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Ductwork sizing, calculation and design for efficiency - HVAC Basics + full worked example
WEBVTT Kind: captions Language: en
00:00:04.820 --> 00:00:05.790 Hey there guys, Paul here 00:00:05.790 --> 00:00:07.370 from TheEngineeringMindset.com. 00:00:07.370 --> 00:00:09.760 In this video, we're going to be looking at ductwork systems 00:00:09.760 --> 00:00:12.130 for mechanical ventilation. 00:00:12.130 --> 00:00:13.610 We're going to look at how to design 00:00:13.610 --> 00:00:16.910 a basic ventilation system with a full worked example. 00:00:16.910 --> 00:00:18.303 We'll also look at how to calculate the losses 00:00:18.303 --> 00:00:21.300 through the bends, the tees, the ducts and the branches, 00:00:21.300 --> 00:00:23.050 we'll consider the shapes in the material 00:00:23.050 --> 00:00:25.580 that the ducts are made from to improve the efficiency, 00:00:25.580 --> 00:00:27.480 and then lastly, we're going to look at how to improve 00:00:27.480 --> 00:00:29.470 the efficiency and optimize the design, 00:00:29.470 --> 00:00:30.780 using a freemium software 00:00:30.780 --> 00:00:33.113 for fluid flow simulation by SimScale. 00:00:34.190 --> 00:00:35.980 Methods of ductwork design, 00:00:35.980 --> 00:00:37.470 there are many different methods used 00:00:37.470 --> 00:00:39.170 to design ventilation systems, 00:00:39.170 --> 00:00:42.160 the most common ways being velocity reduction, 00:00:42.160 --> 00:00:44.640 equal friction and static regain. 00:00:44.640 --> 00:00:46.740 We're going to focus on the equal friction method 00:00:46.740 --> 00:00:49.580 in this example, as it's the most common method used 00:00:49.580 --> 00:00:51.340 for commercial HVAC systems, 00:00:51.340 --> 00:00:52.990 and it's fairly simple to follow. 00:00:53.910 --> 00:00:56.300 So we'll jump straight into designing a system. 00:00:56.300 --> 00:00:58.680 We use a small engineering office as an example 00:00:58.680 --> 00:01:01.040 and we'll want to make a layout drawing for the building 00:01:01.040 --> 00:01:03.300 which we'll use for the design and calculations. 00:01:03.300 --> 00:01:04.790 This is a really simple building. 00:01:04.790 --> 00:01:07.707 It has just four offices, a corridor, and a mechanical room. 00:01:07.707 --> 00:01:08.750 And the mechanical room 00:01:08.750 --> 00:01:10.170 is where we're going to have the fan, 00:01:10.170 --> 00:01:12.970 the filters, the air heaters or the air cooler. 00:01:12.970 --> 00:01:15.540 The first thing we'll need to do is calculate the heating 00:01:15.540 --> 00:01:18.020 and cooling loads for each room. 00:01:18.020 --> 00:01:19.820 I won't cover how to do that in this video, 00:01:19.820 --> 00:01:21.620 we'll have to cover that in a separate tutorial 00:01:21.620 --> 00:01:23.523 as it's a separate subject area. 00:01:24.400 --> 00:01:26.530 Once you have these figures, just tally them together 00:01:26.530 --> 00:01:28.180 to find which is the biggest load 00:01:28.180 --> 00:01:30.670 as we need to size the system to be able to operate 00:01:30.670 --> 00:01:32.070 at the peak demand. 00:01:32.070 --> 00:01:34.290 The cooling load is usually the highest load 00:01:34.290 --> 00:01:35.723 as it is in this case. 00:01:36.800 --> 00:01:38.420 Now we need to convert the cooling loads 00:01:38.420 --> 00:01:40.580 into volume flow rates, but to do that, 00:01:40.580 --> 00:01:43.080 we first need to convert this into mass flow rates, 00:01:43.080 --> 00:01:46.910 we use the formula M dot equals Q divided by CP, 00:01:46.910 --> 00:01:50.520 multiply by delta T, with M dot meaning the mass flow rate, 00:01:50.520 --> 00:01:52.740 the Q being the cooling load of the room, 00:01:52.740 --> 00:01:55.560 CP is the specific heat capacity of the air 00:01:55.560 --> 00:01:57.610 and delta T being the temperature difference 00:01:57.610 --> 00:01:59.650 between the design air temperature 00:01:59.650 --> 00:02:01.700 and the design return temperature, 00:02:01.700 --> 00:02:03.540 just to note that we will use a CP 00:02:03.540 --> 00:02:07.680 of 1.026 kilojoules per kilogram per kelvin as standard, 00:02:07.680 --> 00:02:10.490 and the delta T should be less than 10, and in this case, 00:02:10.490 --> 00:02:12.730 we're going to use eight degrees Celsius. 