What does the word “robot” mean? A.Clone B. Athlete C. Servant D. Actor

Determine the nature of the following cycle (reversible, irreversible, or impossible): a refrigeration cycle draws heat from a c

vlabodo [156]

Answer:

Impossible.

Explanation:

The ideal Coefficient of Performance is:

$COP_{i} = \frac{250\,K}{300\,K-250\,K}$

$COP_{i} = 5$

The real Coefficient of Performance is:

$COP_{r} = \frac{950\,kJ-70\,kJ}{70\,kJ}$

$COP_{r} = 12.571$

Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.

6 0

3 years ago

(3) Calculate the heat flux through a sheet of brass 7.5 mm (0.30 in.) thick if the temperatures at the two faces are 150°Cand 5

bezimeni [28]

Answer:

a.) 1.453MW/m2, b.) 2,477,933.33 BTU/hr c.) 22,733.33 BTU/hr d.) 1,238,966.67 BTU/hr

Explanation:

Heat flux is the rate at which thermal (heat) energy is transferred per unit surface area. It is measured in W/m2

Heat transfer(loss or gain) is unit of energy per unit time. It is measured in W or BTU/hr

1W = 3.41 BTU/hr

Given parameters:

thickness, t = 7.5mm = 7.5/1000 = 0.0075m

Temperatures 150 C = 150 + 273 = 423 K

50 C = 50 + 273 = 323 K

Temperature difference, T = 423 - 323 = 100 K

We are assuming steady heat flow;

a.) Heat flux, Q" = kT/t

K= thermal conductivity of the material

The thermal conductivity of brass, k = 109.0 W/m.K

Heat flux, $Q" = \frac{109 * 100}{0.0075} = 1,453,333.33 W/m^{2} \\ Heat flux, Q" = 1.453MW/m^{2} \\$

b.) Area of sheet, A = 0.5m2

Heat loss, Q = kAT/t

Heat loss, $Q = \frac{109*0.5*100}{0.0075} = 726,666.667W$

Heat loss, Q = 726,666.667 * 3.41 = 2,477,933.33 BTU/hr

c.) Material is now given as soda lime glass.

Thermal conductivity of soda lime glass, k is approximately 1W/m.K

Heat loss, $Q=\frac{1*0.5*100}{0.0075} = 6,666.67W$

Heat loss, Q = 6,666.67 * 3.41 = 22,733.33 BTU/hr

d.) Thickness, t is given as 15mm = 15/1000 = 0.015m

Heat loss, $Q=\frac{109*0.5*100}{0.015} =363,333.33W$

Heat loss, Q = 363,333.33 * 3.41 = 1,238,966.67 BTU/hr

5 0

2 years ago

Freeee Poinntssss 100!!!!!!<br> Hi how you doin?

Andreas93 [3]

Thank u very much , im doin good wby :)

6 0

2 years ago

Read 2 more answers

5.11: Population Write a program that will predict the size of a population of organisms. The program should ask the user for th

fomenos

Answer:

// using c++ language

#include "stdafx.h";

#include <iostream>

#include<cmath>

using namespace std;

//start

int main()

{

//Declaration of variables in the program

double start_organisms;

double daily_increase;

int days;

double updated_organisms;

//The user enters the number of organisms as desired

cout << "Enter the starting number of organisms: ";

cin >> start_organisms;

//Validating input data

while (start_organisms < 2)

{

cout << "The starting number of organisms must be at least 2.\n";

cout << "Enter the starting number of organisms: ";

cin >> start_organisms;

}

//The user enters daily input, here's where we apply the 5.2% given in question

cout << "Enter the daily population increase: ";

cin>> daily_increase;

//Validating the increase

while (daily_increase < 0)

{

cout << "The average daily population increase must be a positive value.\n ";

cout << "Enter the daily population increase: ";

cin >> daily_increase;

}

//The user enters number of days

cout << "Enter the number of days: ";

cin >> days;

//Validating the number of days

while (days<1)

{

cout << "The number of days must be at least 1.\n";

cout << "Enter the number of days: ";

cin >> days;

}

//Final calculation and display of results based on formulas

for (int i = 0; i < days; i++)

{

updated_organisms = start_organisms + (daily_increase*start_organisms);

cout << "On day " << i + 1 << " the population size was " << round(updated_organisms)<<"."<<"\n";

start_organisms = updated_organisms;

}

system("pause");

return 0;

//end

}

4 0

2 years ago

5. A pump operating at steady state receives 1.2 kg/s of liquid water at 50o C, 1.5 MPa. The pressure of the water at the pump e

Andrew [12]

Answer:

the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW

the isentropic pump efficiency is 78%

Explanation:

Given that;

m = 1.2 kg/sec

T = 50 degree Celsius { Vf = 0.001012 m^3/kg}

P1 = 1.5 Mpa

P2 = 15 Mpa

W-actual = 21 kw

W reversible = m*Vf (p2 - p1)

= 1.2 * 0.001012 * ( 15*10^3 - 1.5*10^3)

= 1.2 * 0.001012 * 13500

= 16.39 kW

the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW

Isentropic Pump efficiency = W-reversible / W-actual

= 16.39 / 21 = 0.78

= 78%

the isentropic pump efficiency is 78%

6 0

3 years ago