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Introduction
The preliminary design of a pipeline basically involves the following:
Evaluation of the properties of the fluid to be pumped (density, viscosity etc.)
Selection of pipe material
Using a contour map to determine the elevation encountered along the route of thepipeline and to estimate the length of the pipe required.
Determining of pipe internal diameter and wall thickness of the pipe.
Calculation of pressure drop in the form of head (m), associated with pipe friction,
elevation and miscellaneous fittings. Head at the inlets and outlets of the pipe are
also to be considered.
Selection of appropriate pumps and their positioning along the pipeline.
Selection of appropriate control valves to control flow from the pump. This requires
correction of pump selection and positioning.
All of the above is required, but is certainly not all that is needed to obtain a full detailed
design of the pipeline that workmen can use for construction. The design that is
considered here, needs to be re-evaluated again and again to replace all estimations
with accurate values. Make a good attempt at identifying these shortcomings and
mention them in the report.
The design is simple enough to be done with the use of Microsoft Excel. The
spreadsheet used, must be of high detail, as many changes may be necessary while
doing the design. The better the spreadsheet is set up, the easier it is to perform
corrections. Keep that in mind instead of rushing to be ahead in the design. Strive to be
on schedule.
1. Preliminary step-by-step guide:
1.1Read the document titled “Design 2009.pdf” available from the S drive. This
document must be read preferably before the first design lecture but at least
before the second lecture. Read at least twice for a good understanding.
1.2 Each students design will vary with their student number to avoid plagiarism. The
first step is to determine how the design varies and make sure all values in the
process description and regarding the fluid are known. The route of the pipeline
also varies according to the student number. Make a clear note of this and makesure all of the information is correct according to your student number.
1.3 A topographical map will be provided to determine pipe length and elevation
according to your student number. Follow the guidelines on page 4 of “Design
2009.pdf” in order to obtain a 2 dimensional map of elevation (a graph of
elevation vs distance, drawn to scale). Note that for preliminary design, the
length of the pipe will be regarded as the straight line distance between the
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appropriate depots. Ensure the graph is complete before the due date as this
constitutes a portion of the assessment. For the purpose of assessment, the
graph can be drawn by hand, but it is recommended that the graph be plotted in
Excel as this will be useful throughout the design.
1.4 Specifications on the location of miscellaneous valves, elbows etc are given.
Assume that all these fittings are grouped in single locations and work out were
these locations are. They may be displayed on the graph for further design
purposes. This is not required for the graph to be presented on the due date of
the graph hand-in. An illustrative example is shown below.
Figure 1: Example of an Elevation Graph
1.5 Consider all of the components constituting the mixture to be pumped. At winter
and summer temperatures, find the density, viscosity and vapour pressure of
each individual component in the mixture using data and equations given onpage A4-1 and A4-2 of “Design 2009.pdf”. Also find the molar mass of each of
these components.
1.6 Thereafter, analyze the vapour pressures to determine the order of volatility of
each component and hence determine the composition of the mixture as per the
process description. An example follows:
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A mixture of Decane, Isopropanol and Cycloheptane is considered. 35% most
volatile, 25% least component. At 32oC:
Vapour Pressure:
For Decane: A = 16.0114 B = 3456.8 C = -78.67.
Using the equation on Page A4-2 of “Design 2009.pdf”:
Vapour Pressure = exp(16.0114-(3456.8/(305.15+(-78.67))) x 133.32/1000
= 0.28 kPa
Similarly for Isopropanol: Vapour pressure = 9.12 kPa
Cycloheptane: Vapour pressure = 4.14 kPa
Hence Isopropanol is the most volatile and decane the least volatile. Composition
is 35% Isopropanol, 25% Decane and 40% Cycloheptane (by mole).
