However, at high relative flow rates in the branch, the RMCs change very sharply, therefore, in this area, the tables under consideration are manually interpolated with difficulty and with a significant error. In addition, in the case of using MS Excel spreadsheets, it is again desirable to have formulas for directly calculating the CMR through the ratio of flow rates and sections. Moreover, such formulas should, on the one hand, be quite simple and convenient for mass design and use in the educational process, but, at the same time, should not give an error exceeding normal accuracy engineering calculation. Previously, a similar problem was solved by the author in relation to resistances encountered in water heating systems. Let us now consider this issue for mechanical systems B and HF. Below are the results of data approximation for unified tees (branch nodes) per passage. General form dependencies were chosen based on physical considerations, taking into account the convenience of using the resulting expressions while ensuring permissible deviation from tabular data:
❏ for supply tees, with Loʹ ≤ 0.7 and fnʹ ≥ 0.5: and with Loʹ ≤ 0.4, you can use the simplified formula:
❏ for exhaust tees: 
It is easy to notice that the relative area of the passage fnʹ during discharge or, respectively, the branch foʹ during suction affects the CMR in the same way, namely, with an increase in fnʹ or foʹ the resistance will decrease, and the numerical coefficient for the indicated parameters in all the given formulas is the same, namely (-0.25). In addition, for both supply and exhaust tees, when the air flow rate in the branch changes, the relative minimum KMS occurs at the same level Loʹ = 0.2. These circumstances indicate that the obtained expressions, despite their simplicity, sufficiently reflect the general physical laws underlying the influence of the studied parameters on pressure losses in tees of any type. In particular, the larger fnʹ or foʹ, i.e. the closer they are to unity, the less the flow structure changes when passing resistance, and therefore the less the CMR. For the Loʹ value the dependence is more complex, but here it will be common to both modes of air movement.
An idea of the degree of correspondence between the found relationships and the initial CMR values is given in Fig. 1, which shows the results of processing Table 22.37 for KMS standardized tees (branch assemblies) for the passage of round and rectangular section when pumping. Approximately the same picture is obtained for the approximation of the table. 22.38 using formula (3). Note that, although in the latter case we're talking about O round section, it is easy to see that expression (3) quite well describes the data in table. 22.39, already related to rectangular nodes. 
The error of formulas for CMS is generally 5-10% (maximum up to 15%). Slightly higher deviations may be given by expression (3) for tees during suction, but even here this can be considered satisfactory, taking into account the complexity of changing the resistance in such elements. In any case, the nature of the dependence of the IMR on the factors influencing it is reflected very well here. In this case, the obtained relationships do not require any other initial data other than those already available in the aerodynamic calculation table. In fact, it must explicitly indicate both the air flow rates and the cross sections in the current and adjacent sections included in the listed formulas. This especially simplifies calculations when using MS Excel spreadsheets.
At the same time, the formulas given in this work, are very simple, visual and easily accessible for engineering calculations, especially in MS Excel, as well as in the educational process. Their use makes it possible to abandon the interpolation of tables while maintaining the accuracy required for engineering calculations, and to directly calculate KMS tees for passage with a wide variety of cross-sectional ratios and air flow rates in the trunk and branches. This is quite sufficient for the design of V and HF systems in most residential and public buildings.
1. A.D. Altshul, L.S. Zhivotovsky, L.P. Ivanov. Hydraulics and aerodynamics. - M.: Stroyizdat, 1987.
2. Designer's Handbook. Internal sanitary installations. Part 3. Ventilation and air conditioning. Book 2 / Ed. N.N. Pavlova and Yu.I. Schiller. - M.: Stroyizdat, 1992.
3. O.D. Samarin. On the calculation of pressure losses in elements of water heating systems // Journal S.O.K., No. 2/2007.
The resistance to the passage of air in a ventilation system is mainly determined by the speed of air movement in this system. As speed increases, resistance also increases. This phenomenon is called pressure loss. The static pressure created by the fan causes air movement in the ventilation system, which has a certain resistance. The higher the resistance of such a system, the less consumption air, moved or . Calculation of friction losses for air in air ducts, as well as the resistance of network equipment (filter, silencer, heater, valve, etc.) can be made using the corresponding tables and diagrams indicated in the catalog. The total pressure drop can be calculated by summing the resistance values of all elements ventilation system.
