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Mama L [17]
3 years ago
9

A well-established way of power generation involves the utilization of geothermal energy-the energy of hot water that exists nat

urally underground- as the heat source.If a supply of hot water at 140oC is discovered at a location where the environmental temperature is 20oC, determine the maximum thermal efficiency a geothermal power plant built at that location can have.
Engineering
1 answer:
jeka943 years ago
3 0

Answer:

the maximum thermal efficiency is 29%

Explanation:

the maximum efficiency for a thermal engine that works between a cold source and a hot source is the one of a Carnot engine. Its efficiency is given by

Maximum efficiency= 1 - T2/T1

where

T2= absolute temperature of the cold sink (environment)= 20°C + 273 = 293

T2= absolute temperature of the hot source (hot water supply) = 140°C + 273 = 413

therefore

Maximum efficiency= 1 - T2/T1 = 1 - 293/413 = 0,29 =29%

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Write a C++ program to display yearly calendar. You need to use the array defined below in your program. // the first number is
ddd [48]

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

3 0
3 years ago
For the unity negative feedback system G(s) = K(s+6)/ (s + 1)(s + 2)(s + 5) It's known that the system is operating with a domin
Ad libitum [116K]

Answer:The awnser is 5

Explanation:Just divide all of it

4 0
3 years ago
Find E[x] when x is sum of two fair dice?
Ksenya-84 [330]

Answer:

When two fair dice are rolled, 6×6=36 observations are obtained.

P(X=2)=P(1,1)=

36

1

​

P(X=3)=P(1,2)+P(2,1)=

36

2

​

=

18

1

​

P(X=4)=P(1,3)+P(2,2)+P(3,1)=

36

3

​

=

12

1

​

P(X=5)=P(1,4)+P(2,3)+P(3,2)+P(4,1)=

36

4

​

=

9

1

​

P(X=6)=P(1,5)+P(2,4)+P(3,3)+P(4,2)+P(5,1)=

36

5

​

P(X=7)=P(1,6)+P(2,5)+P(3,4)+P(4,3)+P(5,2)+P(6,1)=

36

6

​

=

6

1

​

P(X=8)=P(2,6)+P(3,5)+P(4,4)+P(5,3)+P(6,2)=

36

5

​

P(X=9)=P(3,6)+P(4,5)+P(5,4)+P(6,3)=

36

4

​

=

9

1

​

P(X=10)=P(4,6)+P(5,5)+P(6,4)=

36

3

​

=

12

1

​

P(X=11)=P(5,6)+P(6,5)=

36

2

​

=

18

1

​

P(X=12)=P(6,6)=

36

1

​

Therefore, the required probability distribution is as follows.

Then, E(X)=∑X

i

​

⋅P(X

i

​

)

=2×

36

1

​

+3×

18

1

​

+4×

12

1

​

+5×

9

1

​

+6×

36

5

​

+7×

6

1

​

+8×

36

5

​

+9×

9

1

​

+10×

12

1

​

+11×

18

1

​

+12×

36

1

​

=

18

1

​

+

6

1

​

+

3

1

​

+

9

5

​

+

6

5

​

+

6

7

​

+

9

10

​

+1+

6

5

​

+

18

11

​

+

3

1

​

=7

E(X

2

)=∑X

i

2

​

⋅P(X

i

​

)

=4×

36

1

​

+9×

18

1

​

+16×

12

1

​

+25×

9

1

​

+36×

36

5

​

+49×

6

1

​

+64×

36

5

​

+81×

9

1

​

+100×

12

1

​

+121×

18

1

​

+144×

36

1

​

=

9

1

​

+

2

1

​

+

3

4

​

+

9

25

​

+5+

6

49

​

+

9

80

​

+9+

3

25

​

+

18

121

​

+4

=

18

987

​

=

6

329

​

=54.833

Then, Var(X)=E(X

2

)−[E(X)]

2

=54.833−(7)

2

=54.833−49

=5.833

∴ Standard deviation =

Var(X)

​

=

5.833

​

=2.415

4 0
2 years ago
Why do wires hang loosely during summer?
Stella [2.4K]

Answer:

due to the expansion process and they contract during winter due to the contraction process. Explanation: Electric cables are the solids which exhibit the property of contraction and expansion.

6 0
2 years ago
A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 2
Furkat [3]

Answer:

P_{1} = 403,708\,kPa\,(58.553\,psi)

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}

The initial pressure is:

P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}

The speed at outlet is:

v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}

v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }

v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )

The initial pressure is:

P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}

P_{1} = 403,708\,kPa\,(58.553\,psi)

7 0
3 years ago
Read 2 more answers
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