1.7 Obtain the overall density, viscosity and vapour pressure of the mixture using
mole weighted averages as determined from step 1.5 above, at winter and
summer temperatures. An example follows:Density of Decane at 30
oC: Use e quation found on page A4-1 of “Design 2009”
For Decane, A = 1084.5; B = -2.8243; C = 6.8997x10 -3; D = -8.7449x10-6
ρ = 1084.5 – 2.8243(305.15) + 6.8997x10-3(305.15)2 – 8.7449x10-6(305.15)3
= 616.66 kg.m-3
Similarly for Isopropanol: ρ = 544.16 kg.m-3
Cycloheptane: ρ = 798.97 kg.m-3
Overall density = ??(544.16) + ??(616.66) + ??(798.97) =
1.8 Note that at higher temperatures (ie. summer), the viscosity and density of
mixtures should be less but the vapour pressures should be higher. The reverse
is true for winter temperatures. The pipe system must be designed to cater forthe worst conditions of operation. Summer or winter conditions can be chosen as
the worst case scenario. Or the worst case scenario might not depend only on a
single season. It may be property specific. Use reasoning and establish the most
unfavourable conditions of operation for the pipeline. This reasoning must be
decided upon at this point before continuing, as it affects further design
considerations. Make certain to explain your choice in the write-up.
1.9 Obtain your unique molar flow rate as described in the process description, and
calculate the molar flow rates of the individual components, using the mole %
specified by the process description.
1.10 Convert the individual molar flow rates into mass flow rates, using the
molar masses of each component. Thereafter, use the density (ie. worst case
density) of each component to convert mass flow rate into volumetric flow rate.
Add up the individual volumetric flow rates to obtain the total volumetric flow rate
of the mixture [m3/s]. Be sure that the vol. flow rate obtained is the worst case
flow rate. (Hint: Lower densities provide higher volumes).
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All of above are relatively simple tasks that need to be done before any actual design of
the pipeline commences. It is recommended that they be done by the due date of the
topographical map (ie. Elevation vs Distance graph).
2. Pipe Sizing
2.1 The pipe material can now be selected. Material selection should be based
mainly on corrosiveness, cost and availability. No exact figures need be given for
these factors for this design. One should have a general knowledge of the
materials that are produced locally and hence would be in abundance.
Corrosivity figures need not be given but a material must be chosen such that
none of the components of the fluid would cause it to corrode. Costs must be
expressed in relative terms. Any suggestions of corrosion inhibitors and
galvanizing etc. will be acknowledged provided that the argument makes sense
and has technically correct evidence. For whatever material that is chosen, themain property that must be available is the pipe roughness (ε, in m).
2.2 The calculation of wall thickness involves an estimation of design stress. This is
not within the scope of this design.
2.3 The pipe diameter needs to be calculated:
2.3.1 Add a 20% safety factor to the volumetric flow rate. This is to cater for
increased demand and error due to estimates.
2.3.2 Refer to [1], page 218, for a suggestions on fluid velocities. For
hydrocarbon mixtures, the velocity is 1-3 m/s.
2.3.3 Refer to Simpsons table on [1], page 219, for suggestions on fluid velocity
based on the density of the fluid. Select a fluid velocity using this table.Interpolation may necessary. Bear in mind that the velocity must lie
between 1-3 m/s. Higher velocities result in decreased pipe diameters and
less costs.
2.3.4 Pipe Area = Q[m3/s] / v[m/s] and D =
A4
3. Pump Selection
Now that the volumetric flow rate is known, the number of possible pumps can be
narrowed down. Refer to the document titled “pump curves.pdf”. The “fish scale”
graphs are used for preliminary pump selection.
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Figure 2: “Fish Scale” Pump Curves
Each “fish scale” is a pump curve. A pump is selected according to the desired flow
rate. It is selected such that the desired flow rate lies on the sloping region of thepump curve, so that the head supplied varies with flow rate. Eg. if the desired flow
rate was 130m3/hr, the series 80 pumps are applicable. Note that the maximum flow
rate is that flow rate which includes the 20% safety factor (In this case 157m 3/hr).
Both the desired and maximum flow rate must lie on the sloping region of a pump
curve. In some cases, 2 or 3 series of pumps are applicable. However, a detailed
analysis using “Pump curves 3.pdf” can be used to show which pumps are suitable.
80 series pump curves are found in “Pump curves 3.pdf”. Refer next page for 80 -315
and 80-250 pumps.
