3

mass kgvolume m

2.1.2. Volume flow, mass flow and the continuity equation2.1.2. Volume flow, mass flow and the continuity equation

…………………………………… …………………………………… (2.1)(2.1)

3

mass flow M kg svolume flow Q s m

andand

………………………………………………………….… (2.2).… (2.2)

givinggiving

/M Q kg s

tan /M Q cons t kg s

32m mQ uA m or

s s

………………………………………………………….… (2.3).… (2.3)

…………………………………………………………………….… (2.4).… (2.4)

Then the continuity equation becomesThen the continuity equation becomes

tan /M uA cons t kg s ……………….…………………….… (2.5).…………………….… (2.5)

3tan /Q uA cons t m s ……………….…………………….… (2.6).…………………….… (2.6)

2.2. FLUID PRESSURE2.2. FLUID PRESSURE

2.2.1. The cause of fluid pressure2.2.1. The cause of fluid pressure

2

force Nparea m

3volume hA m

……………….…………………………….… (2.7).…………………………….… (2.7)

2.2.2. Pressure head2.2.2. Pressure head

Then from the definition of density (mass/volume), the mass of liquid isThen from the definition of density (mass/volume), the mass of liquid is

mass volume x densityhA kg

The weight of the liquid will exert a force, F, on the base of the tube equal to The weight of the liquid will exert a force, F, on the base of the tube equal to mass x gravitational acceleration (g):mass x gravitational acceleration (g):

F hA g N

2

F NP gh or PaA m

However, as pressure = force/area, the pressure on the base of the tube isHowever, as pressure = force/area, the pressure on the base of the tube is

…………………………… …………………………… (2.8)(2.8)

One standard atmosphereOne standard atmosphere 3

3

13.5951 10 9.8066 0.760

101.324 10

x g x h

x x x

x Pa

2.2.3. Atmospheric pressure and gauge pressure2.2.3. Atmospheric pressure and gauge pressure

oror 101.324 kPa101.324 kPa

Absolute pressure = atmospheric pressure + gauge pressureAbsolute pressure = atmospheric pressure + gauge pressure ………… ………… (2.9)(2.9)

The atmospheric pressure could then be calculated as (see equation (2.8))The atmospheric pressure could then be calculated as (see equation (2.8))

1 1 2 2p gh gh Pa

p gh Pa

2.2.4. Measurement of air pressure2.2.4. Measurement of air pressure

Where, in this case, is the density of mercury. Where, in this case, is the density of mercury.

12 1

2

h h m

oror

………………………….……… (2.10).……… (2.10)

2.3. FLUIDS IN MOTION2.3. FLUIDS IN MOTION

2.3.1. Bernoulli’s equation for ideal fluids2.3.1. Bernoulli’s equation for ideal fluids

Kinetic energyKinetic energy

If we commence with the mass, m, at rest and accelerate it to velocity u in t If we commence with the mass, m, at rest and accelerate it to velocity u in t seconds by applying a constant force F, then the acceleration will be uniform and seconds by applying a constant force F, then the acceleration will be uniform and the mean velocity isthe mean velocity is

0 /2 2u u m s

Distance travelled = mean velocity x timeDistance travelled = mean velocity x timeThenThen

2u t m

Furthermore, the acceleration is defined asFurthermore, the acceleration is defined as

2/increaseinvelocity u m stime t

The force is given byThe force is given by

and the work done to accelerate from rest to velocity u isand the work done to accelerate from rest to velocity u is

F mass accelerationum Nt

WD = force x distance N mWD = force x distance N m

2

2

2

u um ttum N m or J

……………….…………………………….… (2.11).…………………………….… (2.11)

The kinetic energy of the mass m is, therefore muThe kinetic energy of the mass m is, therefore mu22/2 joules/2 joules

where g is the gravitational acceleration. In moving upward to the final where g is the gravitational acceleration. In moving upward to the final elevation of Z metres above the datum, the work done iselevation of Z metres above the datum, the work done is

F mass accelerationmg N

Potential energyPotential energy

tanWD force dis cemgZ J

This gives the potential energy of the mass at elevation ZThis gives the potential energy of the mass at elevation Z

……………….……………………………….….… (2.12).……………………………….….… (2.12)

Flow workFlow work

PmWD J

……………….…………………………………..… (2.13).…………………………………..… (2.13)

tanwork done force dis cePAs J

However, the product As is the swept volume v, givingHowever, the product As is the swept volume v, giving