Recommended air speed in air ducts:
Determination of air speed in air ducts:
V= L / 3600*F (m/sec)
Where L- air flow, m 3 / h;
F- channel cross-sectional area, m2.
Recommendation 1.
Pressure loss in a duct system can be reduced by increasing the cross-section of the ducts to ensure relatively uniform air velocity throughout the system. In the image we see how it is possible to ensure relatively uniform air speeds in a duct network with minimal pressure loss.
Recommendation 2.
In systems with long air ducts and a large number of ventilation grilles It is advisable to place the fan in the middle of the ventilation system. This solution has several advantages. On the one hand, pressure losses are reduced, and on the other hand, air ducts of a smaller cross-section can be used.
Example of calculation of a ventilation system:
The calculation must begin by drawing up a sketch of the system indicating the locations of air ducts, ventilation grilles, fans, as well as the lengths of the air duct sections between the tees, then determine the air flow in each section of the network.
Let's find out the pressure loss for sections 1-6, using the graph of pressure loss in round air ducts, determine the required diameters of the air ducts and the pressure loss in them, provided that it is necessary to ensure the permissible air speed.
Section 1: air flow will be 220 m 3 /h. We assume the diameter of the air duct is 200 mm, the speed is 1.95 m/s, the pressure loss is 0.2 Pa/m x 15 m = 3 Pa (see the diagram for determining pressure loss in air ducts).
Section 2: Let's repeat the same calculations, not forgetting that the air flow through this section will already be 220 + 350 = 570 m 3 / h. We assume the diameter of the air duct is 250 mm, the speed is 3.23 m/s. The pressure loss will be 0.9 Pa/m x 20 m = 18 Pa.
Section 3: the air flow through this section will be 1070 m 3 /h.
We take the diameter of the air duct equal to 315 mm, the speed is 3.82 m/s. The pressure loss will be 1.1 Pa/m x 20= 22 Pa.
Section 4: the air flow through this section will be 1570 m 3 /h. We take the diameter of the air duct equal to 315 mm, speed - 5.6 m/s. The pressure loss will be 2.3 Pa x 20 = 46 Pa.
Section 5: the air flow through this section will be 1570 m 3 /h. We take the diameter of the air duct equal to 315 mm, speed 5.6 m/s. The pressure loss will be 2.3 Pa/m x 1= 2.3 Pa.
Section 6: the air flow through this section will be 1570 m 3 /h. We take the diameter of the air duct equal to 315 mm, speed 5.6 m/s. The pressure loss will be 2.3 Pa x 10 = 23 Pa. The total pressure loss in the air ducts will be 114.3 Pa.
When the calculation of the last section is completed, it is necessary to determine the pressure loss in the network elements: in the CP 315/900 silencer (16 Pa) and in check valve KOM 315 (22 Pa). We will also determine the pressure loss in the taps to the grilles (the total resistance of 4 taps will be 8 Pa).
Determination of pressure losses at bends of air ducts
The graph allows you to determine the pressure loss in the outlet based on the bend angle, diameter and air flow.
Example. Let us determine the pressure loss for a 90° outlet with a diameter of 250 mm at an air flow of 500 m3/h. To do this, we find the intersection of the vertical line corresponding to our air flow with the inclined line characterizing the diameter of 250 mm, and on the vertical line on the left for a 90° outlet we find the value of pressure loss, which is 2 Pa.
We accept for installation ceiling diffusers of the PF series, the resistance of which, according to the schedule, will be 26 Pa.
Now let’s sum up all the pressure loss values for straight sections of air ducts, network elements, bends and grilles. The desired value is 186.3 Pa.
We calculated the system and determined that we need a fan that removes 1570 m3/h of air with a network resistance of 186.3 Pa. Taking into account the characteristics required for the operation of the system, we will be satisfied with the fan; the characteristics required for the operation of the system will suit us with the VENTS VKMS 315 fan.
Determination of pressure losses in air ducts.
Determination of pressure loss in a check valve.
Selection of the required fan.
Determination of pressure loss in silencers.
Determination of pressure losses at bends of air ducts.
Determination of pressure loss in diffusers.
When the parameters of the air ducts are known (their length, cross-section, coefficient of air friction on the surface), it is possible to calculate the pressure loss in the system at the designed air flow.