00:02:12.730 --> 00:02:14.380 We know all the values for this formula, 00:02:14.380 --> 00:02:15.927 so we can calculate the mass flow rate, 00:02:15.927 --> 00:02:18.480 and the mass flow rate is really just how many kilograms 00:02:18.480 --> 00:02:20.640 per second of air needs to enter that room. 00:02:20.640 --> 00:02:22.560 If we look at the calculation for room one, 00:02:22.560 --> 00:02:26.150 we see that it requires 0.26 kilograms per second. 00:02:26.150 --> 00:02:28.710 So we just repeat that calculation for the rest of the rooms 00:02:28.710 --> 00:02:30.850 to find all the mass flow rates. 00:02:30.850 --> 00:02:32.720 Now we can convert these mass flow rates 00:02:32.720 --> 00:02:34.030 into volume flow rates. 00:02:34.030 --> 00:02:36.080 To do that we need the specific volume 00:02:36.080 --> 00:02:37.630 or density of the air. 00:02:37.630 --> 00:02:40.660 We'll specify that the air needs to be 21 degrees Celsius, 00:02:40.660 --> 00:02:41.840 and we'll assume that it's going to be 00:02:41.840 --> 00:02:45.600 at atmospheric pressure of 101.325 kPa, 00:02:45.600 --> 00:02:47.670 we can look up the specific volume or density 00:02:47.670 --> 00:02:49.110 from our air properties tables, 00:02:49.110 --> 00:02:50.930 but I like to just use an online calculator 00:02:50.930 --> 00:02:52.130 as it's much quicker. 00:02:52.130 --> 00:02:53.460 So we just drop those numbers in 00:02:53.460 --> 00:02:56.210 and we get the density of air being 1.2 kilograms 00:02:56.210 --> 00:02:57.670 per meter cubed. 00:02:57.670 --> 00:02:59.060 You see that density has the units 00:02:59.060 --> 00:03:02.290 of kilogram per meter cubed, but we need specific volume 00:03:02.290 --> 00:03:04.350 which is meter cube per kilogram. 00:03:04.350 --> 00:03:06.400 So to convert that we just take the inverse 00:03:06.400 --> 00:03:08.840 which means to calculate the density or 1.2 00:03:08.840 --> 00:03:10.460 to the power of minus one, 00:03:10.460 --> 00:03:13.270 you can just do that in Excel very quickly to get the answer 00:03:13.270 --> 00:03:16.920 of 0.83 meters cube per kilogram. 00:03:16.920 --> 00:03:19.440 Now that we have that we can calculate the volume flow rate, 00:03:19.440 --> 00:03:23.140 using the formula V dot equals M dot multiplied by V, 00:03:23.140 --> 00:03:25.150 where V dot equals the volume flow rate, 00:03:25.150 --> 00:03:27.840 M dot equals the mass flow rate of the individual room 00:03:27.840 --> 00:03:31.860 and V equals the specific volume, which we just calculated. 00:03:31.860 --> 00:03:33.880 So if we drop these values in for room one, 00:03:33.880 --> 00:03:38.720 we get the volume flow rate of 0.2158 meters per second. 00:03:38.720 --> 00:03:40.880 That is how much air it needs to enter the room 00:03:40.880 --> 00:03:42.030 to meet the cooling load. 00:03:42.030 --> 00:03:45.310 So just repeat that calculation for all the remaining rooms. 00:03:45.310 --> 00:03:47.420 Now we're going to sketch out our ductwork route 00:03:47.420 --> 00:03:50.450 onto the floor plan so we can start to size it. 00:03:50.450 --> 00:03:53.130 Before we size that we need to consider some things 00:03:53.130 --> 00:03:55.570 which will play a big role in the overall efficiency 00:03:55.570 --> 00:03:56.940 of the system. 00:03:56.940 --> 00:03:58.830 The first point we need to consider is the shape 00:03:58.830 --> 00:03:59.760 of the ductwork, 00:03:59.760 --> 00:04:03.070 Ductwork comes in round, rectangular, and flat oval shape. 00:04:03.070 --> 00:04:05.700 Round duct is by far the most energy efficient type 00:04:05.700 --> 00:04:06.860 and that's what we're going to use 00:04:06.860 --> 00:04:08.440 in our worked example later on. 00:04:08.440 --> 00:04:10.600 If we compare round duct to rectangle duct, 00:04:10.600 --> 00:04:13.140 we see that a round duct with a cross sectional area 00:04:13.140 --> 00:04:17.890 of 0.6 meters squared has a perimeter of 2.75 meters, 00:04:17.890 --> 00:04:20.180 a rectangular duct with equal cross sectional area 00:04:20.180 --> 00:04:24.820 of 0.6 meters squared, has a perimeter of 3.87 meters. 