WD Pv

Now, by definition, the density isNow, by definition, the density is

3

m kgv m

orormv

Hence, the work done in moving the plug of fluid into the pipe isHence, the work done in moving the plug of fluid into the pipe is

/ logor P joules per ki ram

Total mechanical = kinetic + potential + flowTotal mechanical = kinetic + potential + flow energy energy energy work energy energy energy work

2

2mu PmZg m J

………………………………..… (2.14)..… (2.14)

2

tan2u Pm Zg cons t J

………………………………………………………………..… (2.15)..… (2.15)

Another way of expressing this equation is to consider two stations, 1 and Another way of expressing this equation is to consider two stations, 1 and 2 along the pipe, duct or airway. Then2 along the pipe, duct or airway. Then

2 21 1 2 2

1 21 22 2g g

u P u Pm Z m Z

Now as we are still considering the fluid to be incompressible,Now as we are still considering the fluid to be incompressible,

1 2 ( )say

givinggiving

2 21 2 1 2

1 2( ) 02

u u P P JZ Z gkg

………………………………..… (2.16)..… (2.16)

2.3.2. Static, total and velocity pressure2.3.2. Static, total and velocity pressure

22 1 1

2P P u

21

2 1 2vup P P Pa …………………………………………………………..… (2.17)..… (2.17)

v t sp p p

oror

t s vp p p Pa …………………………………………………………..… (2.18)..… (2.18)

1( )dp gh Pa …………………………………………………………..… (2.19)..… (2.19)

1 2( ) ( )d d ap gh gh Pa …………………………………………..… (2.20)..… (2.20)

2.3.3. Viscosity2.3.3. Viscosity

2

F NA m

……………………..……………..…..… (2.21)..……………..…..… (2.21)

2

F du NA dy m

……………………..……………..…..… (2.22)..……………..…..… (2.22)

62(17.0 0.045 ) 10s

airNtm

And for water,And for water,

32

64.72 0.2455 1031.766

swater

Nt m

where t = temperature (where t = temperature (ooC) in the range 0-60 C) in the range 0-60 ooCC

The units of viscosity are derived by transposing equation (2.22): The units of viscosity are derived by transposing equation (2.22):

2 2sNdy N sm or

du m m m

A term which commonly occurs in fluid mechanics is the ratio of dynamic A term which commonly occurs in fluid mechanics is the ratio of dynamic viscosity to fluid density. This is called the kinematic viscosity, v (Greek ‘nu’)viscosity to fluid density. This is called the kinematic viscosity, v (Greek ‘nu’)

3

2sN m sv or N m

m kg kg

As 1 N = 1 kg x 1 (m/sAs 1 N = 1 kg x 1 (m/s22), these units become), these units become

2

2

m ms mkgs kg s

2 21 1 2 2

1 2 122 2g gu P u P JZ Z F

kg …………………………………………..… (2.23)..… (2.23)

2.3.4. Laminar and turbulent flow; Reynolds’ number2.3.4. Laminar and turbulent flow; Reynolds’ number

2 12

inertial force u dyviscous force du

…………………………………………..… (2.24)..… (2.24)

2 1 Luu

oror

ReuL

…………………………………………..… (2.25)..… (2.25)

Re ud

ExampleExample A ventilation shaft of diameter 5 m passes an airflow of 200 m A ventilation shaft of diameter 5 m passes an airflow of 200 m33/s /s at a mean density of 1.2 kg/mat a mean density of 1.2 kg/m33 and a mean temperature of 18 and a mean temperature of 18ooC. Determine C. Determine the Reynolds number for the shaft.the Reynolds number for the shaft.

Solution Solution For air at 18For air at 18ooCC6

6 2

(17.0 0.045 18) 10

17.81 10 /N s m

Air velocityAir velocity

2

66

2005 / 4

10.186 /

Re

1.2 10.186 5 3.432 1017.81 10

QuA

m sud

This Reynolds’ number indicates that the flow will be turbulent.This Reynolds’ number indicates that the flow will be turbulent.

