A 1.6 m³ tank is filled with air at a pressure of 5 bar and a temperature of 100°C. The air is then let off to the atmosphere through a valve. Assuming no heat transfer, determine the work obtainable by utilising the kinetic energy of the discharge air to run a frictionless turbine.

Take : Atmospheric pressure = 1 bar ;

$c_p$ for air = 1 kJ/kg K ;

$c_v$ for air = 0.711 kJ/kg K.

Question

A 1.6 m³ tank is filled with air at a pressure of 5 bar and a temperature of 100°C. The air is then let off to the atmosphere through a valve. Assuming no heat transfer, determine the work obtainable by utilising the kinetic energy of the discharge air to run a frictionless turbine.

Take : Atmospheric pressure = 1 bar ;

$c_p$ for air = 1 kJ/kg K ;

$c_v$ for air = 0.711 kJ/kg K.

Take : Atmospheric pressure = 1 bar ;

[latex]c_p[/latex] for air = 1 kJ/kg K ;

[latex]c_v[/latex] for air = 0.711 kJ/kg K.

Accepted Answer

Initial volume of air, [latex]V_1=1.6[/latex] m³

Initial pressure of air, [latex]p_1=5 ext { bar }=5 imes 10^5[/latex] N/m²

Initial temperature of air, [latex]T_1=100+273=373[/latex] K

Final pressure of air, [latex]p_2=1 ext { bar }=1 imes 10^5[/latex] N/m²

Now, initial quantity of air in the tank before discharge,

[latex]m_1=\frac{p_1 V_1}{R T_1}=\frac{5 imes 10^5 imes 16}{\left(0.287 imes 10^3 ight) imes 373}=7.47[/latex] kg.

Assuming that system undergoes a reversible adiabatic expansion

[latex]\frac{T_2}{T_1}=\left(\frac{p_2}{p_1} ight)^{\frac{\gamma-1}{\gamma}}[/latex]

where [latex]T_2[/latex] is the final temperature of air in the tank.

∴ [latex]\frac{T_2}{373}=\left(\frac{1}{5} ight)^{\frac{1.4-1}{1.4}}=0.631[/latex]

[latex]T_2=373 imes 0.631=235.4[/latex] K (i.e., finally in the line)

The final quantity of air remaining in the tank is

[latex]m_2=\frac{p_2 V_2}{R T_2}=\frac{1 imes 10^5 imes 1.6}{\left(0.287 imes 10^3 ight) imes 235.4}=2.368[/latex] kg.

With Q = 0, kinetic energy is found from,

[latex]\left(m_1-m_2 ight)\left(h^{\prime}+\frac{C^{\prime 2}}{2} ight)=m_1 u_1-m_2 u_2[/latex]

or [latex]\left(m_1-m_2 ight) h^{\prime}+\left(m_1-m_2 ight) \frac{C^{\prime 2}}{2}=m_1 u_1-m_2 u_2[/latex]

∴ Kinetic energy,

[latex]\left(m_1-m_2 ight) \frac{C^{\prime 2}}{2}=\left(m_1 u_1-m_2 u_2 ight)-\left(m_1-m_2 ight) h^{\prime}[/latex]

= [latex]m_1 c_v T_1-m_2 c_v T_2-\left(m_1-m_2 ight) c_p T_2[/latex]

= 7.47 × 0.771 × 373 – 2.368 × 0.711 × 235.4 – (7.47 – 2.368) × 1 × 235.4

= 2148.24 – 396.33 – 1201 = 550.9 kJ.

Question 4.65: A 1.6 m³ tank is filled with air at a pressure of 5 bar and ...

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