Ductwork sizing, calculation and design for efficiency – HVAC Basics + full worked example


Hey there guys, Paul here from TheEngineeringMindset.com. In this video, we’re going to
be looking at ductwork systems for mechanical ventilation. We’re going to look at how to design a basic ventilation system
with a full worked example. We’ll also look at how
to calculate the losses through the bends, the tees,
the ducts and the branches, we’ll consider the shapes in the material that the ducts are made from
to improve the efficiency, and then lastly, we’re going
to look at how to improve the efficiency and optimize the design, using a freemium software for fluid flow simulation by SimScale. Methods of ductwork design, there are many different methods used to design ventilation systems, the most common ways
being velocity reduction, equal friction and static regain. We’re going to focus on
the equal friction method in this example, as it’s
the most common method used for commercial HVAC systems, and it’s fairly simple to follow. So we’ll jump straight
into designing a system. We use a small engineering
office as an example and we’ll want to make a
layout drawing for the building which we’ll use for the
design and calculations. This is a really simple building. It has just four offices, a
corridor, and a mechanical room. And the mechanical room is where we’re going to have the fan, the filters, the air
heaters or the air cooler. The first thing we’ll need to
do is calculate the heating and cooling loads for each room. I won’t cover how to
do that in this video, we’ll have to cover that
in a separate tutorial as it’s a separate subject area. Once you have these figures,
just tally them together to find which is the biggest load as we need to size the
system to be able to operate at the peak demand. The cooling load is
usually the highest load as it is in this case. Now we need to convert the cooling loads into volume flow rates, but to do that, we first need to convert
this into mass flow rates, we use the formula M dot
equals Q divided by CP, multiply by delta T, with M
dot meaning the mass flow rate, the Q being the cooling load of the room, CP is the specific heat
capacity of the air and delta T being the
temperature difference between the design air temperature and the design return temperature, just to note that we will use a CP of 1.026 kilojoules per
kilogram per kelvin as standard, and the delta T should be less
than 10, and in this case, we’re going to use eight degrees Celsius. We know all the values for this formula, so we can calculate the mass flow rate, and the mass flow rate is
really just how many kilograms per second of air needs
to enter that room. If we look at the
calculation for room one, we see that it requires
0.26 kilograms per second. So we just repeat that calculation
for the rest of the rooms to find all the mass flow rates. Now we can convert these mass flow rates into volume flow rates. To do that we need the specific volume or density of the air. We’ll specify that the air
needs to be 21 degrees Celsius, and we’ll assume that it’s going to be at atmospheric pressure of 101.325 kPa, we can look up the
specific volume or density from our air properties tables, but I like to just use
an online calculator as it’s much quicker. So we just drop those numbers in and we get the density of
air being 1.2 kilograms per meter cubed. You see that density has the units of kilogram per meter cubed,
but we need specific volume which is meter cube per kilogram. So to convert that we
just take the inverse which means to calculate
the density or 1.2 to the power of minus one, you can just do that in Excel
very quickly to get the answer of 0.83 meters cube per kilogram. Now that we have that we can
calculate the volume flow rate, using the formula V dot
equals M dot multiplied by V, where V dot equals the volume flow rate, M dot equals the mass flow
rate of the individual room and V equals the specific
volume, which we just calculated. So if we drop these
values in for room one, we get the volume flow rate
of 0.2158 meters per second. That is how much air it
needs to enter the room to meet the cooling load. So just repeat that calculation
for all the remaining rooms. Now we’re going to sketch
out our ductwork route onto the floor plan so
we can start to size it. Before we size that we need
to consider some things which will play a big role
in the overall efficiency of the system. The first point we need
to consider is the shape of the ductwork, Ductwork comes in round,
rectangular, and flat oval shape. Round duct is by far the
most energy efficient type and that’s what we’re going to use in our worked example later on. If we compare round
duct to rectangle duct, we see that a round duct
with a cross sectional area of 0.6 meters squared has
a perimeter of 2.75 meters, a rectangular duct with
equal cross sectional area of 0.6 meters squared, has
a perimeter of 3.87 meters. The rectangular duct
therefore requires more metal for its construction. This adds more weight
and cost to the design. A larger perimeter also
means that more air will come into contact with the material
and this adds friction to the system. Friction in a system means
a fan needs to work harder and this results in
higher operating costs. Always use round duct where possible although in many cases the
rectangular duct needs to be used as space is very limited. The second thing to consider
is the material being used for the ducts. The rougher the material, the more the friction it will cause. For example, if we had two
ducts with equal dimensions, volume, flow rate and velocity, the only difference being the material, one is made from standard galvanized steel the other from fiberglass. The pressure drop over a 10-meter distance for this example is around 11 pascals for the galvanized steel and
16 pascals for the fiberglass. The third thing we have to consider is the dynamic losses
caused by the fittings. We want to use the
smoothest fittings possible for energy efficiency. For example, use long radius
bends rather than right angles as the sudden change in direction, wastes a huge amount of energy. We can compare the performance of different ductwork designs
quickly and easily using CFD or computational fluid dynamics. These simulations on screen were produce using a
revolutionary cloud-based CFD and FEA engineering platform by SimScale who have kindly sponsored this video. You can access this software