00:04:24.820 --> 00:04:27.040 The rectangular duct therefore requires more metal 00:04:27.040 --> 00:04:27.940 for its construction. 00:04:27.940 --> 00:04:30.080 This adds more weight and cost to the design. 00:04:30.080 --> 00:04:32.790 A larger perimeter also means that more air will come 00:04:32.790 --> 00:04:35.170 into contact with the material and this adds friction 00:04:35.170 --> 00:04:36.190 to the system. 00:04:36.190 --> 00:04:38.910 Friction in a system means a fan needs to work harder 00:04:38.910 --> 00:04:41.160 and this results in higher operating costs. 00:04:41.160 --> 00:04:43.000 Always use round duct where possible 00:04:43.000 --> 00:04:45.650 although in many cases the rectangular duct needs to be used 00:04:45.650 --> 00:04:47.960 as space is very limited. 00:04:47.960 --> 00:04:50.620 The second thing to consider is the material being used 00:04:50.620 --> 00:04:51.800 for the ducts. 00:04:51.800 --> 00:04:52.970 The rougher the material, 00:04:52.970 --> 00:04:55.040 the more the friction it will cause. 00:04:55.040 --> 00:04:57.680 For example, if we had two ducts with equal dimensions, 00:04:57.680 --> 00:04:59.030 volume, flow rate and velocity, 00:04:59.030 --> 00:05:01.000 the only difference being the material, 00:05:01.000 --> 00:05:02.930 one is made from standard galvanized steel 00:05:02.930 --> 00:05:04.470 the other from fiberglass. 00:05:04.470 --> 00:05:06.480 The pressure drop over a 10-meter distance 00:05:06.480 --> 00:05:08.710 for this example is around 11 pascals 00:05:08.710 --> 00:05:12.530 for the galvanized steel and 16 pascals for the fiberglass. 00:05:12.530 --> 00:05:13.920 The third thing we have to consider 00:05:13.920 --> 00:05:16.480 is the dynamic losses caused by the fittings. 00:05:16.480 --> 00:05:18.680 We want to use the smoothest fittings possible 00:05:18.680 --> 00:05:20.060 for energy efficiency. 00:05:20.060 --> 00:05:23.410 For example, use long radius bends rather than right angles 00:05:23.410 --> 00:05:25.140 as the sudden change in direction, 00:05:25.140 --> 00:05:27.430 wastes a huge amount of energy. 00:05:27.430 --> 00:05:28.840 We can compare the performance 00:05:28.840 --> 00:05:32.290 of different ductwork designs quickly and easily using CFD 00:05:32.290 --> 00:05:34.200 or computational fluid dynamics. 00:05:34.200 --> 00:05:35.530 These simulations on screen 00:05:35.530 --> 00:05:38.410 were produce using a revolutionary cloud-based CFD 00:05:38.410 --> 00:05:40.220 and FEA engineering platform by SimScale 00:05:40.220 --> 00:05:42.720 who have kindly sponsored this video. 00:05:42.720 --> 00:05:45.420 You can access this software free of charge using the links 00:05:45.420 --> 00:05:46.750 in the video description below, 00:05:46.750 --> 00:05:48.390 and they offer a number of different account types, 00:05:48.390 --> 00:05:50.660 depending on your simulation needs. 00:05:50.660 --> 00:05:52.620 SimScale is not just limited to ductwork design, 00:05:52.620 --> 00:05:55.360 it's also used for data centers, AEC applications, 00:05:55.360 --> 00:05:57.190 electronics design, as well as thermal 00:05:57.190 --> 00:05:58.850 and structural analysis. 00:05:58.850 --> 00:06:00.280 Just a quick look through their site 00:06:00.280 --> 00:06:02.700 and you can find thousands of simulations for everything 00:06:02.700 --> 00:06:05.350 from buildings, HVAC systems, heat exchangers, 00:06:05.350 --> 00:06:07.730 pumps and valves, to race cars and airplanes, 00:06:07.730 --> 00:06:09.930 which can all be copied and used as templates 00:06:09.930 --> 00:06:11.970 for your own design analysis. 00:06:11.970 --> 00:06:14.550 They also offer free webinars, courses and tutorials 00:06:14.550 --> 00:06:17.120 to help you set up and run your own simulations. 