contestada

an air conditioning system operating on reversed carnot cycle is required to remove heat from the house at a rate of 32kj/s to maintain its temperature constant at 20. if the temperature of the outside is 35 the power required to operate this air conditioning system is
(a) 0.58 kW
(b) 3.20 kW
(c) 1.56 kW
(d) 2.26 kW
(e) 1.64 kW

Respuesta :

xero099

Answer:

(e) 1.64 kW

Explanation:

The Coefficient of Performance of the Reverse Carnot's Cycle is:

[tex]COP = \frac{T_{L}}{T_{H}-T_{L}}[/tex]

[tex]COP = \frac{293.15\,K}{308.15\,K-293.15\,K}[/tex]

[tex]COP = 19.543[/tex]

Lastly, the power required to operate the air conditioning system is:

[tex]\dot W = \frac{\dot Q_{L}}{COP}[/tex]

[tex]\dot W = \frac{32\,kW}{19.543}[/tex]

[tex]\dot W = 1.637\,kW[/tex]

Hence, the answer is E.

Lanuel

The power that is required to operate this air-conditioning system is: E. 1.64 kW.

Given the following data:

  • Rate of heat loss = 32 kJ/s = 32000 Joules.
  • House temperature = 20°C to K = 293.15 Kelvin.
  • Outside temperature = 35°C to K = 308.15 Kelvin.

The coefficient of performance (COP).

In Science, the coefficient of performance (COP) is a mathematical expression that is used to show the relationship between the power output of an air-conditioning system and the power input of its compressor.

Mathematically, the coefficient of performance (COP) is given by this formula:

[tex]COP=\frac{T_L}{T_H-T_L} \\\\COP=\frac{293.15}{308.15-293.15}\\\\COP=\frac{293.15}{15}[/tex]

COP = 19.543

For the power input:

[tex]W=\frac{Q_L}{COP} \\\\W=\frac{32000}{19.543}[/tex]

W = 1.64 kW.

Read more on power input here: https://brainly.com/question/15878720

Otras preguntas