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Ductwork sizing, calculation and design for efficiency - HVAC Basics + full worked example - YouTube
Channel: The Engineering Mindset
[4]
Hey there guys, Paul here
[5]
from TheEngineeringMindset.com.
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In this video, we're going to
be looking at ductwork systems
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for mechanical ventilation.
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We're going to look at how to design
[13]
a basic ventilation system
with a full worked example.
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We'll also look at how
to calculate the losses
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through the bends, the tees,
the ducts and the branches,
[21]
we'll consider the shapes in the material
[23]
that the ducts are made from
to improve the efficiency,
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and then lastly, we're going
to look at how to improve
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the efficiency and optimize the design,
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using a freemium software
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for fluid flow simulation by SimScale.
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Methods of ductwork design,
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there are many different methods used
[37]
to design ventilation systems,
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the most common ways
being velocity reduction,
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equal friction and static regain.
[44]
We're going to focus on
the equal friction method
[46]
in this example, as it's
the most common method used
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for commercial HVAC systems,
[51]
and it's fairly simple to follow.
[53]
So we'll jump straight
into designing a system.
[56]
We use a small engineering
office as an example
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and we'll want to make a
layout drawing for the building
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which we'll use for the
design and calculations.
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This is a really simple building.
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It has just four offices, a
corridor, and a mechanical room.
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And the mechanical room
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is where we're going to have the fan,
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the filters, the air
heaters or the air cooler.
[72]
The first thing we'll need to
do is calculate the heating
[75]
and cooling loads for each room.
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I won't cover how to
do that in this video,
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we'll have to cover that
in a separate tutorial
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as it's a separate subject area.
[84]
Once you have these figures,
just tally them together
[86]
to find which is the biggest load
[88]
as we need to size the
system to be able to operate
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at the peak demand.
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The cooling load is
usually the highest load
[94]
as it is in this case.
[96]
Now we need to convert the cooling loads
[98]
into volume flow rates, but to do that,
[100]
we first need to convert
this into mass flow rates,
[103]
we use the formula M dot
equals Q divided by CP,
[106]
multiply by delta T, with M
dot meaning the mass flow rate,
[110]
the Q being the cooling load of the room,
[112]
CP is the specific heat
capacity of the air
[115]
and delta T being the
temperature difference
[117]
between the design air temperature
[119]
and the design return temperature,
[121]
just to note that we will use a CP
[123]
of 1.026 kilojoules per
kilogram per kelvin as standard,
[127]
and the delta T should be less
than 10, and in this case,
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we're going to use eight degrees Celsius.
[132]
We know all the values for this formula,
[134]
so we can calculate the mass flow rate,
[135]
and the mass flow rate is
really just how many kilograms
[138]
per second of air needs
to enter that room.
[140]
If we look at the
calculation for room one,
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we see that it requires
0.26 kilograms per second.
[146]
So we just repeat that calculation
for the rest of the rooms
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to find all the mass flow rates.
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Now we can convert these mass flow rates
[152]
into volume flow rates.
[154]
To do that we need the specific volume
[156]
or density of the air.
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We'll specify that the air
needs to be 21 degrees Celsius,
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and we'll assume that it's going to be
[161]
at atmospheric pressure of 101.325 kPa,
[165]
we can look up the
specific volume or density
[167]
from our air properties tables,
[169]
but I like to just use
an online calculator
[170]
as it's much quicker.
[172]
So we just drop those numbers in
[173]
and we get the density of
air being 1.2 kilograms
[176]
per meter cubed.
[177]
You see that density has the units
[179]
of kilogram per meter cubed,
but we need specific volume
[182]
which is meter cube per kilogram.
[184]
So to convert that we
just take the inverse
[186]
which means to calculate
the density or 1.2
[188]
to the power of minus one,
[190]
you can just do that in Excel
very quickly to get the answer
[193]
of 0.83 meters cube per kilogram.
[196]
Now that we have that we can
calculate the volume flow rate,
[199]
using the formula V dot
equals M dot multiplied by V,
[203]
where V dot equals the volume flow rate,
[205]
M dot equals the mass flow
rate of the individual room
[207]
and V equals the specific
volume, which we just calculated.
