8/27/2022
Lecture 5
Food Engineering Principles
Slide No. 1
Content of the Lecture
Slide No. 2
Topics to be covered:
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FLUID
MECHANICS
Dr. Dang Quoc Tuan
Department of Food Technology
International University - VNU
International University
Dept Food Technology
Food Engineering Principles
Fluid; fluid statics
Hydrostatic pressure, pressure head
Pressure measurement
Fluid deformation: Shear, shear rate, shear stress
Viscosity, Newton Law of viscosity
Momentum transport
Classification of fluids
Flow rate, fluid velocity
Reynold’s Number, Velocity Profile
Viscosity measurement
Reynolds number for non-Newtonian fluid
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Slide No. 3
Slide No. 4
Fluid Statics
Pascal’s Law
Fluid
 Fluid static is a term that is referred to the state of a
fluid where its velocity is zero and this condition is also
called hydrostatic.
• A fluid is defined as a substance that
deforms continuously when acted on by a
shearing stress of any magnitude.
 So, in fluid static, which is the state of fluid in which the
shear stress is zero throughout the fluid volume.
• Fluid tends to flow or to conform to the
outline of its container.
 In a stationary fluid, the most important variable is
pressure.
• They are gases and liquids and mixtures of
solids and liquids capable of flow.
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Force Equilibrium of a Fluid
Element
 Pascal’s Law: For any fluid, the pressure is the same
regardless its direction. As long as there is no shear
stress, the pressure is independent of direction.
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Slide No. 5
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Static Pressure Equation
Slide No. 6
A=Area (m 2)
Fluid surfaces
Fluid density ρ
H = Height
F1
Pressure acting uniformly in all directions
dF
dh
F2
F
Direction of fluid pressures on boundaries
dF = ρgA.dh = F2-F1;
dP = ρgdh
Pressure units:
SI: Pascals (Pa) = N/m2; kPa = 103 Pa
Engineering units: psi = lbf/in2 = 6.89 x 102 Pa
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∆P = ρg.∆h
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Slide No. 7
Slide No. 8
Pressure Calculation
Container Shape and Pressure
Example 1:
Atm.
Pressure
A
H2O
Pressure depends only on depth and not on the cross section area
of the column of fluid.
B
C 0.1 m
If all are filled with fluids of the same density, at any given depth,
the pressure will be the same in each container.
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Slide No. 9
Sol 1: Pressure calculation
H2O
1.2 m
Hg
0.3 m
B
C 0.1 m
0.3 m
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Slide No. 10
Example 2
•An underground gasoline tank is accidentally opened during
raining causing the water to seep in and occupying the bottom
part of the tank as shown in Fig. E2.1. If the specific gravity
for gasoline 0.68, calculate the absolute pressure at the
interface of the gasoline and water and at the bottom of the
tank. Use water = 998 kg/m3 and g = 9.81 m/s2.
Pressure at point C ??
A
Hg
1.2 m
The container is filled to a
depth of 0.3 meters with
mercury (ρ= 13,600 kg/m3).
Above this is 1.2 meters of
H2O (ρ = 1,000 kg/m3).
What is the absolute
pressure at a point 0.1m
above the bottom of the
tank?
Po
P1
P2
P = P + ∆PAB + ∆PBC =
101.4 kPa + 11.8 kPa + 26.7 kPa = 139.9 kPa
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Slide No. 11
Sol 2:
Density for gasoline:
ρg = 0.68(998) = 678.64 kg/m3
At the free surface, take the absolute pressure to be po =
1 atm;
po = 101.300 kPa.
p1 = p0 + ρgghg = 101,300 Pa + (678.64)(9.81)(5.5)
= 137916 Pa = 137.92 kPa
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Pressure measurement
U-tube manometer
Slide No. 12
Pressure is defined as a force per unit area - and the most accurate
way to measure low air pressure is to balance a column of liquid of
known weight against it and measure the height of the liquid column
so balanced. The units of measure commonly used are inches of
mercury (in. Hg), using mercury as the fluid and inches of water (in.
w.c.), using water or oil as the fluid
At the bottom of the tank, the pressure:
p2 = p1 + ρw ghw = 137916 + (998)(9.81)(1)
= 156418 N/m2 = 156.4 kPa
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Slide No. 13
Manometer
1
The left arm of a mercury manometer is connected to the bottom of
tank A and the right arm to the bottom of tank B. Both tanks contain
water and this water extends into and fills both arms above the
manometric fluid. The mercury in the right arm is 16.2 cm above the
level in the left. A pressure gauge at the bottom of tank A reads 59.0
kPa (gauge).
