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Mass and energy equations for CV; nozzle, condenser

Short answer

Mass conservation: 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). Adiabatic nozzle: h₁ + V₁²/2 = h₂ + V₂²/2. Condenser: ṁ_s(h₁ - h₂) = ṁ_w(h₄ - h₃).

Write down general expressions for mass and energy conservation for a control volume. Reduce these equations for an adiabatic nozzle and condenser.

General expressions 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)

For steady state (dm_cv/dt = 0, dE_cv/dt = 0): Σṁ_i = Σṁ_e, and
Q̇ − Ẇ = Σṁ_e (h + V²/2 + gz)_e − Σṁ_i (h + V²/2 + gz)_i

Adiabatic nozzle

One inlet (1), one exit (2); Q̇ = 0, Ẇ = 0, ΔPE = 0.

Mass: ṁ₁ = ṁ₂ ⇒ A₁V₁/v₁ = A₂V₂/v₂

Energy: h₁ + V₁²/2 = h₂ + V₂²/2

V₂ = √[ V₁² + 2 (h₁ − h₂) ]

Condenser

A steam stream (1 → 2) and a cooling-water stream (3 → 4) exchange heat inside; with the whole condenser as CV, Q̇ = 0 (insulated), Ẇ = 0, ΔKE = ΔPE = 0.

Mass (each stream): ṁ_s1 = ṁ_s2 = ṁ_s, ṁ_w3 = ṁ_w4 = ṁ_w

Energy: ṁ_s h₁ + ṁ_w h₃ = ṁ_s h₂ + ṁ_w h₄

ṁ_s (h₁ − h₂) = ṁ_w (h₄ − h₃)

The heat given up by condensing steam equals the heat gained by the cooling water. Cooling water flow needed: ṁ_w = ṁ_s (h₁ − h₂)/(c_w (T₄ − T₃)).

✓ Answer checked by Claude Opus 5.5 (high effort)

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