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General mass and energy conservation for CV; unsteady

Short answer

General mass: dm_cv/dt = Σṁ_i - Σṁ_e. Energy: dE_cv/dt = Q̇ - Ẇ + Σṁ_i(h_i+V_i²/2+gz_i) - Σṁ_e(...). Integrated for unsteady uniform-state: m₂ - m₁ = Σm_i - Σm_e; Q - W = m₂u₂ - m₁u₁ + Σm_e h_e - Σm_i h_i (neglecting KE, PE).

Write down general mass conservation and energy conservation equations for a control volume. Also reduce them for a control volume operating under unsteady state condition.

General conservation equations for a control volume

Mass: rate of change of mass in CV = inflow − outflow

dm_cv/dt = Σṁ_i − Σṁ_e, ṁ = ρ A V = A V/v

Energy: the rate of change of energy in the CV equals net heat, minus work, plus energy carried in and out by mass (including flow work, hence enthalpy):

dE_cv/dt = Q̇_cv − Ẇ_cv + Σṁ_i (h_i + V_i²/2 + g z_i) − Σṁ_e (h_e + V_e²/2 + g z_e)

with E_cv = m_cv (u + V²/2 + g z).

Unsteady (transient) process: uniform-state, uniform-flow

Inlet and exit conditions are constant over the time interval, and the CV state is uniform at the start (1) and end (2). Integrating over time:

Mass: m₂ − m₁ = Σm_i − Σm_e

Energy: Q − W = (m₂e₂ − m₁e₁) + Σm_e (h + V²/2 + gz)_e − Σm_i (h + V²/2 + gz)_i

If KE and PE are neglected (e ≈ u):

Q − W = m₂u₂ − m₁u₁ + Σm_e h_e − Σm_i h_i

Special cases:

  • Charging a tank (no exit, adiabatic, W = 0): m₂u₂ − m₁u₁ = m_i h_i.
  • Discharging: m₂u₂ − m₁u₁ = Q − m_e h_e.

✓ Answer checked by Claude Opus 5.5 (high effort)

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