Chemical_Process_Equipment_2.pptx

Chemical Equipment Design
Lecturer:
Hamid Kazemi Hakki
Autumnsemester 2022-2023
Piping and Instrumentation

The design of piping systems, and the specification of the process
instrumentation and control systems, is usually done by specialist design
groups, and a detailed discussion of piping design and control systems is
beyond the scope of this book. Only general guide rules are given. The piping
handbook edited by Nayyar et al. (2000) is particularly recommended for the
guidance on the detailed design of piping systems and process instrumentation
and control. The references cited in the text and listed at the end of the
chapter should also be consulted.
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2.1. INTRODUCTION

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2.2. THE P AND I DIAGRAM
• The P and I diagram shows the arrangement of the process equipment, piping, pumps, instruments, valves and
other fittings. It should include:
1. All process equipment
identified by an equipment
number. The equipment should be
drawn roughly in proportion, and
the location of nozzles shown.
2. All pipes, identified by a line
number. The pipe size and material
of construction should be shown.
The material may be included as part
of the line identification number.
3. All valves, control and block valves,
with an identification number. The type
and size should be shown. The type
may be shown by the symbol used for
the valve or included in the code used
for the valve number.

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2.2. THE P AND I DIAGRAM
4. Ancillary fittings that are part of the piping
system, such as inline sight-glasses, strainers
and steam traps; with an identification number.
5. Pumps, identified by a suitable
code number.
6. All control loops and instruments,
with an identification number.
For simple processes, the utility (service) lines can be shown on the P and I diagram. For complex processes,
separate diagrams should be used to show the service lines, so

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• The P and I diagram will resemble the process
flow-sheet, but the process information is not
shown. The same equipment identification
numbers should be used on both diagrams.
• the information can be shown
clearly, without cluttering up the
diagram. The service connections
to each unit should, however, be
shown on the P and I diagram.

2.2.1 Symbols and layout
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o The symbols used to show the equipment,
valves, instruments and control loops will
depend on the practice of the particular
design office. The equipment symbols are
usually more detailed than those used for
the process flow-sheet.
o Standard symbols for instruments,
controllers and valves are given in the
British Standard BS 1646.

• Austin (1979) gives a comprehensive summary of the British Standard symbols, and also shows the American standard
symbols (ANSI) and examples of those used by some process plant contracting companies. The German standard symbols
are covered by DIN 28004, DIN (1988).
• In the laying of the diagram ‫و‬it is only
necessary to show the relative elevation
of the process connections to the
equipment where these affect the
process operation.
For example
• the net positive suction head (NPSH) of
pumps, barometric legs, syphons and the
operation of thermosyphon reboilers.
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5.2.2. Basic symbols
Instruments and controllers
Control valve:
represent all types of control
valve, and both pneumatic and
electric actuators.
Failure mode:
means that the controller and display is located out
on the plant near to the sensing instrument location.
means that they are located on a panel in the
control room. Except on small plants, most
controllers would be mounted in the control room.
Chemical Equipment Design 8

This is indicated on the circle representing the instrument-
controller by a letter code.
The first letter indicates the property measured; for
example, F = flow. Subsequent letters indicate the
function; for example,
I = indicating
RC = recorder controller
Type of instrument
5.2.2. Basic symbols
Table 2.1 type of instruments

The valves depending on their primary function
are divided into two broad classes:
1. Shut-off valves (block valves), whose purpose
is to close off the flow.
2. Control valves, both manual and automatic,
used to regulate flow.
5.3. VALVE SELECTION
 The main types of shut – off valves are:
Gate valve Plug valve Ball valve
A valve selected for shut-off purposes should give a positive seal
in the closed position and minimum resistance to flow when open.
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● The careful selection and design of
control valves is important;
● good flow control must be achieved,
● whilst keeping the pressure drop as
low as possible.
● The valve must also be sized to avoid
the flashing of hot liquids and the super-
critical flow of gases and vapors.
• If flow control is required, the valve
should be capable of giving smooth
control over the full range of flow,
from fully open to closed. Globe
valves are normally used, though the
other types can be used.
5.3. VALVE SELECTION
Globe valve
• Butterfly valves are often used for
the control of gas and vapour flows.
Butterfly valve