Total pressure loss (in kg/sq.m.) is calculated using the formula:
P = R*l + z,
where R is friction pressure loss per 1 linear meter of air duct, l is the length of the air duct in meters, z is pressure loss per local resistance(with variable cross section).
1. Friction losses:
In a round air duct, pressure loss due to friction P tr is calculated as follows:
Ptr = (x*l/d) * (v*v*y)/2g,
where x is the friction resistance coefficient, l is the length of the air duct in meters, d is the diameter of the air duct in meters, v is the air flow speed in m/s, y is the air density in kg/cub.m., g is the acceleration of free fall (9 .8 m/s2).
- Note: If the duct has a rectangular rather than a round cross-section, the equivalent diameter must be substituted into the formula, which for an air duct with sides A and B is equal to: deq = 2AB/(A + B)
2. Losses due to local resistance:
Pressure losses due to local resistance are calculated using the formula:
z = Q* (v*v*y)/2g,
where Q is the sum of the local resistance coefficients in the section of the air duct for which the calculation is being made, v is the air flow speed in m/s, y is the air density in kg/cub.m., g is the acceleration of gravity (9.8 m/s2 ). Q values are presented in tabular form.
Permissible speed method
When calculating the air duct network using the permissible speed method, the optimal air speed is taken as the initial data (see table). Then they count required section air duct and pressure loss in it.
Procedure for aerodynamic calculation of air ducts using the permissible speed method:
- Draw a diagram of the air distribution system. For each section of the air duct, indicate the length and amount of air passing in 1 hour.
- We start the calculation from the areas farthest from the fan and the most loaded.
- Knowing the optimal air speed for a given room and the volume of air passing through the air duct in 1 hour, we will determine the appropriate diameter (or cross-section) of the air duct.
- We calculate the pressure loss due to friction P tr.
- Using the tabular data, we determine the sum of local resistances Q and calculate the pressure loss due to local resistances z.
- The available pressure for the following branches of the air distribution network is determined as the sum of pressure losses in the areas located before this branch.
During the calculation process, it is necessary to sequentially link all branches of the network, equating the resistance of each branch to the resistance of the most loaded branch. This is done using diaphragms. They are installed on lightly loaded areas of air ducts, increasing resistance.
Table of maximum air speed depending on duct requirements
|
Purpose |
Basic Requirement |
||||
|
Silence |
Min. head loss |
||||
|
Main channels |
Main channels |
Branches |
|||
|
Inflow |
Hood |
Inflow |
Hood |
||
|
Living spaces |
|||||
|
Hotels |
|||||
|
Institutions |
|||||
|
Restaurants |
|||||
|
The shops |
|||||
Note: speed air flow in the table it is given in meters per second
Constant head loss method
This method assumes a constant loss of pressure per 1 linear meter of air duct. Based on this, the dimensions of the air duct network are determined. The method of constant pressure loss is quite simple and is used at the stage of feasibility study of ventilation systems:
- Depending on the purpose of the room, according to the table of permissible air speeds, select the speed on the main section of the air duct.
- Based on the speed determined in paragraph 1 and based on the design air flow, the initial pressure loss is found (per 1 m of duct length). The diagram below does this.
- The most loaded branch is determined, and its length is taken as the equivalent length of the air distribution system. Most often this is the distance to the farthest diffuser.
- Multiply the equivalent length of the system by the pressure loss from step 2. The pressure loss at the diffusers is added to the resulting value.
Now, using the diagram below, determine the diameter of the initial air duct coming from the fan, and then the diameters of the remaining sections of the network according to the corresponding air flow rates. In this case, the initial pressure loss is assumed to be constant.
Diagram for determining pressure loss and diameter of air ducts
Using rectangular ducts
The pressure loss diagram shows the diameters round ducts. If rectangular ducts are used instead, their equivalent diameters must be found using the table below.
Notes:
- If space allows, it is better to choose round or square air ducts;
- If there is not enough space (for example, during reconstruction), rectangular air ducts are chosen. As a rule, the width of the duct is 2 times the height).
The table shows the height of the air duct in mm along the horizontal line, its width in the vertical line, and the cells of the table contain the equivalent diameters of the air ducts in mm.
