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Travka [436]
3 years ago
9

. A Carnot heat pump is to be used to heat a house and maintain it at 22 °C in winter. When the outdoor temperature remains at 3

°C, the house is estimated to lose heat at a rate of 76,000 kJ/h. If the heat pump consumes 9 kW of power, how long does it need to run in a single day to keep the temperature constant inside the house?
Engineering
1 answer:
max2010maxim [7]3 years ago
8 0

Answer:

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

Explanation:

The net heat daily loss of the house is:

Q_{losses} = \left(76000\,\frac{kJ}{h}\right)\cdot (24\,h)

Q_{losses} = 1.824\times 10^{6}\,kJ

In order to keep the house warm, given heat must be equal to heat losses:

Q_{H} = Q_{losses}

Besides, the Coefficient of Performance for a Carnot heat pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

Where,

T_{L} - Temperature of the cold reservoir (Outdoors), measured in Kelvin.

T_{H} - Temperature of the hot reservoir (House), measured in Kelvin.

Given that T_{L} = 276.15\,K and T_{H} = 295.15\,K, the Coefficient of Performance is:

COP_{HP} = \frac{295.15\,K}{295.15\,K-276.15\,K}

COP_{HP} = 15.534

For a real heat machine, the Coefficient of Performance is determined by the following expression:

COP_{HP} = \frac{Q_{H}}{W}

Where:

Q_{H} - Heat received by the house, measured in kilojoules.

W - Work consumed by the Carnot heat pump, measured in kilojoules.

The daily work consumed is now cleared in the previous expression:

W = \frac{Q_{H}}{COP_{HP}}

W = \frac{1.824\times 10^{6}\,kJ}{15.534}

W = 117419.853\,kJ

The working time is calculated by dividing this result by input power. That is:

\Delta t = \frac{W}{\dot W}

\Delta t = \left(\frac{117419.853\,kJ}{9\,kW} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)

\Delta t = 3.624\,h

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

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gulaghasi [49]

Answer:

<em>181 °C</em>

<em></em>

Explanation:

Initial pressure P_{1} = 100 kPa

Initial temperature T_{1} = 30 °C = 30 + 273 K = 303 K

Final pressure P_{2} = 1200 kPa

Final temperature T_{2} = ?

n = 1.2

For a polytropic process, we use the relationship

(T_{2}/T_{1} ) = (P_{2}/P_{1})^γ

where γ = (n-1)/n

γ = (1.2-1)/1.2 = 0.1667

substituting into the equation, we have

(T_{2}/303) = (1200/100)^0.1667

T_{2}/303 = 12^0.1667

T_{2}/303 = 1.513

T_{2} = 300 x 1.513 = 453.9 K

==> 453.9 - 273 = 180.9 ≅ <em>181 °C</em>

5 0
3 years ago
A binary liquid system exhibits LLE at 25°C. Determine from each of the following sets of miscibility data estimates for paramet
Phoenix [80]

Answer:

(a) - A12 = A21 = 2.747

(b) - A12 = 2.148; A21 = 2.781

(c)-  A12 = 2.781; A21 = 2.148

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(a) - x1(a) = 0.1 |  x2(a) = 0.9 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

  • x1(a)*γ1(a) = x1(b)γ1(b)

        x2(a)*γ2(a) = x2(b)γ2(b)

  • A12 = A21 = 2.747

(b) -  x1(a) = 0.2 |  x2(a) = 0.8 | x1(b) = 0.9 | x2(b) = 0.1

LLE equations:

  • x1(a)*γ1(a) = x1(b)γ1(b)

        x2(a)*γ2(a) = x2(b)γ2(b)

  • A12 = 2.148; A21 = 2.781

(c) -  x1(a) = 0.1 |  x2(a) = 0.9 | x1(b) = 0.8 | x2(b) = 0.2

LLE equations:

  • x1(a)*γ1(a) = x1(b)γ1(b)

        x2(a)*γ2(a) = x2(b)γ2(b)

  • A12 = 2.781; A21 = 2.148
7 0
3 years ago
21. How long can food that requires time-temperature control be left in the danger zone?
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Answer: A maximum of 1 hour

Explanation:

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8 0
3 years ago
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CS3733: Homework/Practice 05 Suppose we would like to write a program called monitor which allows two other programs to communic
valina [46]

Answer:

#include<stdio.h>

#include<stdlib.h>

#include<unistd.h>

#include<sys/types.h>

#include<string.h>

#include<pthread.h>

//#include<sys/wait.h>

int main(int argc, char** argv)

{

int fd1[2];

int fd2[2];

int fd3[2];

int fd4[2];

char message[] = "abcd";

char input_str[100];

pid_t p,q;

if (pipe(fd1)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd2)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd3)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd4)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

p = fork();

if (p < 0)

{

 fprintf(stderr, "fork Failed" );

return 1;

}

// child process-1

else if (p == 0)

{

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 printf("\n\tEnter meaaage:"):

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 int k = strlen(concat_str);

 int i;

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 {

  concat_str[k++] = fixed_str[i];

 }

 concat_str[k] = '\0';//string ends with '\0'

 // Close both writting ends

 close(fd1[1]);

 wait(NULL);

//.......................................................................

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 read(fd2[0], concat_str, 100);

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 {

 printf("\n\tmessage not send");

 }

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 {

  printf("\n\tmessage send to prog_2(child_2).");

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 close(fd2[0]);//close reading end of pipe 2

 exit(0);

}

else

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 close(fd1[1]);//Close writting end of first pipe

 char concat_str[100];

 read(fd1[0], concal_str, strlen(concat_str)+1);

 close(fd1[0]);

 close(fd2[0]);//Close writing end of second pipe

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 {

  //if not then

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  return 0;

 }

 else

 {

  //if yes then

  write(fd2[1], "valid",sizeof(concat_str));

  close(fd2[1]);

  q=fork();//create chile process 2

  if(q>0)

  {

   close(fd3[0]);/*close read head offd3[] */

   write(fd3[1],concat_str,sizeof(concat_str);//write message by monitor(main process) using fd3[1]

   close(fd3[1]);

   wait(NULL);//wait till child_process_2 send ACK

   //...........................................................

   close(fd4[1]);

   read(fd4[0],concat_str,100);

   close(fd4[0]);

   if(sctcmp(concat_str,"ack")==0)

   {

    printf("Messageof child process_1 is received by child process_2");

   }

   else

   {

    printf("Messageof child process_1 is not received by child process_2");

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   {

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   else

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    close(fd3[1]);//Close writing end of first pipe

    char concat_str[100];

    read(fd3[0], concal_str, strlen(concat_str)+1);

    close(fd3[0]);

    close(fd4[0]);//Close writing end of second pipe

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  }

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8 0
3 years ago
A geothermal heat pump absorbs 15 KJ/s of heat from the Earth 15 m below a house. This heat pump uses a 7.45 kJ/s compressor.
Anna007 [38]

Answer:

COP of the heat pump is 3.013

OP of the cycle is  1.124

Explanation:

W = Q₂ - Q₁

Given

a)

Q₂ = Q₁ + W

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COP = Q₂ / W = 22.45 / 7.45 = 3.013

b)

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

Q₁ = 15.825 - 7.45 = 8.375

∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124

4 0
3 years ago
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