[211]
So if we drop these
values in for room one,
[213]
we get the volume flow rate
of 0.2158 meters per second.
[218]
That is how much air it
needs to enter the room
[220]
to meet the cooling load.
[222]
So just repeat that calculation
for all the remaining rooms.
[225]
Now we're going to sketch
out our ductwork route
[227]
onto the floor plan so
we can start to size it.
[230]
Before we size that we need
to consider some things
[233]
which will play a big role
in the overall efficiency
[235]
of the system.
[236]
The first point we need
to consider is the shape
[238]
of the ductwork,
[239]
Ductwork comes in round,
rectangular, and flat oval shape.
[243]
Round duct is by far the
most energy efficient type
[245]
and that's what we're going to use
[246]
in our worked example later on.
[248]
If we compare round
duct to rectangle duct,
[250]
we see that a round duct
with a cross sectional area
[253]
of 0.6 meters squared has
a perimeter of 2.75 meters,
[257]
a rectangular duct with
equal cross sectional area
[260]
of 0.6 meters squared, has
a perimeter of 3.87 meters.
[264]
The rectangular duct
therefore requires more metal
[267]
for its construction.
[267]
This adds more weight
and cost to the design.
[270]
A larger perimeter also
means that more air will come
[272]
into contact with the material
and this adds friction
[275]
to the system.
[276]
Friction in a system means
a fan needs to work harder
[278]
and this results in
higher operating costs.
[281]
Always use round duct where possible
[283]
although in many cases the
rectangular duct needs to be used
[285]
as space is very limited.
[287]
The second thing to consider
is the material being used
[290]
for the ducts.
[291]
The rougher the material,
[292]
the more the friction it will cause.
[295]
For example, if we had two
ducts with equal dimensions,
[297]
volume, flow rate and velocity,
[299]
the only difference being the material,
[301]
one is made from standard galvanized steel
[302]
the other from fiberglass.
[304]
The pressure drop over a 10-meter distance
[306]
for this example is around 11 pascals
[308]
for the galvanized steel and
16 pascals for the fiberglass.
[312]
The third thing we have to consider
[313]
is the dynamic losses
caused by the fittings.
[316]
We want to use the
smoothest fittings possible
[318]
for energy efficiency.
[320]
For example, use long radius
bends rather than right angles
[323]
as the sudden change in direction,
[325]
wastes a huge amount of energy.
[327]
We can compare the performance
[328]
of different ductwork designs
quickly and easily using CFD
[332]
or computational fluid dynamics.
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These simulations on screen
[335]
were produce using a
revolutionary cloud-based CFD
[338]
and FEA engineering platform by SimScale
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who have kindly sponsored this video.
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You can access this software
free of charge using the links
[345]
in the video description below,
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and they offer a number of
different account types,
[348]
depending on your simulation needs.
[350]
SimScale is not just
limited to ductwork design,
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it's also used for data
centers, AEC applications,
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electronics design, as well as thermal
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and structural analysis.
[358]
Just a quick look through their site
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and you can find thousands
of simulations for everything
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from buildings, HVAC
systems, heat exchangers,
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pumps and valves, to
race cars and airplanes,
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which can all be copied
and used as templates
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for your own design analysis.
[371]
They also offer free webinars,
courses and tutorials
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to help you set up and
run your own simulations.
[377]
If like me,
[377]
you have some experience
creating CFD simulations,
[380]
then you'll know that
this type of software
[381]
is usually very expensive,
[383]
and you would need a
powerful computer to run it.
[385]
With SimScale however, it can all be done
[388]
from a web browser, as the
platform is cloud-based,
[391]
their servers do all the work
[392]
and we can access and design
simulations from anywhere
[395]
which makes our lives as
engineers a lot easier.
[398]
So if you're an engineer,
a designer, or an architect
[400]
or just someone interested
[401]
in trying out simulation technology,
[403]
then I highly recommend you
check out this software,
[405]
get your free account
by following the links
[407]
in the video description below.
[409]
Now, if we look at the
comparison for the two designs,
[412]
we have a standard design on the left
[413]
and a more efficient design on the right
[415]
which has been optimized using SimScale.
