Fluid Flow of Food Processing
Production Line for milk processing
Fluid Mechanics Overview
Fluid Mechanics
Gas
Liquids
Statics
F  0
i
Air, He, Ar,
N2, etc.
Compressibility Density
Introduction
Water, Oils,
Alcohols,
etc.
Viscosity
Dynamics
 F  0 , Flows
Stability
Pressure Buoyancy
i
Compressible/
Incompressible
Surface
Laminar/
Tension
Turbulent
Steady/Unsteady
Vapor
Viscous/Inviscid
Pressure
Fluid Statics
Fluid Dynamics
Characteristics of Fluids
•
•
•
•
Gas or liquid state
“Large” molecular spacing relative to a solid
“Weak” intermolecular cohesive forces
Can not resist a shear stress in a stationary
state
• Will take the shape of its container
• Generally considered a continuum
• Viscosity distinguishes different types of fluids
Measures of Fluid Mass and Weight: Density
The density of a fluid is defined as mass per unit volume.
m

v
m = mass, and v = volume.
•Different fluids can vary greatly in density
•Liquids densities do not vary much with pressure and temperature
•Gas densities can vary quite a bit with pressure and temperature
•Density of water at 4° C : 1000 kg/m3
•Density of Air at 4° C : 1.20 kg/m3
Alternatively, Specific Volume:

1

Measures of Fluid Mass and Weight: Specific Weight
The specific weight of fluid is its weight per unit volume.
  g
g = local acceleration of gravity, 9.807 m/s2
•Specific weight characterizes the weight of the fluid system
•Specific weight of water at 4° C : 9.80 kN/m3
•Specific weight of air at 4° C : 11.9 N/m3
Measures of Fluid Mass and Weight: Specific Gravity
The specific gravity of fluid is the ratio of the density of the fluid
to the density of water @ 4° C.
SG 

H O
2
•Gases have low specific gravities
•A liquid such as Mercury has a high specific gravity, 13.2
•The ratio is unitless.
•Density of water at 4° C : 1000 kg/m3
Viscosity: Kinematic Viscosity



