Fluid Flow.pptx

UNIT OPERATION OF
CHEMICAL ENGINEERING I
Hikmatun Ni’mah, S.T., M.Sc., Ph.D
Fluid Flow

1. Types of Fluid Flow and Reynold Number
A. Introduction and Types of Fluid Flow
• The principles of the static of fluids are almost an exact science.
• On the other hand, the principles of the motion of fluids are quite
complex. The basic relations describing the motions of a fluid are the
equations for the overall balances of mass, energy, and momentum.
• Two types of fluid flow that can occur: laminar and turbulent flow.

B. Laminar and Turbulent Flow
• When fluids move through a closed channel of any cross section,
either of two distinct types of flow can be observed according to the
conditions present. These two types of flow can be commonly seen in
a flowing open stream or river.
• When the velocity of flow is slow, the flow patterns are smooth and
without eddies or swirls present, which is called laminar flow or
viscous flow.
• However, when the velocity is quite high, an unstable pattern is
observed in which eddies or small packets of fluid particles are
present moving in all direction and at all angles. This flow pattern is
called turbulent flow.

Laminar and Turbulent Flow
4
Reynolds’ experiment for different types of flow
Turbulent flow
Laminar flow
B. Laminar and Turbulent Flow
• The existence of laminar and
turbulent flow is most easily
visualized by experiment of
Reynold.
• Water was allowed to flow at
steady state through a
transparent pipe with the flow
rate controlled by a valve at the
end of the pipe.
• A fine steady stream of dye-
colored water was introduced
from a fine jet and its flow
pattern observed.
• The velocity at which the flow changes is
known as the critical velocity.

C. Reynold Number
• Studies have shown that the transition from laminar to turbulent flow
in tubes is not only a function of velocity but also of density and
viscosity of the fluid and the tube diameter.
• These variables are combined into the Reynold number, which is
dimensionless.
• The instability of the flow that leads to disturbed or turbulent flow is
determined by the ratio of the kinetic or inertial forces to the viscous
forces in the fluid stream.
• The inertial force are proportional to ρv2 and the viscous forces to
μv/D, and the ratio is the Reynold number.

Reynolds Number


Dv
N 
Re
Where:
D= Diameter (m)
= fluid density (kg/m3)
µ=fluid viscosity (Pa.s)
v= average velocity (m/s), v=Q/A
D
ce
viscousFor
ce
kineticFor
N
/
2
Re




NRe < 2100  Laminar
NRe > 4000  turbulent
6
For a straight circular pipe:
• In between, which is called the transition region,
the flow can be viscous or turbulent, depending
upon the apparatus details, which cannot be
predicted.

Water at 303 K is flowing at the rate of 10 gal/min
in a pipe having an inside diameter (ID) of 2.067 in.
Calculate the Reynolds number!
Example:
Solution:
7

2. Overall Mass Balance and Continuity Equation
Introduction and Simple Mass Balances
• In fluid dynamics fluids are in motion. Generally, they are moved from
place to place by means of mechanical devices such as pumps or
blowers, by gravity head, or by pressure, and flow through systems of
piping and/or pressure equipment.
• Simple mass or material balances is:
input = out + accumulation
• Since, in fluid flow, we are usually working with the rates of flow and
usually at steady state, the rate of accumulation is zero and we
obtain:
rate of input = rate of out (steady state)

2. Overall Mass Balance and Continuity Equation
Introduction and Simple Mass Balances
• In Fig.2.6-1 a simple flow system is shown where the fluid enters section 1
with an average velocity v1 m/s and density ρ1 kg/m3. The cross-sectional
area is A1 m2. The fluid leaves section 2 with average velocity v2.
• The mass balance is:
Where,
m = kg/s
vρ is expressed as G = vρ , where
G is mass velocity or mass flux in
kg/s.m2.

Example: Flow of Crude Oil and Mass Balance
11

3. Overall Energy Balance
A. Introduction
• The energy-conservation equation is combined with the first law of
thermodynamics to obtain the final overall energy-balance equation.