What is the absolute pressure at the bottom of tank B?
5
h5-h4
B
h1-h2
A
4
h4-h3=R
3
2
Slide No. 14
Example 3:
P2 = P 1 + ρA.g.(h1 -h2)
P3 = P 5 + ρB.g.(h5-h4) + ρM .g.(h4-h3)
139 kPa
Set P2 = P3, then
A
B
P1 + ρA.g.(h1-h2) = P 5 + ρB.g.(h5-h4) + ρM .(h4-h3)
P1 - P5= g [ρB.(h5-h4) + ρM .R - ρA.(h1-h2)]
If ρA=ρB
16.2 cm
P1 - P5= g{ρA.[(h5-h4) - (h1-h2)]+ ρM .R }
P1 - P5 = (ρM – ρA).g.R
(h5-h4) – (h1-h2) = -R
P1 - P5 = ρM.g.R
If ρA < < ρM
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Inclined Manometer
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Slide No. 15
Slide No. 16
Piezometer Tube
It is difficult to read small changes in pressure on an
ordinary manometer. The inclined manometer is designed
to make small changes easier to read.
Open
h
B
A
θ
Rm
P abs
R1
=
P gauges+ P
atm
Rm = R1 sinθ
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Slide No. 17
Bourdon Tubes
• The Bourdon tube is a long flat tube made with a curved
shape.
• The tube is connected to the container to measure
pressure.
• As pressure increases inside the Bourdon tube it tends to
straighten out.
• The tube is linked to a dial through a set of levers so that,
as the tube straightens, the dial moves and displays the
pressure.
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Piezometer tube – a device consists
of a vertical tube.
One end is connected to the
pressure to be measured while the
other end is open to the atmosphere.
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Slide No. 18
STRAIN GAUGE
If the metal is stretched or distorted, its resistance will
change and this will be reflected in a change in the
measured voltage or current
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Slide No. 19
Pressure Head
Slide No. 20
Pressure Head
Example 3:
A water system has a pressure of 31.6
psia. Verify that this is equivalent to a
pressure head of 3.9 ft of water.
1 PSI = 689 Pa
g= 9.81 m/s2
1 ft = 0.3048 m
Water density: 1000kg/m 3
Pg = 11,644 Pa
1 psi = 689 Pa
1 m = 3.28 ft
• Pressure (in N/m2) is computed from the height of manometric
fluid that the pressure supports (in units of length).
• We can report pressure by giving the height supported.
• When pressure is given this way we call it the pressure head.
• The HEAD is always against 1 atm ( so it is a gauge pressure)
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Slide No. 21
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Liquid Transport Systems
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FLUID DYNAMICS
Slide No. 22
Fluid dynamics will see what happens when fluids move.
This information is needed to answer such questions
as:
• How long must a holding tube of a pasteurizer be in
order to kill all pathogens.
• How much mixing will take place to achieve certain
degree of homogeneity of a food product.
• How much power is required to achieve such a level of
mixing.
• How rapidly will cells separate from fermentation broth
in a centrifuge or separator.
• At what rate can we obtain heat from steam flowing
through a pipe.
Viscosity
Fluid Flow
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Slide No. 23
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Measurement of Flow
Slide No. 24
Mass Flow Rate =
Mass Flux =
Volumetric Flow Rate =
Volumetric Flux =
SI Units:
Engineering Units:
Production line for fruit juice (TH true juice)
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Slide No. 25
Slide No. 26
Average Velocity
Local Velocity
 The layer of molecules in immediate contact with the
pipe wall will have 0 velocity.