•Kinematic viscosity is another way of representing
viscosity
•Used in the flow equations
•The units are of L2/T or m2/s and ft2/s
Surface Tension
At the interface between a liquid and a gas or two immiscible liquids, forces
develop forming an analogous “skin” or “membrane” stretched over the
fluid mass which can support weight.
This “skin” is due to an imbalance of cohesive forces. The interior of the fluid is
in balance as molecules of the like fluid are attracting each other while on the
interface there is a net inward pulling force.
Surface tension is the intensity of the molecular attraction per unit length along
any line in the surface.
Surface tension is a property of the liquid type, the temperature, and the other
fluid at the interface.
This membrane can be “broken” with a surfactant which reduces the surface
tension.
Surface Tension: Liquid Drop
The pressure inside a drop of fluid can be calculated using a free-body diagram:
Real Fluid Drops
Mathematical Model
R is the radius of the droplet, s is the surface tension, Dp is the pressure
difference between the inside and outside pressure.
The force developed around the edge due to surface tension along the line:
Applied to Circumference
F
 2Rs
surface
This force is balanced by the pressure difference Dp:
Fpressure  DpR
2
Applied to Area
Surface Tension: Liquid Drop
Now, equating the Surface Tension Force to
the Pressure Force, we can estimate
Dp = pi – pe:
2s
Dp 
R
This indicates that the internal pressure in the
droplet is greater that the external pressure since
the right hand side is entirely positive.
Surface Tension: Capillary Action
Capillary action in small tubes which involve a liquid-gas-solid
interface is caused by surface tension. The fluid is either drawn up
the tube or pushed down.
“Wetted”
“Non-Wetted”
Adhesion
Cohesion
Adhesion
Cohesion
Adhesion > Cohesion
Cohesion > Adhesion
h is the height, R is the radius of the tube, q is the angle of contact.
The weight of the fluid is balanced with the vertical force caused by surface
tension.
Surface Tension: Capillary Action
Free Body Diagram for Capillary Action for a Wetted Surface:
Fsurface  2Rs cosq
W  R2h
Equating the two and solving for h:
2s cosq
h
R
For clean glass in contact with water, q  0°, and thus as R decreases, h
increases, giving a higher rise.
For a clean glass in contact with Mercury, q  130°, and thus h is negative
or there is a push down of the fluid.
Pressure
Pressure is the force on an object that is spread over a surface area. The
equation for pressure is the force divided by the area where the force is applied.
Although this measurement is straightforward when a solid is pushing on a
solid, the case of a solid pushing on a liquid or gas requires that the fluid be
confined in a container. The force can also be created by the weight of an
object.
Pressure =
F  mg
F
A
Unit of pressure is Pa
Force Equilibrium of a Fluid
Element
• 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.
• So, in fluid static, which is the state of fluid in which
the shear stress is zero throughout the fluid volume.
• In a stationary fluid, the most important variable is
pressure.
• 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. This statement is
known as Pascal’s law
Force Equilibrium of a Fluid
Element
Fluid surfaces
Figure 2.1 Pressure acting uniformly in all directions
Figure 2.2: Direction of fluid pressures on boundaries
Standard Atmosphere
Patm
Patm
1 atm
P-vapor
h = 76 cm
Mercury
Mercury barometer
=
101325 Pa
=
760 mmHg
=
760 torr
=
1 bar
Pabs = Pgages + P
atm
Pgages
Patm
 A pressure is quoted in its gauge value, it usually refers to a
standard atmospheric pressure p0. A standard atmosphere is
an idealised representation of mean conditions in the earth’s
atmosphere.
 Pressure can be read in two different ways; the first is to
quote the value in form of absolute pressure, and the second
to quote relative to the local atmospheric pressure as
reference.
 The relationship between the absolute pressure and the
gauge pressure is illustrated in Figure 2.6.
 The pressure quoted by the latter approach (relative to
the local atmospheric pressure) is called gauge
pressure, which indicates the ‘sensible’ pressure since
this is the amount of pressure experienced by our
senses or sensed by many pressure transducers.
 If the gauge pressure is negative, it usually represent
suction or partially vacuum. The condition of absolute
vacuum is reached when only the pressure reduces to
absolute zero.
Pressure Measurement
Based on the principle of hydrostatic pressure distribution, we can
develop an apparatus that can measure pressure through a column
of fluid (Fig. 2.7)
Pressure Measurement
We can calculate the pressure at the bottom surface which has to
withstand the weight of four fluid columns as well as the
atmospheric pressure, or any additional pressure, at the free
surface. Thus, to find p5,
Total fluid columns = (p2 – p1) + (p3 – p2) + (p4 – p3) + (p5 – p4)
p5 – p1 = og (h2 – h1) + wg (h3 – h2) +gg (h4 – h3) + mg (h5 – h4)
The p1 can be the atmospheric pressure p0 if the free surface at z1 is
exposed to atmosphere. Hence, for this case, if we want the
value in gauge pressure (taking p1=p0=0), the formula for p5
becomes
p5 = og (h2 – h1) + wg (h3 – h2) + gg (h4 – h3) + mg (h5 – h4)
The apparatus which can measure the atmospheric pressure is called
barometer (Fig 2.8).
For mercury (or Hg — the chemical symbol for mercury), the height
formed is 760 mm and for water 10.3 m.
patm = 760 mm Hg (abs) = 10.3 m water (abs)
By comparing point A and point B, the atmospheric pressure in the SI
unit, Pascal,
pB = pA + gh
pacm = pv + gh
= 0.1586 + 13550 (9.807)(0.760)
 101 kPa
This concept can be extended to general pressure measurement using
an apparatus known as manometer. Several common manometers
are given in Fig. 2.9. The simplest type of manometer is the
piezometer tube, which is also known as ‘open’ manometer as
shown in Fig. 2.9(a). For this apparatus, the pressure in bulb A can
be calculated as:
pA = p 1 + p0
= 1gh1 + p0
 Here, p0 is the atmospheric
pressure. If a known local
atmospheric pressure value is
used for p0, the reading for pA
is in absolute pressure. If only
the gauge pressure is required,
then p0 can be taken as zero.
Although this apparatus (Piezometer) is simple, it has limitations,
i.e.
It cannot measure suction pressure which is lower than the
atmospheric pressure,
The pressure measured is limited by available column height,
It can only deal with liquids, not gases.
The restriction possessed by the piezometer tube can be overcome
by the U-tube manometer, as shown in Fig. 2.9(b). The U-tube
manometer is also an open manometer and the pressure pA can be
calculated as followed:
p2 = p3
pA + 1gh1 = 2gh2 + p0
 pA = 2gh2 - 1gh1 + p0
If fluid 1 is gas, further simplification can be made since it can be
assumed that 1   2, thus the term 1gh1 is relatively very small
compared to 2gh2 and can be omitted with negligible error.
Hence, the gas pressure is:
pA  p2
= 2gh2 - p0
There is also a ‘closed’ type of manometer as shown in Fig. 2.9(c),
which can measure pressure difference between two points, A and
B. This apparatus is known as the differential U-tube manometer.
For this case, the formula for pressure difference can be derived
as followed:
p2 = p3
pA + 1gh1 = pB + 3gh3 + 2gh2
 pA - pB = 3gh3 + 2gh2 - 1gh1
Piezometer tube
Open
h
P  gh
  Density
(kg / m 3 )
g  9.81
m2 / s
h  higth level of fluid (m)
Pabs = Pgages + P
atm
U-tube manometer
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
P  gh
  Density
(kg / m 3 )
g  9.81
m2 / s
h  higth level of fluid (m)
FB
=
Wfluid
Example
•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 gauge
pressure at the interface of the gasoline and water and
at the bottom of the tank. Express the pressure in Pascal
and as a pressure head in metres of water. Use
water = 998 kg/m3 and g = 9.81 m/s2.
For gasoline:
g = 0.68(998) = 678.64kg/m3
At the free surface, take the atmospheric pressure to be zero, or p0 = 0
(gauge pressure).
p1 = p0 + pgghg = 0 + (678.64)(9.81)(5.5)
= 36616.02 N/m2 = 36.6 kPa
The pressure head in metres of water is:
h1 = p1 – p0 = 36616.02 - 0
pwg
(998)(9.81)
= 3.74 m of water
At the bottom of the tank, the pressure:
p2 = p1 + pgghg = 36616.02 + (998)(9.81)(1)
= 46406.4 N/m2 = 46.6 kPa
And, the pressure head in meters of water is:
h2 = p1 – p0 = 46406.4 - 0
pwg
(998)(9.81)
= 4.74 m of water
Strain and Strain (Shear) Rate
• Strain
– a dimensionless quantity representing the
relative deformation of a material
Normal Strain
Shear Strain
Shear Stress is the intensity of force per unit area, acting tang
Simple Shear Flow
Solids: Elastic and Shear Moduli
• When a solid material is exposed to a
stress, it experiences an amount of
deformation or strain proportional to the
magnitude of the stress
• Stress (s)  Strain ( or )
• Stress = Modulus  Strain
• Normal stress: elastic modulus (E)
• Shear stress: shear modulus (G)
Fluid Viscosity
• Newtonian fluids
– viscosity is constant (Newtonian viscosity, )
s  
• Non-Newtonian fluids
– shear-dependent viscosity (apparent
viscosity,  or a)
s
  f ( ) 

