How to design a duct system. In this article we’ll be learning how to size and design a ductwork system for efficiency. We’ll include a full worked example as well as using CFD simulations to optimise the performance and efficiency using SimScale. **Scroll to the bottom to watch the FREE YouTube video tutorial!**

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## Methods of ductwork design

There are many different methods used to design ventilation systems, the most common ways being:

- Velocity reduction method: (Residential or small commercial installations)
- Equal friction method: (Medium to large sized commercial installations)
- Static regain: Very large installations (concert halls, airports and industrial)

Weâ€™re going to focus on the equal friction method in this example as itâ€™s the most common method used for commercial HVAC systems and its fairly simple to follow.

## Design example

So weâ€™ll jump straight into designing a system. Weâ€™ll use a small engineering office as an example and we want to make a layout drawing of the building which weâ€™ll use for the design and calculations. This is a really simple building it has just 4 offices a corridor and a mechanical room which is where the fan, filters and air heater or cooler will be located.

The first thing we need to do is calculate the heating and cooling loads for each room. I wonâ€™t cover how to do that in this article weâ€™ll have to cover that in a separate tutorial as itâ€™s a separate subject area.

Once you have these, 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, 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 to mass flow rate so we use the formula:

mdot = Q / (cp x Î”t)

Where mdot means mass flow rate (kg/s), the Q being the cooling load of the room (kW), cp is the specific heat capacity of the air (kJ/kg.K) and Î”t being the temperature difference between the designed air temperature and the design return temperature. Just to note that we will use a cp of 1.026 kJ/kg.k as standard and the delta T should be less than 10*C so weâ€™ll use 8*c.

We know all the values for this so we can calculate the mass flow rate (how many kilograms per second of air needs to enter the room). If we look at the calculation for room 1, we see it requires 0.26 kg/s. So we just repeat that calculation for the rest of the room to find all the mass flow rates.

Now we can convert these into volume flow rates. To do that we need the specific volume or density of the air. Weâ€™ll specify 21*c and assume atmospheric pressure of 101.325 kPa. We can look this up in our air properties tables but I like to just use an online calculator http://bit.ly/2tyT8yp as its quicker. So we just drop those numbers in and we get the density of air being 1.2 kg/m3.

You see that density has the units of kg/m3 but we need specific volume which is m3/kg so to convert that we just take the inverse which means to calculate 1.2 to the power of -1. You can just do that in excel very quickly (copy paste this =1.2^-1) to get the answer of 0.83m3/kg.

Now that we have that we can calculate the volume flow rate using the formula:

vdot = mdot multiplied by v.

where vdot equals the volume flow rate, mdot equals the mass flow rate of the room and v equals the specific volume which we just calculated.

So if we drop those values in for room 1 we get a volume flow rate of 0.2158m3/s that is how much air needs to enter the room to meet the cooling load. So just repeat that calculation for all the rooms.

Now weâ€™re going to sketch out our ductwork route onto the floor plan so we can start to size it.

Before we go any further we need to consider some things which will play a big role in the overall efficiency of the system.

## Design considerations

The first one being 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 thatâ€™s what weâ€™ll use in our worked example later on. If we compare round duct to rectangular duct we see that:

A round duct with a cross sectional area of 0.6m2 has a perimeter of 2.75m

A rectangular duct with an equal cross sectional area has a perimeter of 3.87m

The rectangular duct therefore requires more metal for its construction, this adds more weight and costs to the design. The larger perimeter also means more air will come into contact wit the material and this adds friction to the system. Friction in a system means the fan needs to work harder and this results in higher operating costs. Always use round duct where possible although in many cases rectangular duct needs to be used as space is limited.

The second thing to consider is the material being used for the ducts, and the roughness of this material as this causes friction. For example, if we had two ducts, with equal dimensions, volume flow rate and velocity, the only difference is the material. One is made from standard galvanised steel the other from fibreglass, the pressure drop over a 10m distance for this example, is around 11 Pa for the galvanised steel and 16 Pa for the Fibreglass.

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 were produced using a revolutionary cloud based CFD and FEA engineering platform, by SimScale, who have kindly sponsored this article.

You can access this software free of charge by clicking here, and they offer a number of different account types depending on your simulation needs.

SimScale is not just limited to ductwork design, itâ€™s also used for data centers, AEC applications, electronics design, as well as thermal and structural analysis.

Just a quick look on their site and you can find thousands of simulations for everything from buildings, HVAC systems, heat exchangers, pumps and valves to race cars and air planes, 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 youâ€™ll know that this type of software is usually very expensive and you would also need a powerful computer to run it.

