# ENGR2000 Lecture Notes - Lecture 5: A.D. Vision, Continuity Equation, Allis-Chalmers D Series

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2nd-Year Fluid Mechanics, Faculty of Science & Engineering, Curtin University

ENGR2000: FLUID MECHANICS

For Second-year Chemical, Petroleum, Civil & Mechanical Engineering

FLUID MECHANICS LECTURE NOTES

CHAPTER 5: APPLICATIONS OF MASS CONSERVATION,

MOMENTUM EQUATION AND BERNOULLI EQUATION

5.1 Introduction

The fundamental principles of ﬂuid behaviour covered in Chapters 3 & 4 are

applied to a number of engineering problems. Unless otherwise stated the

ﬂow will be assumed to be incompressible and inviscid. As a consequence of

these assumptions, the examples that follow only give approximate solutions.

Nevertheless, they yield results that are useful in engineering applications

(orders of magnitude, physical understanding etc.) in situations that would

make solutions of the full (ﬂuid) equations very complicated and, in some

cases, impossible for the present state-of-the-art.

Chapter 5 −Page 1

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2nd-Year Fluid Mechanics, Faculty of Science & Engineering, Curtin University

5.2 Example E1: Pipe ﬂow: Force on a nozzle

A pipe of circular cross-section is bolted to a nozzle which feeds into another

pipe of narrower cross-section than the upstream pipe. The joint is held by

a set of ﬂange bolts around the circumference of the bigger pipe. The set-up

is shown in Fig. E1.1.

FIGURE E1.1

Assuming that the ﬂow passes smoothly from upstream to downstream pipes,

ﬁnd the force which the ﬂange bolts resist in terms of the upstream ﬂow speed,

V1, upstream pressure, p1, and pipe cross-sectional areas, A1and A2.

Solution: Set up a suitable CV as shown in Fig. E1.1 and apply mass

conservation, momentum equation and the Bernoulli equation.

(i) Mass continuity:

V1A1=V2A2(E1.1)

(ii) Momentum equation: (in x-direction)

Chapter 5 −Page 2

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2nd-Year Fluid Mechanics, Faculty of Science & Engineering, Curtin University

For steady ﬂow, there is no storage term. Thus:

XF=mom. flux out −mom. flux in

Both pressures and the force, Ff(in the x-direction), of the nozzle wall on

the ﬂuid act. Thus: XF=p1A1−p2A2+Ff(E1.2)

mom. flux in =ρ(V1A1)V1(E1.3a)

mom. flux out =ρ(V2A2)V2=ρ(V1A1)V2(E1.3b)

having used Eqn E1.1 for the second relation above.

Substituting Eqns. E1.2 and E1.3 into the momentum equation and re-

arranging gives:

Ff=ρV1A1(V2−V1)−(p1A1−p2A2) (E1.4)

The force experienced by the walls of the nozzle (and which the bolts must

resist) is −Ff. The equation above solves the problem except that we must

ﬁnd expressions for V2and p2.V2is found easily from Eqn. E1.1; V2=

(A1/A2)V1. To obtain the downstream pressure use the B.E.

(iii) Bernoulli equation

Applied along a central (horizontal) streamline from upstream to downstream

of the CV. 1

2ρV 2

1+p1=1

2ρV 2

2+p2

so:

p2=p1−1

2ρV 2

1((A1/A2)2−1) (E1.5)

The force on the bolts is thus:

−Ff=−ρA1V2

1((A1/A2)−1)+p1(A1−A2)+ 1

2ρV 2

1A2((A1/A2)2−1) (E1.6)

Note: Have we included all the pressure forces on the CV in Eqn. E1.2?

Chapter 5 −Page 3

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