00:06:17.120 --> 00:06:17.953 If like me, 00:06:17.953 --> 00:06:20.430 you have some experience creating CFD simulations, 00:06:20.430 --> 00:06:21.990 then you'll know that this type of software 00:06:21.990 --> 00:06:23.420 is usually very expensive, 00:06:23.420 --> 00:06:25.980 and you would need a powerful computer to run it. 00:06:25.980 --> 00:06:28.170 With SimScale however, it can all be done 00:06:28.170 --> 00:06:31.110 from a web browser, as the platform is cloud-based, 00:06:31.110 --> 00:06:32.700 their servers do all the work 00:06:32.700 --> 00:06:35.550 and we can access and design simulations from anywhere 00:06:35.550 --> 00:06:38.190 which makes our lives as engineers a lot easier. 00:06:38.190 --> 00:06:40.410 So if you're an engineer, a designer, or an architect 00:06:40.410 --> 00:06:41.400 or just someone interested 00:06:41.400 --> 00:06:43.400 in trying out simulation technology, 00:06:43.400 --> 00:06:45.620 then I highly recommend you check out this software, 00:06:45.620 --> 00:06:47.420 get your free account by following the links 00:06:47.420 --> 00:06:49.170 in the video description below. 00:06:49.170 --> 00:06:52.040 Now, if we look at the comparison for the two designs, 00:06:52.040 --> 00:06:53.730 we have a standard design on the left 00:06:53.730 --> 00:06:55.450 and a more efficient design on the right 00:06:55.450 --> 00:06:57.600 which has been optimized using SimScale. 00:06:57.600 --> 00:07:00.620 Both designs use an air velocity of five meters per second, 00:07:00.620 --> 00:07:02.310 the color represents the velocity 00:07:02.310 --> 00:07:04.820 with blue meaning low velocity and red representing 00:07:04.820 --> 00:07:06.660 the higher velocity regions. 00:07:06.660 --> 00:07:09.510 We can see from the velocity color scale and the streamlines 00:07:09.510 --> 00:07:11.030 that in the design on the left, 00:07:11.030 --> 00:07:13.760 the inlet air directly strikes the sharp turns 00:07:13.760 --> 00:07:15.200 that are present in the system, 00:07:15.200 --> 00:07:17.940 which causes an increase in the static pressure. 00:07:17.940 --> 00:07:19.730 The sharp turns cause a large amount 00:07:19.730 --> 00:07:21.740 of recirculation regions when the duct's 00:07:21.740 --> 00:07:24.210 preventing the air from moving smoothly, 00:07:24.210 --> 00:07:27.260 the T section at the far end of the main duct causes the air 00:07:27.260 --> 00:07:29.620 to suddenly divide and change direction, 00:07:29.620 --> 00:07:31.320 there is a high amount of backflow here 00:07:31.320 --> 00:07:33.430 which again increases the static pressure 00:07:33.430 --> 00:07:35.600 and reduces the amount of air delivery. 00:07:35.600 --> 00:07:37.720 The higher velocity in the main duct which is caused 00:07:37.720 --> 00:07:40.610 by the sharp turns and sudden bends reduces the flow 00:07:40.610 --> 00:07:43.230 into the three branches on the left. 00:07:43.230 --> 00:07:45.920 If we now focus on the optimized design on the right, 00:07:45.920 --> 00:07:49.330 we see that the fittings used follow a much smoother profile 00:07:49.330 --> 00:07:52.780 with no sudden obstructions, recirculation or backflow, 00:07:52.780 --> 00:07:55.110 which significantly improves the air flow rate 00:07:55.110 --> 00:07:55.943 within the system. 00:07:55.943 --> 00:07:58.470 At the far end of the main duct, the air is divided 00:07:58.470 --> 00:08:02.010 into the two branches through gentle curved T section. 00:08:02.010 --> 00:08:04.380 This allows the air to smoothly change direction 00:08:04.380 --> 00:08:07.170 and thus there is no sudden increase in static pressure 00:08:07.170 --> 00:08:08.220 and the air flow rate 00:08:08.220 --> 00:08:10.890 to these rooms has dramatically increased. 00:08:10.890 --> 00:08:12.960 The three branches within the main duct, 00:08:12.960 --> 00:08:14.610 now receive equal airflow, 00:08:14.610 --> 00:08:17.110 making a significant improvement to the design. 00:08:17.110 --> 00:08:19.380 This is because an additional branch now feeds 00:08:19.380 --> 00:08:20.740 the three smaller branches, 00:08:20.740 --> 00:08:23.110 allowing some of the air to smoothly break away 00:08:23.110 --> 00:08:26.093 from the main flow and feed into these smaller branches. 