Table of equivalent duct diameters
where R is the pressure loss due to friction per 1 linear meter of the air duct, l is the length of the air duct in meters, z is the pressure loss due to local resistance (with a variable cross-section).
1. Friction losses:
Ptr = (x*l/d) * (v*v*y)/2g,
z = Q* (v*v*y)/2g,
Permissible speed method
When calculating the air duct network using the permissible speed method, the optimal air speed is taken as the initial data (see table). Then the required cross-section of the air duct and the pressure loss in it are calculated.
This method assumes a constant loss of pressure per 1 linear meter of air duct. Based on this, the dimensions of the air duct network are determined. The method of constant pressure loss is quite simple and is used at the stage of feasibility study of ventilation systems:
The head loss diagram shows the diameters of round ducts. If rectangular ducts are used instead, their equivalent diameters must be found using the table below.
Notes:
If there is not enough space (for example, during reconstruction), rectangular air ducts are chosen. As a rule, the width of the duct is 2 times the height).
With this material, the editors of the magazine “Climate World” continue the publication of chapters from the book “Ventilation and air conditioning systems. Design guidelines for production
water and public buildings." Author Krasnov Yu.S.
The aerodynamic calculation of air ducts begins with drawing an axonometric diagram (M 1: 100), putting down the numbers of sections, their loads L (m 3 / h) and lengths I (m). The direction of the aerodynamic calculation is determined - from the most distant and loaded area to the fan. When in doubt when determining a direction, consider all possible options.
The calculation begins from a remote section: the diameter D (m) of a round or the cross-sectional area F (m2) of a rectangular air duct is determined:
The speed increases as you approach the fan.
According to Appendix N, the nearest ones are accepted standard values: D CT or (a x b) st (m).
Hydraulic radius of rectangular ducts (m):
 |
where is the sum of the local resistance coefficients in the air duct section.
Local resistances at the border of two sections (tees, crosses) are assigned to the section with lower flow.
Local resistance coefficients are given in the appendices.
Diagram of the supply ventilation system serving a 3-story administrative building
Calculation example
Initial data:
| No. of plots | flow L, m 3 / h | length L, m | υ rivers, m/s | section a × b, m |
υ f, m/s | D l,m | Re | λ | Kmc | losses in the area Δр, pa |
| PP grid at the outlet | 0.2 × 0.4 | 3,1 | - | - | - | 1,8 | 10,4 | |||
| 1 | 720 | 4,2 | 4 | 0.2 × 0.25 | 4,0 | 0,222 | 56900 | 0,0205 | 0,48 | 8,4 |
| 2 | 1030 | 3,0 | 5 | 0.25×0.25 | 4,6 | 0,25 | 73700 | 0,0195 | 0,4 | 8,1 |
| 3 | 2130 | 2,7 | 6 | 0.4 × 0.25 | 5,92 | 0,308 | 116900 | 0,0180 | 0,48 | 13,4 |
| 4 | 3480 | 14,8 | 7 | 0.4 × 0.4 | 6,04 | 0,40 | 154900 | 0,0172 | 1,44 | 45,5 |
| 5 | 6830 | 1,2 | 8 | 0.5 × 0.5 | 7,6 | 0,50 | 234000 | 0,0159 | 0,2 | 8,3 |
| 6 | 10420 | 6,4 | 10 | 0.6 × 0.5 | 9,65 | 0,545 | 337000 | 0,0151 | 0,64 | 45,7 |
| 6a | 10420 | 0,8 | Yu. | Ø0.64 | 8,99 | 0,64 | 369000 | 0,0149 | 0 | 0,9 |
| 7 | 10420 | 3,2 | 5 | 0.53 × 1.06 | 5,15 | 0,707 | 234000 | 0.0312×n | 2,5 | 44,2 |
| Total losses: 185 | ||||||||||
| Table 1. Aerodynamic calculation | ||||||||||
The air ducts are made of galvanized sheet steel, the thickness and size of which correspond to approx. N from. The material of the air intake shaft is brick. Adjustable grilles of the PP type with possible sections: 100 x 200; 200 x 200; 400 x 200 and 600 x 200 mm, shading coefficient 0.8 and maximum air outlet speed up to 3 m/s.
