---
title: "General mass and energy conservation for CV; unsteady"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:mass-energy-conservation-cv-unsteady"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/mass-energy-conservation-cv-unsteady
site: BCT Engineering
---

# General mass and energy conservation for CV; unsteady

Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics

- Asked in: 2071 Bhadra, 6 marks
- 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/mass-energy-conservation-cv-unsteady
- 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. Also reduce them for a control volume operating under unsteady state condition.

## 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).

## Answer

#### 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Uniform-state, uniform-flow assumes:
   A. Inlet/exit conditions constant over time and CV state uniform at start and end
   B. Steady state
   C. No mass flow
   D. Adiabatic only
   Answer: A. Allows integration over the interval. (Key point: Uniform state)

2. Integrated mass balance:
   A. m2 + m1 = Σm_i
   B. m2 - m1 = Σm_i - Σm_e
   C. m2 - m1 = 0
   D. m2 = Σm_e
   Answer: B. Accumulation equals net inflow. (Key point: Mass)

3. Energy equation neglecting KE and PE:
   A. Q - W = Δh
   B. Q - W = m2h2 - m1h1
   C. Q - W = m2u2 - m1u1 + Σm_e h_e - Σm_i h_i
   D. Q = W
   Answer: C. e ≈ u inside CV; h with flows. (Key point: Energy neglect KE PE)

4. Charging an adiabatic tank with W = 0:
   A. m2u2 - m1u1 = m_i u_i
   B. m2h2 = m1h1
   C. m2u2 = Q
   D. m2u2 - m1u1 = m_i h_i
   Answer: D. Flow work raises the energy via enthalpy. (Key point: Charging)

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