Engineering, 28.07.2020 15:01 alanflores40

The data listed below are claimed for power cycles operating between hot and cold reservoirs at 1000 K and 400 K, respectively. For each case determine whether such a cycle is in keeping with the first and second laws of thermodynamics. a) Qh=300 kJ, W(cycle)=160kJ, Qc=140 kJ
b) Qh=300 kJ, W(cycle)=180kJ, Qc=120 kJ
c) Qh=300 kJ, W(cycle)=170kJ, Qc=140 kJ
d) Qh=300 kJ, W(cycle)=200kJ, Qc=100 kJ

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Answer from: Elazjah

a) This cycle observes the First and Second Laws of Thermodynamics, b) This cycle observes the First and Second Laws of Thermodynamics, c) This cycle does not observe the First Law of Thermodynamics, d) This cycle does not observe the Second Law of Thermodynamics.

Explanation:

The Carnot cycle offers a reliable criterion to determine the maximum theoretical efficiency ( $\eta_{th,max}$ ) for a power cycle in term of the temperatures of hot and cold reservoirs and expressed in percentage:

$\eta_{th, max} = \left(1-\frac{T_{L}}{T_{H}} \right)\times 100\,\%$

Where:

$T_{L}$ , $T_{H}$ - Temperatures of cold and hot reservoirs, measured in kelvins.

If $T_{L} = 400\,K$ and $T_{H} = 1000\,K$ , the maximum theoretical thermal efficiency is:

$\eta_{th, max} = \left(1-\frac{400\,K}{1000\,K} \right)\times 100\,\%$

$\eta_{th,max} = 60\,\%$

In addition, the real efficiency of the heat engine is described by the following formula:

$\eta_{th} = \left(1-\frac{Q_{L}}{Q_{H}} \right)\times 100\,\%$

Where:

$Q_{H}$ - Heat absorbed by the heat engine from hot reservoir, measured in kilojoules.

$Q_{L}$ - Heat released by the heat engine to the cold reservoir, measured in kilojoules.

The following conditions must be observed by all heat engines:

First Law of Thermodynamics

$W = Q_{H}-Q_{L}$

Second Law of Thermodynamics

$\eta_{th} \leq \eta_{th,max}$

Now, each cycle is checked:

a) $Q_{H} = 300\,kJ$ , $Q_{L} = 140\,kJ$ and $W = 160\,kJ$

First Law of Thermodynamics

$W = 300\,kJ-140\,kJ$

$W = 160\,kJ$

This cycle observes the First Law of Thermodynamics.

Second Law of Thermodynamics

$\eta_{th} = \left(1-\frac{140\,kJ}{300\,kJ} \right)\times 100\,\%$

$\eta_{th} = 53.333\,\%$

This cycle observes the Second Law of Thermodynamics.

b) $Q_{H} = 300\,kJ$ , $Q_{L} = 120\,kJ$ and $W = 180\,kJ$

First Law of Thermodynamics

$W = 300\,kJ-120\,kJ$

$W = 180\,kJ$

This cycle observes the First Law of Thermodynamics.

Second Law of Thermodynamics

$\eta_{th} = \left(1-\frac{120\,kJ}{300\,kJ} \right)\times 100\,\%$

$\eta_{th} = 60\,\%$

This cycle observes the Second Law of Thermodynamics.

c) $Q_{H} = 300\,kJ$ , $Q_{L} = 140\,kJ$ and $W = 170\,kJ$

First Law of Thermodynamics

$W = 300\,kJ-140\,kJ$

$W = 160\,kJ$

This cycle does not observe the First Law of Thermodynamics.

Second Law of Thermodynamics

$\eta_{th} = \left(1-\frac{140\,kJ}{300\,kJ} \right)\times 100\,\%$

$\eta_{th} = 53.333\,\%$

This cycle observes the Second Law of Thermodynamics.

d) $Q_{H} = 300\,kJ$ , $Q_{L} = 200\,kJ$ and $W = 100\,kJ$

First Law of Thermodynamics

$W = 300\,kJ-100\,kJ$

$W = 200\,kJ$

This cycle observes the First Law of Thermodynamics.

Second Law of Thermodynamics

$\eta_{th} = \left(1-\frac{100\,kJ}{300\,kJ} \right)\times 100\,\%$

$\eta_{th} = 66.667\,\%$

This cycle does not observe the Second Law of Thermodynamics.

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Answer from: Quest

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