = 38%
• Power output = 800 MW

• Power input =?

Power input = Power input / efficiency

Power input = 800 / 38%

Power input = 800 / 0.38

Power input = 2105.26 MW

Thus, the heat transfer from natural gas is 2105.26 MW

B. How to determine the heat transfer to the river

• Total heat = 2105.26 MW
• Heat used by plant = 800 MW
• Heat to the river =?

Heat to the river = 2105.26 – 800

Heat to the river = 1305.26 MW

Learn more about efficiency:

brainly.com/question/2009210

Answer:

heat transfer from natural gas is 2105.26 MW

heat transfer to river is 1305.26 MW

Explanation:

given data

power output Wn = 800 MW

efficiency = 38%

solution

we know that efficiency is express as

   ......................1

put here value we get

38% =  

Qin  = 2105.26 MW

so heat supply is 2105.26

so we can say

Wn = Qin - Qout

800 = 2105.26 - Qout

Qout = 2105.26 - 800

Qout = 1305.26 MW

so heat transfer from natural gas is 2105.26 MW

and heat transfer to river is 1305.26 MW

Answer:

// Program is written in C++

// Comments are used to explain some lines

// Only the required function is written. The main method is excluded.

#include<bits/stdc++.h>

#include<iostream>

using namespace std;

int divSum(int num)

{

// The next line declares the final result of summation of divisors. The variable declared is also

//initialised to 0

int result = 0;

// find all numbers which divide 'num'

for (int i=2; i<=(num/2); i++)

{

// if 'i' is divisor of 'num'

if (num%i==0)

{

if (i==(num/i))

result += i; //add divisor to result

else

result += (i + num/i); //add divisor to result

}

}

cout<<result+1;

}

In this exercise, using the knowledge of computational language in C++, we have that this code will be written as:

The code is in the attached image.

We can write the C++  as:

#include<bits/stdc++.h>

#include<iostream>

using namespace std;

int divSum(int num)

{

int result = 0;

for (int i=2; i<=(num/2); i++)

{

if (num%i==0)

{

if (i==(num/i))

result += i; //add divisor to result

else

result += (i + num/i); //add divisor to result

}

}

cout<<result+1;

}

See more about C++ at brainly.com/question/19705654

Answer:

Va / Vb = 0.5934

Explanation:

First step is to determine total head losses at each pipe

at Pipe A

For 1/4 open gate valve head loss = 17 *Va^2 / 2g

elbow loss = 0.75 Va^2 / 2g

at Pipe B

For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g

elbow loss = 0.75 * Vb^2 / 2g

Given that both pipes are parallel

17 *Va^2/2g +  0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g  + 0.75 * Vb^2 / 2g

∴ Va / Vb = 0.5934

Answer:

//Annual calendar

#include <iostream>

#include <string>

#include <iomanip>

void month(int numDays, int day)

{

int i;

string weekDays[] = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

// Header print

      cout << "\n----------------------\n";

      for(i=0; i<7; i++)

{

cout << left << setw(1) << weekDays[i];

cout << left << setw(1) << "|";

}

cout << left << setw(1) << "|";

      cout << "\n----------------------\n";

      int firstDay = day-1;

      //Space print

      for(int i=1; i< firstDay; i++)

          cout << left << setw(1) << "|" << setw(2) << " ";

      int cellCnt = 0;

      // Iteration of days

      for(int i=1; i<=numDays; i++)

      {

          //Output days

          cout << left << setw(1) << "|" << setw(2) << i;

          cellCnt += 1;

          // New line

          if ((i + firstDay-1) % 7 == 0)

          {

              cout << left << setw(1) << "|";

              cout << "\n----------------------\n";

              cellCnt = 0;

          }

      }

      // Empty cell print

      if (cellCnt != 0)

      {

          // For printing spaces

          for(int i=1; i<7-cellCnt+2; i++)

              cout << left << setw(1) << "|" << setw(2) << " ";

          cout << "\n----------------------\n";

      }

}

int main()

{

int i, day=1;

int yearly[12][2] = {{1,31},{2,28},{3,31},{4,30},{5,31},{6,30},{7,31},{8,31},{9,30},{10,31},{11,30},{12,31}};

string months[] = {"January",

"February",

"March",

"April",

"May",

"June",

"July",

"August",

"September",

"October",

"November",

"December"};

for(i=0; i<12; i++)

{

//Monthly printing

cout << "\n Month: " << months[i] << "\n";

month(yearly[i][1], day);

if(day==7)

{

day = 1;

}

else

{

day = day + 1;

}

cout << "\n";

}

return 0;

}

//end