CHAPTER ONE
1. Introduction to Fluid Machines
In this chapter the purpose of fluid machines, their application in process industries and
the role of the chemical engineer with respect to fluid machines is discussed. In addition
some basic relevant concepts and terminologies are explained.
Objectives: At end of this chapter students’ shall be able to:
 Identify the purpose of fluid machines and their applications in process industries,
 Recognize the role of chemical engineers with respect to fluid machines,
 Determine the specific work, total head, total pressure and useful power for fluid
machines.
1.1
Fluid Machines in Chemical Process Industries
In chemical and process industries it is usually required to increase the mechanical energy
of fluids. In moving fluids from one place to another it may be required to increase the
mechanical energy to account for increase in geodetic energy (potential energy), velocity
energy, pressure energy and losses due to fluid friction. In gasses the energy may be
related to moving the gas or compress it. The machines used for this purpose are called
fluid machines.
Fluid machines that are used to transport liquid are known as pumps. Fans, blowers and
compressors are used to increase the mechanical energy of gasses.
Pumps are used to transport process and service liquids. Fans are used in ventilating
buildings, aerating workstations, exhausting or introducing air or other gasses into
process reactors, dryers, cooling towers, kilns, etc., at relatively low pressure. Fans are
also used in some air-cooled heat exchangers.
Blowers: Fluid machines that are used to compress gases at low pressure relative to
compressors. Compressors are used to compress process gas, supplying plant with
1
compressed air, convey solid material in suspension, exhausting or introducing air into
process reactor at higher pressures.
2
The chemical engineer is involved in selecting, installing, operating, testing and maintaining
fluid machines depending on his/ her assignment. To do this effectively the engineer has to
know the system where the fluid machine is to be used and the operating principles,
capability and limitations of the different types of fluid machines. In moving fluids the most
important system parameters are the properties of the fluid (viscosity, density, composition,
vapor pressure, etc.), the flow rate, the specific energy requirement and the suction condition.
In case of compression the important parameters are the required flow rate, the property of the
gas (Molecular weight, specific heat capacity, and inlet and discharge temperature,
viscosity) and the compression ratio. The determination of these parameters is dealt with in
Fluid Mechanics, Thermodynamics and Substance Parameters.
1.2
Basic Concepts and Terminologies
Mechanical Energy: Energy of a total energy of the flowing fluid that can be converted to
work. It is the sum of the potential, kinetic pressure energy and loss energy of the flow
medium. Heat and internal energy cannot be completely converted to work. Energy that is
converted to heat or internal energy is lost work or loss in mechanical energy. Mechanical
energy is also known as useful energy.
ME=PE+KE+EP+EL
Where: ME, PE, KE Ep and EL are mechanical energy, potential energy, kinetic energy pressure
energy and loss energy, respectively. SI unit of energy is J or kg m2/s2.
Specific work: The specific work of a fluid machine is the useful energy (work) that the
machine transfers to the flow medium per unit mass of the fluid. The SI unit of specific energy is
J/kg or m2/s2. Specific work is represented by Y.
Total head: The total head transferred to a flow medium is the specific energy transferred to the
fluid divided by the gravitational acceleration. It is the measure of the amount of useful energy
(mechanical energy or work) of the flow medium. The term head is more commonly used in
centrifugal pumps. The SI unit of head is meter.
H 
Y
g
(1.1)
H=Head,
3
Y= Specific Work
g=gravitational acceleration.
The head of a pump is the specific energy that the pump can transfer to the flow liquid under
specified condition divided by the gravitational acceleration.
Example 1.1: The rate of energy transfer from a fluid machine to a flow medium is 40kW and
the mass flow rate of the flow medium is 2kg/s. Calculate the specific energy transferred to the
fluid.
Solution
N= 40kW= 40 kJ/s,
Specific Energy 
mass flow rate= 2kg/s
Total energy transfered per second 40 kJ/s

 20 kJ/kg
Massflow per second
2 kg/s
Example 1.2: Determine the head of the fluid machine in Example 1.1.
Solution
Y=20,000 J/kg=20,000 m2/s2
H
Y 20,000 m 2 /s 2

