---
title: "Mass and energy equations for non-flow; cyclic process"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:mass-energy-nonflow-cyclic-closed"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/mass-energy-nonflow-cyclic-closed
site: BCT Engineering
---

# Mass and energy equations for non-flow; cyclic process

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

- Asked in: 2069 Poush, 6 marks
- Video: [Non flow and flow process | Control volume | Control mass | Thermodynamics | Mechanical engineering](https://www.youtube.com/watch?v=1j4YBwRokzc) — Mechanical Studies
- 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-nonflow-cyclic-closed
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4

## Question

State the equations of conservation of mass and conservation of energy for non flow process and deduce it for cyclic process undergoing in a closed system.

## Short answer

For a closed system, mass is constant and Q - W = ΔU + ΔKE + ΔPE. In a cyclic process, ΔE =0, hence ∮ δQ = ∮ δW, i.e. net heat added equals net work done.

## Answer

#### Non-flow process (closed system, no mass crosses the boundary)
**Conservation of mass:** the mass of the system stays constant:  
m = constant, i.e. dm = 0 (m₁ = m₂)

**Conservation of energy (first law for a closed system):** the net energy supplied as heat and work equals the change in the energy of the system:

**Q₁₂ − W₁₂ = E₂ − E₁ = ΔU + ΔKE + ΔPE**

Differential form: δQ − δW = dE; for a stationary system, δQ − δW = dU. Sign convention: heat into the system +ve, work by the system +ve.

#### Reduction for a cyclic process in a closed system
In a cycle the system returns to its initial state: m, E and all other properties regain initial values, so  
m₂ = m₁ and E₂ − E₁ = 0 (∮ dE = 0).

Substituting in the first law:

∮ δQ − ∮ δW = 0

**∮ δQ = ∮ δW**   i.e. **Q_net = W_net**

So in a cycle, the net heat transfer equals the net work transfer. For a power cycle: W_net = Q_H − Q_L; for a refrigeration cycle: W_in = Q_H − Q_L.

## Check yourself

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

1. What does conservation of mass state for a non-flow (closed) process?
   A. Mass decreases as energy is added
   B. Mass flowing in equals mass flowing out
   C. Mass of the system remains constant (m1 = m2)
   D. Mass changes by the amount of heat added
   Answer: C. No mass crosses the boundary of a closed system, so m is constant. (Key point: Mass in a non-flow process)

2. The first law for a closed system process 1-2 is written as:
   A. Q12 + W12 = ΔU only, always
   B. Q12 − W12 = E2 − E1 = ΔU + ΔKE + ΔPE
   C. Q12 = W12 + m
   D. Q12 − W12 = 0
   Answer: B. Net energy supplied as heat minus work done by the system equals the change in total energy of the system. (Key point: First law for closed system)

3. Why does E2 − E1 = 0 for a cyclic process?
   A. Because no heat is transferred
   B. Because no work is done
   C. Because the system returns to its initial state so all properties regain their original values
   D. Because the process is adiabatic
   Answer: C. Energy is a property, so after the system returns to the initial state its change is zero. (Key point: Cycle property change)

4. Applying the first law to a cycle gives:
   A. ∮δQ = ∮δW, i.e. Q_net = W_net
   B. ∮δQ = ΔU
   C. ∮δQ = 0 and ∮δW = 0 separately
   D. ∮δQ = −∮δU
   Answer: A. Since ∮dE = 0, the first law becomes ∮δQ − ∮δW = 0, so net heat equals net work. (Key point: Heat and work in a cycle)

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