 The fluid in the center of the pipe will travel faster
than the average velocity.
Average Velocity = <u> =
 This
is identical to the formula for volumetric flux.
 Volumetric
flux and average flow velocity are two ways
of looking at the same thing.
 When
we speak of flux, we are focusing on a fixed location
and measuring the volume that passes through a unit area
in unit time at that location.
 When
we speak of velocity, we are focusing a moving
portion of water measuring its speed.
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Slide No. 27
Examples 4
The volumetric flow rate of beer flowing in a pipe is 1.8
L/s. The inside diameter of the pipe is 3 cm. The
density of beer is 1100 kg/m3.
(a). Calculate the average velocity of beer and its mass
flow rate in kg/s.
(b). If another pipe with a diameter of 1.5 cm is used,
what will be the velocity for the same volumetric flow
rate
Ans:
(a)
(b)
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0.007707 m2
2.55 m/s
1.98 kg/s
10.2 m/s
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Slide No. 28
Shear Stress



A=
<u> =
m=ρG=
<u>new =
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

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A deck of cards on a table and the top card pushed
sideways
The cards in the deck will slide over each other
Each card moves a little faster than the one below it
When a fluid moves, usually some portions of the fluid
move faster than others
Fluid is distorted as it undergoes shear
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Slide No. 29
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Slide No. 30
Shear Stress
Shear Stress and Shear Rate
A
F
F
Liquid Foods
Area

Force


Opposing forces needed to produce shear
They are parallel to the direction of shear, but not in line
with each other
By friction, the applied force is transmitted from layer
to layer throughout area A
Shear Stress = τ = F/A
U
y
SI Units:
Engineering Units:
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Slide No. 31
Shear Rate
Newton’s Law of Viscosity
Slide No. 32
• Newton observed that if the shear stress is increased
(by increasing force, F), then the shear rate will also
increase in direct proportion.
3.1 m/s
0.02m
2.4 m/s
Shear rate =
More friction within thick oil than within water.
At the same shear stress, water will shear (deform) faster
than oil.
 The ratio: shear force/shear rate  an indicator of flowability.


Shear rate is the relative change in velocity divided
by the distance between the plates.
SI Units:
SI Units:
Conversion:
1 cP = 1 mPa.s
Engineering Units:
cgs Units:
Engineering units:
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Slide No. 33
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Slide No. 34
Viscosity
Newton’s Law of Viscosity
 m is the coefficient of viscosity, or viscosity of
the fluid. It is also called “absolute” or
“dynamic” viscosity.
 Fluids that exhibit a direct proportionality
between shear stress and shear rate, are called
Newtonian Fluids.
 Viscosity is a physical property of the fluid and
it describes the resistance of the material to
shear-induced flow. Furthermore, it depends on
the physico-chemical nature of the material and
the temperature.
 The shear rate is proportional to the shear stress
 Viscosity is the proportionality constant
Shear stress F/A
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µ= Viscosity
of oil
µ= Viscosity of
water
Shear rate dVx/dy
Water, ethanol,.., obey Newton’s law
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Slide No. 36
Slide No. 35
Momentum transport
Viscosity

Viscosity is the measure of the internal friction of
a fluid

The greater the friction, the greater the amount of
force required to cause this movement, which is
called shear

Shearing occurs whenever the fluid is physically
moved or distributed, as in pouring, spreading,
spraying, mixing, etc.

Highly viscous fluids, therefore, require more
force to move than less viscous materials
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The bottom plate remains
stationary, the top plate slides
sideways at a constant velocity
Assuming:
 The temperature remains
constant
 The fluid between the plates
undergoes laminar flow
 Movement is in one direction
only
 This process continues for a
long time so that the system
reaches a steady state
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Slide No. 37
Momentum transport
• The fluid layer next to the moving plate has acquired a certain
momentum in the X- direction.
• Because of friction within the fluid, some of this momentum is
transferred to the adjoining layer.