B. Derivation of Overall Energy-Balance Equation
• The entity balance:
rate of entity output – rate of entity input + rate of entity accumulation = 0

     
 
     
     
 
 
 
 
 
 
rate of energy rate of energy in rate of energy out
accumulation by convection by convection
net rate of heat addition net rate of work
to the system from performed on the sys
the surroundings
 
 
 
 
 
tem (2-8)
by the surroundings

• The rate of energy input and output associated with mass in the control
volume.
• The mass added or removed from the system carries internal, kinetic, and
potential energy. In addition, energy is transferred when mass flows into
and out of the control volume.
• Net work is done by the fluid as it flows into and out of the control volume.
This pressure-volume work per unit mass of fluid is pV.
• The pV term and U term are combined using the definition of enthalpy, H.

C. Overall Energy Balance for Steady-State Flow System
17

Overall Energy Balance
Where: m1=m2=m=v1avA1  Eq. 1 is divided by m, thus:
1)


2
2
av
α= kinetic energy velocity
correction factor
 av
av
v3
3

 
Laminer flow  α=0.5
Turbulent flow  close to 1
2)
18

Solution:
0
(no external work)
1 (turbulent flow)
21

Example2.7-2:
25
H2 + v2
2/2α + z2g + Q = H1 + v1
2/2α + z1g + Ws
H2 - H1 + v2
2/2α - v1
2/2α + z2g - z1g = Ws - Q

H2 + v2
2/2α + z2g + Q = H1 + v1
2/2α + z1g + Ws
H2 - H1 + v2
2/2α - v1
2/2α + z2g - z1g = Ws - Q
• H1 water at 93.3 oC (from interpolation) = 390.8 kJ/kg.
• Density water at 93.3 oC (from interpolation), ρ = 963.04 kg/m3.
• Mass rate, m = (0.189 m3/min)(963.04 Kg/m3)(1 min/60 s) = 3.03 kg/s.
• Q = (704) (1/3.30) = 232.11 kJ/kg
• Ws = (1.49) (1/3.30) = 0.491 kJ/kg
• Potential energy: (z2-z1)g = (15.24) (9.8) = 149.352 m2/s2 = 149.352 J/kg = 0.149 kJ/kg
• Overall energy balance:
H2 - H1 + v2
2/2α - v1
2/2α + z2g - z1g = Ws - Q
H2 - H1 + 0 + (z2 - z1)g = Ws - Q
H2 – 390.8 kJ/kg + 0.149 kJ/kg = 0.491 kJ/kg – 232.11 kJ/kg
H2 = 159.032 kJ/kg  From steam table, t2 = (between 36 and 40) oC  by interpolation
SOLUTION:

4. Overall Mechanical-Energy Balance
1. A more useful type of energy balance for flowing fluids, especially
liquids, is a modification of the total energy balance to deal with
mechanical energy. Engineers are often concerned with this special type
of energy, called mechanical energy, which includes the work term,
kinetic energy, potential energy, and the flow work part of the enthalpy
term.
2. Mechanical energy  a form of energy that is either work or a form
that can be directly converted into work. Heat (Q) and internal energy
terms (U or H) (in energy-balance equation) cannot be simply converted
into work.
3. Energy converted to heat or internal energy is lost work or a loss in
mechanical energy which is caused by frictional resistance to flow.

It is convenient to write an energy balance in term of this loss, which is
the sum of all frictional losses per unit mass.
For the case of steady-state flow, when a unit mass of fluid passes from
inlet to outlet, the batch work done by the fluid, W’, is expressed as
This work W’ differs from the W in equation of first law of
thermodynamics, , which also includes kinetic and
potential-energy effect. Writing the first law of thermodynamics for this
case, where ΔE becomes ΔU,

The equation defining enthalpy, , can be written as
Substitution Eq. (2.7-23) into (2.7-24) and the combining the resultant with
Eq. (2.7-25), we obtain

Finally, we substitute Eq. (2.7-26) into (2.7-10, Eq. overall energy balance)
and 1/ρ for V, to obtain the overall mechanical-energy balance equation:
For English units, the kinetic- and potential-energy terms of Eq. (2.7-27)
are divided by gc.
If the fluid is an incompressible liquid, the integral becomes

