Question

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

Answer 1

Answer:

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.

Answer 2

Based on the first parameters, the power cycle obeys both the First and Second Law of Thermodynamics.

Given the following data:

• Temperature of hot reservoir = 1000 K.

• Temperature of cold reservoir = 400 K.

How to verify which law a power cycle obeys.

In order to verify a power cycle obeys the first and second laws of thermodynamics, we would use the Carnot cycle.

Mathematically, the maximum theoretical efficiency for a power cycle in terms of the temperature is given by this formula:

$\eta_{th, max}=(1-\frac{T_c}{T_h} ) \times 100$

Where:

• $T_c$ is the temperature of cold reservoir.

• $T_h$ is the temperature of hot reservoir.

Substituting the given parameters into the formula, we have;

$\eta_{th, max}=(1-\frac{400}{1000} ) \times 100\\\\\eta_{th, max}=60\%$

Similarly, the real efficiency of a power cycle, we have:

$\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100$

The conditions for a power cycle.

According to the First Law of Thermodynamics, the following condition must be met:

$W_{cycle}=Q_h-Q_c$

According to the Second Law of Thermodynamics, the following condition must be met:

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

Next, we would determine whether or not each obeys the first and second laws of thermodynamics:

When $Q_h=300\; kJ, \;W_{(cycle)}=160kJ, \;and\;Q_c=140\; kJ$

For the First Law:

$W_{cycle}=Q_h-Q_c\\\\160=300-140\\\\160=160$

Therefore, the power cycle obeys the First Law of Thermodynamics.

For the Second Law:

$\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{140}{300} ) \times 100\\\\\eta_{th}=53.33\%\\\\\eta_{th }\leq \eta_{th, max } =53.33\%\leq 60\%$

Therefore, the power cycle obeys the Second Law of Thermodynamics.

Using the second parameters.

When $Q_h=300\; kJ, \;W_{(cycle)}=180kJ, \;and\;Q_c=120\; kJ$

For the First Law:

$W_{cycle}=Q_h-Q_c\\\\180=300-120\\\\180=180$

Therefore, the power cycle obeys the First Law of Thermodynamics.

For the Second Law:

$\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{120}{300} ) \times 100\\\\\eta_{th}=60\%\\\\\eta_{th }\leq \eta_{th, max } =60\%\leq 60\%$

Therefore, the power cycle obeys the Second Law of Thermodynamics.

Using the third parameters.

When $Q_h=300\; kJ, \;W_{(cycle)}=170kJ, \;and\;Q_c=140\; kJ$

For the First Law:

$W_{cycle}=Q_h-Q_c\\\\170=300-140\\\\170\neq 160$

Therefore, the power cycle doesn't obey the First Law of Thermodynamics.

For the Second Law:

$\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{140}{300} ) \times 100\\\\\eta_{th}=53.33\%\\\\\eta_{th }\leq \eta_{th, max } =53.33\%\leq 60\%$

Therefore, the power cycle obeys the Second Law of Thermodynamics.

Using the fourth parameters.

When $Q_h=300\; kJ, \;W_{(cycle)}=200kJ, \;and\;Q_c=100\; kJ$

For the First Law:

$W_{cycle}=Q_h-Q_c\\\\200=300-100\\\\200=200$

Therefore, the power cycle obeys the First Law of Thermodynamics.

For the Second Law:

$\eta_{th}=(1-\frac{Q_c}{Q_h} ) \times 100\\\\\eta_{th}=(1-\frac{100}{300} ) \times 100\\\\\eta_{th}=66.67\%\\\\\eta_{th }\leq \eta_{th, max } \neq 63.33\%\geq 60\%$

Therefore, the power cycle doesn't obey the Second Law of Thermodynamics.

Read more on thermodynamics here: brainly.com/question/11628413