2.4. PUMPS
2.4.1. Pump selection
Pumps
Positive displacement
pumps
Dynamic pumps
Dynamic pumps
Based on the direction of fluid
axial flow radial flow Mixed flow
Based on the vacuum
Single stage
Multi stage
Based on the Mechanical structure
Open - without cap
Closed - with cap
Half open
Reciprocating pumps Rotary pumps
Piston pump
Plunger pump
Diaphragm pump
Single stage
Double stage pump
Screw
Piston
Vane
Flexible membrane
Peristaltic
Gear
Lobe
Circumferential pump
Screw
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Centrifugal pump selection guide. Single-stage >1750 rpm, multi-stage 1750 rpm

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(2-1)
(2-2)

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2.4.2. Pressure drop in pipelines
(2-3)

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The friction factor is a dependent on the Reynolds number and pipe roughness. The friction factor for use in
equation 2.3 can be found from Figure 5.7.
Table 2.2 Pipe roughness
(2-4)
Values for the absolute surface roughness of commonly used pipes are given in Table 2.2.
The parameter to use with Figure 2.1 is the relative roughness, given by:
relative roughness, e = absolute roughness/pipe inside diameter

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Fig. 2.1. Pipe friction versus Reynolds number and relative roughness

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o Non-Newtonian fluids
In equation 2.3, and when calculating the Reynolds number for use with Figure 2.1, the fluid viscosity and density are
taken to be constant. This will be true for Newtonian liquids but not for non-Newtonian liquids, where the apparent
viscosity will be a function of the shear stress.
o Gases
When a gas flows through a pipe the gas density is a
function of the pressure and so is determined by the
pressure drop. Equation 2.3 and Figure 2.1 can be used to
estimate the pressure drop, but it may be necessary to
divide the pipeline into short sections and sum the results.
o Miscellaneous pressure losses
Any obstruction to flow will generate turbulence and
cause a pressure drop. So, pipe fittings, such as: bends,
elbows, reducing or enlargement sections, and tee
junctions, will increase the pressure drop in a pipeline.

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There will also be a pressure drop due to the valves used to isolate equipment and control the fluid flow. The
pressure drop due to these miscellaneous losses can be estimated using either of two methods:
1. As the number of velocity heads, K, lost at each fitting
or valve. A velocity head is u2/2g, metres of the fluid,
equivalent to u2/2, N/m2. The total number of velocity
heads lost due to all the fittings and valves is added to the
pressure drop due to pipe friction. 2. As a length of pipe that would cause the same pressure loss as
the fitting or valve. As this will be a function of the pipe
diameter, it is expressed as the number of equivalent pipe
diameters. The length of pipe to add to the actual pipe length is
found by multiplying the total number of equivalent pipe
diameters by the diameter of the pipe being used.

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Table 2.2 Pressure loss in pipe fittings and valves (for turbulent flow)

Example 5.1
A pipeline connecting two tanks contains four
standard elbows, a plug valve that is fully open and a
gate valve that is half open. The line is commercial
steel pipe, 25 mm internal diameter, length 120 m.
The properties of the fluid are: viscosity 0.99 mNM-
2s, density 998 kg/m3. Calculate the total pressure
drop due to friction when the flow rate is 3500 kg/h.
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Solution:
Step 1. Calculation of Reynolds no.
Re = ρud/μ
Pipe area= πD2/4
Pipe area= π×(0.025)2/4 = 0.00049
Re = (998 × 1.98 × 0.025) / 0.00099 = 49900
= 5×104

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Step 2. Finding frisction factor
Absolute roughness commercial steel pipe, Table 2.2 = 0.046 mm Relative roughness = 0.046/0.025 D= 0.0018≈0.002
From friction
factor chart,
fig. 2.1,
f = 0.0032

Step 3.Miscellaneous losses
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Method 1, velocity heads
A velocity head = u2/2g = 1.982/2 × 9.8 = 0.20 m of liquid.
Head loss d 0.20 × 14.7 = 2.94 m
as pressure = 2.94 × 998 × 9.8 = 28,754 N/m2
Total pressure = 28754 + 240388 = 269142 N/m2 = 270 kN/m2
Step 4. pressure calculation

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Method 2, equivalent pipe diameters
Extra length of pipe to allow for miscellaneous losses = 735 × 0.025 =18.4 m
So, total length for ∆P calculation = 120 + 18.4 = 138.4 m
Note: the two methods will not give exactly the same result. The method using velocity heads is the
more fundamentally correct approach, but the use of equivalent diameters is easier to apply and
sufficiently accurate for use in design calculations.