Overall Mechanical-Energy Balance
Sum of all frictional losses
per unit mass
35
Overall energy balance

5. Bernoulli Eq. for Mechanical Energy Balance
0
(no external work)
0
(no friction)
1 (turbulent flow)
Bernoulli Eq. 45
In the special case where no mechanical energy is added and for no friction, then
Eq. (2.7-28) becomes the Bernoulli equation, Eq (2.7-32)

172.4 kN/m2
103.4 kN/m2
3.35 m

Solution:
Ws = - 209.2 J/kg
P1 = 103.4 kN/m2 = 103400 N/m2
P2 = 172.4 kN/m2 = 172400 kN/m2
Z1 as datum/reference point, so Z1 = 0
Z2 = 3.35 m
Flow is turbulent, hence α = 1
T = 303 K, ρ soybean oil = 919 kg/m3 (from App. A.4)
Mechanical energy balance:
(0 because exit and entrance pipes are the same diameter)

Solution:
9.8 (3.35) J/kg + (172400 – 103400)/919 J/kg + - 209.2 J/kg = 0
= 209.2 J/kg – 32.83 J/kg – 75.08 J/kg = 101.29 J/kg
(0 because exit and entrance pipes are the same diameter)

• One of the most important applications of fluid flow is the flow
inside circular conduits, pipes, and tubes.
• When fluid is flowing in a circular pipe and the velocities are
measured at different distances from the pipe wall to the center
of the pipe, it has been shown that in both laminar and turbulent
flow, the fluid in the center of the pipe is moving faster than the
fluid near the walls.
A. Velocity Profiles in Pipes
57
6. Design Equations for Laminar and Turbulent Flow in Pipes

58
Laminar Flow in Pipes
Velocity and momentum flux profiles for laminar flow in pipe

Laminar Flow in Pipes
• The picture above illustrates the state of laminar flow in a pipe with radius r.
• In laminar flow, there is a velocity gradient of flow at radius equal to r to
radius equal to 0, where at radius equal to 0 the velocity of flow is at its
maximum, whereas at radius equal to r velocity is at the smallest state, even
said v = 0.
• This is because at a radius equal to r, there is the greatest friction force, and
the effect of friction force decreases from the radius r to r = 0.
59

60
Velocity distribution of fluid across the pipe

Previous figure show the plot of the relative distance from the center of the
pipe versus the fraction of maximum velocity v’/vmax, where v’ is local
velocity at given position and vmax is the maximum velocity at the center of
the pipe.
For viscous or laminar flow, the velocity profile is a true parabola. The
velocity at the wall is zero.
Whereas in turbulent flow, the velocity distribution is not always 0 on the
pipe wall, it but can reach 0.4v from vmax.
61
Differences in velocity distribution in laminar and
turbulent flow

62
B. Pressure Drop and Friction Loss in Laminar Flow
Pressure drop and loss due to friction
1
When the fluid is in steady-state laminar flow in a pipe, the for a
Newtonian fluid the shear stress can be described by the following
equation:
Using this relationship and making a shell momentum balance on the
fluid over a cylindrical cell, by deriving it from the Hagen-Poiseuille
equation it can be derived to:
Pressure Loss

63
Friction Loss
The quantity (p1-p2)f or ∆pf is the pressure loss due to skin friction,
the for constant density, the friction loss Ff is:
This is the mechanical-energy loss due to skin friction for the pipe in
N.m/kg of fluid and part of total friction term for frictional losses in
the mechanical-energy-balance equation.

66
Penggunaan Friction Factor untuk Menghitung
Friction Loss pada Laminar Flow
 Previously we have known the equations for calculating pressure drop and
friction loss, using the difference in pressure from point 1 and point 2.
 In general, the pressure-drop and friction loss can also be calculated using
another parameter known as the Fanning friction factor (f).
 The friction factor is defined as the drag force per wetted surface unit
area (shear stress at the surface) divided by the product of density times
velocity head.

