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First law for cyclic process; CV mass and energy equations

Short answer

For a control mass cycle, ∮δQ = ∮δW. Mass conservation for CV: dm_cv/dt = Σṁ_i - Σṁ_e. Energy: dE_cv/dt = Q̇ - Ẇ + Σṁ_i(h_i+V_i²/2+gz_i) - Σṁ_e(h_e+V_e²/2+gz_e). Steady state: Σṁ_i=Σṁ_e; Q̇ - Ẇ = Σṁ_e(...) - Σṁ_i(...).

State first law of thermodynamics for a control mass undergoing cyclic process. Write the mass and energy equations for a control volume and reduce them for steady state process.

First law for a control mass undergoing a cycle

For any process of a closed system, Q − W = ΔE. In a cycle the system returns to its initial state, so ΔE = 0:

∮ δQ = ∮ δW (Q_net = W_net)

The first law states: when a closed system undergoes a cycle, the net heat transferred to it equals the net work done by it.

Mass and energy equations for a control volume

Mass: dm_cv/dt = Σṁ_i − Σṁ_e, with ṁ = ρAV = AV/v.

Energy: dE_cv/dt = Q̇ − Ẇ + Σṁ_i (h_i + V_i²/2 + g z_i) − Σṁ_e (h_e + V_e²/2 + g z_e)

Steady-state reduction

Properties do not change with time, so dm_cv/dt = 0 and dE_cv/dt = 0:

Σṁ_i = Σṁ_e

Q̇ − Ẇ = Σṁ_e (h_e + V_e²/2 + g z_e) − Σṁ_i (h_i + V_i²/2 + g z_i)

For a single inlet and exit (ṁ₁ = ṁ₂ = ṁ), per unit mass:
q − w = (h₂ − h₁) + (V₂² − V₁²)/2 + g (z₂ − z₁)

✓ Answer checked by Claude Opus 5.5 (high effort)

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