With SimScale, however, all can be done from a web browser. As the platform is cloud-based, their servers do all the work and we can access our design simulations from anywhere, which makes our lives as engineers a lot easier.

So if youâ€™re an engineer, designer, architect or just someone interested in trying out simulation technology, then I highly recommend you check this software out, get your free account by following this link.

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 optimised using simscale. Both designs use an air velocity of 5m/s, the colours represent the velocity with blue meaning low velocity and red representing the high velocity regions.

We can see from the velocity colour 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 within the ducts, preventing the air from moving smoothly.

The tee 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 high velocity in the main duct which is caused by the sharp turns and sudden bends, reduces the flow into the 3 branches on the left.

If we now focus on the optimised design on the right, we see 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 two branches through a gentle, curved tee 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 the rooms has dramatically increased.

The three branches within the main duct now receive equal air flow 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.

With these considerations in place we can come back to the duct design.

Now we need to label every section of ductwork as well as the fittings with a letter. Notice we are only designing a very simple system here so Iâ€™ve only included ducts and basic fittings, Iâ€™ve not included things such as grilles, inlets, flexible connections, fire dampers etc.

Now we want to make a table with the rows labelled as per the example. Each duct and fitting needs its own row, if the air stream splits such as with a Tee section, then we need to include a line for each direction, weâ€™ll see that later in the article.

Just add in the letters to separate rows then declare what type of fitting or duct 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 as its just the volume flow rate for the room which it serves. You can see on the chart Iâ€™ve filled that in.

Then we can start to size the main ducts. To do this make sure you start at the main duct which is furthest away. Then we just add up the volume flow rates for all the branches downstream of this. For the main duct G we just sum branches L and I. For D thatâ€™s just the sum of L I and F and for duct A its then the sum of L, I, F and C. so just enter those into the table.

From the rough drawing we measure out the length of each duct section and enter this into the chart.

## Duct sizing – How to size ductwork

To size the ducts you’re going to need a duct sizing chart. You can obtain these from ductwork manufacturers or from industry bodies such as CIBSE and ASHRAE. If you don’t have one, you can find them in the following links. Link 1 and Link 2

These charts hold a lot of information. We 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, the upward diagonal lines are for duct diameter.

We start sizing from the first main duct which is section A. To limit the noise in this section weâ€™ll specify that it can only have a maximum velocity of 5m/s. We know that this duct also requires a volume flow rate of 0.79m3/s so we can use the velocity and volume flow rate to find the missing data.

We take the chart and scroll up from the bottom left until we hit the volume flow rate of 0.79m3/s. Then we locate where the velocity line is of 5m/s 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 pa per meter. So add this figure into the chart. As weâ€™re using the equal pressure drop method we can use this pressure drop for all the duct lengths so fill those in too. Then we scroll up again and align our intersection with the upward diagonal lines to see this requires a duct with a diameter of 0.45m so we add that into the table also.

We know the volume flow rate and pressure drop so we can now calculate the values for section C and then the remaining ducts.

For the remainder of the ducts we use the same method.

On the chart we start by drawing a line from 0.65 pa/m all the way up and then draw a line across from our required volume flow rate, in this case for section C we need 0.21m3/s. At this intersection we draw a line to find the velocity and we can see that it falls within the lines of 3 and 4m/s so we need to estimate the value, in this case it seems to be about 3.6m/s 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 about 0.27m and weâ€™ll add that to the table too.

Repeat that last process for all the remaining ducts and branches until the table is complete.

Now find the total duct losses for each duct and branch, thatâ€™s very easy to do simply multiply the duct length by the pressure drop per meter, in our example we found it to be 0.65pa/m. Do that for all the ducts and branches on the table.

## Sizing ductwork fittings

The first fitting weâ€™ll look at is the 90* bend between ducts J and L

For this we look up our loss coefficient for the bend from the manufacturer or the industry body, you can find that by clicking this link.

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 2 where co is our coefficient, rho is the density of the air and v is the velocity.

We already know all these values so if we drop the figures in we get an answer of 0.718 pascals. So just add that to the table. (Watch the video at the bottom of the page to see how to calculate that).

The next fitting weâ€™ll look at is the tee which connects the main duct to the branches, weâ€™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 a calculation for both directions.

If we look at the air travelling straight though first, we find the velocity ratio first using the formula velocity out divided by velocity in. In this example the air out is 3.3m/s and the air in is 4m/s which gives us 0.83

Then we perform another calculation to find the area ratio, this uses the formula diameter out squared divided by diameter in squared. In this example the diameter out is 0.24m and the diameter in is 0.33m so if we square them and then divide we get 0.53

Now we look up the fitting weâ€™re using from the manufacturer or the industry body, again link here for that.

