The aim of this article is to find the final temperature of the gas after it has gone through a polytropic process of compression from lower to higher pressure.
The basic concept of this article is the Polytropic process and Ideal Gas Law.
The polytropic process is a thermodynamic process involving the expansion or compression of a gas resulting in heat transfer. It is expressed as follows:
\[PV^n\ =\ C\]
Where:
$P\ =$ The pressure of the gas
$V\ =$ The volume of the gas
$n\ =$ Polytropic Index
$C\ =$ Constant
Expert Answer
Given that:
Polytropic Index $n\ =\ 1.2$
Initial Pressure $P_1\ =\ 120\ kPa$
Initial Temperature $T_1\ =\ 30°C$
Final Pressure $P_2\ =\ 1200\ kPa$
Final Temperature $T_2\ =\ ?$
First, we will convert the given temperature from Celsius to Kelvin.
\[K\ =\ ^{\circ}C+273\ =\ 30+273\ =\ 303K\]
Hence:
Initial Temperature $T_1\ =\ 303K$
We know that as per the Polytropic process:
\[PV^n\ =\ C\]
For a polytropic process between two states:
\[P_1{V_1}^n\ =\ P_2{V_2}^n\]
By rearranging the equation, we get:
\[\frac{P_2}{P_1}\ =\ \frac{{V_1}^n}{{V_2}^n}\ =\ \left(\frac{V_1}{V_2}\right)^n\]
As per Idea Gas Law:
\[PV\ =\ nRT\]
For two states of gas:
\[P_1V_1\ =\ nRT_{1\ }\]
\[V_1\ =\ \frac{nRT_{1\ }}{P_1}\]
And:
\[P_2V_2\ =\ nRT_2\]
\[V_2\ =\ \frac{nRT_2}{P_2}\]
Substituting the values from Idea Gas law into Polytropic process relation:
\[\frac{P_2}{P_1}\ =\ \left(\frac{\dfrac{nRT_{1\ }}{P_1}}{\dfrac{nRT_2}{P_2}}\right)^n\]
Cancelling $nR$ from numerator and denominator, we get:
\[\frac{P_2}{P_1}\ =\ \left(\frac{\dfrac{T_{1\ }}{P_1}}{\dfrac{T_2}{P_2}}\right)^n\]
\[\frac{P_2}{P_1}\ =\ \left(\frac{T_{1\ }}{P_1}\times\frac{P_2}{T_2}\right)^n\]
\[\frac{P_2}{P_1}\ =\ \left(\frac{P_{2\ }}{P_1}\times\frac{T_{1\ }}{T_2}\right)^n\]
\[\frac{P_2}{P_1}\ =\ \left(\frac{P_{2\ }}{P_1}\right)^n\times\left(\frac{T_{1\ }}{T_2}\right)^n\]
\[\left(\frac{T_{1\ }}{T_2}\right)^n\ =\ \left(\frac{P_{2\ }}{P_1}\right)^{1-n}\]
\[\frac{T_{1\ }}{T_2}\ =\ \left(\frac{P_{2\ }}{P_1}\right)^\dfrac{1-n}{n}\ or\ \ \frac{T_{2\ }}{T_1}\ =\ \left(\frac{P_{2\ }}{P_1}\right)^\dfrac{n-1}{n}\]
Now substituting the given values of pressures and temperatures of argon gas in two states, we get:
\[\frac{T_{2\ }}{303K}\ =\ \left(\frac{1200}{120}\right)^\dfrac{1.2-1}{1.2}\]
\[T_{2\ }\ =\ {303K\left(\frac{1200\ kPa}{120\ kPa}\right)}^\dfrac{1.2-1}{1.2}\]
\[T_{2\ }\ =\ {303K\times10}^{0.16667}\]
\[T_{2\ }\ =\ 444.74K\]
Converting the Final Temperature $T_{2\ }$ from Kelvin to Celsius, we get:
\[K\ =\ ^{\circ}C+273\]
\[444.74\ =\ ^{\circ}C+273\]
\[T_{2\ }\ =\ 444.74-273\ =171.74\ ^{\circ}C\]
Numerical Result
The Final Temperature $T_{2\ }$ of the argon gas after it has gone through a polytropic process of compression from $120$ $kPa$ at $30^{\circ}C$ to $1200$ $kPa$ in a piston-cylinder device:
\[T_{2\ }=171.74\ ^{\circ}C\]
Example
Determine the final temperature of hydrogen gas after it has gone through a polytropic process of compression with $n=1.5$ from $50$ $kPa$ and $80^{\circ}C$ to $1500$ $kPa$ in a screw compressor.
Solution
Given that:
Polytropic Index $n\ =\ 1.5$
Initial Pressure $P_1\ =\ 50\ kPa$
Initial Temperature $T_1\ =\ 80°C$
Final Pressure $P_2\ =\ 1500\ kPa$
Final Temperature $T_2\ =\ ?$
First, we will convert the given temperature from Celsius to Kelvin.
\[K\ =\ ^{\circ}C+273\ =\ 80+273\ =\ 353K\]
Hence:
Initial Temperature $T_1\ =\ 303K$
As per polytropic process expressions in term of pressure and temperature:
\[\frac{T_{2\ }}{T_1}\ =\ \left(\frac{P_{2\ }}{P_1}\right)^\dfrac{n-1}{n}\]
\[T_{2\ }\ =\ T_1\left(\frac{P_{2\ }}{P_1}\right)^\dfrac{n-1}{n}\]
Substituting the given values:
\[T_{2\ }\ =\ 353K\left(\frac{1500\ kPa}{50\ kPa}\right)^\dfrac{1.5-1}{1.5}\]
\[T_{2\ }\ =\ 353K\left(\frac{1500\ kPa}{50\ kPa}\right)^\dfrac{1.5-1}{1.5}\]
\[T_{2\ }\ =\ 1096.85K\]
Converting the Final Temperature $T_{2\ }$ from Kelvin to Celsius:
\[T_{2\ }\ =\ 1096.85-273\ =\ 823.85^{\circ}C \]