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A system executes a power cycle while receiving 1000 kJ by heat transfer at a temperature of 500 K and discharging 700 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Determine the cycle efficiency. Use the Clausius Inequality to determine , in kJ/K. Determine if this cycle is internally reversible, irreversible, or impossible.

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Answer:

[tex]\eta_{th} = 30\,\%[/tex], [tex]\eta_{th,max} = 40\,\%[/tex], [tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex], The cycle is irreversible.

Explanation:

The real cycle efficiency is:

[tex]\eta_{th} = \frac{1000\,kJ-700\,kJ}{1000\,kJ} \times 100\,\%[/tex]

[tex]\eta_{th} = 30\,\%[/tex]

The theoretical cycle efficiency is:

[tex]\eta_{th,max} = \frac{500\,K-300\,K}{500\,K} \times 100\,\%[/tex]

[tex]\eta_{th,max} = 40\,\%[/tex]

The reversible and real versions of the power cycle are described by the Clausius Inequalty:

Reversible Unit

[tex]\frac{1000\,kJ - 600kJ}{300\,K}= 0[/tex]

Real Unit

[tex]\Delta S = \frac{1000\,kJ-600\,kJ}{300\,K} -\frac{1000\,kJ-700\,kJ}{300\,K}[/tex]

[tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex]

The cycle is irreversible.

The cycle efficiency using clausius inequality is;

σ_cycle = 0.333 kJ/kg and is internally irreversible

  • For the cycle, we know that efficiency is;

η = 1 - Q_c/Q_h

Thus;

Q_c = (1 - η)Q_h

  • Now, the cycle efficiency is derived from the integral;

σ_cycle = -∫(dQ/dt)ₐ

Thus; σ_cycle = -[(Q_h/T_h) - (Q_c/T_c)]

We are given;

Q_h = 1000 kJ

T_h = 500 k

T_c = 300 k

Q_c = 700 kJ

  • Thus;

σ_cycle = -[(1000/500) - (700/300)]

σ_cycle = -(2 - 2.333)

σ_cycle = 0.333 kJ/kg

Since σ_cycle > 0, then the cycle is internally irreversible

Read more about cycle efficiency at; https://brainly.com/question/16014998

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