[417]
Both designs use an air velocity
of five meters per second,
[420]
the color represents the velocity
[422]
with blue meaning low
velocity and red representing
[424]
the higher velocity regions.
[426]
We can see from the velocity
color scale and the streamlines
[429]
that in the design on the left,
[431]
the inlet air directly
strikes the sharp turns
[433]
that are present in the system,
[435]
which causes an increase
in the static pressure.
[437]
The sharp turns cause a large amount
[439]
of recirculation regions when the duct's
[441]
preventing the air from moving smoothly,
[444]
the T section at the far end
of the main duct causes the air
[447]
to suddenly divide and change direction,
[449]
there is a high amount of backflow here
[451]
which again increases the static pressure
[453]
and reduces the amount of air delivery.
[455]
The higher velocity in the
main duct which is caused
[457]
by the sharp turns and
sudden bends reduces the flow
[460]
into the three branches on the left.
[463]
If we now focus on the
optimized design on the right,
[465]
we see that the fittings used
follow a much smoother profile
[469]
with no sudden obstructions,
recirculation or backflow,
[472]
which significantly
improves the air flow rate
[475]
within the system.
[475]
At the far end of the main
duct, the air is divided
[478]
into the two branches through
gentle curved T section.
[482]
This allows the air to
smoothly change direction
[484]
and thus there is no sudden
increase in static pressure
[487]
and the air flow rate
[488]
to these rooms has dramatically increased.
[490]
The three branches within the main duct,
[492]
now receive equal airflow,
[494]
making a significant
improvement to the design.
[497]
This is because an
additional branch now feeds
[499]
the three smaller branches,
[500]
allowing some of the air
to smoothly break away
[503]
from the main flow and feed
into these smaller branches.
[507]
Now that we have decided
to use circular ducts,
[509]
made from galvanized steel, we
can continue with the design.
[513]
Label every section of ductwork
and fitting with a letter.
[516]
Notice we are only designing
a very simple system here
[519]
so I've only included
ducts and basic fittings,
[521]
I've not included things
such as grills, inlets,
[524]
flexible connections,
fire dampers, et cetera.
[527]
Now we want to make a table
[529]
with the rows and columns
labeled as per the example
[531]
on screen now, each duct and
fitting needs its own row.
[535]
If the air stream splits
such as with a T section,
[538]
then we need to include a
line for each direction.
[540]
So just add in the letters
to separate the rows
[542]
then declare what type of fittings
[544]
or ducts that corresponds to.
[546]
We can start to fill some of the data in.
[548]
We can first include the volume flow rates
[549]
for each of the branches.
[551]
This is easy, it's just the
volume flow rates for the rooms
[553]
which the branch serves.
[555]
You can see on the chart
I've filled that in now,
[557]
then we can start to size the main ducts.
[559]
To do this, make sure you
start with the main duct
[561]
which is furthest away,
[563]
then we just add up the volume flow rates
[565]
for all the branches downstream from this.
[567]
For the main duct G, we just
sum the branches L and I,
[570]
for D, that's just the sum of L, I and F,
[574]
and for duct A then it's
the sum of L, I, F and C,
[578]
so just enter those into the table.
[580]
Now from the rough drawing,
[581]
we measure out the lengths
of each of the duct sections
[584]
and the branches and
enter this into the chart,
[587]
and now we can begin to calculate
the size of the ductwork.
[590]
To do that we need a
duct pressure loss chart,
[592]
you can obtain these from
ductwork manufacturers
[594]
or from industry bodies
such as CIBSE and ASHRAE.
[597]
I've included links to these guides
[599]
in the video description
below so do check those out.
[601]
These charts hold a lot of information,
[604]
you can use them to find
the pressure drop per meter,
[606]
the air velocity, the volume flow rate,
[608]
and also the size of the ductwork.
[610]
The layout of the chart
does vary a little,
[612]
depending on the manufacturer,
but in this example,
[615]
the vertical lines are for
pressure drop per meter of duct,
[618]
the horizontal lines are
for volume flow rate,
[620]
the downward diagonal
lines are for velocity,
[623]
and the upward diagonal
lines are for duct diameter.
