Question

Carnot cycle works with isentropic compression ratio of 5bar and heat capacity ratio of 2. The volume of air at the beginning of the isothermal expansion is$0.3{{m}^{3}}$. If the temperature and pressure is limited to 550K and 21bar, determine minimum temperature in the cycle.

Answer 1

Hint: You could first recall the expression for efficiency in terms of isentropic compression ratio and the heat capacity ratio of a Carnot cycle. And then you could substitute the given values in it. Now you could recall the other expression for efficiency in terms of temperatures of the reservoir and then substitute the maximum possible limit of temperature given. After that, you could equate the above two expressions and thus find the minimum temperature in the cycle.

Formula used:
Expressions for efficiency,
$\eta =1-\dfrac{1}{{{r}^{\gamma -1}}}$
$\eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$

Complete step by step answer:
In the question, we are discussing a Carnot cycle. We are given the isentropic ratio of the Carnot cycle as 5bar and it has the heat capacity ratio as 2. The volume of air at the beginning of the isothermal expansion is given as$0.3{{m}^{3}}$. Also, it is said that the temperature and pressure is limited to 550K and 21bar.
Now let us recall the expression for efficiency of a Carnot’s engine which given by,
$\eta =1-\dfrac{1}{{{r}^{\gamma -1}}}$
Where, r is the isentropic compression ratio and $\gamma $ is the heat capacity ratio. Substituting the given values we get,
$\eta =1-\dfrac{1}{{{\left( 5 \right)}^{2-1}}}$
$\Rightarrow \eta =1-\dfrac{1}{5}$
$\therefore \eta =\dfrac{4}{5}$ ………………………… (1)
Now we have another expression for efficiency of Carnot cycle which is given by,
$\eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$
Where, ${{T}_{1}}$ is the high temperature at which the reservoir operates and ${{T}_{2}}$ is the low temperature at which the reservoir operates.
Since the maximum temperature at which the reservoir operates is limited to 550K, we can take that as${{T}_{1}}$, so,
${{T}_{1}}=550K$
And let the minimum temperature of the cycle that we are asked to find be${{T}_{2}}$, then, the efficiency could be given as,
$\eta =1-\dfrac{{{T}_{2}}}{550K}$ ………………………………… (2)
Now, we could equate both (1) and (2) as both expressions give the efficiency of the cycle.
$\dfrac{550-{{T}_{2}}}{550}=\dfrac{4}{5}$
$\Rightarrow 550-{{T}_{2}}=440$
$\therefore {{T}_{2}}=110K$
Therefore, we found the minimum temperature in the cycle to be 110K.

Note:
From the expression for efficiency, we know that, higher the temperature of the engine can get the more efficient the engine will be. However, the automobiles are operated at low frequencies. This is not because we can’t achieve such high temperatures, but because, no known engine materials can withstand such high temperatures.