Viscosity: Introduction
The viscosity is measure of the “fluidity” of the fluid which is not
captured simply by density or specific weight. A fluid can not resist a
shear and under shear begins to flow. The shearing stress and
shearing strain can be related with a relationship of the following form
for common fluids such as water, air, oil, and gasoline:
 or s  
du
dy
 is the absolute viscosity or dynamics viscosity of the fluid, u is the
velocity of the fluid and y is the vertical coordinate as shown in the
schematic below:
“No Slip
Condition”
Viscosity
Viscosity is a property of fluids that indicates resistance to flow. When a force is applied to a
volume of material then a displacement (deformation) occurs. If two plates (area, A),
separated by fluid distance apart, are moved (at velocity V by a force, F) relative to each
other,
dv
s  
dy
y=0
s  
dv
dy
y = y-max

= Coefficient Viscosity ( Pa s)
Newton's law states that the shear stress (the force divided by area parallel to the force, F/A)
is proportional to the shear strain rate . The proportionality constant is known as the
(dynamic) viscosity
Shear stress

dv
s  
dy
The unit of viscosity in the SI system of units is
pascal-second (Pa s)
Shear rate
In cgs unit , the unit of viscosity is expressed
as poise
1 poise = 0.1 Pa s
1 cP = 1 m Pa s
Example : 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
• 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?
Kinematics Viscosity
Dynamic viscosity
Kinematics Viscosity =
Density