free of charge using the links in the video description below, and they offer a number of
different account types, depending on your simulation needs. SimScale is not just
limited to ductwork design, it’s also used for data
centers, AEC applications, electronics design, as well as thermal and structural analysis. Just a quick look through their site and you can find thousands
of simulations for everything from buildings, HVAC
systems, heat exchangers, pumps and valves, to
race cars and airplanes, which can all be copied
and used as templates for your own design analysis. They also offer free webinars,
courses and tutorials to help you set up and
run your own simulations. If like me, you have some experience
creating CFD simulations, then you’ll know that
this type of software is usually very expensive, and you would need a
powerful computer to run it. With SimScale however, it can all be done from a web browser, as the
platform is cloud-based, their servers do all the work and we can access and design
simulations from anywhere which makes our lives as
engineers a lot easier. So if you’re an engineer,
a designer, or an architect or just someone interested in trying out simulation technology, then I highly recommend you
check out this software, get your free account
by following the links in the video description below. Now, if we look at the
comparison for the two designs, we have a standard design on the left and a more efficient design on the right which has been optimized using SimScale. Both designs use an air velocity
of five meters per second, the color represents the velocity with blue meaning low
velocity and red representing the higher velocity regions. We can see from the velocity
color scale and the streamlines that in the design on the left, the inlet air directly
strikes the sharp turns that are present in the system, which causes an increase
in the static pressure. The sharp turns cause a large amount of recirculation regions when the duct’s preventing the air from moving smoothly, the T section at the far end
of the main duct causes the air to suddenly divide and change direction, there is a high amount of backflow here which again increases the static pressure and reduces the amount of air delivery. The higher velocity in the
main duct which is caused by the sharp turns and
sudden bends reduces the flow into the three branches on the left. If we now focus on the
optimized design on the right, we see that the fittings used
follow a much smoother profile with no sudden obstructions,
recirculation or backflow, which significantly
improves the air flow rate within the system. At the far end of the main
duct, the air is divided into the two branches through
gentle curved T section. This allows the air to
smoothly change direction and thus there is no sudden
increase in static pressure and the air flow rate to these rooms has dramatically increased. The three branches within the main duct, now receive equal airflow, making a significant
improvement to the design. This is because an
additional branch now feeds the three smaller branches, allowing some of the air
to smoothly break away from the main flow and feed
into these smaller branches. Now that we have decided
to use circular ducts, made from galvanized steel, we
can continue with the design. Label every section of ductwork
and fitting with a letter. Notice we are only designing
a very simple system here so I’ve only included
ducts and basic fittings, I’ve not included things
such as grills, inlets, flexible connections,
fire dampers, et cetera. Now we want to make a table with the rows and columns
labeled as per the example on screen now, each duct and
fitting needs its own row. If the air stream splits
such as with a T section, then we need to include a
line for each direction. So just add in the letters
to separate the rows then declare what type of fittings or ducts that corresponds to. We can start to fill some of the data in. We can first include the volume flow rates for each of the branches. This is easy, it’s just the
volume flow rates for the rooms which the branch serves. You can see on the chart
I’ve filled that in now, then we can start to size the main ducts. To do this, make sure you
start with the main duct which is furthest away, then we just add up the volume flow rates for all the branches downstream from this. For the main duct G, we just
sum the branches L and I, for D, that’s just the sum of L, I and F, and for duct A then it’s
the sum of L, I, F and C, so just enter those into the table. Now from the rough drawing, we measure out the lengths
of each of the duct sections and the branches and
enter this into the chart, and now we can begin to calculate
the size of the ductwork. To do that we need a
duct pressure loss chart, you can obtain these from
ductwork manufacturers or from industry bodies
such as CIBSE and ASHRAE. I’ve included links to these guides in the video description
below so do check those out. These charts hold a lot of information, you can use them to find
the pressure drop per meter, the air velocity, the volume flow rate, and also the size of the ductwork. The layout of the chart
does vary a little, depending on the manufacturer,
but in this example, the vertical lines are for
pressure drop per meter of duct, the horizontal lines are
for volume flow rate, the downward diagonal
lines are for velocity, and the upward diagonal
lines are for duct diameter. We start sizing from the first main duct, which in this example is section A. To limit the noise in this section, we’ll specify that you can
only have a maximum velocity of five meters per second. We know that this duct also
requires a volume flow rate of 0.79 meters cubed per second,
so we can use the velocity and volume flow rate to find
the missing data on the chart. We take the chart and scroll
up from the bottom left until we find the volume flow rate of 0.79 meters cubed per second, then we locate where the velocity line is of five meters per second
and we draw a line across until we hit that, then
to find the pressure drop, we draw a vertical line
down from this intersection. In this instance, we see it comes out at 0.65 pascals per meter, so add this figure to our table. As we are using the equal
pressure drop method, we can also use this pressure
drop for all the duct lengths, so fill these in too. Then coming back to the
chart, we scroll up again and align our intersection