2.4.3 Power requirements for pumping liquids
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To transport a liquid from one vessel to another through a pipeline, energy has to be supplied to:
1. overcome the friction losses in the pipes;
2. overcome the miscellaneous losses in the pipe fittings (e.g. bends), valves, instruments etc.;
3. overcome the losses in process equipment (e.g. heat exchangers);
4. overcome any difference in elevation from end to end of the pipe;
5. overcome any difference in pressure between the vessels at each end of the pipeline.
The total energy required can be calculated from the equation:
Fig. 2.2. Piping system
(2-5)

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2.4.3 Power requirements for pumping liquids
If W is negative a pump is required; if it is positive a turbine could be installed to extract energy from the system.
The head required from the pump =
The power is given by:
Power = (W × m) / ɳ, for a pump
and = (W × m) × ɳ for a turbine
(2-5a)
(2-6a)
(2-6b)
where m = mass flow rate, kg/s,
ɳ = efficiency = power out/power in.
The efficiency will depend on the type of pump used and
the operating conditions. For preliminary design
calculations, the efficiency of centrifugal pumps can be
determined using Figure. 2.3.
Fig. 2.3 Centrifugal pump efficiency

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Example 5.2
A tanker carrying toluene is unloaded, using the ship’s pumps, to an on-shore storage tank. The pipeline is 225 mm internal
diameter and 900 m long. Miscellaneous losses due to fittings, valves, etc., amount to 600 equivalent pipe diameters. The
maximum liquid level in the storage tank is 30 m above the lowest level in the ship’s tanks. The ship’s tanks are nitrogen
blanketed and maintained at a pressure of 1.05 bar. The storage tank has a floating roof, which exerts a pressure of 1.1 bar on
the liquid. The ship must unload 1000 tonne within 5 hours to avoid demurrage charges. Estimate
the power required by the pump. Take the pump efficiency as 70%. Physical properties of toluene: density 874 kg/m3,
viscosity 0.62 mNm2 s.

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Pipe area= π×(0.225)2/4 = 0.0398
Solution:
Absolute roughness commercial steel pipe, Table 2.2 = 0.046 mm
Relative roughness = 0.046/225 = 0.0002
Friction factor from Fig. 2.1, f = 0.0019
Total length of pipeline, including miscellaneous losses,
= 900 + 600 × 0.225 =1035 m

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Solution:

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2.4.4. Characteristic curves for centrifugal pumps
The performance of a centrifugal pump is characterised by plotting the head developed against the flow-rate.
The pump efficiency can be shown on the same curve. A typical plot is shown in Figure 2.4. The head
developed by the pump falls as the flow-rate is increased. The efficiency rises to a maximum and then falls. For
a given type and design of pump, the performance will depend on the impeller diameter, the pump speed, and
the number of stages. Pump manufacturers publish families of operating curves for the range of pumps they
sell. These can be used to select the best pump for a given duty. A typical set of curves is shown in Figure 2.5.

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Fig. 2.4 Pump characteristic for a range of impeller sizes
(a) 250 mm (b) 225 mm (c) 200 (d) 175 mm (e) 150 mm.
Fig. 2.5 Family of pump curves

2.4.5. System curve (operating line)
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There are two components to the pressure head that has to be supplied by the pump in a piping system:
1. The static pressure, to overcome the differences in head (height) and pressure.
2. The dynamic loss due to friction in the pipe, the miscellaneous losses, and the pressure loss through equipment.
The static pressure difference will be independent of the fluid flow-
rate. The dynamic loss will increase as the flow-rate is increased.
The system curve, or operating line, is a plot of the total pressure head
versus the liquid flow-rate. The operating point of a centrifugal pump can
be found by plotting the system curve on the pump’s characteristic curve.

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Example 2.3
A process liquid is pumped from a storage tank to a distillation column, using a centrifugal pump. The
pipeline is 80 mm internal diameter commercial steel pipe, 100 m long. Miscellaneous losses are
equivalent to 600 pipe diameters. The storage tank operates at atmospheric pressure and the column at
1.7 bara. The lowest liquid level in the tank will be 1.5 m above the pump inlet, and the feed point to the
column is 3 m above the pump inlet. Plot the system curve on the pump characteristic given in Figure A
and determine the operating point and pump efficiency.
Properties of the fluid: density 900 kg/m3, viscosity 1.36 mN m2s.

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Difference in elevation, Δz= 3.0 - 1.5 = 1.5 m
Difference in pressure, ΔP= (1.7 - 1.013)105 = 0.7×105 N/m2
as head of liquid = (0.7 × 105) / (900 × 9.8) = 7.9 m
Total static head = 1.5 + 7.9 = 9.4 m
Solution:
Friction factor from Fig. 2.1, f = 0.0027

To find the system curve the calculations were
repeated for the velocities shown in the table below:
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Chemical_Process_Equipment_2.pptx

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