• This layer in turn, transfers some of its momentum to the next layer,
etc.
• Momentum in the X-direction is transferred from layer to layer in the
Y-direction. The result is a gradual change in momentum from top to
bottom along a momentum gradient. We call this process momentum
transport.
Moving plate
Y
Momentum
Flux
y4
y3
y2
y1
Stationary plate
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Slide No. 39
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Kinematic Viscosity
Kinematic Viscosity, n , is defined as ratio of
dynamic viscosity to fluid density
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Slide No. 40
Approximate Viscosities of Common Materials
(At Room Temperature-70°F) *
Material
Water
Milk
SAE 10 Motor Oil
SAE 20 Motor Oil
SAE 30 Motor Oil
SAE 40 Motor Oil
Castrol Oil
Karo Syrup
Honey
Chocolate
Ketchup
Mustard
Sour Cream
Peanut Butter
Viscosity (Pa.s)
1.9 x 10-5 (0.000019)
1 x 10-3 (0.001)
0.1
1
8
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where τyx represents shear stress in the y direction due to movement in the x
direction. But shear stress is
(Momentum Flux) = (Viscosity)*(Velocity Gradient)
The viscosities of common
substances
Substance
Air (at 18 oC)
Water (at 20 oC)
Canola Oil at room temp.
Motor Oil at room temp.
Corn syrup at room temp.
According to Newton's law, the shear stress between any two vectors is:
In other words, τyx represents the transfer (in the y direction) of momentum (in
the x direction) per unit time per unit area or the rate of movement of momentum
per unit area.
This is the momentum flux of the system. Therefore, Newton’s Law tells us that
Momentum
Gradient
Vx1
X
Slide No. 38
Vm
Vx4
Vx3
Vx2
Momentum transport
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Viscosity in Centipoise
1 cps
3 cps
85-140 cps
140-420 cps
420-650 cps
650-900 cps
1,000 cps
5,000 cps
10,000 cps
25,000 cps
50,000 cps
70,000 cps
100,000 cps
250,000 cps
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Slide No. 41
Slide No. 42
Problem: Shear stress in soybean oil
The distance between the two parallel plates is 0.00914 m
and the lower plate is being pulled at a relative velocity of
0.366 m/s greater than the top plate. The fluid used is
soybean oil with viscosity of 0.004 Pa.s at 303 K
• Calculate the shear stress and the shear rate
Unit:
Kinematic Viscosity = Momentum Diffusivity
• If water having a viscosity of 880x10-6 Pa.s is used
instead of soybean oil, what relative velocity in m/s
needed using the same distance between plates so that
the same shear stress is obtained? Also, what is the new
shear rate?
5A
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Slide No. 43
Fluid Classification
(Casson fluids)
(Catsup, molten chocolate, toothpaste)
(Pseudoplastics)
(Salad dressings, concentrated fruit juices,
dairy cream, fruit purees, biological fluids)
(Dilatent)
(Corn flour and sugar solutions,
wet beach sand)
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Slide No. 45
Slide No. 44
• If the increase in shear rate results in an
increase in apparent viscosity, then the liquid is
called a shear-thickening liquid or dialatant
liquid
• Examples of shear-thickening liquids include
60% suspension of corn starch in water;
Suspensions of sand in water
• With shear-thickening liquids, the apparent
viscosity increases with increasing shear rate
• The liquids become “stiffer” at higher shear
rates
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Herschel-Bulkley Model
Shear-Thickening Liquid
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Power Law Fluid
Apparent viscosity
Slide No. 46
• For power law fluids, the relationship between shear
stress and shear rate by equation:
where:
τ = m. γn
m = the coefficient of consistency
n = the flow behavior index.