00:08:27.270 --> 00:08:29.500 Now that we have decided to use circular ducts, 00:08:29.500 --> 00:08:33.260 made from galvanized steel, we can continue with the design. 00:08:33.260 --> 00:08:36.740 Label every section of ductwork and fitting with a letter. 00:08:36.740 --> 00:08:39.460 Notice we are only designing a very simple system here 00:08:39.460 --> 00:08:41.980 so I've only included ducts and basic fittings, 00:08:41.980 --> 00:08:44.500 I've not included things such as grills, inlets, 00:08:44.500 --> 00:08:47.700 flexible connections, fire dampers, et cetera. 00:08:47.700 --> 00:08:49.180 Now we want to make a table 00:08:49.180 --> 00:08:51.850 with the rows and columns labeled as per the example 00:08:51.850 --> 00:08:55.360 on screen now, each duct and fitting needs its own row. 00:08:55.360 --> 00:08:58.060 If the air stream splits such as with a T section, 00:08:58.060 --> 00:09:00.600 then we need to include a line for each direction. 00:09:00.600 --> 00:09:02.690 So just add in the letters to separate the rows 00:09:02.690 --> 00:09:04.230 then declare what type of fittings 00:09:04.230 --> 00:09:06.010 or ducts that corresponds to. 00:09:06.010 --> 00:09:08.030 We can start to fill some of the data in. 00:09:08.030 --> 00:09:09.950 We can first include the volume flow rates 00:09:09.950 --> 00:09:11.110 for each of the branches. 00:09:11.110 --> 00:09:13.470 This is easy, it's just the volume flow rates for the rooms 00:09:13.470 --> 00:09:15.030 which the branch serves. 00:09:15.030 --> 00:09:17.530 You can see on the chart I've filled that in now, 00:09:17.530 --> 00:09:19.690 then we can start to size the main ducts. 00:09:19.690 --> 00:09:21.950 To do this, make sure you start with the main duct 00:09:21.950 --> 00:09:23.610 which is furthest away, 00:09:23.610 --> 00:09:25.350 then we just add up the volume flow rates 00:09:25.350 --> 00:09:27.430 for all the branches downstream from this. 00:09:27.430 --> 00:09:30.980 For the main duct G, we just sum the branches L and I, 00:09:30.980 --> 00:09:34.170 for D, that's just the sum of L, I and F, 00:09:34.170 --> 00:09:38.120 and for duct A then it's the sum of L, I, F and C, 00:09:38.120 --> 00:09:40.390 so just enter those into the table. 00:09:40.390 --> 00:09:41.740 Now from the rough drawing, 00:09:41.740 --> 00:09:44.460 we measure out the lengths of each of the duct sections 00:09:44.460 --> 00:09:47.130 and the branches and enter this into the chart, 00:09:47.130 --> 00:09:50.190 and now we can begin to calculate the size of the ductwork. 00:09:50.190 --> 00:09:52.650 To do that we need a duct pressure loss chart, 00:09:52.650 --> 00:09:54.850 you can obtain these from ductwork manufacturers 00:09:54.850 --> 00:09:57.770 or from industry bodies such as CIBSE and ASHRAE. 00:09:57.770 --> 00:09:59.140 I've included links to these guides 00:09:59.140 --> 00:10:01.870 in the video description below so do check those out. 00:10:01.870 --> 00:10:04.320 These charts hold a lot of information, 00:10:04.320 --> 00:10:06.630 you can use them to find the pressure drop per meter, 00:10:06.630 --> 00:10:08.640 the air velocity, the volume flow rate, 00:10:08.640 --> 00:10:10.550 and also the size of the ductwork. 00:10:10.550 --> 00:10:12.650 The layout of the chart does vary a little, 00:10:12.650 --> 00:10:15.340 depending on the manufacturer, but in this example, 00:10:15.340 --> 00:10:18.420 the vertical lines are for pressure drop per meter of duct, 00:10:18.420 --> 00:10:20.840 the horizontal lines are for volume flow rate, 00:10:20.840 --> 00:10:23.470 the downward diagonal lines are for velocity, 00:10:23.470 --> 00:10:26.440 and the upward diagonal lines are for duct diameter. 00:10:26.440 --> 00:10:28.720 We start sizing from the first main duct, 00:10:28.720 --> 00:10:30.770 which in this example is section A. 00:10:30.770 --> 00:10:32.170 To limit the noise in this section, 00:10:32.170 --> 00:10:34.810 we'll specify that you can only have a maximum velocity 00:10:34.810 --> 00:10:36.490 of five meters per second. 