reading materials: chapter 6

CHEM ENG 1007 1

Reading Materials: Chapter 6

LECTURES 16-18

CHEM ENG 1007 2

What is a Fluid

Material that continually deforms under a shear stress

Divided into two groupsLiquid

Gases

CHEM ENG 1007 3

Characterized as loosely-associated molecules which are normally not close together and which travel through space for long distances before colliding with each other. The velocity of their travel depends on the temperature of the gas.Characteristic of gases:

Readily compressibleExpand quickly and fill a container

To convert from volume to mass/mole use PV=nRT

GASES

CHEM ENG 1007 4

Characterized by molecules which are very close together and which are in collision with each other very frequently as they move around each other. The velocity of that motion and the rate of that collision depend on the temperature of the liquid.Characteristic of liquids:

Slightly compressibleTakes shape of container sides and bottom

To convert from volume to mass/mole use

Density = mass/volume

LIQUIDS

CHEM ENG 1007 5

Fluid-Like Systems

Solids present in fluids Slurries

fine particles suspended in liquid

Solids in a fluidized bed particles moving with the

fluid in a tall reactor.

CHEM ENG 1007 6

Variables associated with fluids

Density

Flow rate

Pressure

Viscosity

Surface tension

Others

Thermal conductivity

Electrical conductivity

Boiling point

Freezing point

Heat capacity

Enthalpy

Chapter 7 CHEM ENG 1007 7

Flow Rate

Rate at which a material is transported through a process line.

Mass flow rate:

Molar flow rate:

Volume flow rate:

where:

mass mm

time t

mole nn

time t

Volume VV

time t

avem V MW nρ: average density

aveMW : average molecular weight


Pressure

The pressure of the fluid is defined as the total force (exerted on the boundary by the fluid molecules) divided by the surface area of the boundary it is acting upon.

{F} {A} {P}

SI N m2 Nm-2 (Pa)

cgs dyne cm2 dynecm-2

American lbf in2 lbfin-2 (psi)


A woman’s high heels sink

into the soft ground,

but the larger shoes of the

much bigger man do not.

Pressure = force/area

Puzzle


PressureWe express pressure in two ways: absolute and gauge pressures

Gauge Pressure: fluid pressure that is measured relative to the atmospheric pressure. Absolute Pressure: the total magnitude of force exerted per unit area

In process calculations:Pabs = Pgauge + Patmosphere

Pabs = 0 in complete vacuumLetter “a” or “g” is added to designate absolute or gauge, thus psia or psig


Standard Atmospheric Pressure

At sea level, 0oC and 45o latitude:

Patm = 1 atm

= 101,325 Pa

= 14.7 psi

= 1.01325 bars

= 760 mmHg

= 10.333 m H2O = 33. 9 ft H2O


Torricelli filled a tube with mercury and inverted it into an open

container of mercury. Air pressure acting on the mercury in the dish

can support a column of mercury 76 cm in height.

How much does the atmosphere heigh?

Answer: the same as 76 cm of mercury. How?

Standard Atmospheric Pressure


Example 7.1

A man pumps his automobile tire until the tire gauge reads 34.0 psi. If the atmosphere in his community is 14.2 psia, what is the absolute pressure of the air in the tire?

Solution:

Pg = 34.0 psig

Patm = 14.2 psia

Pabs = 34.0 + 14.2 = 48.2 psia


Hydrostatic Pressure

It is the pressure (P) of the fluid at the base of the column. That is the force (F) exerted on the base divided by the base area (A). F thus equals the force on the top surface plus the weight of the fluid in the column.P = P0+ gh

h = height of a hypothetical column of the fluidA pressure may also be expressedas a head of a particular fluid (Ph)

Ph = P0 + h


Example 7.2

For the tank depicted in Fig. 7.2, if the NaOH solution is 8 ft high, what is the pressure at the bottom of the tank? Assume that the density of the NaOH solution is the same as that of water. Perform the calculation in metric units.

Solution:2 1

2 1 3

1000f

f

P P p ghkg

P P p ghm

9.80 m2

8 ft

s

0.3048 m1ft

2 1 223,896 23,896 23.896 kkg

msP P Pa Pa

P1 = 0 Pa (gauge), so P2 = 23,896 Pa (gauge)

or P2 = 23,896 + 101,325 = 125,221 Pa (absolute)


Quick Quiz 1

What is the pressure 30.0 m below the surface of a lake? Assume the atmospheric pressure (the pressure at the surface) is 10.4 m H2O? Express your answer in atm.

Solution:

Ph = P0 + h

= 10.4 + 30.0 = 40.4 m H2O

2

2

40.4 1 3.91

10.333 mH O atm

atmmH O


Figure 3.4-3 (p. 57) of Felder and RousseauBourdon gauge. It is used to measured fluid pressure from nearly perfect vacuums to about 7000 atm.