 2038.7m
g
9.81 m/s 2
Total pressure: Total pressure of a fluid machine is the specific energy that a fluid machine
transfers to a flow medium multiplied by the density of the flow medium. It is commonly used
for fans and positive displacement pumps and compressors.
Pt  ρY
(1.2)
Example 1.3: The static pressure, geodetic (potential) and kinetic specific energy transferred by
a fan to a gas are 500 J/kg, 22.5 J/kg and 25 J/kg, respectively. Determine the increase in total
pressure of the flow medium. The average density of the flow medium is 1.2 kg/m3.
Solution
Y=Ypr +Ygeo + Yvel
Y=500 + 22.5+ 25 =547.5 J/kg
Pt=  Y = 1.2  547.5 = 657 Pa
Useful power (N): The useful power is the rate at which useful energy is transferred to the flow
medium. The SI unit of useful power is kW which is (kJ/s). The useful power is calculated using
Equations 1.3 to 1.5.
4
N  m Y
(1.3)
Since mass flow rate is the product of density and volume flow rate.
N  QY
(1.4)
N  QPt
(1.5)
Using (1.2) in (1.4)
Pulsation: The capacity of some fluid machines is not uniform, it varies with time. Pulsation is this
non uniformity of the capacity fluid machines.
Priming: Some pumps require that the air in the suction line should be replaced by liquid
before they start pumping. The process of replacing the air in the suction pipe with liquid is
known as priming.
Absolute pressure: Absolute pressure (static) of a fluid on a surface is the normal force exerted
by the fluid per unit area of the surface.
Gauge Pressure: It is the pressure above the atmospheric pressure. Hence,
Absolute Pressure = Gauge Pressure + Atmospheric Pressure
Vacuum Pressure: It is the pressure below the atmospheric pressure.
Absolute Pressure = Atmospheric Pressure - Vacuum Pressure
Continuity Equation
A1
Input
A2
Process
Output
Figure 1.1 The continuity equation
For a steady state process (no accumulation)
Rate of mass input = Rate of mass output
  1Q1   2 Q2
m
(1.6a)
m   A c   A c
1
1
1
2
2
2
(1.6b)
For incompressible fluid
5
A1c1  A2 c2
(1.7)
Where
m =mass flow rate [kg/s]
Q=Volume flow rate [m3/s]
c= velocity of the flow medium [m/s]
A= Flow Area [m2]
Head Loss: It is loss of the useful head of the flow medium due to fluid friction or the
turbulence that occurs when the fluid passes an obstruction, sudden contraction or sudden
expansion, etc.
Temperature scales
There are two temperature scales in common use. These are the degree Fahrenheit and degree
Celsius.
(T ) 0 C 
5
((T ) 0 F  32)
9
(1.8)
(T ) 0 F 
9
((T ) 0 C )  32
5
(1.9)
Absolute Temperatures
Degree Kelvin (K)
(T ) K  (T ) 0C  273.15
(1.10)
(T ) 0R  (T ) 0F  460
(1.11)
Degree Rankine (0R)
Ideal Gas Law
PV  nRT
(1.12)
Where
P= Absolute Pressure [Pa]
V=Volume of Gas[m3]
n=number of moles of the gas [kmol]
R= universal gas constant =8314.3 J/kmol K
T=absoutr temperature[K]
1.3 Application of Fluid Machines
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Application of Pumps: Production processes in a process plant usually involve a number of unit
operations that are carried out at different places in the plant. Due to this, it is a common practice
to move process liquids and utilities from one place to another in which different unit operations
are involved in between.
In a wine factory for example, fresh grape is crushed and collected below the crushing machine
and sent to fermentation tanks. Wine is moved from tank to tank to separate the clear wine from
the unwanted settled mass. Fermented wine from the cellar is filtered and sent to tanks in the
filling room. From this temporary storage tank it should be pumped to the filling machine. This
movement of liquid requires energy at each stage. Figure 1.2 shows a schematic diagram that
illustrates a typical pumping system for moving liquid from Tank 1 to Tank 2 with a heat
exchanger in between.
P2
Hgeo
P1
P
Figure 1.2 A pumping system
The amount of energy required by a flow medium is the sum of the potential energy, kinetic
energy and pressure energy differences of the liquid between the suction and discharge points
and the energy loss due to friction loss in the pipe line and unit operations like the heat
exchanger in the mentioned example.
Y
P2  P1


