Question

Argon is compressed in a polytropic process with = 1:3 (i.e., PV = costant) from 120 kPa and 10 C until it reaches a temperature of 120 C in a piston{cylinder device. Model the system as an ideal gas and determine the speci c work (work per unit mass) done by the gas and the heat per unit mass transferred during this the process.

Answer 1

Answer:

The work done per unit kg of mass is -76.08 kJ/kg while the heat transferred per unit kg of mass is -42.01 kJ/kg.

Explanation:

For Argon the molar mass is given as

$M=40$

Now the gas constant for Argon is given as

$R=\frac{\bar{R}}{M}\\R=\frac{8.314}{40}\\R=0.208 kJ/K kg$

Initial Temperature is T_1=10 C=10+273=283 K

Final Temperature is T_2=120 C=120+273=393 K

Initial Pressure is P_1=120 kPa

The initial volume is given as

$v_1=\frac{RT_1}{P_1}\\v_1=\frac{0.208\times 283}{120}\\v_1=0.491 m^3/kg$

For the polytropic process with γ=1.3 is given as

             $\frac{P_2}{P_1}=(\frac{T_2}{T_1})^{\frac{\gamma}{\gamma-1}}\\\frac{P_2}{120}=(\frac{393}{283})^{\frac{1.3}{1.3-1}}\\\frac{P_2}{120}=(\frac{393}{283})^{\frac{1.3}{0.3}}\\\frac{P_2}{120}=(\frac{393}{283})^{\frac{1.3}{0.3}}\\{P_2}=(\frac{393}{283})^{\frac{1.3}{0.3} }\times 120\\P_2=497.9 kPA$

Now the volume at the second stage is given as

               $v_2=\frac{RT_2}{P_2}\\v_2=\frac{0.208\times 393}{497.9}\\v_2=0.164 m^3/kg$

Now the work done per kg mass is given as

$w=\frac{p_1v_1-p_2v_2}{\gamma-1}\\w=\frac{120\times 0.491-497.9 \times 0.164}{1.3-1}\\w=-76.08 kJ/kg$

So the work done per unit kg of mass is -76.08 kJ/kg.

The heat per unit mass is given as

$Q_{poly}=\frac{\gamma_{adi}-\gamma_{poly}}{\gamma_{adi}-1} \times w_{poly}$

As the Argon gas is monotonic which gives γ_adiabatic=1.67

$Q_{poly}=\frac{\gamma_{adi}-\gamma_{poly}}{\gamma_{adi}-1} \times w_{poly}\\Q_{poly}=\frac{1.67-1.3}{1.67-1} \times (-76.08 kJ/kg)\\Q_{poly}=-42.01 kJ/kg$

The heat transferred per unit kg of mass is -42.01 kJ/kg