The first chart shows pump curves. Each curve represents a 80-315 pump of aspecific impeller (blade) diameter. All of the curves are based at a standard pump
impeller speed of 2900 min-1
. Any speed increase will cause a pump curve to shift
up or down. Pump efficiency curves are also found on the chart. As would be
noticeable, different impeller diameters and different speeds would result in different
efficiencies.
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The second chart shows the Net Positive Suction Head required (NPSHr) by a pump
operating at the desired flow rate. Although a pump supplies head, a pressure drop
occurs at the inlet of the pump. If the pressure drops to below the liquids vapour
pressure, then cavitation occurs. The NPSHr is the minimum amount of head (ie.
pressure) at which the fluid must enter the pump. If this head is not met then
cavitation in the pump may occur. Cavitation is the vaporization of liquid under low
pressure, in the pump. This results in the formation of air cavities in the pump, which
ultimately leads to excessive pump damage. The NPSHa (NPSH available) must be
higher than the NPSHr for safe, smooth operation.
The third chart shows the power output that the pump would supply. A higher
impeller diameter provides a higher power output and in turn needs a higher power
input.
Power Input required = Power output/η where n is the efficiency of the pump, foundon the first chart at desired operating conditions.
Note that smaller pumps possess higher efficiencies. The 80-250 have higher
efficiencies than the 80-315 pumps. For short distances, small pumps are ideal as
they are more efficient and save power. But for long distances, it is better to use a
few large pumps instead of many small pumps. For your particular case, you would
have to try to justify your pump choice quantitatively. Make sure your pump choice
is definite and your spreadsheet is set up in detail for easy alteration, as any
errors and uncertainties may result in the design being repeated again and
again.
4. Calculation of Head losses:
In reality, there is a loss of energy associated with fluid moving through a pipe. A
pump imparts energy to the fluid to maintain its motion through the pipe, but the fluid
loses energy due to friction against pipe walls, change in direction (sometimes even
just on a molecular level) when fluid encounters elbows, tees, pipe expansions, pipe
constrictions and various valves in the pipe, and energy losses due to gravitational
forces when pumping uphill.
When the energy losses begin to overcome the energy imparted, the fluid velocity
decreases and the desired volumetric flow rate is not maintained. This manifests
itself as a loss of pressure. Pipe friction, fittings and elevation are factors that cause
pressure losses. In piping, it is convenient to represent pressure losses in metres
instead of Pascals. This is referred to as head losses. Pumps supply head in (m),
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4.1.3 Note that an uphill slope can only be compensated by a downhill slope
which came before it. If the topography was such that position F and B
was lower than position A, there would still be head lost due to elevation
A-C.
4.1.4 If pumping occurred between A-E, then the head lost due to overcoming
elevation, would be 60m.
4.1.5 As shown in Figure 3, elevation head loss also includes the height on tank
inlet, pump height above ground and pump inlet position, as well as liquid
level. Liquid level generally acts as a supplier of head due to the pressure
it exerts as a result of gravity acting on it. Be wary of these cases.
4.1.6 All elevation can be added using the above considerations to obtain the
total head loss due to elevation h∆z.
4.2 Pipe Friction and miscellaneous Fittings
4.2.1 Friction head loss is represented by means of a friction factor. This factor
is calculated using the Colebrook equation below:
Where D = pipe diameter (m) ; ε = pipe roughness (m) and
NRE = Reynolds number =
Du
where u = fluid velocity (m); ρ = fluid density (kg/m3)
and μ = fluid viscosity (N.s.m-2
)
4.2.2 The actions of fittings are represented as K values. The K values forvarious fittings are given on page 17-18 of “Fluid flow.pdf”. Certain valves
and fittings have constant K values. Some are a factor of the friction
factor.
4.2.3 The K value for Pipe enlargements is as follows:
And that of pipe contractions:
where d1 = smaller diameter and d2=larger diameter; θ < 45o
4.2.4 Once the friction factor and all K values are found, they are combined in a
single equation to represent the head lost by friction, valves and fittings:
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Where K1,K2 etc. are K values of all miscellaneous valves and fittings
um = fluid velocity (m/s) and g = acceleration due to gravity (m/s 2)
D
fL represents the head lost due to friction, which is obviously dependent
on the length of pipe L(m).
