unit4 fluid dynamics

FLUID DYNAMICS J3008/4/1

FLUID DYNAMICS

OBJECTIVES

General Objective : To know, understand and apply the mechanism of flow to simple pipes.

Specific Objectives : At the end of the unit you should be able to :

define types of flow

define discharge, continuity equation and mass flowrate in pipes

solve problems related to the use of continuity equation

UNIT 4


4.1 TYPES OF FLOW

4.1.1 Steady flow The cross-sectional area and velocity of the stream may vary from cross-section, but for each cross-section they do not change with time. Example: a wave travelling along a channel.

4.1.2 Uniform flow The cross-sectional area and velocity of the stream of fluid are the same at each successive cross-section. Example: flow through a pipe of uniform bore running completely full.

4.1.3 Laminar flowAlso known as streamline or viscous flow, in which the particles of the fluid move in an orderly manner and retain the same relative positions in successive cross-sections.

4.1.4 Turbulent flowTurbulent flow is a non steady flow in which the particles of fluid move in a disorderly manner, occupying different relative positions in successive cross-sections.

4.2 Discharge and Mass Flowrate

4.2.1 DischargeThe volume of liquid passing through a given cross-section in unit time is called the discharge. It is measured in cubic meter per second, or similar units and denoted by Q.

vAQ .=

INPUTINPUT


Example 4.1

If the diameter d = 15 cm and the mean velocity, v = 3 m/s, calculate the actual discharge in the pipe.

Solution to Example 4.1

AvQ =

vd ×=4

2π

( )

34

15.0 2

×= π

sm /053.0 3=

4.2.2 Mass Flowrate

The mass of fluid passing through a given cross section in unit time is called the mass flow rate. It is measured in kilogram per second, or similar units and denoted by

•m .

vAm ××=•

ρ

••= 21 mm

222111 vAvA ρρ =

`Example 4.2

A1 v

1A

2 v

2

inout


Oil flows through a pipe at a velocity of 1.6 m/s. The diameter of the pipe is 8 cm. Calculate discharge and mass flowrate of oil. Take into consideration soil = 0.85.


111 vAQ =

( ) ( )6.1

4

08.0 2π=

sm /10042.8 33−×=

Qm ρ=•

( )( )310042.8100085.0 −×= skg /836.6=

A very simple way to measure the rate at which water is flowing along the pipe is by catching all the water that is coming out of the pipe in a bucket over a fixed time period. We can obtain the rate of accumulation of mass by measuring the weight of the water in the bucket and dividing this by the time taken to collect this water. This is known as the mass flowrate.

Example 4.3

The weight of an empty bucket is 2.0 kg. After 7 seconds of collecting water the weight of the bucket is 8.0 kg. Calculate the mass flowrate of the fluid.


fluidthecollecttotakentime

bucketinflowratemassmflowratemass =•

,

7

0.20.8 −=

skg /857.0=

ACTIVITY 4A


TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

4.1 List down four types of flow. Define any three types of flow that you have listed.

FEEDBACK ON ACTIVITY 4A


4.11. Steady flow

The cross-sectional area and velocity of the stream may vary from cross-section, but for each cross-section they do not change with time. Example: a wave travelling along a channel.

2. Uniform flow The cross-sectional area and velocity of the stream of fluid are the same at each successive cross-section. Example: flow through a pipe of uniform bore running completely full.

3. Laminar flowAlso known as streamline or viscous flow, in which the particles of the fluid move in an orderly manner and retain the same relative positions in successive cross-sections.

4. Turbulent flowTurbulent flow is a non steady flow in which the particles of fluid move in a disorderly manner, occupying different relative positions in successive cross-sections.

INPUTINPUT


4.3 Continuity EquationFor continuity of flow in any system of fluid flow, the total amount of fluid entering the system must equal the amount leaving the system. This occurs in the case of uniform flow and steady flow.

QP = discharge through cross-section P-P AP = cross-sectional area through P-P vp = fluid mean velocity through P-PQR = discharge through cross-section R-RAR = cross-sectional area through R-RvR = fluid mean velocity through R-R

Discharge at section P = Discharge at section R QP = QR

AP vP = AR vR

Application

We can apply the principle of continuity to pipes with cross sections that have changes along their length. Consider the diagram below of a pipe with a contraction.