The resistance of the insulated intake valve with fully open blades is 10 Pa. The hydraulic resistance of the heating unit is 100 Pa (according to a separate calculation). Filter resistance G-4 250 Pa. Hydraulic resistance of the muffler 36 Pa (according to acoustic calculation). Based on architectural requirements, rectangular air ducts are designed.
The cross-sections of brick channels are taken according to table. 22.7.
Local resistance coefficients
Section 1. PP grid at the outlet with a cross section of 200×400 mm (calculated separately):
| No. of plots | Type of local resistance | Sketch | Angle α, deg. | Attitude | Rationale | KMS | ||
| F 0 /F 1 | L 0 /L st | f pass /f stv | ||||||
| 1 | Diffuser |
 |
20 | 0,62 | - | - | Table 25.1 | 0,09 |
| Retraction |
 |
90 | - | - | - | Table 25.11 | 0,19 | |
| Tee-pass |
 |
- | - | 0,3 | 0,8 | Adj. 25.8 | 0,2 | |
| ∑ = | 0,48 | |||||||
| 2 | Tee-pass |
 |
- | - | 0,48 | 0,63 | Adj. 25.8 | 0,4 |
| 3 | Branch tee |
 |
- | 0,63 | 0,61 | - | Adj. 25.9 | 0,48 |
| 4 | 2 bends | 250×400 | 90 | - | - | - | Adj. 25.11 | |
| Retraction | 400×250 | 90 | - | - | - | Adj. 25.11 | 0,22 | |
| Tee-pass |
 |
- | - | 0,49 | 0,64 | Table 25.8 | 0,4 | |
| ∑ = | 1,44 | |||||||
| 5 | Tee-pass |
 |
- | - | 0,34 | 0,83 | Adj. 25.8 | 0,2 |
| 6 | Diffuser after fan |
 |
h=0.6 | 1,53 | - | - | Adj. 25.13 | 0,14 |
| Retraction | 600×500 | 90 | - | - | - | Adj. 25.11 | 0,5 | |
| ∑= | 0,64 | |||||||
| 6a | Confusion in front of the fan |
 |
D g =0.42 m | Table 25.12 | 0 | |||
| 7 | Knee | 90 | - | - | - | Table 25.1 | 1,2 | |
| Louvre grille | Table 25.1 | 1,3 | ||||||
| ∑ = | 1,44 | |||||||
| Table 2. Determination of local resistances | ||||||||
Krasnov Yu.S.,
When the parameters of the air ducts are known (their length, cross-section, coefficient of air friction on the surface), it is possible to calculate the pressure loss in the system at the designed air flow.
Total pressure loss (in kg/sq.m.) is calculated using the formula:
where R is the pressure loss due to friction per 1 linear meter of the air duct, l is the length of the air duct in meters, z is the pressure loss due to local resistance (with a variable cross-section).
1. Friction losses:
In a round air duct, pressure loss due to friction P tr is calculated as follows:
Ptr = (x*l/d) * (v*v*y)/2g,
where x is the friction resistance coefficient, l is the length of the air duct in meters, d is the diameter of the air duct in meters, v is the air flow speed in m/s, y is the air density in kg/cub.m., g is the acceleration of free fall (9 .8 m/s2).
- Note: If the duct has a rectangular rather than a round cross-section, the equivalent diameter must be substituted into the formula, which for an air duct with sides A and B is equal to: deq = 2AB/(A + B)
2. Losses due to local resistance:
Pressure losses due to local resistance are calculated using the formula:
z = Q* (v*v*y)/2g,
where Q is the sum of the local resistance coefficients in the section of the air duct for which the calculation is being made, v is the air flow speed in m/s, y is the air density in kg/cub.m., g is the acceleration of gravity (9.8 m/s2 ). Q values are presented in tabular form.
Permissible speed method
When calculating the air duct network using the permissible speed method, the optimal air speed is taken as the initial data (see table). Then the required cross-section of the air duct and the pressure loss in it are calculated.
Procedure for aerodynamic calculation of air ducts using the permissible speed method:
- Draw a diagram of the air distribution system. For each section of the air duct, indicate the length and amount of air passing in 1 hour.
- We start the calculation from the areas farthest from the fan and the most loaded.
- Knowing the optimal air speed for a given room and the volume of air passing through the air duct in 1 hour, we will determine the appropriate diameter (or cross-section) of the air duct.
