Engineering College

A completely reversible heat pump produces heat at a rate of 300 kW to warm a house maintained at 24°C. The exterior air, which is at 7°C, serves as the source. Calculate the rate of entropy change of the two reservoirs and determine if this heat pump satisfies the second law according to the increase of entropy principle

Answers

Answer 1

Answer:

Entropy generation rate of the two reservoirs is approximately zero ([tex]\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}[/tex]) and system satisfies the Second Law of Thermodynamics.

Explanation:

Reversible heat pumps can be modelled by Inverse Carnot's Cycle, whose key indicator is the cooling Coefficient of Performance, which is the ratio of heat supplied to hot reservoir to input work to keep the system working. That is:

[tex]COP_{H} = \frac{\dot Q_{H}}{\dot W}[/tex]

The following simplification can be used in the case of reversible heat pumps:

[tex]COP_{H,rev} = \frac{T_{H}}{T_{H} - T_{L}}[/tex]

Where temperature must written at absolute scale, that is, Kelvin scale for SI Units:

[tex]COP_{H, rev} = \frac{297.15\,K}{297.15\,K-280.15\,K}[/tex]

[tex]COP_{H, rev} = 17.479[/tex]

Then, input power needed for the heat pump is:

[tex]\dot W = \frac{\dot Q}{COP_{H,rev}}[/tex]

[tex]\dot W = \frac{300\,kW}{17.749}[/tex]

[tex]\dot W = 16.902\,kW[/tex]

By the First Law of Thermodynamics, heat pump works at steady state and likewise, the heat released from cold reservoir is now computed:

[tex]-\dot Q_{H} + \dot W + \dot Q_{L} = 0[/tex]

[tex]\dot Q_{L} = \dot Q_{H} - \dot W[/tex]

[tex]\dot Q_{L} = 300\,kW - 16.902\,kW[/tex]

[tex]\dot Q_{L} = 283.098\,kW[/tex]

According to the Second Law of Thermodynamics, a reversible heat pump should have an entropy generation rate equal to zero. The Second-Law model for the system is:

[tex]\dot S_{in} - \dot S_{out} - \dot S_{gen} = 0[/tex]

[tex]\dot S_{gen} = \dot S_{in} - \dot S_{out}[/tex]

[tex]\dot S_{gen} = \frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}}[/tex]

[tex]\dot S_{gen} = \frac{283.098\,kW}{280.15\,K} - \frac{300\,kW}{297.15\,K}[/tex]

[tex]\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}[/tex]

Albeit entropy generation rate is positive, it is also really insignificant and therefore means that such heat pump satisfies the Second Law of Thermodynamics. Furthermore, [tex]\dot S_{in} = \dot S_{out}[/tex].

Answer 2

The rate of entropy change of the two reservoirs is; 9.318 * 10⁻⁴ kW/K and it satisfies second law of thermodynamics

What is the rate of entropy?

The formula for Coefficient of Performance is;

COP = T_H/(T_H - T_L)

Where;

T_H = 24°C = 297.15 K

T_L = 7°C = 280.15 K

Thus;

COP = 297.15/(297.15 - 280.15)

COP = 17.479

Input power is;

Input power needed for the heat pump is:

W' = Q'/COP

We are given; Q' = 300 kW

Thus;

W' = 300/17.479

W' = 16.902 kW

From first law of thermodynamics, we can deduce that;

Q_L = Q_H - W'

Thus;

Q_L = 300 - 16.902

Q_L = 283.098 kW

From second law of thermodynamics, the rate of entropy generation is;

S_gen = (Q_L/T_L) - (Q_H/T_H)

S_gen = (283.098/280.15) - (300/297.15)

S_gen = 9.318 * 10⁻⁴ kW/K

Read more about Entropy at; https://brainly.com/question/15022152

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Explanation:

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