with the upward diagonal lines to see that this requires
a duct with a diameter of 0.45 meters, so we add
that to the table also. We know the volume flow
rates and the pressure drop so we can now calculate
the values for section C and then also the remaining ducts. For the remainder of the ducts, we need to use the same method. On the chart we start by drawing a line from 0.65 pascals per meter. We draw this line all the way up. Then we draw another line across from where our required
volume flow rate is, in this case for section C, we require 0.21 cubic meters per second. At this intersection, we draw a line to find the velocity and we
can see that it falls between the lines of three and
four meters per second so we need to estimate the value. In this case it seems to be
around 3.6 meters per second, so we add that to the chart. Then we draw another line
on the other diagonal grid to find our duct diameter, which in this case is around 0.27 meters, and we’ll add that to the table also. So just repeat that last process for all the remaining ducts and branches until the table is complete. Now find the total duct
losses for each of the ducts and branches, it’s very
easy and simple to do, just multiply the duct length
by the pressure drop per meter of 0.65 pascals per meter and do that for all the ducts
and branches on the table. Now we can start on the fittings. The first fitting we’ll look
at is the 90-degree bend between ducts J and L. For this we look up our lost
coefficient from the bend from the manufacturer or
from the industry body. Again link’s in the video
description below for these. In this example, we can see the coefficient
comes out at 0.11. We then need to calculate
the dynamic loss caused by the bend changing
the direction of flow. For that we use the formula
Co multiplied by rho, multiplied by V squared divided by two, where Co is our coefficient,
rho is the density of the air and V is the velocity. We already know these values,
so if we drop these figures in then we get an answer of 0.718 pascals, so just add that to the table. The next fitting will look at is the Tee which connects the main
duct to the branches. We’ll use the example of
the tee with the ID letter H between G and J in the system. Now for this we need to
consider that the air is moving in two directions, straight
through and also turning off into the branch. So we need to perform calculations for both of these directions. If we look at the air traveling
straight through first, we find the velocity ratio first, using the formula velocity
out, divided by velocity in. In this example, the air out
is 3.3 meters per second, and the air in is four meters per second, which gives us answer of 0.83. Then we perform another
calculation to find the area ratio. This uses the formula
diameter out squared, divided diameter in squared. In this example, the
diameter out is 0.24 meters, and the diameter in is 0.33 meters. So if we square them, we would get 0.53. Now we look up the fitting we’re
using from the manufacturer or the industry body, again, link’s in the video
description below for those. In the guides, we find two tables, the one you use depends
on the direction of flow, we’re using the straight
direction, so we’ll okay that one and then we look up each ratio
to find our last coefficient. Here you can see both values we calculated for between values listed in the table, so we need to perform a
bilinear interpolation. To save time, we’ll just use an online
calculator to find that, links to the site are in
the video description below. So we fill out our values and
we find the answer of 0.143. Now we calculate the dynamic
loss for the straight path with a tee using the formula
Co, multiplied by rho, multiplied by V squared divided by two. If we drop our values in, we
get the answer of 0.934 pascals so add that to the table. Then we can calculate the
dynamic loss for the air which turns into the bend. For this we use the
same formulas as before, velocity out, divided by velocity in, to find our velocity ratio. We take our values from our table and use 3.5 meters per second, divided by four meters
per second to get 0.875, for the velocity ratio,
then we find the area ratio, using the formula diameter out squared, divided by diameter in squared, and we use 0.26 meters squared, divided by 0.33 meters squared to get 0.62 for the area ratio, then we use the bend
table for the T section. Again, it’s between the
values listed in the table, so we have to find the numbers
using bilinear interpolation. We drop the values in to get
the answer of 0.3645 pascals, so just add that to the table too. Now repeat that calculation
for the other tees and fittings until the table is complete. Next, we need to find the index run. which is the run with the
largest pressure drop. It’s usually the longest run,
but it could also be the run with the most fittings. We find it easily by adding
up all the pressure losses from start to the exit of each branch. For example, to get from A
to C, we lose 5.04 pascals, for A to F, we lose 8.8 pascals, for A to I we lose 10.56 pascals, and for A to L we lose 12.5 pascal. Therefore the fan we use,
must overcome the run with the highest loss that
being A to L with 12.5 pascals, this being the index run. To balance the system,
we need to add dampers to each of the branches to
ensure equal pressure drop through all to achieve the
design flow rates to each room. We can calculate how much pressure drop each damper needs to provide
simply by subtracting the loss of the run from the index run. A to C is 5.04 pascals, which means that branch
C would need a damper, providing 7.46 pascals. A to F is 8.8 pascals which means that branch F would require a damper providing 3.7 pascals, and A to I is 10.56 pascals,
meaning that duct I, would require a damper
providing 1.94 pascals and that is our ductwork system. We’ll do another video
covering additional ways to improve efficiency in ductwork systems, but unfortunately we’ve run
out of time in this video. Okay guys, that’s it for this video. Thank you very much for watching. I hope you’ve enjoyed this
and it has helped you. If so, please don’t forget
to like, subscribe and share and also check out SimScale software. You can follow us on
Facebook, Twitter, Instagram, Google Plus, as well as our website, TheEngineeringMindset.com. Once again, thanks for watching.

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