c c
2
2
2
1
2
 eg  F
(1.13)
Y= The specific energy that should be transferred from the fluid machine to the
flow medium.
P2 - P1= The static pressure difference between the suction and discharge end of
the pumping system
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c1, c2 = The average flow velocities at point 1 and 2 respectively.
e = the elevation difference between 1 and 2.
F = Energy loss due to fluid friction
 = Density of the flow medium at the flow condition
The energy loss due to fluid friction in simple pipes is the sum of the friction losses in the
straight pipe and minor losses in pipe fittings like elbows, valves and losses due to sudden
contraction and sudden expansion. Equations 1.14 and 1.15 give the formulas for the straight
pipe friction loss and the minor losses respectively.
L c2
Fst , pipe  f
D 2
(1.14)
c2
Fst , pipe   k i 
2
(1.15)
It is not only the energy that is a requirement of the pump. In most processes it is required to
transport a defined amount of liquid per unit time. Hence, the capacity, i.e., the volume the pump
delivers per unit time is the other important performance characteristic of the pump. The rate of
energy transfer is known as power and the power that should be transferred from the fluid
machine to a flow medium is the product of the mass flow rate and the specific energy.
There are various types of pumps. Their difference is mainly due to the fact that different
liquid properties and different performance requirements result in different types of design.
For example, screw pumps are more effective in pumping viscous liquids than centrifugal
pumps. Similarly, temperature, vapor and gas content and solid content are important fluid
properties in this respect. On the other hand, the capacity and head, uniformity of delivery, price,
operating cost, maintenance cost, space requirement, and weight are some of the important
process requirements that may lead to different choices of pumps.
Example 1.4: Water at 200C should be pumped from Tank 1 to Tank 2 (Figure 1.3) at the rate of
120m3/hr. All pipes in the system are 6-in. schedule 40 commercial steel pipes. The total length
of the straight pipe is 45m. Determine the specific mechanical energy, head and power that
should be transferred to the flow medium to move the water.
8
900 regular elbow
Patm
15m
Patm
900 regular elbow
Figure 1.3 A pumping system for Example 1.1
Solution
Flow medium: water at 200C
Q=120m3/hr =0.0333 m3/s
=998.2 kg/m3
=1.005 10-3 Pa s
(viscosity)
D=0.1541m ( Internal Diameter of 6-in Schedule 40 pipe)
c
Q
0.0333

 1.79 m/s
2
D / 4  (0.15412 ) / 4
Re 
cD 998.2(1.79)(0.1541)

 2.74  10 5
3

1.005  10
The flow is in turbulent region
a) The specific energy requirement of the pumping system
1) Friction loss in the pumping system
i)
Contraction loss at the tank exit of Tank 1
Ap 

k c  0.421 
  0.42(1  0)  0.42
At 

The specific energy loss due to contraction loss at the exit of Tank 1
F1  k c
ii)
c2
1.79 2
 0.42
 0.67 J / kg
2
2
Friction loss in the straight pipe
The relative friction of commercial steel pipe
9

D

4.6  10 5
 0.00029
0.1541
Re  2.74  105
Calculated above
The friction coefficient from Moody chart with /D=0.00029 and Re=2.74105 ,
f=0.0171
F2  f
iii)
L c2
45 1.79 2
 0.0171
 8.0 J/kg
D 2
0.1541 2
Friction loss in the two elbows
F3  2 K
iv)
c2
1.79 2
 2(0.64)
 2.05 J/kg
2
2
Expansion loss at the entrance of Tank 2
2
k Ex
Ap 

2
 1 
  (1  0)  1
At 

F4  k Ex
c2
1.79 2
 1
 1.60 J/kg
2
2
The total energy loss due to friction
F  F1  F2  F3  F4  0.67  8.0  2.05  1.60  12.32 J/kg
2) The static pressure difference P2-P1=0 since both are at the same pressure
3) The geodetic energy difference
Ygeo  eg  15(9.81)  147.15 J/kg
4) The velocity energy difference
Yvel

c c
2
2
2
1
2

1.79  1.79
0
2
Therefore the total specific energy requirement of the pump is
Y  12.35  147.15  159.5 J/kg
b) The total head requirement of the pump is
H
Y 159.5

 16.26 m
g
9.81
c) N  QY  998.2(0.0333)(159.5)  5301.8W  5.3 kW
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Fans: There are various industrial processes that involve movement of air and other gasses
without significant increase in the static pressure. Ventilation and air conditioning, feeding of
reactors, removal of exhaust gases, feeding of drying air are some of these. The movement of
gasses may be through pipes, tunnels and equipment or may be just in a room for moving air
without piping as in a simple ventilation. To achieve such movement of gasses energy should be
added to the gas in the form of mechanical energy to cover for the net increases in energy due
to difference in potential energy, kinetic energy, pressure energy and energy loss in the form
of fluid friction. The machines used to transfer mechanical energy to gases at low discharge
pressure are known as fans. The calculation of total pressure (specific energy requirement) and
power requirement of systems for moving gasses with fans is similar to pumps.
Example 1.5: A fan is used to deliver 1300m3/min (measured at the inlet) of methane. The
specific energy requirement of the system is calculated to be 5.4 kJ/kg. The inlet temperature and
pressure are 12oC and 100 kPa, respectively. Determine the total pressure and the useful power
that should be transferred to the flow medium.
Solution
Flow Medium: Methane
M=16kg/kmol
Q=1300m3/min
Y=5.4 kJ/kg
T1=120C=285.15K
P1=100 kPa
Density of the gas at the inlet condition assuming ideal gas