Fluid Pressure Measurement

A hollow tube closed at one end and bent into a C configuration. The open end of the tube is exposed to the fluid whose pressure is to be measured. As the pressure increases, the tube tends to straighten, causing a pointer attached to the tube to rotate.


Figure 3.4-4 (p. 58)Manometers. It is used for more accurate measurements of pressure (below about 3 atm)

Fluid Pressure MeasurementGauge pressure Absolute pressurePressure difference

Chapter 7 CHEM ENG 100719

Figure 3.4-5 (p. 58)Differential Manometer variables.

Fluid Pressure Measurement

1 2

1 2

1 2 1 2

f f

f

if p p p

P P p p gh p gh

P P p p g d d

1 1 1 2 2 2

( ) ( )

f

P a P b

P p g d P p g d p g h

If is gas then may be neglected


Illustration 1

A manometer reading gives 100 mmHg, calculate the absolute pressure.

Solution:

Measured gauge pressure:

Pabs = 13,328 + 101,325 = 114,653 Pa

= 115 kPa

3 2

13,600 9.80 0.1 13,328 ag

kg mP gh m P

m s


How does pressure relate to flow?

In the absence of other forces, fluids tend to flow from regions of high pressure to regions where the pressure is lower. Therefore, pressure differences provide a driving force for fluid flow.

Example: when a tire is punctured, air flows out of the high pressure tire to the atmosphere, which is a low pressure.


Viscosity ()

Characterises its resistance to flow A measure of “stickiness” of a fluid A “frictional force”

Less energy required for mixing

More energy required for mixing

Low viscosity fluid High viscosity fluid


Measurement - “Shear stress/shear rate”

Viscosity ()

dv

dyviscosity

Units SI: kg/(m.s) = N.s/m2 = Pa.s cgs: cp (centipoise) 1 cP = 10-2 poise = 10-3 Pa.s = 1 mPa.s


Viscosity of various fluids

Fluid Temperature (oC) ViscosityPa.s

Water 15.6 1.1x10-3

Gasoline 15.6 0.3x10-3

SAE 30 oil 15.6 383x10-3

Air 15 1.8x10-5

Methane 20 1.1x10-5


Substance (25°C) Viscosity (Pa.s)

Water 1 x 10-3

Mercury 1.5 x 10-3

Air 1.8 x 10-5

Castor oil 0.99

Viscosity of various fluids


Kinematic viscosity

2 -1 m .s

The property viscosity may also be combined with the fluid’s density to give the property kinematic viscosity


Types of viscous fluids

1

2

3

4

She

ar s

tres

s

Velocity gradient dv

dy

dv

dyviscosity


1. Newtonian fluid

e.g. water, air, other gases.

2-4 Non-Newtonian fluids

2. Bingham-plastic.

e.g. toothpaste, margarine, soap

3. Pseudo-plastic (shear thinning)

e.g. mayonnaise, polymer melts, paints.

4. Dilatant (shear thickening)

e.g. wet beach sand, starch in water

1

2

3

4

She

ar s

tres

s

Velocity gradient dv

dy

Types of viscous fluids


Newtonian fluids

With Newtonian fluids, shear stress increases proportionately with the shear rate

Question:

How would you determine the viscosity from a plot of shear stress against shear rate?

Shear rateVi

scos

ity

ExamplesWaterGlycerineAlcoholAir

Shear rate

Shea

r stre

ss


Other types of Non-Newtonian fluids

Pseudoplastic

Shear rate

Vis

cosi

ty

Dilatant

Shear rateV

isco

sity

Viscosity changes with power input


Thixotropic

Time

Vis

cosi

ty

Rheopetic

Time

Vis

cosi

tyViscosity changes with time at constant

shear rate

Other types of Non-Newtonian fluids


Non-Newtonian fluids can have more than one property

Example:

Damp gypsum Pseudoplastic, Thixotropic and Viscoelastic

Cream Dilatant, Rheopectic and not Viscoelastic

Viscoelastic: Fluid returns to original viscosity after power input ceases

Non-Newtonian fluids


Ideal / Inviscid Fluid

Hypothetical fluid which is incompressible and has zero viscosity.


Fluid Flow Principles

Flow patterns vary with:velocitygeometry of surface, andfluid properties such as viscosity, density.

Classic experiment by Osborne Reynolds (1883) observed two types of fluid flow:

Laminar flow – low flow ratesTurbulent flow – higher flow rates


Reynolds’ experiment

(i) Low flow rates: fluid moves in parallel layers.