[626]
We start sizing from the first main duct,
[628]
which in this example is section A.
[630]
To limit the noise in this section,
[632]
we'll specify that you can
only have a maximum velocity
[634]
of five meters per second.
[636]
We know that this duct also
requires a volume flow rate
[638]
of 0.79 meters cubed per second,
so we can use the velocity
[642]
and volume flow rate to find
the missing data on the chart.
[646]
We take the chart and scroll
up from the bottom left
[648]
until we find the volume flow rate
[650]
of 0.79 meters cubed per second,
[653]
then we locate where the velocity line is
[655]
of five meters per second
and we draw a line across
[657]
until we hit that, then
to find the pressure drop,
[660]
we draw a vertical line
down from this intersection.
[663]
In this instance, we see it comes out
[665]
at 0.65 pascals per meter,
[667]
so add this figure to our table.
[669]
As we are using the equal
pressure drop method,
[671]
we can also use this pressure
drop for all the duct lengths,
[674]
so fill these in too.
[675]
Then coming back to the
chart, we scroll up again
[677]
and align our intersection
with the upward diagonal lines
[680]
to see that this requires
a duct with a diameter
[683]
of 0.45 meters, so we add
that to the table also.
[687]
We know the volume flow
rates and the pressure drop
[689]
so we can now calculate
the values for section C
[691]
and then also the remaining ducts.
[694]
For the remainder of the ducts,
[695]
we need to use the same method.
[697]
On the chart we start by drawing a line
[699]
from 0.65 pascals per meter.
[701]
We draw this line all the way up.
[703]
Then we draw another line across
[705]
from where our required
volume flow rate is,
[707]
in this case for section C,
[709]
we require 0.21 cubic meters per second.
[712]
At this intersection, we draw a line
[714]
to find the velocity and we
can see that it falls between
[716]
the lines of three and
four meters per second
[719]
so we need to estimate the value.
[721]
In this case it seems to be
around 3.6 meters per second,
[725]
so we add that to the chart.
[726]
Then we draw another line
on the other diagonal grid
[729]
to find our duct diameter,
[730]
which in this case is around 0.27 meters,
[733]
and we'll add that to the table also.
[735]
So just repeat that last process
[737]
for all the remaining ducts and branches
[738]
until the table is complete.
[740]
Now find the total duct
losses for each of the ducts
[743]
and branches, it's very
easy and simple to do,
[745]
just multiply the duct length
by the pressure drop per meter
[748]
of 0.65 pascals per meter
[751]
and do that for all the ducts
and branches on the table.
[753]
Now we can start on the fittings.
[755]
The first fitting we'll look
at is the 90-degree bend
[758]
between ducts J and L.
[759]
For this we look up our lost
coefficient from the bend
[762]
from the manufacturer or
from the industry body.
[764]
Again link's in the video
description below for these.
[767]
In this example,
[768]
we can see the coefficient
comes out at 0.11.
[772]
We then need to calculate
the dynamic loss caused
[774]
by the bend changing
the direction of flow.
[776]
For that we use the formula
Co multiplied by rho,
[780]
multiplied by V squared divided by two,
[783]
where Co is our coefficient,
rho is the density of the air
[786]
and V is the velocity.
[788]
We already know these values,
so if we drop these figures in
[790]
then we get an answer of 0.718 pascals,
[794]
so just add that to the table.
[796]
The next fitting will look at is the Tee
[798]
which connects the main
duct to the branches.
[800]
We'll use the example of
the tee with the ID letter H
[803]
between G and J in the system.
[805]
Now for this we need to
consider that the air is moving
[808]
in two directions, straight
through and also turning off
[811]
into the branch.
[812]
So we need to perform calculations
[814]
for both of these directions.
[815]
If we look at the air traveling
straight through first,
[818]
we find the velocity ratio first,
[820]
using the formula velocity
out, divided by velocity in.
[823]
In this example, the air out
is 3.3 meters per second,
[828]
and the air in is four meters per second,
[830]
which gives us answer of 0.83.
[833]
Then we perform another
calculation to find the area ratio.
