L5 slide 11: Filling a partly evacuated tank, find T2 and m2 (ENGR5003)

L5 slide 11: Filling a partly evacuated tank, find T2 and m2From lectureCurrent specFirst10 marks14 min

ENGR5003 Lecture 5, slides 11-15 (Flow of air into an evacuated rigid tank - "you should know how to derive this for the exam")

A rigid tank of volume $V$ initially holds air at a low pressure $p_1$ and at the ambient temperature $T_1 = T_0$ . The valve is opened and atmospheric air (at $p_0$ , $T_0$ , stationary far from the tank) flows in through a throttle valve until the tank pressure reaches a sub-atmospheric value $p_2$ . The tank is adiabatic, no work is done, and kinetic and potential energy are negligible. Air is an ideal gas with constant $c_p$ , $c_v$ .

The lecturer asks: what are the mass $m_2$ and temperature $T_2$ of the gas in the tank at the end?

Formulas you may need

Unsteady-flow energy equation: $\delta Q - \delta W + \left(h + \tfrac12 C^2 + gZ\right)_{in}\delta m_{in} - \left(h + \tfrac12 C^2 + gZ\right)_{out}\delta m_{out} = d(mu)_{CV}$ (on the formula sheet)
Ideal gas: $pV = mR_gT$ , $u = c_vT$ , $h = c_pT$ , $\gamma = c_p/c_v$ (on the formula sheet)
Result: $\dfrac{T_2}{T_0} = \dfrac{\gamma}{1 + (\gamma - 1)p_1/p_2}$ (NOT given: "you should know how to derive this for the exam")

(a)
Starting from the unsteady-flow energy equation, show that $\dfrac{T_2}{T_0} = \gamma - \dfrac{m_1}{m_2}(\gamma - 1)$ .
[4]
(b)
Hence show that $\dfrac{T_2}{T_0} = \dfrac{\gamma}{1 + (\gamma - 1)\dfrac{p_1}{p_2}}$ , and say how $m_2$ then follows.
[3]
(c)
What does the result give for an initially fully evacuated tank ( $p_1 = 0$ )? Why is this limit not reached in practice?
[1]
(d)
[Added] Take $V = 0.5\ \mathrm{m^3}$ , $T_0 = 290$ K, $p_1 = 0.2$ bar, $p_2 = 0.8$ bar, $\gamma = 1.4$ and $R_g = 0.287\ \mathrm{kJ/(kg\,K)}$ . Find $T_2$ , $m_2$ and the mass of air that entered.
$T_{2}$ =
K
$m_{2}$ =
kg
$m_{\mathrm{in}}$ =
kg
[2]