00:10:36.490 --> 00:10:38.950 We know that this duct also requires a volume flow rate 00:10:38.950 --> 00:10:42.530 of 0.79 meters cubed per second, so we can use the velocity 00:10:42.530 --> 00:10:46.210 and volume flow rate to find the missing data on the chart. 00:10:46.210 --> 00:10:48.700 We take the chart and scroll up from the bottom left 00:10:48.700 --> 00:10:50.530 until we find the volume flow rate 00:10:50.530 --> 00:10:53.110 00:10:53.110 --> 00:10:55.060 then we locate where the velocity line is 00:10:55.060 --> 00:10:57.880 of five meters per second and we draw a line across 00:10:57.880 --> 00:11:00.170 until we hit that, then to find the pressure drop, 00:11:00.170 --> 00:11:03.370 we draw a vertical line down from this intersection. 00:11:03.370 --> 00:11:05.060 In this instance, we see it comes out 00:11:05.060 --> 00:11:07.550 at 0.65 pascals per meter, 00:11:07.550 --> 00:11:09.330 so add this figure to our table. 00:11:09.330 --> 00:11:11.430 As we are using the equal pressure drop method, 00:11:11.430 --> 00:11:14.120 we can also use this pressure drop for all the duct lengths, 00:11:14.120 --> 00:11:15.240 so fill these in too. 00:11:15.240 --> 00:11:17.710 Then coming back to the chart, we scroll up again 00:11:17.710 --> 00:11:20.460 and align our intersection with the upward diagonal lines 00:11:20.460 --> 00:11:23.020 to see that this requires a duct with a diameter 00:11:23.020 --> 00:11:27.080 of 0.45 meters, so we add that to the table also. 00:11:27.080 --> 00:11:29.210 We know the volume flow rates and the pressure drop 00:11:29.210 --> 00:11:31.870 so we can now calculate the values for section C 00:11:31.870 --> 00:11:34.070 and then also the remaining ducts. 00:11:34.070 --> 00:11:35.320 For the remainder of the ducts, 00:11:35.320 --> 00:11:37.170 we need to use the same method. 00:11:37.170 --> 00:11:39.080 On the chart we start by drawing a line 00:11:39.080 --> 00:11:41.520 from 0.65 pascals per meter. 00:11:41.520 --> 00:11:43.660 We draw this line all the way up. 00:11:43.660 --> 00:11:45.410 Then we draw another line across 00:11:45.410 --> 00:11:47.830 from where our required volume flow rate is, 00:11:47.830 --> 00:11:49.520 in this case for section C, 00:11:49.520 --> 00:11:52.610 we require 0.21 cubic meters per second. 00:11:52.610 --> 00:11:54.370 At this intersection, we draw a line 00:11:54.370 --> 00:11:56.930 to find the velocity and we can see that it falls between 00:11:56.930 --> 00:11:59.820 the lines of three and four meters per second 00:11:59.820 --> 00:12:01.420 so we need to estimate the value. 00:12:01.420 --> 00:12:05.260 In this case it seems to be around 3.6 meters per second, 00:12:05.260 --> 00:12:06.650 so we add that to the chart. 00:12:06.650 --> 00:12:09.180 Then we draw another line on the other diagonal grid 00:12:09.180 --> 00:12:10.570 to find our duct diameter, 00:12:10.570 --> 00:12:13.790 which in this case is around 0.27 meters, 00:12:13.790 --> 00:12:15.700 and we'll add that to the table also. 00:12:15.700 --> 00:12:17.210 So just repeat that last process 00:12:17.210 --> 00:12:18.930 for all the remaining ducts and branches 00:12:18.930 --> 00:12:20.740 until the table is complete. 00:12:20.740 --> 00:12:23.360 Now find the total duct losses for each of the ducts 00:12:23.360 --> 00:12:25.650 and branches, it's very easy and simple to do, 00:12:25.650 --> 00:12:28.730 just multiply the duct length by the pressure drop per meter 00:12:28.730 --> 00:12:31.170 of 0.65 pascals per meter 00:12:31.170 --> 00:12:33.810 and do that for all the ducts and branches on the table. 00:12:33.810 --> 00:12:35.490 Now we can start on the fittings. 00:12:35.490 --> 00:12:38.110 The first fitting we'll look at is the 90-degree bend 00:12:38.110 --> 00:12:39.840 between ducts J and L. 00:12:39.840 --> 00:12:42.093 For this we look up our lost coefficient from the bend 00:12:42.093 --> 00:12:44.860 from the manufacturer or from the industry body. 00:12:44.860 --> 00:12:47.920 Again link's in the video description below for these. 