--- title: "Steady/unsteady system; mass conservation for CV" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 4 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steady-unsteady-mass-conservation-cv" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steady-unsteady-mass-conservation-cv site: BCT Engineering --- # Steady/unsteady system; mass conservation for CV Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics - Asked in: 2072 Magh, 6 marks; 2067 Mangsir, 6 marks (asked 2 times) - Video: [FLOW WORK & Energy Conservation in Open Systems in 11 Minutes!](https://www.youtube.com/watch?v=HleudqDDd9w) — Less Boring Lectures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steady-unsteady-mass-conservation-cv - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4 ## Question Define steady and unsteady state system. Derive the expression of conservation of mass and conservation of energy for control volume having unsteady process. ## Short answer Steady state: properties and flow rates are constant in time; unsteady: they change. Conservation of mass: dm_cv/dt = Σṁ_i − Σṁ_e. Energy: dE_cv/dt = Q̇ − Ẇ_shaft + Σṁ_i(h+ke+pe)_i − Σṁ_e(h+ke+pe)_e; for uniform-state unsteady process, integrate over time. ## Answer #### Steady and unsteady state - **Steady state:** the properties (P, T, v, u, h) and flow rates at any point of the system or control volume do not change with time. The mass and energy in the control volume are constant: dm_cv/dt = 0, dE_cv/dt = 0. Example: a turbine running at constant load. - **Unsteady (transient) state:** properties or flow rates at a point change with time, so the mass or energy inside the control volume changes. Example: filling or emptying a gas cylinder. #### Conservation of mass for a control volume Mass entering − mass leaving = rate of change of mass inside: **dm_cv/dt = Σṁ_i − Σṁ_e** with ṁ = ρ A V = A V/v. For an unsteady process over a time interval: **m₂ − m₁ = Σm_i − Σm_e** #### Conservation of energy for a control volume Energy enters with heat, with mass (specific energy plus flow work P v), and leaves as work and with mass: dE_cv/dt = Q̇ − Ẇ_shaft + Σṁ_i (h_i + V_i²/2 + g z_i) − Σṁ_e (h_e + V_e²/2 + g z_e) where E_cv = m u + m V²/2 + m g z of the contents. Here h = u + Pv includes the flow work. For the unsteady process (uniform state, uniform flow) integrated over time: **Q − W = (m₂e₂ − m₁e₁) + Σm_e (h_e + V_e²/2 + g z_e) − Σm_i (h_i + V_i²/2 + g z_i)** Neglecting KE and PE: Q − W = m₂u₂ − m₁u₁ + Σm_e h_e − Σm_i h_i. For steady state the right-hand storage terms vanish: Σṁ_i = Σṁ_e and Q̇ − Ẇ = Σṁ_e(h + V²/2 + gz)_e − Σṁ_i(h + V²/2 + gz)_i. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In steady state for a control volume: A. Properties and flow rates do not change with time, so dm_cv/dt = 0 and dE_cv/dt = 0 B. Only temperature is constant C. Mass in CV changes at a constant rate D. Inlet flow is zero Answer: A. Example: a turbine at constant load. (Key point: Steady state) 2. Which is an unsteady process? A. Turbine at constant load B. Filling or emptying a gas cylinder C. Nozzle with constant flow D. Steady heat exchanger Answer: B. The contents' mass and energy change with time. (Key point: Unsteady example) 3. The rate form of the CV mass balance is: A. dm_cv/dt = Σṁ_e - Σṁ_i B. dm_cv/dt = 0 always C. dm_cv/dt = Σṁ_i - Σṁ_e D. dm_cv/dt = Σṁ_i + Σṁ_e Answer: C. Accumulation equals inflow minus outflow. (Key point: Mass conservation) 4. Mass flow carries energy h + V²/2 + gz, where h appears because: A. h includes heat transfer B. h replaces kinetic energy C. u alone is carried by the fluid D. h = u + Pv includes flow work to push the fluid in or out Answer: D. Flow work Pv is added to internal energy u. 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