The unit of kinematics viscosity are m2 /s
Reynolds Number
Inertial Forces
Kinematics =
Viscous Forces
vD
Re 

o
Re 
4m
 D

= Density of liquid kg / m

= Density of liquid Pa s
V
= Flow Velocity
D
= Diameter of the pipe
m/s
o
m
3
= mass flow rate kg /sec
m
Experiment for find Re
Laminar flow
Turbulent flow
Re  2100
Re  10,000
Example
At what velocity does water flow convert
from laminar to transition flow in a 5 cm
diameter pipe at 20 C
Properties of water @ 20 C
Milk is flowing at 0.12 m 3 min -1 in a 2.5-cm diameter pipe. If the temperature
of the milk is 21°C, is the flow turbulent or streamline
Given
Density of milk
=
Viscosity
=
1030
2.12
kg /m 3
cP
Velocity Profile in a Liquid Flowing under Fully Developed Flow Condition
Circumferential area A  2  rL
(2 rL)s
2
r
 r 2 ( P1 )  r ( P2 )
 r 2 ( P2 )
 r 2 (P2  P1 )  (2 rL)s
DP x  r 2
 s (2 rL)
The velocity profile for a larminar , fully develop flow in horizontal pipe is
v(r ) 
DP R 2
4 L
  r 2 
1    
  R  
Measurement of Viscosity
Viscosity of a liquid can be measurement
Capillary Tube Viscosity
Rotational Viscometer
Viscosity: Measurements
A Capillary Tube Viscosimeter is one method of measuring
the viscosity of the fluid.
Viscosity Varies from Fluid to Fluid and is dependent on
temperature, thus temperature is measured as well.
Units of Viscosity are N·s/m2 or lb·s/ft2
Movie Example using a Viscosimeter:
Capillary Tube Viscometer
Rotational Viscometer
Influence of Temperature on Viscosity
There is considerable evidence that the influence of temperature on viscosity for
liquid food may be described by an Arrhenius type relationship
ln 
 ln BA 
Ea
RgTA
Energy Balance
In addition to the mass balance, the other important quantity we must consider in
the analysis of fluid flow, is the energy balance. Referring again to this picture ,
we shall consider the changes in the total energy of unit mass of fluid, one
kilogram, between Section 1 and Section 2
There may be energy interchange with the
surroundings including:
(4) Energy lost to the surroundings due to friction.
(5) Mechanical energy added by pumps.
(6) Heat energy in heating or cooling the fluid.
Firstly, there are the changes in the intrinsic
energy of the fluid itself which include changes
in:
(1) Potential energy.
(2) Kinetic energy.
(3) Pressure energy.
Bernoulli Equation
Bernoulli’s equation is a consequence of Conservation of Energy applied to an
ideal fluid Assumes the fluid is incompressible and non-viscous, and flows in a
non-turbulent, steady-state manner
P
v2

 z  Constant
 g 2g
States that the sum of the pressure, kinetic energy per unit volume, and the
potential energy per unit volume has the same value at all points along a
streamline
A 3 m diameter stainless steel tank contains wine. In the tank , the wine is filled to
5 cm depth. A discharge port, 10 cm diameter , is opened to drain the wine.
Calculate the discharge velocity of wine ,assuming the flow is steady and frictionless
and the time required in emptying it
3m
2
Location 1
P2
v
 2  z2
 g 2g
v2

2g ( z2  z1 )
v2

2 x 9.81x (5  0)
D = 10 cm
5m
 9.9 m / s
Location 2
Volume of tank

4
D L 
2

2
x3 x5
4
 35.3 m 2
2
P1
v
 1  z1 
 g 2g
Then volumetric flow rate from the discharge port using
Q =AV =
Time to empty the tank =

4
x 0.12 x 9.9 m / s = 0.0078 m 3 /s
35.3 m
3
3
0.0078 m / s
=
452.6 sec or 7.5 min
Water flows at the rate of 0.4 m3 / min in a 7.5 cm diameter pipe at a pressure of 70
kPa. If the pipe reduces to 5 cm diameter calculate the new pressure in the pipe
P = 70 kPa
P = ? kPa
D = 7.5 cm
D = 5 cm
A = 4.42 x 10 -3 m 2
A = 1.90 x 10 -3 m 2
2
2
P1
v1

 z1 
 g 2g
P2
P2
 70 kPa 
 P1 


P2
v2

 z2
 g 2g

v
2
2
1
1000
1.512  3.54 2
2
 v22


P2 = 65.3 k Pa.
Forces due to Friction
When a fluid moves through a pipe or through fittings, it encounters frictional
resistance and energy can only come from energy contained in the fluid and so
frictional losses provide a drain on the energy resources of the fluid. The actual
magnitude of the losses depends upon the nature of the flow and of the system through
which the flow takes place. In the system, let the energy lost by 1 kg fluid between
section 1 and section 2, due to friction, be equal to Eƒ (J).
Common parameter used in laminar and turbulent flow is
Fanning friction factor (ƒ)
ƒ is defined as
drag force per wetted surface unit area
product of density times velocity head
=
s
½v2
Major loss
DPf
u2L
Ef 
 2f

