---
title: "Integral continuity for a compressing duct"
subject: "Applied Computational Fluid Dynamics"
chapter: 2
key: "semester_7/applied_computational_fluid_dynamics:integral-continuity-duct-numerical"
source: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/integral-continuity-duct-numerical
site: BCT Engineering
---

# Integral continuity for a compressing duct

Applied Computational Fluid Dynamics · Chapter 2: Governing Equations and Numerical Foundations

- Asked in: year not given, 5 marks
- Video: [Conservation of mass (a.k.a., continuity) [Fluid Mechanics #2]](https://www.youtube.com/watch?v=PFgiW7P6omE) — Prof. Van Buren
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/applied_computational_fluid_dynamics/integral-continuity-duct-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/applied_computational_fluid_dynamics/chapter-2

## Question

State the integral form of the continuity equation for a control volume. Air flows steadily through a duct. At the inlet the area is 0.020 m², density 1.2 kg/m³ and velocity 20 m/s. At the outlet the area is 0.010 m² and the density is 1.5 kg/m³. Find the mass flow rate and the outlet velocity.

## Short answer

The integral continuity equation states time rate of mass stored plus net outflow equals zero. Mass flow is 1.2*0.020*20=0.48 kg/s; outlet velocity = 0.48/(1.5*0.010)=32 m/s.

## Answer

#### Integral form
For a fixed control volume (CV) with control surface (CS):
$$
\frac{\partial}{\partial t}\int_{CV}\rho\,dV+\int_{CS}\rho\,(\vec V\cdot\hat n)\,dA=0
$$
The first term is the rate of change of mass stored in the CV; the second is the net mass outflow through the surface. For steady flow the first term is zero, so total mass inflow equals total outflow. For one inlet and one outlet:
$$
\rho_1A_1V_1=\rho_2A_2V_2=\dot m
$$
The integral form is the basis of the finite volume method, since it is applied to each cell. The differential form follows by applying the divergence theorem to a very small volume.

#### Numerical
Mass flow rate:
$$
\dot m=\rho_1A_1V_1=1.2\times0.020\times20=0.48\ \text{kg/s}
$$
Outlet velocity:
$$
V_2=\frac{\dot m}{\rho_2A_2}=\frac{0.48}{1.5\times0.010}=32\ \text{m/s}
$$
Check: halving the area doubles the velocity, while the 1.25 times higher density reduces it by a factor of 1.25, giving a net factor of 1.6: $V_2=V_1\times\dfrac{\rho_1A_1}{\rho_2A_2}=20\times\dfrac{0.024}{0.015}=32$ m/s.

**Answer:** $\dot m=0.48$ kg/s; $V_2=32$ m/s.

## Check yourself

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

1. In the integral form for a fixed control volume, the second term represents:
   A. Rate of change of mass stored in the CV
   B. Net mass outflow through the control surface
   C. Pressure force on the surface
   D. Viscous work
   Answer: B. The first term is storage; the surface integral of rho(V.n) dA is the net outflow. (Key point: Integral continuity)

2. For steady flow with one inlet and one outlet:
   A. V1 = V2 always
   B. A1 = A2 always
   C. rho1 A1 V1 = rho2 A2 V2
   D. rho1 = rho2 always
   Answer: C. Steady flow means storage is zero, so inflow mass rate equals outflow. (Key point: Steady mass balance)

3. The mass flow rate in the duct is:
   A. 0.24 kg/s
   B. 0.48 kg/s
   C. 0.03 kg/s
   D. 0.72 kg/s
   Answer: B. m = 1.2 x 0.020 x 20 = 0.48 kg/s. (Key point: Mass flow rate)

4. The outlet velocity is:
   A. 10 m/s
   B. 20 m/s
   C. 40 m/s
   D. 32 m/s
   Answer: D. V2 = m/(rho2 A2) = 0.48/(1.5 x 0.010) = 32 m/s. (Key point: Outlet velocity)

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