RP

SYSTEM

P R

QR

QP

QP =QR

Figure 4.1


A liquid is flowing from left to right and the pipe is narrowing in the same direction. By the continuity principle, the discharge must be the same at each section. The mass going into the pipe is equal to the mass going out of the pipe.

Discharge at section 1 = Discharge at section 2 21 QQ =

2211 vAvA =

Example 4.4

If the area A1 = 10 × 10-3 m2 and A2 = 3 × 10-3 m2 and the upstream mean velocity, v1=2.1 m/s, calculate the downstream mean velocity.


2

112 A

vAv =

( )3

3

103

1.21010−

−

××=

sm /0.7=

Now try this on a diffuser, a pipe which expands or diverges as in the figure below.

Section 1 Section 2

Figure 4.2

Section 1 Section 2

Figure 4.3


Example 4.5

Referring to the Figure the diameter at section 1 is d1 = 30 mm and at section 2 is d2=40 mm and the mean velocity at section 2 is v2 = 3.0 m/s. Calculate the velocity entering the diffuser.


2

2

1

21 v

d

dv

=

0.330

402

×

=

sm /3.5=

Another example in the use of the continuity principle is to determine the velocities in pipes coming from a junction.

The downstream velocity only changes from the upstream by the ratio of the two areas of the pipe. As the area of the circular pipe is a function of the diameter, we can reduce the calculation further. Thus,

1

3

2

Figure 4.4


Total discharge into the junction = Total discharge out of the junction Q1 = Q2 + Q3

A1v1 = A2v2 + A3v3

Example 4.6A pipe is split into 2 pipes which are BC and BD as shown in the Figure 4.5. The following information is given:

diameter pipe AB at A = 0.45 mdiameter pipe AB at B = 0.3 mdiameter pipe BC = 0.2 mdiameter pipe BD = 0.15 m

Calculate:

a) discharge at section A if vA = 2 m/sb) velocity at section B and section D if velocity at section C = 4 m/s


a) Discharge at section A

AAA vAQ ×=

( )

24

45.0 2

×= π

A

D

C

Figure 4.5

B


sm /318.0 3=

b) Discharge at section A = Discharge at section B

BA QQ = BBAA vAvA =

B

AAB A

vAv =

( )

( ) 23.0

4318.0

π=

sm /5.4=

For continuity of flow

DCB QQQ +=

CBD QQQ −=

( ) ( )CCBB vAvA −×=

( ) ( )

×−

×= 4

4

2.05.4

4

3.0 22 ππ

sm /192.0 3=

For pipe BD

DDD vAQ ×=

sm /192.0 3=

D

DD A

Qv =

( )215.0

4192.0

π×= sm /86.10=

ACTIVITY 4B

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

4.2 State the actual discharge equation for the following pipes.

6

5

4

3

2

7

8

1


Q1 = _______________

Q2 = _______________

Q7 = _______________

FEEDBACK ON ACTIVITY 4B

4.2

6

5

4

3

2

7

8

1


Q1 = _Q2 +Q3_

Q2 = _Q4 +Q5 +Q6_

Q7 = _Q3 –Q8_

SELF-ASSESSMENT

You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback on Self-Assessment. If you face any problems, discuss it with your lecturer. Good luck.


4.1 Water flows through a pipe AB of diameter d1 = 50 mm, which is in series with a pipe BC of diameter d2 = 75 mm in which the mean velocity v2 = 2 m/s. At C the pipe forks and one branch CD is of diameter d3 such that the mean velocity v3 is 1.5 m/s. The other branch CE is of diameter d4 = 30 mm and conditions are such that the discharge Q2 from BC divides so that Q4 = ½ Q3. Calculate the values of Q1,v1,Q2,Q3,D3,Q4 and v4..

FEEDBACK ON SELF-ASSESSMENT

Answers:

B

E

DC

A


4.1 Q1 = 8.836 × 10-3 m3/s

v1 = 4.50 m/s

Q3 = 5.891 × 10-3 m3/s

Q4 = 2.945 × 10-3 m3/s

d3 = 71 mm

v4 = 4.17 m/s

unit4 fluid dynamics

Engineering

types of flow

uniform flow

steady flow

viscous flow

turbulent flow turbulent

laminar flow

mass flow rate

mechanism of flow