(ii) High flow rates: cross currents (eddies) develop


Reynolds Number

Reynolds’ key variables

pipe diameter (D)mean velocity (v)fluid density ( )fluid viscosity ( )

Arrange into single “dimensionless group”:

vD

Re


Reynolds Number

For pipe flow:

Re < 2,100 - laminar flow

2,100 < Re < 10,000 - transition region

Re > 10,000 - turbulent flow

Pipe-flow systems with the same Re are said to be dynamically similar.


Fluid flow through a pipe

(i) Is the flow laminar flow or turbulent?

(ii) What is the effect of reducing the velocity by a factor of 10?

Water at 25oC flows through a pipe of internal diameter 0.1 m at a velocity 0.2 m/s.

Illustration 2


Solution

3

1000 0.2 0.1vD(i) Re 20,000

1x10

Hence flow is turbulent

(ii) Re v20,000

Re = 2,00010

Hence flow is laminar


Velocity Profiles: Laminar Flow

(i) Velocity profile is parabolic with the maximum velocity occurring in the centre of the pipe (r = 0)

2

max

2

x max 2

R pv

4 L

At other radial positions:

rv v 1

R

(i) Laminar Flow


Rr

L

(i) Velocities of a fluid in laminar flow through a circular pipe

2

x max 2

rv v 1

R



(ii) Volumetric flow rate:

4R p

V8 L

Hagen Poiseuille equation.

(iii) Mean velocity:

maxV 1

v vA 2


22

where

A cross-sectional area

DA R

4


Velocity Profiles: Turbulent Flow

Difficult to mathematically model due to its complex and rapidly changing flow patterns.

Experimental measurements show for time-averaged velocity and mean velocity

1/7

max

max

rv v 1

Rv 0.8v (ii) Turbulent Flow


(iii) Plug Flow

Velocity Profiles: Plug Flow

Common assumption for highly turbulent flow is that velocity does not vary over cross-section:

v v


Mass Conservation in Fluid Flow

Consider steady-state, one-dimensional fluid through pipe

A1

v

2

2

1

1

1

A

v2

2

Flow In Flow out

1 2

1 1 2 2

1 1 1 2 2 2

1 2

2 21

1

1 2

m m m

V V

A v A v

for the same fluid

A vv

A

v v

22

where

A cross-sectional area

DA R

4


Mass Conservation in Fluid Flow

Is v2 in (a) less than v2 in (b)? No, they are both the same.

1 1 1 2 2 2

1 2 1 2

1 2

A v A v

for the same fluid and A A

v v


Fluid Friction

Frictional force cause pressure drop during fluid flow through a constant diameter horizontal pipe.

Consider flow situation in pipe below.

L

v

1 2

D

p

w

w

p1 2


Fluid FrictionMomentum balance:

2

w

w

2

D0 p DL

4

L p 4

DL

p 2f vD

f = friction factor; L = length of pipe;

D = diameter of pipe;


Fluid Friction

2f

2f

2f

p L J2f v e kgD

Lfrictional energy e 2f v

D

e 2f Lfrictional head loss v

g g D


Example 7.7What value of the friction per mass of fluid (ef) is necessary to cause a decrease in pressure equal to

a) 10 psi (answer in Btu/lbm)?

f

3 2 2f

2 2m f

m

pe

10 lb ft 1 Btu 12 in

62.4 lb 778.1 ft lbin 1 ft

Btu0.030

lb


Example 7.7What value of the friction per mass of fluid (ef) is necessary to cause a decrease in pressure equal to

b) 68,900 Pa (answer in J/kg)?

f

pe

68,900 Pa 21 kg/ms

1 Pa

3 2

2

m 1 J s

1000 kg 1 kg m

J68.9

kg


Laminar Flow

f determined analytically from velocity profile.

16

i.e. fRe

Experimental data confirmed for Re < 2,100 (see Figure 2.10-3)


Figure 2.10-3: Friction Factor Chart


Turbulent Flowf cannot be determined analytically.

Experimental curves have been devised.

f also depends on surface roughness factor,

Pipe (mm)

Concrete 0.3-3.0

Cast iron 0.26

Galvanized iron 0.15

Plastic 0.0 (smooth)


0.25 f 0.079Re

Turbulent Flow

For smooth pipe & 2,100 < Re < 105

Blassius equation may be used:


Frictional losses in a pipe

Pipeline 5 km long & 30 cm internal diameter

Conveys water at 25oC at a rate of 180 kg/s.

Roughness factor /D = 0.001

Estimate the pressure drop across the pipe due to friction.

Illustration 3


Solution

For water

3

3

24

53

1,000 kg/m

10 Pa.s

m V vA

m 180v 2.55 m / s

A 1,000 0.3

1,000 2.55 0.3vDRe 7.65x10

10


From Figure 2.10-3, f = 0.005

Hence,

Solution

2 2L 5000p 2f v 2 0.005 1000 2.55

D 0.3

1,083,750 Pa

1084 kPa

reading materials: chapter 6

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