Example: A section of pipe 1180m long has 2 gate valves, 4 globe valves,
12 elbows and 3 branched tees. The pipe used is commercial steel of
diameter 0.1525m and roughness (ε) = 4.6E-5m. Assume 80-315 pump is
used with impeller diameter 300. The flow rate is 130.88 m3/hr. Fluid
properties are as follows:
Density = 577.66 kg.m-3
; Viscosity = 2.06e-3 N.s.m-2
;
Vapour pressure = 17.62 kPaPump outlet diameter (from “Pump curves.pdf) = 80mm = 0.08m
Pump inlet diameter = 125mm = 0.125m
u = (130.88/3600)/(π(0.1525)2/4) = 1.99 m/s
NRE =306.2
)66.577)(99.1)(1525.0(
e= 85003.1
and ε/D = (4.6x10-5) / 0.1525 = 3.02 x 10
-4.
Using the Colebrook equation (4.2.1), f = 0.005
For Globe Valves K = 6 For gate valves K = 8f = 0.04For Elbows K = 30f = 0.15 For Tees K = 20f = 0.1
K(enlargement at pump exit):
d1 = 0.08 ; d2 = 0.1525 ; θ = 45o = π/4 radians d1/d2 = 0.525
Using equation for enlargement (4.2.3) K = 6.44
K(constriction for next pump inlet):
d1 = 0.125 ; d2 = 0.1525 ; θ = 45o = π/4 radians d1/d2 = 0.820
Using equation for enlargement (4.2.3) K = 0.151
Hence
hL=)81.9(2
99.1
1525.0
)1180)(005.0()1.0(3)15.0(12)6(4)04.0(2151.044.6
2
=14.42m
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Note that depending on the pump to be used, pipe enlargements and
constrictions will have to be considered due to the pump inlet and outlet
having diameters different to that of the pipeline.
4.3 Control Valve Head Loss
4.3.1 Control valves need to be given special attention. They are electronically
controlled valves that are continuously active and are most frequently
used to control flow, unlike other miscellaneous valves which are used for
other reasons such as occasional maintenance.
4.3.2 Control valves regulate flow. For every pump that is installed, a control
valve is installed immediately after that pump, in the line of flow.
4.3.3 It is most common for control valves to operate at positions ranging from
10% open to 80% open. They are seldom used as shut off valves and
usually not used in the fully open position for safety reasons.
4.3.4 For this design, assume that normal flow corresponds to a valve position
of 60% open and maximum flow corresponds to 80%. In this design, equalpercentage valves are used.
4.3.5 To choose the correct valve, one makes use of the table of control valves
on page 2 of “Control valves.pdf”. Two important constraints on valve
choice, is that a valves size must not be bigger than the pipe diameter,
and it must not be smaller than half of the pipe diameter.
4.3.6 The table of control valves has Cv values for up flow and down flow. So Cv
will have to be chosen based on the position of the valve on the pipeline.
4.3.7 For 60% valve position, interpolate between the 30% and 70% Cv. For
80% flow, interpolate between 70% and 100% Cv. Not much attention
needs to payed to the XF values on the bottom of the table.4.3.8 Once the Cv values are obtained, the pressure drop can be found by
∆P =2)/( QC
SG
V
where SG = specific gravity of fluid; Q = flow rate of fluid (gpm)
and ∆P = Pressure drop (Psi).
Note that the above equation is only valid in the units stated above.
4.3.9 ∆P can be converted to S.I. units and then divided by ρ.g to obtain the
pressure drop in head form hcv(m).
4.4 NPSHr4.4.1 NPSHr is explained under section 3. Consider pumping in figure 4 and
assume for example, that pumps were positioned strategically at points A
and E. Suppose that pump E has a NPSHr of 4m.
4.4.2 Pump A has to supply enough head not only to overcome the various
head losses along line AE, but also to provide enough NPSHa for the next
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pump (Pump E) to operate safely and efficiently. NPSHa must be greater
than or at least equal to NPSHr (in this case, 4m).
4.4.3 NPSHa = Head supplied by pump – line head loss (friction, fittings,
elevation, control valve) – vapour pressure of liquid at worst conditions.