- We calculate the pressure loss due to friction P tr.
- Using the tabular data, we determine the sum of local resistances Q and calculate the pressure loss due to local resistances z.
- The available pressure for the following branches of the air distribution network is determined as the sum of pressure losses in the areas located before this branch.
During the calculation process, it is necessary to sequentially link all branches of the network, equating the resistance of each branch to the resistance of the most loaded branch. This is done using diaphragms. They are installed on lightly loaded areas of air ducts, increasing resistance.
Table of maximum air speed depending on duct requirements
Note: air flow speed in the table is given in meters per second
Constant head loss method
This method assumes a constant loss of pressure per 1 linear meter of air duct. Based on this, the dimensions of the air duct network are determined. The method of constant pressure loss is quite simple and is used at the stage of feasibility study of ventilation systems:
- Depending on the purpose of the room, according to the table of permissible air speeds, select the speed on the main section of the air duct.
- Based on the speed determined in paragraph 1 and based on the design air flow, the initial pressure loss is found (per 1 m of duct length). The diagram below does this.
- The most loaded branch is determined, and its length is taken as the equivalent length of the air distribution system. Most often this is the distance to the farthest diffuser.
- Multiply the equivalent length of the system by the pressure loss from step 2. The pressure loss at the diffusers is added to the resulting value.
Now, using the diagram below, determine the diameter of the initial air duct coming from the fan, and then the diameters of the remaining sections of the network according to the corresponding air flow rates. In this case, the initial pressure loss is assumed to be constant.
Diagram for determining pressure loss and diameter of air ducts
Using rectangular ducts
The pressure loss diagram shows the diameters of round ducts. If rectangular ducts are used instead, their equivalent diameters must be found using the table below.
Notes:
- If space allows, it is better to choose round or square air ducts;
- If there is not enough space (for example, during reconstruction), rectangular air ducts are chosen. As a rule, the width of the duct is 2 times the height).
The table shows the height of the air duct in mm along the horizontal line, its width in the vertical line, and the cells of the table contain the equivalent diameters of the air ducts in mm.
Table of equivalent duct diameters
- System performance serving up to 4 rooms.
- Dimensions of air ducts and air distribution grilles.
- Resistance of the air network.
- Heater power and estimated energy costs (when using an electric heater).
If you need to choose a model with humidification, cooling or recovery, use the calculator on the Breezart website.
An example of calculating ventilation using a calculator
In this example we will show how to calculate supply ventilation for 3 room apartment, in which a family of three lives (two adults and a child). During the day, sometimes relatives come to visit them, so up to 5 people can stay in the living room for a long time. The ceiling height of the apartment is 2.8 meters. Room parameters:
We will set the consumption rates for the bedroom and nursery in accordance with the recommendations of SNiP - 60 m³/h per person. For the living room we will limit ourselves to 30 m³/h, since a large number of There are rarely people in this room. According to SNiP, such air flow is permissible for rooms with natural ventilation (you can open a window for ventilation). If we set the air flow rate of 60 m³/h per person for the living room, then the required productivity for this room would be 300 m³/h. The cost of electricity to heat this amount of air would be very high, so we made a compromise between comfort and efficiency. To calculate air exchange by multiplicity for all rooms, we will select a comfortable double air exchange.
The main air duct will be rectangular, rigid, and the branches will be flexible, sound-insulated (this combination of duct types is not the most common, but we chose it for demonstration purposes). For additional cleaning supply air A fine dust filter of class EU5 will be installed (we will calculate the network resistance with dirty filters). Air velocities in air ducts and permissible level We will leave the noise on the grilles equal to the recommended values, which are set by default.
We begin the calculation by drawing up a diagram of the air distribution network. This diagram will allow us to determine the length of the air ducts and the number of turns that can be both horizontal and vertical plane(we need to count all right angle turns). So, our scheme:
The resistance of the air distribution network is equal to the resistance of the longest section. This section can be divided into two parts: the main air duct and the longest branch. If you have two branches of approximately the same length, then you need to determine which one has more resistance. To do this, we can assume that the resistance of one turn is equal to the resistance of 2.5 meters of the air duct, then the greatest resistance will be the branch whose value (2.5 * number of turns + length of the air duct) is maximum. It is necessary to select two parts from the route in order to be able to specify different type air ducts and different speed air for the main section and branches.