Casson Model for Chocolate
Apparent Viscosity:
ηa = t/γ =(m γn-1)
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Slide No. 47
If n < 1, the curve bends
downward and the fluid is
pseudoplastic
n=1
Shear Stress
If n> 1, the curve bends
upward and the fluid is
dilatent
n>1
n<1
• τo = the yield stress, i.e.
the stress required
before any flow takes
place
• m = the slope of the
line after this threshold
has been exceeded
• Non-Bingham:
τ = τo + m.γ
Shear Stress
If n = 1, the fluid obeys
Newton's law; m = µ or
viscosity
Slide No. 48
Equations for Plastics
The power law
τo
Shear Rate
Shear Rate
τ = τo + m.γn
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General Equation for Viscosity
Slide No. 50
Slide No. 49
Time dependent behavior
τ = τo + m.γ
Time 0
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Thixotropic Materials
• Time-dependent non-Newtonian liquids obtain a
constant value of apparent viscosity only after a
certain finite time has elapsed after the
application of shear stress.
• Apparent viscosity decreases with duration of
stress
• Examples include certain types of starch pastes,
some clays, some drilling mud, many paints,
honey under certain conditions.
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Rheological classification of materials
Time 2
Shear Rate
Slide No. 51
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Time 1
Shear Stress
Newtonian
Shear Thinning
Shear Thickening
Bingham
Herschel-Bulkley
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n
1
<1
>1
1
<1
t0
0
0
0
>0
>0
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Rheopectic Materials
Slide No. 52
• Rheopecty or rheopexy is the rare property
of some non-Newtonian fluids to show a
time-dependent change in viscosity; the
longer the fluid undergoes shearing force,
the higher its viscosity.
• Apparent viscosity increases with duration of
stress.
• Rheopectic fluids, such as some lubricants,
thicken or solidify when shaken, whipped
cream.
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Slide No. 54
Slide No. 53
FLOW PATTERNS
Materials
Ideal Fluids
Viscoelastic Solids
Non-Newtonian
Newtonian
Time Independent
Pseudo-Plastic
Dilatant
Time Dependent
Bingham Plastic
Thixotropic
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Laminar flow: fluid flows in
parallel layers, with no
disruption between the layers.
Ideal Solids
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Turbulent flow: fluid undergoes
irregular fluctuations and mixing
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Slide No. 55
FLOW PATTERNS
Eddies and Cylinder Wakes
The Reynold’s Number
Slide No. 56
• Laminar flow: at low velocities ;
• Turbulent flow: at higher velocities.
• The transition: at lower velocities with fluids of
higher density.
• The transition: at lower velocities in pipes of
greater diameter. Larger diameter pipes provide
more room for lateral movement and is more
conducive to turbulent flow.
• The transition: at higher velocities with fluids of
greater viscosity. High viscosity fluids such as
thick oils stay in laminar flow longer than low
viscosity fluids such a water .
Re = 30
Re = 40
Re = 47
Re = 55
Re = 67
Re = 100
Tritton, D.J. Physical Fluid Dynamics, 2nd Ed.
Oxford University Press, Oxford. 519 pp.
Re = 41
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The Reynold’s Number
for Newtonian fluids
•
•
•
•
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Slide No. 57
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The Reynold’s Number
Slide No. 58
Example 5:
ρ = density of the fluid
D = the inside diameter of the pipe
<V> = the average velocity of the fluid
µ = the viscosity of the fluid.
A pipe with an inside diameter of 3.2 cm is carrying water at
20oC at a rate of 0.005 cubic meters per minute. What type
of flow is it exhibiting?
Solution 5:
The average velocity of the water is:
Re < 2100: laminar flow
Re > 4000: turbulent flow.
2100 < Re < 4000: unstable, the transition region
The Reynold's number is:
Re [=]
the water flow is in transition region
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The Reynold’s Number
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The viscosity:
µ = 0.469 cP = 0.460x10-3 Pa.s (at 60 °C)
The Reynold's number is:
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Slide No. 59
Example 6: A pipe with an inside diameter of 3.2 cm is
carrying water at 60oC at a rate of 0.005 cubic meters per
minute. What type of flow is it exhibiting?
Sol 6:
The average velocity of the water is:
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Slide No. 60
Example 7:
Calculate the Reynolds number for water flowing at 5 m3/h in
a tube with 2 in inside diameter if the viscosity and density
of water are 1 cP and 0.998 g/mL,respectively.