4.4.4 Vapour pressure is found using steps as in section 1.6. It is used in the
above equation, in head form:
Vapour pressure (m) = g
PaessureVapour
][Pr
4.4.5 Ensure NPSHa≥NPSHr
5. Total head loss and plotting of the system curve
5.1 Total head loss hT (m) = hL + h∆z + hcv …….(1)
hL = friction and fittings head lossh∆z = head loss due to an increase in elevation (Inc. special cases)
hcv = head loss across control valve5.2 For any section of pipe, the total head loss can be found. The elevation
associated with this section of pipe accounts for a particular head loss which is
constant regardless of the fluids velocity or properties.
5.3 The head loss due to friction and fittings depends on the fluid velocity and hence
volumetric flow rate.
5.4 Consider the example on 4.2.4. Suppose the elevation is 60 m and Cv for the
control valve is 81.58. Compute the total head loss:
SG = 0.58, Q = 130.88 m3/hr = 576.22 gpm
∆P = Psi94.28)22.576/58.81(
58.02
hcv= 28.94 x 6894.76/ (577.66 x 9.81) = 35.07 m
hT (m) = hL + h∆z + hcv = 14.42 + 60 + 35.07 = 109.49 m
5.5 The total head loss for a section of pipe can be computed for different flow rates
and a curve of Head loss (m) vs Volumetric flow rate (m3/hr) can be plotted. This
is known as a system curve. An example is shown in figure below.
Figure 5:
System Curve
for Entire
Length of Pipe
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5.6 Figure 5 above shows a system curve when the entire pipeline is considered,
including elevation, all fittings and a single control valve. It is plotted based on the
assumption that only 1 pump will be needed for the entire line.
5.7 Note that the control valve head loss is dependent on the valve position. For max
flow, all control valves would have to be 80% open, hence less head loss occurs
and the system curve is lower. The less open the control valve is, the higher the
head loss across that valve.
5.8 At zero flow rate the system still shows a minimum head loss of 300m. This is
due to the elevation which is independent of fluid flow.
5.9 When comparing the magnitude of head loss on figure 5 with head that pumps
supply (as shown on “pump curves.pdf”), it is easy to conclude that more than 1
pump would be required as no pump can supply such large magnitudes of head.
5.10 The system will have to be divided and many pumps will be used in series
at specific positions to overcome the head loss of separate sections of piping.
5.11 Note that for the example on 5.4, NPSH is not taken into account. This isdone as a separate calculation.
NPSHa = 114-109.49 – 17.62 x 1000 / (9.81 x 577.62) = 1.4m < NPSHr.
For this case, NPSH is not satisfied. One can either move the next pump forward
or change the current pump’s speed. This is explained in section 6.4.
6. Sectioning piping, positioning pumps and using affinity laws
6.1 By now, the choice of pump (ie. the series, exact head magnitude and impeller
diameter) should be known to the design engineer. For the desired and
maximum flow rates, the head supplied by the chosen pump should be known.
Generally, try to stick to using the same type of pump along the pipeline.6.2 It is recommended that the curve of the chosen pump be replicated in Excel. This
would allow for easy shifting of the pump curve when changing pump speed
(explained in 6.4). Other likely candidates for use as pumps on the pipeline
should also be plotted in Excel.
6.3 The logic of the procedure is as follows:
6.3.1 Consult the Height vs Distance map that you have already drawn and
ensure the location of miscellaneous fittings and each depot is present on
the graph.
6.3.2 Unless specific information is given, set the first pumps position to be at
the zero point of distance.
6.3.3 The head provided by the pump at the desired conditions and at standard
are already known. The question is, how far along the pipeline can the
pump drive the fluid before another pump is needed.
6.3.4 Assume the pump can drive fluid to a certain length and then compute the
head loss associated with that length. Find the friction head loss, consider
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the elevation associated with that chosen length, fittings if they are present
in the length, and consider the control valve.
6.3.5 Compute the NPSHa at the end of that chosen length of pipe. Compare it
to the NPSHr, which is known for the chosen pump at prevailing
conditions.
