--- title: "Mass and energy equations for CV; gas cylinder in kitchen" subject: "Fundamental of Thermodynamics & Heat Transfer" chapter: 4 key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:mass-energy-cv-gas-cylinder-kitchen" source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/mass-energy-cv-gas-cylinder-kitchen site: BCT Engineering --- # Mass and energy equations for CV; gas cylinder in kitchen Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics - Asked in: 2068 Magh, 6 marks - Video: [Energy Conservation Equation & FIRST LAW of Thermodynamics in 12 Minutes!](https://www.youtube.com/watch?v=PKk3N32XKmU) — 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/mass-energy-cv-gas-cylinder-kitchen - Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4 ## Question Write down general mass conservation and energy conservation equations for a control volume system. Also derive mass and energy conservation equations for a gas cylinder being used in a kitchen. ## Short answer General mass: dm_cv/dt = Σṁ_i − Σṁ_e; energy: dE_cv/dt = Q̇−Ẇ + Σṁ_i(h_i+V²/2+gz)_i − Σṁ_e(...). For kitchen cylinder: no inlet, no work, mass m₂−m₁=−m_e, energy m₂u₂−m₁u₁=Q −∫ h dm_e. Adiabatic quick discharge (ideal gas): T₂/T₁ = (P₂/P₁)^((k−1)/k). Slow isothermal: Q = m_e(h−u) = m_e R T. ## Answer 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a gas cylinder discharging (no inlet), the mass balance is: A. dm_cv/dt = +ṁ_e B. dm_cv/dt = 0 C. dm_cv/dt = ṁ_i + ṁ_e D. dm_cv/dt = −ṁ_e, so m2 = m1 − m_e Answer: D. Mass only leaves, so the mass in the cylinder decreases by the mass used. (Key point: Mass equation for discharge) 2. In the energy equation for the discharging cylinder, the enthalpy carried out h_e is: A. the enthalpy of the gas inside the cylinder, h B. the enthalpy of the burner flame C. a constant equal to the initial h D. equal to u Answer: A. The gas leaving is the gas in the cylinder (uniform state), so h_e = h = u + Pv. (Key point: Exit enthalpy) 3. For adiabatic quick discharge of an ideal gas, the temperature and pressure are related by: A. T2/T1 = P2/P1 B. T2 = T1 C. T2/T1 = (P2/P1)^((k−1)/k) D. T2/T1 = (P1/P2)^((k−1)/k) Answer: C. Using m du = RT dm and PV = mRT gives dT/T = ((k−1)/k) dP/P, so temperature falls as pressure falls. (Key point: Adiabatic quick discharge) 4. For slow use at constant temperature in thermal equilibrium with the kitchen, the heat taken from surroundings is: A. Q = 0 B. Q = m_e (h − u) = m_e RT C. Q = m_e cv T D. Q = m_e u Answer: B. u is constant for an ideal gas at constant T, so Q = m_e h − m_e u = m_e RT, the flow work of the leaving gas. (Key point: Slow discharge heat) ## Related in Fundamental of Thermodynamics & Heat Transfer - [Mass and energy equations for CV; gas filling process](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Amass-energy-cv-gas-filling-station) — Fundamental of Thermodynamics & Heat Transfer, chapter 4 - [General mass and energy conservation for CV; unsteady](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Amass-energy-conservation-cv-unsteady) — Fundamental of Thermodynamics & Heat Transfer, chapter 4 - [Mass and energy equations for CV; nozzle, condenser](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Amass-energy-cv-nozzle-condenser) — Fundamental of Thermodynamics & Heat Transfer, chapter 4 - [Steady/unsteady system; mass conservation for CV](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Asteady-unsteady-mass-conservation-cv) — Fundamental of Thermodynamics & Heat Transfer, chapter 4, asked 2× - [First law for cyclic process; CV mass and energy equations](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Afirst-law-cyclic-mass-energy-cv) — Fundamental of Thermodynamics & Heat Transfer, chapter 4 - [General energy equation for control volume](https://www.bctengineer.com/question.md?key=semester_2%2Ffundamental_of_thermodynamics_and_heat_transfer%3Ageneral-energy-equation-control-volume) — Fundamental of Thermodynamics & Heat Transfer, chapter 4 ## Related in other subjects - [Two-mass system equations, X2/F, F-I and F-V analogies](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ade-tf-fi-fv-m1-m2) — Control Systems, chapter 2 - [General block diagram of a control system](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Ageneral-bd-control-system) — Control Systems, chapter 1 - [Derive resultant force, variable mass system](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Avariable-mass-resultant-force) — Applied Mechanics (Dynamics), chapter 3, asked 2×