2.3.5. Frictional losses in laminar flow; Poiseuille’s equation2.3.5. Frictional losses in laminar flow; Poiseuille’s equation

22 rtL r p

However , (equation (2.22) with a negative du), givingHowever , (equation (2.22) with a negative du), giving/du dr

2du r pdr L

oror

/2

p rdu dr m sL

…………………………………………..… (2.26)..… (2.26)

212 2

p ru CL

…………………………………………..… (2.27)..… (2.27)

2

4p RCL

2 21 ( ) /4

pu R r m sL

…………………………………………..… (2.28)..… (2.28)

2max

14

p mu RL s

…………………………………………..… (2.29)..… (2.29)

/ /mu Q A m s …………………………………………..… (2.30)..… (2.30)

2dQ u r dr

Substituting for u from equation (2.28) givesSubstituting for u from equation (2.28) gives

2 2

2 3

0

2 ( )42 ( )4

R

pdQ R r r drLpQ R r r drL

Integrating givesIntegrating gives

43 /

8R pQ m s

L

…………………………………………..… (2.31)..… (2.31)

4

8 Lp QR

oror

Lp R Q Pa …………………………………………..… (2.32)..… (2.32)

wherewhere

54

8 /LLR N s mR

4

2

2

18

/8

mR pu

L RR p m s

L

…………………………………………..… (2.33)..… (2.33)

oror

2

8 mup L PaR

…………………………………………..… (2.34)..… (2.34)

2 21 2 1 2

1 2 12( )2 g

u u P P JZ Z Fkg

1 212

P P JFkg

…………………………………………………….....… (2.35).....… (2.35)

12 2

8 mu JF LR kg

…………………………………………………….....… (2.36).....… (2.36)

2 21 2 1 2

1 2 2

8( )

2m

guu u P P JZ ZR kg

……………………. (2.37). (2.37)

2.3.6. Frictional losses in turbulent flow2.3.6. Frictional losses in turbulent flow

A huper L

/A hu c m sper L

…………………………………………………………………….. (2.38).. (2.38)

per L

per L Ap N …………………………………………………………………….. (2.39).. (2.39)

(A similar equation was used in section 2.3.5 for a circular pipe.) However,(A similar equation was used in section 2.3.5 for a circular pipe.) However,

p gh Pa

(equation (2.8) ) giving(equation (2.8) ) giving

2

A h Ngper L m

…………………………………………………………………….. (2.40).. (2.40)

2

3 3 22u J Nm Nor

m m m

oror

2

22u Nf

m …………………………………………………………………….. (2.41).. (2.41)

2

2u A hf g

per L

oror

2 /g A hu m sf per L

……………………………….. (2.42).. (2.42)

1/ 22 /gc m sf

………………..…………………….... (2.43)..…………………….... (2.43)

22 2 1

4g d huf d L

oror

242fLuh metres of fluidgd

………………………….….... (2.44).….... (2.44)

242

fL pup Pad

………………..…………………….... (2.45)..…………………….... (2.45)

or a frictional work termor a frictional work term

2

124

2p fL u JF

d kg ………………..……...……………..... (2.46)..……...……………..... (2.46)

The Bernoulli equation for frictional and turbulent flow becomesThe Bernoulli equation for frictional and turbulent flow becomes

2 2 21 2 1 2

1 24( )

2 2gu u P P fL u JZ Z

d kg

……..... (2.47)..... (2.47)

2

4 4

hAr mper

d dd

………………..……...……………..... (2.48)..……...……………..... (2.48)

2

2per up fL PaA

………………..……...……………... (2.49)..……...……………... (2.49)

32fL perp Q Pa

A

oror

2tp R Q Pa ………………..……...……………... (2.50)..……...……………... (2.50)

wherewhere

432t

fL perR mA

………………..……...………..... (2.51)..……...………..... (2.51)

Combining equation (2.34) and (2.45) givesCombining equation (2.34) and (2.45) gives

2

2

8 42

uL fL up PaR d

Substituting R = d/2 givesSubstituting R = d/2 gives

16fud

oror

16 dimRe

f ensionless ………………..……...………..... (2.52)..……...………..... (2.52)

Smooth pipe turbulent curveSmooth pipe turbulent curve

101 4log (Re ) 0.4ff

………………..……...………..... (2.53)..……...………..... (2.53)

0.25

0.0791Re

f ………………..……...…………......................... (2.54)..……...…………......................... (2.54)

210

14[2log ( / ) 1.14]

fd e

…………………….................................. (2.55).................................. (2.55)

101 18.71.74 2log 24 Re 4

edf f

.................................. (2.56).................................. (2.56)

andand

101 / 1.2554log

3.7 Ree d

f f

.................................... (2.57).................................... (2.57)

2

10/4 log

3.7e df

..................................... (2.58)..................................... (2.58)