00:12:47.920 --> 00:12:48.820 In this example, 00:12:48.820 --> 00:12:52.350 we can see the coefficient comes out at 0.11. 00:12:52.350 --> 00:12:54.710 We then need to calculate the dynamic loss caused 00:12:54.710 --> 00:12:56.980 by the bend changing the direction of flow. 00:12:56.980 --> 00:13:00.130 For that we use the formula Co multiplied by rho, 00:13:00.130 --> 00:13:03.170 multiplied by V squared divided by two, 00:13:03.170 --> 00:13:06.890 where Co is our coefficient, rho is the density of the air 00:13:06.890 --> 00:13:08.760 and V is the velocity. 00:13:08.760 --> 00:13:10.983 We already know these values, so if we drop these figures in 00:13:10.983 --> 00:13:14.090 then we get an answer of 0.718 pascals, 00:13:14.090 --> 00:13:15.590 so just add that to the table. 00:13:16.530 --> 00:13:18.530 The next fitting will look at is the Tee 00:13:18.530 --> 00:13:20.910 which connects the main duct to the branches. 00:13:20.910 --> 00:13:23.780 We'll use the example of the tee with the ID letter H 00:13:23.780 --> 00:13:25.760 between G and J in the system. 00:13:25.760 --> 00:13:28.140 Now for this we need to consider that the air is moving 00:13:28.140 --> 00:13:31.250 in two directions, straight through and also turning off 00:13:31.250 --> 00:13:32.520 into the branch. 00:13:32.520 --> 00:13:34.080 So we need to perform calculations 00:13:34.080 --> 00:13:35.820 for both of these directions. 00:13:35.820 --> 00:13:38.270 If we look at the air traveling straight through first, 00:13:38.270 --> 00:13:40.300 we find the velocity ratio first, 00:13:40.300 --> 00:13:43.880 using the formula velocity out, divided by velocity in. 00:13:43.880 --> 00:13:48.120 In this example, the air out is 3.3 meters per second, 00:13:48.120 --> 00:13:50.370 and the air in is four meters per second, 00:13:50.370 --> 00:13:53.260 which gives us answer of 0.83. 00:13:53.260 --> 00:13:57.020 Then we perform another calculation to find the area ratio. 00:13:57.020 --> 00:13:59.700 This uses the formula diameter out squared, 00:13:59.700 --> 00:14:02.050 divided diameter in squared. 00:14:02.050 --> 00:14:05.580 In this example, the diameter out is 0.24 meters, 00:14:05.580 --> 00:14:08.960 and the diameter in is 0.33 meters. 00:14:08.960 --> 00:14:12.260 So if we square them, we would get 0.53. 00:14:12.260 --> 00:14:14.740 Now we look up the fitting we're using from the manufacturer 00:14:14.740 --> 00:14:15.750 or the industry body, 00:14:15.750 --> 00:14:18.300 again, link's in the video description below for those. 00:14:18.300 --> 00:14:20.110 In the guides, we find two tables, 00:14:20.110 --> 00:14:22.700 the one you use depends on the direction of flow, 00:14:22.700 --> 00:14:25.310 we're using the straight direction, so we'll okay that one 00:14:25.310 --> 00:14:28.190 and then we look up each ratio to find our last coefficient. 00:14:28.190 --> 00:14:30.540 Here you can see both values we calculated 00:14:30.540 --> 00:14:32.680 for between values listed in the table, 00:14:32.680 --> 00:14:34.940 so we need to perform a bilinear interpolation. 00:14:34.940 --> 00:14:35.773 To save time, 00:14:35.773 --> 00:14:38.360 we'll just use an online calculator to find that, 00:14:38.360 --> 00:14:40.450 links to the site are in the video description below. 00:14:40.450 --> 00:14:44.940 So we fill out our values and we find the answer of 0.143. 00:14:44.940 --> 00:14:47.510 Now we calculate the dynamic loss for the straight path 00:14:47.510 --> 00:14:49.913 with a tee using the formula Co, multiplied by rho, 00:14:49.913 --> 00:14:52.800 multiplied by V squared divided by two. 00:14:52.800 --> 00:14:57.580 If we drop our values in, we get the answer of 0.934 pascals 00:14:57.580 --> 00:14:58.853 so add that to the table. 00:15:00.000 --> 00:15:02.240 Then we can calculate the dynamic loss for the air 00:15:02.240 --> 00:15:03.820 which turns into the bend. 00:15:03.820 --> 00:15:06.290 For this we use the same formulas as before, 00:15:06.290 --> 00:15:08.530 velocity out, divided by velocity in, 00:15:08.530 --> 00:15:10.020 to find our velocity ratio. 