A 125 m3 rigid tank initially contains air at 101 kPa and 25

A 1.25 m3 rigid tank initially contains air at 101 kPa and 25 °C (State 1). The tank is connected to an air supply line at 325 kPa and 51 °C. Air from the supply line is added to the tank until the mass of air inside the tank doubles (State 2). Assuming this process occurs adiabatically and air has Cp = 3.5*R , calculate the following:

Solution

Given: P₁ = 101 kPa, T₁ = 25^oC, V= 1.25 m³, P₂ = 325 kPa, T₁ = 51^oC, M₂ =2*M₁

R_air = 287 J/kgK

M₁ = P₁*V/(R_air*T₁) = 1.4762 kg

M₂ = P₂*V/(R_air*T₂) = 4.36885 kg

M₃ =M₁ +M₂ = 5.84505 kg

Now,

M₃C_vT₃ = M₁C_vT₁ +M₂C_vT₂

Therefore, T₃ = 44.43^oC

Now, P₃ = M₃*R_air*T₃/V

P₃ = 425.98 kPa