Convert the units
Q = 5 m3/h =
D = 2 in =
μ = 1cP =
ρ = 0.998 g/mL
to SI:
1.39x10-3
0.0508
1.0x10-3
998
m3/s
m
kg/m.sec
kg/m3
Re = 34,000
the water flow is in turbulent region
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Slide No. 61
Problem:
Milk is flowing at 0.12 m3 min-1 in a 2.5-cm
diameter pipe. If the temperature of the milk is
21°C, is the flow turbulent or streamline?
Slide No. 62
Problem:
A 3 cm inside diameter pipe is being used to pump
liquid food into a buffer tank. The tank is 1.5 m
diameter and 3 m high. The density of the liquid is
1040 kg/m3 and viscosity is 1600 x 10-6 Pa.s.
Given
Density of milk
=
Viscosity
=
1030
kg /m 3
2.12
cP
a. What is the minimum time to fill the tank with this
liquid food if it is flowing under laminar conditions in
the pipe?
b. What will be the maximum time to fill the tank if the
flow in the pipe is turbulent?
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Physical meanings of the Reynold’s Number
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Slide No. 63
Viscous Transport. When flow is laminar, momentum is passed from
layer to layer in the fluid by the frictional forces between the layers.
We call this viscous transport of momentum. We can also say that
viscous forces are transporting momentum.
Molecular Transport. When flow is turbulent, molecules move laterally
and in so doing carry momentum to different layers. We call this a
molecular transport of momentum. Since the molecules are
carrying kinetic energy, this can also say that kinetic or inertial
forces are transporting of momentum.
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Slide No. 64
Reynold’s Number calculation for different geometries
Equivalent Diameter:
For pipes with noncircular cross sections, the Reynold's
number can be computed with the equation:
De = the equivalent diameter = 4rh
rh = the hydraulic radius = Cross Section Area / Wetted
Primeter
inertial force:
kinetic force:
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Slide No. 65
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For Non-circular pipe
D =
e
4 Hydraulic radius
=
Slide No. 66
4(Cross-sectional area)
Wetted Perimeter
Example:
b
a
h
De = 4
L
De =
axb
2 ( a + b)
 D22 - ( D12 ) 
 = D2 - D1
 D2 + D1 
= 
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Reynold’s Number for
non-Newtonian fluids
Slide No. 67
Slide No. 68
Power Law Fluids:
d 1 = 10 cm
m = the consistency coefficient
De =
n = the flow behavior index
Generalized Reynolds number:
De =
D 2 = 1.2 m
De =
=
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0.875 m
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Slide No. 69
Power Law Fluids
m = the consistency coefficient
m
n
Egg
2.2
0.62
1.0
5x10-3
0.60
0.94
Pudding
Whey
Sauce
12.0
0.55
Tomato
10.5
0.45
Peach puree
7.2
0.35
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n=1
Newtonian
Fluid
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Blunt
Flow
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n >1
applesauce is a non-Newtonian fluid
m = 13
n = 0.3
D = 0.0508 m; A=0.00203 m2
<u> = 0.684 m/s
ρ = 1100 kg/m3
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Slide No. 71
Parabolic
Flow
V = constant
n<1
Solution 8:
Answer GRe = 67.8
As a function of Flow Behavior Index
Plug
Flow
Slide No. 70
Example 8
•
•
•
•
•
•
Velocity Profile
n=0
Food Engineering Principles
•Calculate the Reynolds number for applesauce flowing
at 5 m3/h in a tube with 2 in ( 1 inch = 2.54 cm) inside
diameter if the consistency index is 13 [Pa.s0.3], the flow
behavior index is 0.3, and the density is 1100 kg/m3.