In our system, balancing throttle valves are installed on all branches, allowing you to adjust the air flow in each room in accordance with the project. Their resistance (in open state) has already been taken into account, since it is a standard element of the ventilation system.
The length of the main air duct (from the air intake grille to the branch to room No. 1) is 15 meters; there are 4 turns at right angles in this section. The length of the air supply unit and air filter can be ignored (their resistance will be taken into account separately), and the resistance of the silencer can be taken equal to the resistance of the air duct of the same length, that is, simply consider it part of the main air duct. The longest branch is 7 meters long and has 3 right angle turns (one at the branch, one at the duct and one at the adapter). Thus, we have specified all the necessary initial data and can now begin calculations (screenshot). The calculation results are summarized in tables:
Calculation results for premisesResults of calculation of general parameters
| Ventilation system type | Regular | VAV |
| Performance | 365 m³/h | 243 m³/h |
| Cross-sectional area of the main air duct | 253 cm² | 169 cm² |
| Recommended dimensions of the main air duct | 160x160 mm 90x315 mm 125x250 mm |
125x140 mm 90x200 mm 140x140 mm |
| Air network resistance | 219 Pa | 228 Pa |
| Heater power | 5.40 kW | 3.59 kW |
| Recommended Supply unit | Breezart 550 Lux (in 550 m³/h configuration) |
Breezart 550 Lux (VAV) |
| Maximum performance recommended PU |
438 m³/h | 433 m³/h |
| Electric power heater PU | 4.8 kW | 4.8 kW |
| Average monthly electricity costs | 2698 rubles | 1619 rubles |
Air duct network calculation
- For each room (subsection 1.2), the performance is calculated, the cross-section of the air duct is determined and a suitable air duct of standard diameter is selected. Using the Arktos catalog, the dimensions of distribution grilles with a given noise level are determined (data for the AMN, ADN, AMP, ADR series is used). You can use other grilles with the same dimensions - in this case, there may be a slight change in the noise level and network resistance. In our case, the grilles for all rooms turned out to be the same, since at a noise level of 25 dB(A) the permissible air flow through them is 180 m³/h (there are no smaller grilles in these series).
- The sum of the air flow rates for all three rooms gives us the overall system performance (subsection 1.3). When using a VAV system, the system performance will be one third lower due to separate adjustment of air flow in each room. Next, the cross-section of the main air duct is calculated (in the right column - for VAV systems) and appropriately sized rectangular air ducts are selected (usually several options are given with different aspect ratios). At the end of the section, the resistance of the air network is calculated, which turns out to be quite large - this is due to the use of a fine filter in the ventilation system, which has a high resistance.
- We have received all the necessary data to complete the air distribution network, with the exception of the size of the main air duct between branches 1 and 3 (this parameter is not calculated in the calculator, since the network configuration is unknown in advance). However, the cross-sectional area of this section can be easily calculated manually: from the cross-sectional area of the main air duct, you need to subtract the cross-sectional area of branch No. 3. Having obtained the cross-sectional area of the air duct, its size can be determined by.
Calculation of heater power and selection of air handling unit
The recommended model Breezart 550 Lux has software-configurable parameters (performance and heater power), so the performance that should be selected when setting up the control unit is indicated in brackets. It can be noted that the maximum possible heater power of this unit is 11% lower than the calculated value. The lack of power will be noticeable only when the outside temperature is below -22°C, and this does not happen often. In such cases, the air handling unit will automatically switch to a lower speed to maintain the set outlet temperature (“Comfort” function).
The calculation results, in addition to the required performance of the ventilation system, indicate the maximum performance of the control unit at a given network resistance. If this performance turns out to be significantly higher than the required value, you can use the ability to programmatically limit the maximum performance, which is available for all Breezart ventilation units. For a VAV system, the maximum capacity is provided for reference only, as performance is adjusted automatically while the system is running.
Operating cost calculation
This section calculates the cost of electricity spent on heating air in cold period of the year. The costs for a VAV system depend on its configuration and operating mode, therefore they are assumed to be equal to the average value: 60% of costs conventional system ventilation. In our case, you can save money by reducing air consumption in the living room at night and in the bedroom during the day.
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