D
2
u L
h f  2f
Dg
16
16

f =
 uD N Re
D

Type of pipe
Roughness ()
Ft
m
Riveted steel
0.003-0.03
0.0009-0.009
Concrete
0.001-0.01
0.0003-0.003
0.0006-0.003
0.0002-0.0009
Cast iron
0.00085
0.00026
Galvanized iron
0.0005
0.00015
Asphalted cast iron
0.0004
0.0001
Commercial steel or wrought
iron
0.00015
0.000046
Drawn brass or copper tubing
0.000005
“smooth”
0.0000015
“smooth”
Wood stave
Glass and plastic
Moody Diagram for the Fanning friction factor
Moody Diagram for the Moody friction factor
Laminar Zone
Turbulent Zone
Laminar
Lamina Zone
f
Turbulent Zone

16
Re
Transition
f 
0.33137
  
5.74
ln

  3.7 D Re 2 

 
2
Water at 30 C is being pumped through a 30 m section of 2.5 cm. diameter
steel pipe at a mass flow rate 2.5 kg/s. Compute the pressure loss due to
friction in pipe section
o
m = 2.5 kg/s
30 m
3
Density and viscosity of water@30 C are 997.5 kg/m and 0.000792 Pa s
m
2.5kg / s

 4.092 m / s
A
3 
2
(995.7kg / m )( 0.025 )
4
0
4m
4 x 2 kg / s

 128,550
Re =
 D
(0.000792Pa s)( x 0.0252 )
v 
D

45.7 x106
0.025
 1.828x103
1.828 x 10

f = 0.006
-3
DP  2 f
L
 v2
D
 2 x0.006 x(30 / 0.025 )( 995 .7)( 4.092 )
= 240.08 kPa
128550
A liquid is flowing through a horizontal straight commercial
steel pipe at 4.57 m/s. The inside diameter of the pipe is
2.067 in. The viscosity of the liquid is 4.46 cP and the
density 801 kg/m3. Calculate the mechanical-energy friction
loss in J/kg for a 36.6 m section of pipe.
Given : roughness of commercial steel pipe is 4.6 X 10-5 m.
For Non-circular pipe
D =
e
4 Hydraulic radius
=
4(Cross-sectional area)
Wetted Perimeter
b
a
De = 4
axb
2 ( a + b)
 2  2 
 4 D2  4 D1 
De = 4 


D


D
1
2 



 D22 - (D12 ) 
= 
 = D2 - D1
 D2 + D1 
d 1 = 10 cm
 2 
 2
D

4
(
d1 ) 
4 2
4
De = 4 


D

4

d
2
1 



De =
 D22  4(d12 ) 


D

4
d
1 
 2
De =
1.2 2  4 (0.12 ) 


 1.2  4 (0.1) 
D 2 = 1.2 m
=
0.875 m
Water at 30 C is being pumped through a 30 m section of annular area of steel
pipe at a mass flow rate 2 kg/s. Compute the pressure loss due to friction in
pipe section
2.5 cm
Determine Equivalent Diameter
De
=
=
=
 D22  (d12 ) 


D

d
 2 1 
 0.052  (0.0252 ) 


0
.
05

0
.
025


0.025 m
5 cm
Determine Reynolds Number and
Re 

D

vD
4 x2 kg / s

 128,550

(0.000792Pa s)( x 0.0252 )
45.7 x106
0.025
 1.828x103
Determine Friction factor
1.828 x 10
f = 0.006
-3
128550
DP  2 f
L
 v2
De
 2 x0.006x(30 / 0.025)(995.7)(4.092)
= 240.08 kPa
Energy Equation for Steady Flow of Fluids
elevation head
2
1
2
2
v
p1 v
p2
 y1  z1  
 y2  z2   hL,12
2g
γ 2g
γ
velocity head
pressure head
Energy loss between
sections 1 and 2
Frictional Energy Loss
The frictional energy loss for a liquid flowing in pipe is composed of major and
minor loses
Ef
=
E f , Major + E f , Minor
The major losses are due to the flow of viscous
liquid in the straight portions of a pipe
The minor losses are due to various components
used in pipeline system such as values , tees and
elbow and contraction of fluid
E Minor = E contraction + E expansion + E fitting
General Equation for Minor Losses
hL m 
hL m
KV
2
2g
2 fLeV 2