[837]
This uses the formula
diameter out squared,
[839]
divided diameter in squared.
[842]
In this example, the
diameter out is 0.24 meters,
[845]
and the diameter in is 0.33 meters.
[848]
So if we square them, we would get 0.53.
[852]
Now we look up the fitting we're
using from the manufacturer
[854]
or the industry body,
[855]
again, link's in the video
description below for those.
[858]
In the guides, we find two tables,
[860]
the one you use depends
on the direction of flow,
[862]
we're using the straight
direction, so we'll okay that one
[865]
and then we look up each ratio
to find our last coefficient.
[868]
Here you can see both values we calculated
[870]
for between values listed in the table,
[872]
so we need to perform a
bilinear interpolation.
[874]
To save time,
[875]
we'll just use an online
calculator to find that,
[878]
links to the site are in
the video description below.
[880]
So we fill out our values and
we find the answer of 0.143.
[884]
Now we calculate the dynamic
loss for the straight path
[887]
with a tee using the formula
Co, multiplied by rho,
[889]
multiplied by V squared divided by two.
[892]
If we drop our values in, we
get the answer of 0.934 pascals
[897]
so add that to the table.
[900]
Then we can calculate the
dynamic loss for the air
[902]
which turns into the bend.
[903]
For this we use the
same formulas as before,
[906]
velocity out, divided by velocity in,
[908]
to find our velocity ratio.
[910]
We take our values from our table
[911]
and use 3.5 meters per second,
[913]
divided by four meters
per second to get 0.875,
[917]
for the velocity ratio,
then we find the area ratio,
[920]
using the formula diameter out squared,
[922]
divided by diameter in squared,
[925]
and we use 0.26 meters squared,
[927]
divided by 0.33 meters squared to get 0.62
[931]
for the area ratio,
[933]
then we use the bend
table for the T section.
[936]
Again, it's between the
values listed in the table,
[938]
so we have to find the numbers
using bilinear interpolation.
[941]
We drop the values in to get
the answer of 0.3645 pascals,
[946]
so just add that to the table too.
[948]
Now repeat that calculation
for the other tees and fittings
[951]
until the table is complete.
[953]
Next, we need to find the index run.
[955]
which is the run with the
largest pressure drop.
[957]
It's usually the longest run,
but it could also be the run
[960]
with the most fittings.
[961]
We find it easily by adding
up all the pressure losses
[964]
from start to the exit of each branch.
[967]
For example, to get from A
to C, we lose 5.04 pascals,
[972]
for A to F, we lose 8.8 pascals,
[975]
for A to I we lose 10.56 pascals,
[978]
and for A to L we lose 12.5 pascal.
[981]
Therefore the fan we use,
must overcome the run
[984]
with the highest loss that
being A to L with 12.5 pascals,
[988]
this being the index run.
[989]
To balance the system,
we need to add dampers
[991]
to each of the branches to
ensure equal pressure drop
[994]
through all to achieve the
design flow rates to each room.
[998]
We can calculate how much pressure drop
[1000]
each damper needs to provide
simply by subtracting
[1003]
the loss of the run from the index run.
[1005]
A to C is 5.04 pascals,
[1008]
which means that branch
C would need a damper,
[1011]
providing 7.46 pascals.
[1013]
A to F is 8.8 pascals
[1015]
which means that branch F would require
[1018]
a damper providing 3.7 pascals,
[1021]
and A to I is 10.56 pascals,
meaning that duct I,
[1026]
would require a damper
providing 1.94 pascals
[1030]
and that is our ductwork system.
[1032]
We'll do another video
covering additional ways
[1034]
to improve efficiency in ductwork systems,
[1036]
but unfortunately we've run
out of time in this video.
[1039]
Okay guys, that's it for this video.
[1040]
Thank you very much for watching.
[1041]
I hope you've enjoyed this
and it has helped you.
[1043]
If so, please don't forget
to like, subscribe and share
[1046]
and also check out SimScale software.
[1048]
You can follow us on
Facebook, Twitter, Instagram,
[1050]
Google Plus, as well as our website,
[1051]
TheEngineeringMindset.com.
[1053]
Once again, thanks for watching.
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