67
Pressure Drop, Friction Loss, and Friction Factor
in Laminar and Turbulent Flow
Laminar Flow Turbulent Flow
To determine the value of f in
turbulent flow, figure 2.10-3 is
used.

69
In Figure 2.10-3, we can know the value of the fanning
friction factor, both for laminar and turbulent flow.
First, we have to know the NRe of a fluid flow, then choose the
type of pipe and see its roughness value.
Then calculate the relative roughness value so that it can be
determined on which line we will determine the friction factor
value.

73
In problem involving the friction loss in pipes
Case A
Known:
1. D
2. v
3. DL
Unknown: Ff
+
1.Calculate NRe
2.Calculate e/D
3.With known NRe and e/D
value, read f value from
Figure 2.10-3
4.Calculate Ff by using Eq.
Case B
Known:
1. Head of liquid
2. Q
3. DL
Unknown: D ???
+

74
Case B
Known:
1. Head of liquid
2. Q
3. DL
Unknown: D
1. Calculate Ff=g.h
2. Calculate velocity with unknown D
3. Calculate NRe with unknown D
4. Trial value of D
5. Substitute D value (Trial) to calculate Nre
6. Calculate e/D
7. Read f from Figure 2.10-3
8. Substitute f value to Eq Ff=gh=4fDL v2/(2D)
9. Check/Calculate D value
10. If the D value is not the same with D trial, we need to trial D
value again
+

C. Pressure Drop and Friction Factor in Flow of Gases
For gases, Eq. (2.10-5) can be written as follows for laminar
and turbulent flow:

80
D. Friction Loss
Losses in
straight pipe
Sudden
enlargement
losses
Sudden
contraction
losses
Losses in
fittings and
valves

Losses in Straight Pipe
81
1
Friction loss in straight pipe is the friction that occurs due to fluid
movement in a straight pipe. Friction loss can be calculated using the
fanning friction factor.
Fluid flow
Friction loss
In a straight pipe there is a friction force that can cause heat or
reduce speed.

Losses in Sudden Enlargement
2
If the cross section of a pipe grows gradually, no extra losses will occur. However, if there is a sudden
change, it will produce a separate eddy due to the expanding jet, resulting in extra losses.
The friction loss on enlarged suction can be written as:
1 2
Due to a sudden change in the size of
the pipe area (becoming bigger) there is
a change in speed to be smaller as well
as the force.

Losses in Sudden Contraction
3
If the cross section of a pipe shrinks suddenly, the flow cannot flow following
the sharp corner so that it will cause additional friction loss because of this.
The friction loss on enlarged suction can be written as:
α = 1 untuk aliran turbulen
α = 0.5 untuk aliran laminar

84
Sudden contraction Losses
When there is a narrowing of the pipe’s cross-sectional
area, there is an increase in speed and a reduction in
pressure.
1 2

Losses in Fittings and Valves
4
Pipe fittings and valves also interfere with the flow of the
pipe and cause additional losses. If the short pipe has a lot
of fittings, the friction loss caused by the fitting will be
greater than that caused by the straight pipe.
Friction loss on fittings and valves follows the equation:
The value of Kf can be seen
at following table:

88
Fitting and valve
Elbow 45O

89
Fitting and valve
Elbow 90O

91
Fitting and valve
Return Bend

94
Fitting and valve
Gate Valve

95
Fitting and valve
Globe Valve

96
Fitting and valve
Angle Valve

97
Fitting and valve
Check Valve

98
Fitting and valve
Water meter

100
Total Friction loss F

1.Contraction loss at tank exit
2.Friction in the 4-in, straight
pipe
3.Friction in 4-in elbow
4. Contraction loss from 4-in to 2-
in pipe
5. Friction in the 2-in, straight
pipe
6. Friction in the two 2-in elbows

101
fittings and valves that should be calculated

List of Groups
Group 1:
Presentation Topic: Pump
(centrifugal and positive
displacement pumps)
Group 2:
Presentation Topic: Compressor
Group 3:
Presentation Topic: Blower
Group 4:
Presentation Topic: Fan

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