Dg
hLm = minor loss
K = minor loss coefficient
Le = equivalent length
Expansion and Contraction
b1
width 1
b2
1
Contraction
b1 > b 2
Expansion
b1 < b 2
width 2
2
Transition Loss Equations
2
2
V2
A2 V2
• Contraction: h  Ki
 0.55(1  )
2g
A1 2 g
• Expansion:
2
1
2
1
V
A1 2 V
h  Ke
 (1  )
2g
A2 2 g
Pipe Bends
Friction loss of turbulent flow through valves
and fittings
Example : Friction losses and mechanicalenergy balance
An elevated storage tank contains water at 82.2ºC as
shown in figure. It is desired to have a discharge rate at
point 2 of 6.315 X 10-3 m3/s. What must be the height H
in m of the surface of the water in the tank relative to the
discharge point? The pipe used is commercial steel pipe.
15.2 m
6.1 m
3m
38.1 m
Pump
Power and work:
• Using the total mechanical-energy-balance equation on
pump and piping system, the actual or theoretical
mechanical energy (Ws) added to fluid by the pump can be
calculated.
• If  = fractional efficiency and Wp = the shaft work
delivered to the pump
Wp = Ws/ 
• The mechanical energy Ws in J/kg added to fluid often
expressed as the developed head (H) of pump in m of fluid
Wp = H.g.mº/ 
Suction lift and cavitation
• Power can be calculated from difference in pressure
between discharge and suction.
• Practically, lower limit of suction pressure is fixed by
vapor pressure (corresponding to temperature at suction). If
suction pressure is equal to vapor pressure, liquid flashes
into vapor. This process call cavitation that can occur in
suction line and no liquid can be drawn into pump. This
process can cause severe erosion and mechanical damage to
pump.
• Therefore, it should have net positive suction head
(NPSH)
Net positive suction head (NPSH)
• NPSH = positive difference between pressure at pump
inlet and vapor pressure of liquid being pumped to prevent
cavitation.
• NPSH :
• required NPSH : function of impeller design (its value
is provided by manufacturer)
• available NPSH : function of suction system.
NPSHa = ha – hvp – hs – hf
Where
ha
= absolute pressure
hvp
= vapor pressure
hs
= static head of liquid above center line of pump
hf
= friction loss
Centrifugal pump
• Pressure developed by rotating impeller.
• Impeller impact a centrifugal force on liquid entering the
center of impeller.
• Affinity laws : govern performance of centrifugal pumps at
various impeller speeds
Vº2 = Vº1(N2 /N1)
h2 = h1(N2 /N1)2
P`2 = P`1(N2 /N1)3
Where Vº = volumetric flow rate
h = total head
P` = power
Positive displacement
• Constant discharge pressure at difference flow rates.
• Regulation of flow rate is done by changing displacement
or capacity of intake chamber
Characteristic curve
• plot head, power consumption and efficiency with respect
to volumetric flow rate (capacity)
• use for rating pumps
Peak Efficiency
Head
Efficiency
Power requirement
Volumetric flow rate
(capacity)
Pipe flow problems
• Basically, there are 3 types of problems
– Direct solution : finding h or DP for given L, D,
v, f
– Finding v or Q for given L, D, h or DP, /D
• Trial and error type solution
• Use Re.f1/2 = constant (not depend on v or Q)
– Finding D for given L, Q, h or DP, /D
• Trial and error type solution
• Use Re.f1/5 = constant (not depend on D)
Assumption V
Trial and error
type solution
Determine Re
For finding v
Determine “ f ”
Determine “ V ”
Vnew
 Vold
NO
Vold
YES

Vnew
Example :Trial and error type solution
For finding v
2
Constant level
5m
1
12 m
P
Constant level
2
P1
v
 1  z1  h
 g 2g
2
P

