Mass and energy equations for CV; gas cylinder in kitchen

A control volume (CV) is a fixed region of space across whose boundary mass, heat and work may cross.

General mass conservation

dm_cv/dt = Σ ṁ_i − Σ ṁ_e

General energy conservation

dE_cv/dt = Q̇ − Ẇ + Σ ṁ_i (h_i + V_i²/2 + g z_i) − Σ ṁ_e (h_e + V_e²/2 + g z_e)

with h = u + Pv and E_cv = m(u + V²/2 + gz).

Gas cylinder discharging in a kitchen

Take the cylinder (LPG or other gas, volume V, rigid) as the CV. Gas leaves through the regulator and valve to the burner. There is no inlet (ṁ_i = 0) and no work (W = 0). Assume uniform state inside the cylinder and negligible kinetic and potential energy.

Mass equation.

dm_cv/dt = −ṁ_e, so m₂ − m₁ = −m_e

where m₁ and m₂ are the initial and final masses in the cylinder (m₂ < m₁) and m_e is the mass used.

Energy equation.

d(m u)/dt = Q̇ − ṁ_e h_e

The exit enthalpy equals the enthalpy of the gas inside the cylinder, h_e = h. Integrating over the time of use:

m₂u₂ − m₁u₁ = Q − ∫ h dm_e

Case 1: adiabatic, quick discharge (Q = 0). Write d(mu) = h dm (since dm = −dm_e):

m du + u dm = h dm, so m du = (h − u) dm = Pv dm = RT dm

For an ideal gas, m cv dT = RT dm, so dT/T = (R/cv) dm/m. From PV = mRT: dm/m = dP/P − dT/T. Hence

dT/T (1 + R/cv) = (R/cv) dP/P, so dT/T = ((k−1)/k) dP/P

T₂/T₁ = (P₂/P₁)^((k−1)/k)

The gas left in the cylinder cools as pressure falls (this is why the cylinder feels cold when used fast).

Case 2: slow use, cylinder in thermal equilibrium with the kitchen. The heat Q from the surroundings keeps T constant, so for an ideal gas u is constant and m₂u − m₁u = −m_e u. The energy equation gives Q = m_e (h − u) = m_e RT (the flow work of the gas leaving).

The mass left is m₂ = P₂V/(RT₂), and the gas used is m_e = m₁ − m₂.