---
title: "First law for cyclic process; CV mass and energy equations"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:first-law-cyclic-mass-energy-cv"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/first-law-cyclic-mass-energy-cv
site: BCT Engineering
---

# First law for cyclic process; CV mass and energy equations

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

- Asked in: 2072 Asoj, 6 marks
- Video: [First Law of Thermodynamics 003 | Heat transfer | Cyclic process](https://www.youtube.com/watch?v=gRyTAPwAIc0) — Think Solve Pleasure
- 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/first-law-cyclic-mass-energy-cv
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4

## Question

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.

## 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(...).

## Answer

#### 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)

## Check yourself

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

1. In a cycle, the net heat to the system equals:
   A. Net work done by the system
   B. Zero
   C. Net work done on the system, with opposite sign
   D. Change in enthalpy
   Answer: A. ∮δQ = ∮δW. (Key point: Cyclic first law)

2. The CV energy rate equation includes mass terms with:
   A. u + V²/2 + gz
   B. h + V²/2 + gz
   C. h only
   D. P v
   Answer: B. Enthalpy includes flow work. (Key point: CV energy)

3. At steady state:
   A. dm_cv/dt = Σṁ_i
   B. Q̇ = 0
   C. dm_cv/dt = 0 and dE_cv/dt = 0, so Σṁ_i = Σṁ_e
   D. Ẇ = 0
   Answer: C. Storage terms vanish. (Key point: Steady reduction)

4. For one inlet and exit:
   A. q + w = Δh
   B. q - w = Δu
   C. q = w
   D. q - w = Δh + ΔKE + ΔPE
   Answer: D. Divide the energy equation by ṁ. (Key point: Per kg form)

## Related in Fundamental of Thermodynamics & Heat Transfer

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