P2
v2
L v2

 z2  2 f
 g 2g
D g

v2
L v2
 z2  2 f
2g
D g
2
h
P
2
12 
v2
30
 5 2 f
v22
2(9.81)
0.05x9.81
7  (0.05  122.5 f )v 2
v 
7
0.05  122.5 f
1 Equation 2 Variable cannot determine
Assumption V
Determine Re
Determine “ f ”
Determine “ V ”
v 
Vold
7
0.05  122.5 f
Vnew

Vold
NO
YES

Vnew
Guess V =3.059
Guess V = 1.5
Guess V = 3.164
Re =75361
Re =153678
Re = 158964
f =0.0057
f =0.0053
f = 0.0053
V new =3.059
1.5 =3.059
NO
v = 3.059
YES
V new =3.164
3.059 = 3.164
NO
v = 3.164
V new = 3.164
YES
3.164= 3.164
NO
V =3.164 m/s
YES
Whole milk flows through a horizontal stainless steel pipe ( = 4.2 x 10-6 m) , 30 m
long and having inside diameter D = 2.54 cm . Determine the flow rate of milk
when a 2 HP motor with 75 % efficiency is used.
Given
Density of whole milk = 1030 kg /m
Viscosity of whole milk = 0.00202 Pa s
Whole Milk
30 m
1
P
Diameter of pipe = 2.54 cm
Power = 2 HP
75 % efficiency
2
2
P1
v
 1  z1  hP
 g 2g
HP 
 g hp
xQ
746
0.75 x 2 HP = 1030 x 9.81 x hp  (0.0254 2 ) v
4
746
h p = 218.86
hP
2

P2
v2
L v2

 z2  2 f
 g 2g
D g
L v2
 2f
D g
2
218.86 = 2 x f x 30 x v
v
0.0254 x 9.81
218.86 = 240.8 f
v3
v
1/3
v =
0.91
f
Guess V = 5.06
Guess V = 1
Guess V = 5.59
Re = 12951.5
Re = 65534
Re = 72399
f = 0.007
f = 0.0052
f = 0.005
V new = 5.06
1 = 5.06
NO
v = 5.06
YES
V new = 5.59
5.06 = 5.59
NO
v = 5.59
V new = 5.66
YES
5.59 ~ 5.66
NO
V = 5.66 m/s
YES
Flow Measurement
Flow measurement is essential in many industries such as the oil, power,
chemical, food, water, and waste treatment industries. These industries require the
determination of the quantity of a fluid, either gas, liquid, or steam, that passes
through a check point, either a closed conduit or an open channel, in their daily
processing or operating. The quantity to be determined may be volume flow rate,
mass flow rate, flow velocity, or other quantities related to the previous three.
These methods include (a) Pitot tube, (b) Orifice plate and (c) Venturi tube are the
measurement involves pressure difference. Differential pressure flow meters
employ the Bernoulli equation that describes the relationship between pressure
and velocity of a flow. These devices guide the flow into a section with difference
cross section areas (different pipe diameters) that causes variations in flow velocity
and pressure. By measuring the changes in pressure, the flow velocity can then be
calculated. Many types of differential pressure flow meters are used in the industry.
The Pitot Tube
The Pitot tube is a widely used sensor to measure velocity of fluid
Total pressure
The principle is based on the Bernoulli
Equation where each term can be
interpreted as a form of pressure
Air
2
Static Pressure
Stagnation
point (v = 0)
Pa
v
 a  z1 
 g 2g
Va 
2 ( Pb  Pa )
Water
C d = Discharge Coefficient
2
Pb
v
 b  z2
 g 2g
 Cd
 air
2 ( Pb  Pa )
 air
V1  Cd
2 ( P2  P1 )
 air
 Cd
2  water g DH water
 air
 129.11Cd DH water
The Orifice Meter
A flat plate with an opening is inserted into the
pipe and placed perpendicular to the flow
stream. As the flowing fluid passes through the
orifice plate, the restricted cross section area
causes an increase in velocity and decrease in
pressure. The pressure difference before and
after orifice plate is used to calculate the flow
velocity.
V  Cd
D1
D2
PA - PB
2( PA  PB )
  D 4 
 f 1   2  
  D1  
The Venturi Meter
To reduce energy loss due to friction created by the sudden contraction in flow in
an orifice meter
  f

 2 g 
 1 z f
  

V  Cd 
4
 1   D2 
D 

 1






1/ 2
Variable-Area Meter
Variable area flow meter 's cross section area available to the flow varies with the
flow rate. Under a (nearly) constant pressure drop, the higher the volume flow rate,
the higher the flow path area.