A rigid insulated tank is initially evacuated. It is connected through a valve to a supply line that carries air at a constant pressure and temperature of 250 kPa and 400 K respectively. Now the valve is opened and air is allowed to flow into the tank until the pressure inside the tank reaches to 250 kPa at which point the valve is closed. Assume that the air behaves as a perfect gas with constant properties (C_p = 1.005 kJ/kg.K, C_v = 0.718 kJ/kg.K, R = 0.287 kJ/kg.K). Final temperature of the air inside the tank is______ K (round off to one decimal place).

The correct answer is 560.

Step-by-step Explanation:

This problem represents a classic transient (unsteady) flow process where a rigid, insulated tank is filled (charged) from a supply line. We can analyze this system using the conservation of mass and the first law of thermodynamics.

1. Mass Balance:
Let the initial mass in the tank be _m1 and the final mass be _m2. Let the mass entering the tank from the supply line be _min.
Since the tank is initially evacuated:
$m_1=0$
Since no mass leaves the tank, the conservation of mass simplifies to:
$m_{\text{in}}=m_2$

2. Energy Balance:
For an unsteady flow system, the first law of thermodynamics is written as:
$Q-W+m_{\text{in}}{h}_{\text{in}}-m_{\text{out}}{h}_{\text{out}}=E_2-E_1$
Where:
- _{Q = 0} because the tank is insulated (adiabatic).
- _{W = 0} because the tank is rigid (no boundary work or shaft work).
- _{mout = 0} as no mass leaves the control volume.
- _{E1 = m1u1 = 0} because the tank is initially evacuated.
- _{E2 = m2u2} represents the final internal energy of the air in the tank.
Substituting these parameters into the energy balance equation yields:
$m_{\text{in}}{h}_{\text{in}}=m_2{u}_2$

Since _{min = m2}, the mass terms cancel out, leaving:
$h_{\text{in}}=u_2$

3. Relating Enthalpy and Internal Energy:
For a perfect gas with constant specific heats, the specific enthalpy _h and specific internal energy _u are related to temperature by:
$h_{\text{in}}=C_p{T}_{\text{in}}$
and
$u_2=C_v{T}_2$
Where _Tin is the supply line temperature (400 K) and _T2 is the final temperature inside the tank. Substituting these relations back into the energy equation gives:
$C_p{T}_{\text{in}}=C_v{T}_2$
Solving for the final temperature _T2:
$T_2=\frac{C_p}{C_v}{T}_{\text{in}}$

4. Final Calculation:
We first calculate the ratio of specific heats, _γ:
$\gamma =\frac{C_p}{C_v}=\frac{1.005}{0.718}\approx 1.4$
Now, substitute the value of _γ and _Tin into the expression for _T2:
$T_2=1.4\times 400\text{K}$
$T_2=560\text{K}$

Thus, the final temperature of the air inside the tank is 560 K.