P1 M
(100  10 3 )16

 0.675 kg/m 3
RT1 (8314.3)(285.15)
The total pressure, from (1.2)
Pt  Y  0.675(5.4)  3.645kPa
The useful power transferred to the flow medium is
11
N  QPt 
1300
(3645)  78975W  79.0 kW
60
Compressors: Compressed air is one of the most common utility in process industries. Among
other purposes it is used for most automatic control systems and for cleaning, pneumatic
conveying. In liquefaction of gasses and process that depend on them (like separation)
compression is a very important step. Various gas phase reactions that take place at high
pressures are also core in some chemical production, like ammonia production. Compression
process requires large amount of energy. The machines used to transfer mechanical energy in
compression process to gasses are known as compressors.
The energy requirement in compression process depends on the type of compression. There are
two types of ideal compressions: adiabatic compression and isothermal compression.
Adiabatic compression is carried out with no heat transfer between the gas and the surrounding.
The adiabatic compression specific energy for compressing a gas from a suction temperature T1
and pressure P1 is given by Equation 1.16.
( k 1) / k

RT1  k   P2 


Yad 
 1

  

M  k  1   P1 


(1.16)
Yad= the adiabatic compression specific energy requirement
R= Universal gas constant = 8314.3 J/kmol K
T1= The suction temperature
M= Molecular weight of the gas to be compressed
P2= Final pressure
P1=Initial pressure
P1/P2= compression ratio
k=ratio of specific heats k=Cp/Cv
The power requirement in adiabatic compression is given by:
 Yad
N ad  m
(1.17)
Isothermal compression is carried out under constant temperature, k is 1. The specific energy
requirement for compressing gas under isothermal compression is given by Equation 1.18. For
the same compression ratio and flow medium isothermal compression requires less
compression energy than adiabatic compression.
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Yiso 
RT 1  P2 
ln  
M  P1 
(1.18)
The power requirement for isothermal compression of a gas is given by Equation 1.19
 Yiso
Niso  m
(1.19)
Example 1.6: It is required to compress 0.02 k mol/s of air from 1 atm and 250C to 6 atm.
Calculate the specific energy requirement and the compression power (i) for adiabatic
compression
(ii) for isothermal compression.
Solution
Flow medium: Air
T1=250C =298.15K
P1=1 atm
P2=6 atm
M=28.9 kg/kmol
Molar flow rate =0.02 kmol/s
i)
Adiabatic compression
( k 1) / k
 8314.3(298.15)  1.44   6 (1.441) /1.44 
RT1  k   P2 
Yad 
 1 
 1

  

  


M  k  1   P1 
28.9
1.44  1   1 




= 204615 J / kg
The mass flow rate is given by
  M  molar flow rate  28.9(0.02)  0.578 kg/s
m
The power requirement for adiabatic compression
 Yad  0.578(204615)  118267.5 W  118.3 kW
Nad  m
ii)
Isothermal compression
Yiso 
RT1  P2  8314.3(298.15)  6 
ln   
ln    153689 J/kg
M  P1 
28.9
1
The power requirement for isothermal compression
 Yiso  0.578(153689)  88832W  88.83 kW
Niso  m
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Based on the property of the gas to be compressed, performance and safety requirements there
are various types of compressors that are effective. The most important gas properties are
composition, temperature, molecular weight, specific heat ratio. Flow rate, compression ratio
(Discharge Pressure/ Suction Pressure), discharge temperature are some of the most important
performance requirements. Compression processes involve increase in temperature and if the
compression ratio is very high the increase in temperature becomes unsafe hence the design of
compressors in such cases must ensure that the safety limit is not exceeded.
Review Exercises for Chapter One
1.
What is the main purpose of fluid machines?
2.
What is the role of the chemical engineer in tasks related to fluid machines?
3.
What is the difference between the mechanical energy and the total energy transferred
to a flow medium?
4.
The inlet temperature of a flowing gas is measured to be 82.40C. What is the
temperature in K,0F and 0R.
5.
It is required to ventilate a workshop using fan. The total pressure requirement of the
designed flow system is calculated to be 100mm of water (density 1000kg/m3). The
volume flow rate required is 28,000m3/min measured at inlet. The inlet temperature
and pressure are 200C and 1.01bar respectively. Determine the useful power that the
fan has to deliver to the air.
6.
A pump draws 320m3/hr of liquid solution having a density of 1402 Kg/m3 from an
open storage tank of large cross-sectional area trough a 8-in schedule 40 pipe. The
end of the discharge line is 6-in schedule 40 pipe to an open overhead tank. The end
of the discharge line is 16m above the level of the liquid. The friction losses in the
piping system are 6.4 m. calculate the useful head and power that should be
transferred to the liquid by the pump.
7.
A compressor is required to compress 1000m3/min (at the inlet condition) of air to 4
atm. The inlet temperature and pressure are 150C and 1atm respectively. Calculate the
useful specific energy and power that should be transferred to the gas assuming (i)
adiabatic compression (ii) isothermal compression.
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