00:15:10.020 --> 00:15:11.600 We take our values from our table 00:15:11.600 --> 00:15:13.960 and use 3.5 meters per second, 00:15:13.960 --> 00:15:17.650 divided by four meters per second to get 0.875, 00:15:17.650 --> 00:15:20.520 for the velocity ratio, then we find the area ratio, 00:15:20.520 --> 00:15:22.960 using the formula diameter out squared, 00:15:22.960 --> 00:15:25.060 divided by diameter in squared, 00:15:25.060 --> 00:15:27.590 and we use 0.26 meters squared, 00:15:27.590 --> 00:15:31.890 divided by 0.33 meters squared to get 0.62 00:15:31.890 --> 00:15:33.740 for the area ratio, 00:15:33.740 --> 00:15:36.180 then we use the bend table for the T section. 00:15:36.180 --> 00:15:38.670 Again, it's between the values listed in the table, 00:15:38.670 --> 00:15:41.530 so we have to find the numbers using bilinear interpolation. 00:15:41.530 --> 00:15:46.100 We drop the values in to get the answer of 0.3645 pascals, 00:15:46.100 --> 00:15:47.800 so just add that to the table too. 00:15:48.710 --> 00:15:51.320 Now repeat that calculation for the other tees and fittings 00:15:51.320 --> 00:15:53.280 00:15:53.280 --> 00:15:55.310 Next, we need to find the index run. 00:15:55.310 --> 00:15:57.640 which is the run with the largest pressure drop. 00:15:57.640 --> 00:16:00.150 It's usually the longest run, but it could also be the run 00:16:00.150 --> 00:16:01.890 with the most fittings. 00:16:01.890 --> 00:16:04.550 We find it easily by adding up all the pressure losses 00:16:04.550 --> 00:16:07.920 from start to the exit of each branch. 00:16:07.920 --> 00:16:12.400 For example, to get from A to C, we lose 5.04 pascals, 00:16:12.400 --> 00:16:15.040 for A to F, we lose 8.8 pascals, 00:16:15.040 --> 00:16:18.240 for A to I we lose 10.56 pascals, 00:16:18.240 --> 00:16:21.550 and for A to L we lose 12.5 pascal. 00:16:21.550 --> 00:16:24.190 Therefore the fan we use, must overcome the run 00:16:24.190 --> 00:16:28.160 with the highest loss that being A to L with 12.5 pascals, 00:16:28.160 --> 00:16:29.520 this being the index run. 00:16:29.520 --> 00:16:31.910 To balance the system, we need to add dampers 00:16:31.910 --> 00:16:34.203 to each of the branches to ensure equal pressure drop 00:16:34.203 --> 00:16:38.460 through all to achieve the design flow rates to each room. 00:16:38.460 --> 00:16:40.290 We can calculate how much pressure drop 00:16:40.290 --> 00:16:43.070 each damper needs to provide simply by subtracting 00:16:43.070 --> 00:16:45.630 the loss of the run from the index run. 00:16:45.630 --> 00:16:48.080 A to C is 5.04 pascals, 00:16:48.080 --> 00:16:51.120 which means that branch C would need a damper, 00:16:51.120 --> 00:16:53.410 providing 7.46 pascals. 00:16:53.410 --> 00:16:55.620 A to F is 8.8 pascals 00:16:55.620 --> 00:16:58.150 which means that branch F would require 00:16:58.150 --> 00:17:01.160 a damper providing 3.7 pascals, 00:17:01.160 --> 00:17:06.160 and A to I is 10.56 pascals, meaning that duct I, 00:17:06.660 --> 00:17:10.440 would require a damper providing 1.94 pascals 00:17:10.440 --> 00:17:12.400 and that is our ductwork system. 00:17:12.400 --> 00:17:14.370 We'll do another video covering additional ways 00:17:14.370 --> 00:17:16.420 to improve efficiency in ductwork systems, 00:17:16.420 --> 00:17:19.260 but unfortunately we've run out of time in this video. 00:17:19.260 --> 00:17:20.810 Okay guys, that's it for this video. 00:17:20.810 --> 00:17:21.990 Thank you very much for watching. 00:17:21.990 --> 00:17:23.780 I hope you've enjoyed this and it has helped you. 00:17:23.780 --> 00:17:26.180 If so, please don't forget to like, subscribe and share 00:17:26.180 --> 00:17:28.160 and also check out SimScale software. 00:17:28.160 --> 00:17:30.140 You can follow us on Facebook, Twitter, Instagram, 00:17:30.140 --> 00:17:31.942 Google Plus, as well as our website, 00:17:31.942 --> 00:17:33.720 TheEngineeringMindset.com. 00:17:33.720 --> 00:17:35.320 Once again, thanks for watching.
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