6.3.6 If the chosen length of pipe requires more head than what the pump can
supply, or if NPSHa is not larger than NPSHr, then shorten the length of
pipe and repeat the calculation. If the head is not largely more, then the
speed of the pump can be adjusted.
6.3.7 If the head loss is lower than what the pump can supply, then simply
increase the assumed length and repeat the calculation.
6.3.8 The calculation must be done for all magnitudes of flow rate and a system
curve must be plotted to represent the section of piping.
6.3.9 The calculation can be repeated for the maximum and minimum flow rate.
It is important to do the calculation particularly for the maximum flow rateas the NPSH requirements are more stringent.
6.3.10 Once a section is catered for, place the next pump and repeat the
procedure until you reach the end of the pipeline.
6.3.11 Note that pumping cannot be done across tanks. Fluid has to be pumped
to a tank and then pumped from a tank. Hence the total route will have to
be divided into their individual routes between tanks.
6.4 Using affinity laws:
6.4.1 An important equation is
n1 = original pump speed (usually the standard pump speed – 2900min-1)H1 = head supplied by pump at speed n1
n2 = new pump speed (min-1
)
H2 = head supplied by pump at speed n2
Increasing pump speed allows the head supplied by a pump to be
increased. This often leads to a decrease in efficiency however, as shown
in the pump curves.
6.4.2 Another important equation is
Q1 = Flowrate at speed n1
Q2 = Flowrate at speed n2
This equation is normally used simply to obtain new plots of the pump
curve. A change in flow rate only occurs if head is constant. And for a
constant flow rate, it is the head that changes. Observe the pump curve to
understand this.
6.4.3 The above equations can be used to perform minor changes on pump
speed to alter the head it provides.
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6.4.4 Example: An 80-315 pump with impeller diameter of 300mm can provide a
head of 114 m when driving a liquid at a flow rate of 130.88 m3/hr at a
standard speed of 2900min-1
. Suppose the head loss in a section of pipe
amounts to 116m (incl. satisfying NPSH requirements). At what speed
should the pump operate in order to service this section of pipe and
maintain the flow rate at 130.88?
n2 = 114
1162900
1
2
1 H
H n 2925.33min
-1
6.4.5 The affinity laws can be used to move the entire pump curve to meet the
operating point. This is illustrated on the next page. In this case the speed
had to be lessened.
Standard Speed 2900 New Speed 2831.052
Q (m3/hr) H (m) Q(m3/hr) H (m)
0 110 0 104.83
20 110 19.52 104.83
40 108.5 39.05 103.40
60 108 58.57 102.93
80 105.5 78.10 100.54
100 103 97.62 98.16
120 95.5 117.15 91.01
140 86.5 136.67 82.44
160 75 156.20 71.48
180 59.5 175.72 56.70
200 43 195.24 40.98
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Notes: All of the information above should be sufficient in helping one understand the
design of a piping system. Just apply the logic with the necessary equations.
Important details for the report write-up:
o A map showing all fittings, depots and pump positions.
o System and pump curves for each section of pipe.
o Pump type, impeller diameter, pump speed, distance from zero point, height
above sea level, head provided, entrance and exit diameter of pumps.
o Pipe length for each section, pipe diameter and material of construction
Note that this manual covers the generic design of piping systems. The design
project undergoes minor changes over the years. Some constraints are removed
and different ones are added. Read the “Design 2009.pdf” document carefully to find
out if any other details are necessary for the report.
For system curves, simply form increments of flow rate and perform calculations on
each flow rate in order to plot the curve. Use the pump curves as an indication of the
range of flow rates to calculate and plot.
If you are stuck on any aspect of the design and pending the lecturer’s response, do
not waste time waiting. In the mean time, work on your report. Much of the theory
and introduction can be done without calculation. Certain aspects of the discussion
can also be done. Lastly, it must be reiterated that the efficiency and success at which you do this
design depends largely on your understanding of all the concepts and also the detail
in which you set up your spreadsheet. Much time is used repeating choices and
redoing calculations. Make sure your spreadsheet is flexible and easily changeable.
If anything is unclear here, consult “Fluid flow.pdf” for further information.