In the guides we find two tables the one you use depends on the direction of flow, weâ€™re using the straight direction so we locate that one and then look up each ratio to find our loss coefficient. Here you can see both of the values we calculated fall between vales listed in the table so we need to perform a bilinear interpolation. To save time weâ€™ll just use an online calculator to find that, link here (watch the video to learn how to perform a bilinear interpolation).

We fill out our values and we find the answer of 0.143

Now we calculate the dynamic loss for the straight path through the tee, using the formula co multiplied by rho multiplied by v squared divided by 2. 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 didived by velocity in to find our velocity ratio. Then we find the area ratio using the formula diameter out squared divided by diameter in squared. We take our values from our table and use 3.5m/s divided by 4m/s to get 0.875 for the velocity ratio and we use 0.26m squared divided by 0.33m squared to get 0.62 for the area ratio.

Then we use the bend table for the tee section, again its 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 pascales. So just add that to the table too.

Now repeat that calculation for the other tees and fittings until the table to complete.

## Finding the index run – duct sizing

Next we need to find the index run which is the run with the largest pressure drop. Itâ€™s usually the longest run but could also be the run with the most fittings.

We find it easily by adding up all the pressure losses from the start to the exit of each branch .

For example to get from A to C we lose 5.04pa

A (1.3pa) + B (1.79pa) + C (1.95pa)

For A to F we lose 8.8pa

A (1.3pa) + B (1.7pa) + D (1.3pa) + E (2.55pa) + F (1.95)

For A to I we lose 10.56

A (1.3pa) + B (1.7pa) + D (1.3pa) + E (1.34pa) + G (2.6pa) + H (0.36pa) + I (1.95pa)

For A to L we lose 12.5pa

A(1.3pa) + B (1.7pa) + D (1.3pa) + E (1.34pa) + G (2.6pa) + H (0.93pa) + J (0.65pa) + K (0.72pa) + L (1.95pa)

Therefore the fan we use must overcome the run with the highest loss, that being A â€“ L with 12.5pa, this is the index run.

## Ductwork dampers – system balancing

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 12.5pa â€“ 5.04pa = 7.46pa

A to F is 12.5pa â€“ 8.8pa = 3.7pa

A to I is 12.5pa â€“ 10.56pa = 1.94pa

And that is our ducting system. Weâ€™ll do another tutorial covering additional ways to improve efficiency in ductwork system.

Thank you very much fo this info guys. May you do more lessons on ducting. especially on dumpers, flexible duct calculations, duct risers and toilets ventilation.

This is a great work, it is really well explained and truly beneficial. thanks a lot for this piece. However, I have a question, in calculating the overall pressure drop per duct length. Are we considering the effective length of the duct or just the length of the duct?

Thank you.

thanks

Hi there Paul

Thank you for this informative explanation and its really helpful. My question is on the selection of the unit size Do I use the total cooling load to select the size of my machine or I have to use the also the data on the duct friction loss to select the unit. Please I need help as I am trying to understand clear how its done.

Also can you do another illustration on the return duct size and location.

I would like it as well if you can finish the illustration with diffusers, grills and flexible ducts.

Thank you so much with your info

Thanks a lot for this is a useful work, it is well explained and really helpful. thanks a lot for this great work. May you do more lessons similar to this series on BMS, SCADA, DCS and its interaction with HVAC systems in the smart buildings.

Thank you very much for your great work.

I appreciate that you explain to us how the equal friction method is used for medium to large size commercial installation. Personally, I don know much about ductwork, but I think is great to learn more about it. Thank you for helping me learn more about the methods used for ductwork designs.

Hi,

Thank you for the video, it is well explained. However, please I will like see how grilles and filters affect the external static pressure of a system in your next video.

Thanks,

Newton

thanks for the great work

Hi, thanks for this article. I used this method with the same pressure in all ducts from the main to all the branches with the quantity of air needed, I found out that some duct and branches fall below 2 m/sec. what will happen if I adjust the velocity and the pressure?

So for attaining designed cooling for Room 1 ( 2.4KW OR 0.68 TR) , do we need a flow rate of 457 CFM.

Pls help me team on this , do we need this much CFM for less than a ton cooling.

This is a good article. It is a combination of fun and informative. Thanks for sharing this!

Excellent post. I want to thank you for this informative, I really appreciate sharing this great post. Thanks for sharing this