Questions: Transient Filling and Emptying Processes

Question 1

An evacuated rigid tank is filled with air (γ = 1.4) from a large supply reservoir at 300 K, with no heat transfer to the surroundings. What is the final equilibrium temperature inside the tank?

• A. 300 K — the gas equilibrates to the supply temperature
• B. 214 K — the gas cools as it expands into the vacuum
• C. 420 K — the final temperature is γ × T_supply = 1.4 × 300 K
• D. 600 K — the tank doubles in temperature because of the compression work

Answer: (C)

For adiabatic filling of an evacuated rigid tank, the unsteady energy balance integrates to u_final = h_in (specific internal energy equals inlet enthalpy). For an ideal gas, u = c_v T and h = c_p T, so T_final = (c_p/c_v) × T_supply = γ × T_supply = 1.4 × 300 = 420 K. The extra energy comes from flow work — the supply line pushes the gas into the tank, doing P·v work on each parcel entering. This counterintuitive temperature rise has no external heat source — it is purely a consequence of the enthalpy accounting in the unsteady first law.

Question 2

Why does the gas in a tank filled from a supply line at constant temperature end up hotter than the supply, even with no external heat transfer?

• A. Friction between incoming gas molecules and the tank walls converts kinetic energy to heat
• B. The compression of already-present gas by incoming gas raises the temperature through the ideal gas law
• C. Each parcel of gas entering the tank carries not just internal energy but also flow work (P·v), which is part of its enthalpy. This extra energy is deposited in the tank as the supply line pushes gas through the inlet
• D. The tank walls reflect thermal radiation back into the gas, heating it above supply temperature

Answer: (C)

The key is that flowing streams carry enthalpy h = u + Pv, not just internal energy u. The supply line does Pv work pushing each unit mass of gas into the tank, and this work becomes internal energy of the gas in the tank. The energy balance d(mu)/dt = ṁ_in × h_in integrates to u_final = h_in = u_in + P_in v_in — the extra Pv term is the flow work. This is not friction or compression of existing gas; it is the thermodynamic accounting of what it costs to push mass through a boundary.

Question 3

The energy delivered to a rigid tank by an entering gas stream is proportional to the stream's enthalpy per unit mass, not its internal energy per unit mass.

• T. True
• F. False

Answer: True

This is the central insight of the transient filling problem. The unsteady open-system first law reads dU/dt = Q̇ − Ẇ + ṁ_in h_in − ṁ_out h_out. Inlet streams contribute their enthalpy h = u + Pv, not just u, because the flow work Pv is done by the upstream fluid pushing the parcel through the boundary. In a rigid tank (no shaft work, often adiabatic), the accumulating internal energy equals the enthalpy of all mass that entered — hence u_final = h_in for an initially empty tank.

Question 4

An evacuated rigid tank filled adiabatically from a constant-temperature supply will reach exactly the supply temperature at equilibrium.

• T. True
• F. False

Answer: False

The final temperature is γ × T_supply (1.4× for air), not T_supply. The supply temperature is the temperature of the incoming gas at the source, but once the gas enters the tank, the flow work from the supply line deposits additional energy. The final specific internal energy equals the inlet enthalpy: u_final = h_in = c_p T_supply, so T_final = (c_p/c_v) T_supply = γ T_supply. Reaching exactly the supply temperature would only occur if the filling were reversible and isothermal — which requires heat rejection during filling.

Question 5

Explain why u_final = h_in (not u_in) when an evacuated rigid tank is filled adiabatically, and what physical phenomenon causes this result.

Model answer: The unsteady energy balance for a rigid adiabatic control volume is d(mu)/dt = ṁ_in × h_in. Integrating from empty to full: m_final × u_final = m_final × h_in, so u_final = h_in. The reason h_in rather than u_in appears is that the supply line must do flow work (Pv work per unit mass) to push each parcel of gas through the inlet against the pressure inside the tank. This flow work is part of the enthalpy h = u + Pv. It is deposited as internal energy in the tank, raising the temperature above the supply temperature.

The distinction between u and h is the essential thermodynamic point: internal energy is what a mass stores in place, but enthalpy is what a flowing stream carries — because moving mass through a pressure boundary always involves flow work. Every open-system energy analysis (steady or unsteady) accounts for enthalpy at flow boundaries, never internal energy alone. Students who apply closed-system intuition (where u matters) to open systems make systematic errors in filling/emptying and steady-flow problems alike.