n = the flow behavior index
Food
Dept Food Technology
Elongate
Flow
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Slide No. 72
Velocity Profile
• Plug flow: extremely viscous fluid. Internal friction
is high compared to friction between the fluid and
the wall
• Parabolic flow: newtonian fluids, parabolic shape
• Elongated flow: pseudoplastic fluids. apparent
viscosity decreases with increasing velocity, shear
rate in the center increases
• Blunt flow: dilatent fluid. apparent viscosity
increase with increasing velocity so that the shear
rate in the center increases
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Slide No. 73
Measurement of Viscosity
• Standard laboratory viscometers for
liquids
U-tube viscometers (Capillary Tube
Viscometer)
• Rotational viscometers
• Miscellaneous viscometer types
(Stabinger, Stormer, Bubble)
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Capillary Tube Viscometer
• U-tube viscometers
• These devices also are known as glass
capillary viscometers or Ostwald
viscometers, named after Wilhelm
Ostwald.
• Another version is the Ubbelohde
viscometer, which consists of a Ushaped glass tube held vertically in a
controlled temperature bath.
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Slide No. 75
Volumetric Flow Rate
Volume flow rate
m=
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p R 4D P
8m L
Slide No. 76
Cannon-Fenske Type
Capillary Viscometer
Pressure drop (DP) is sufficient to overcome
the shear forces within the liquid and
produce flow of a given rate.
V=
(Hagen-Poiseuille Eqn )
V=
p D P R4
Volume of Bulb V
=
Discharge Time t
measured
8 LV
Dept Food Technology
Slide No. 74
known
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Slide No. 77
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Operation of Capillary
Viscometer
Slide No. 78
• In one arm of the U is a vertical section of precise
narrow bore (the capillary).
• Above this is a bulb, with is another bulb lower
down on the other arm.
• In use, liquid is drawn into the upper bulb by
suction, then allowed to flow down through the
capillary into the lower bulb.
• Two marks (one above and one below the upper
bulb) indicate a known volume.
• The time taken for the level of the liquid to pass
between these marks is proportional to the
kinematic viscosity.
(Kinematic viscosity = dynamic viscosity / fluid density)
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Rotational Viscometer
Slide No. 79
Slide No. 80
Rotational Viscometer
where W = Torque, Nm
 s = Shear Stress, Pa

u = Radial Velocity, m/s
 ω = Angular Velocity, rad/s

L = spindle length
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Slide No. 81
Slide No. 82
Shear rate
Measured
Shear stress
where w = angular velocity
N = revolution per sec
Ri = the radius of inner cylinder
Ro = the radius of outer cylinder
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Slide No. 83
A variation of the coaxial-cylinder viscometer
is the single-cylinder viscometer. In this device,
a single cylinder of radius Ri is immersed in a
container with the test sample. Then the outer
cylinder radius Ro approaches infinity.
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Slide No. 84
Impeller Viscometer
- Turbine or other Impellers
- Torque and rotational speed measured
- Laminar conditions
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Slide No. 85
Effect of Temperature on Viscosity
Ln (µ)
As the temperature of a gas increases, the viscosity of
the gas increases linearly.
As the pressure of a gas increases, the viscosity of the
gas also increases linearly
For Liquids:
Ea/R
Pressure has a negligible effect on the viscosity of liquids
up to about 40 atmospheres.
Unlike gases, liquids decrease in viscosity as the
temperature increases. Remember that hot fudge
sauces flows better than the cold sauce.
During measurements of viscosity, extra care must be
taken to avoid temperature fluctuations.
1/T
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Slide No. 87
Viscosity of air and water
ToC
Slide No. 86
For Gases:
Arrhenius type equation:
A is a constant;
Ea = activation energy;
R = the gas constant;
T = the absolute
temperature.
Effect of Temperature on Viscosity
0
Water
(μ -cP)
1.787
Air
(μ -cP)
0.01716
20
1.0019
0.01813
40
0.6530
0.01908
60
0.4665
0.01999
80
0.3548
0.02087
100
0.2821
0.02173
Slide No. 88
Further Readings:
1. R. Paul Singh, Dennis R. Heldman. 2013.
Introduction to food engineering. Academic
Press. 5th Edition. (Ch. 2)
HW:
S&H: 2.1; 2.5
TO : 6.1
5
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