Calculation of inflow and exhaust systems air ducts comes down to determining the dimensions cross section channels, their resistance to air movement and pressure balancing in parallel connections. Calculation of pressure losses should be carried out using the method of specific pressure losses due to friction.
An axonometric diagram of the ventilation system is constructed, the system is divided into sections into which the length and flow rate are plotted. The calculation scheme is presented in Figure 1.
The main (main) direction is selected, which represents the longest chain of successively located sections.
3. The sections of the highway are numbered, starting with the section with the lowest flow rate.
4. The cross-sectional dimensions of the air ducts in the design sections of the main are determined. Determine the cross-sectional area, m2:
F p =L p /3600V p ,
where L p is the estimated air flow rate in the area, m 3 / h;
Based on the found values of F p ], the dimensions of the air ducts are taken, i.e. is F f.
5. The actual speed V f, m/s is determined:
V f = L p / F f,
where L p is the estimated air flow rate in the area, m 3 / h;
F f – actual cross-sectional area of the air duct, m2.
We determine the equivalent diameter using the formula:
d eq = 2·α·b/(α+b) ,
where α and b are the transverse dimensions of the air duct, m.
6. Based on the values of d eq and V f, the values of specific pressure loss due to friction R are determined.
Pressure loss due to friction at settlement area will amount to
P t =R l β w,
where R – specific pressure loss due to friction, Pa/m;
l – length of the air duct section, m;
β sh – roughness coefficient.
7. Local resistance coefficients are determined and pressure losses in local resistances in the area are calculated:
z = ∑ζ·P d,
where P d – dynamic pressure:
Pd=ρV f 2 /2,
where ρ – air density, kg/m3;
V f – actual air speed in the area, m/s;
∑ζ – sum of CMR on the site,
8. Total losses by area are calculated:
ΔР = R l β w + z,
l – length of the section, m;
z - pressure loss in local resistance in the area, Pa.
9. Pressure loss in the system is determined:
ΔР p = ∑(R l β w + z) ,
where R is the specific pressure loss due to friction, Pa/m;
l – length of the section, m;
β sh – roughness coefficient;
z- pressure loss in local resistance in the area, Pa.
10. Linking of branches is carried out. Linking is done starting with the longest branches. It is similar to the calculation of the main direction. The resistances in all parallel sections must be equal: the discrepancy is no more than 10%:
where Δр 1 and Δр 2 are losses in branches with higher and lower pressure losses, Pa. If the discrepancy exceeds the specified value, then a throttle valve is installed.
Figure 1 – Design diagram supply system P1.
Sequence of calculation of the supply system P1
Section 1-2, 12-13, 14-15,2-2',3-3',4-4',5-5',6-6',13-13',15-15',16- 16':
Section 2 -3, 7-13, 15-16:
Section 3-4, 8-16:
Section 4-5:
Section 5-6:
Section 6-7:
Section 7-8:
Section 8-9:
Local resistance
Section 1-2:
a) to the output: ξ = 1.4
b) 90° bend: ξ = 0.17
c) tee for straight passage:
Section 2-2’:
a) branch tee
Section 2-3:
a) 90° bend: ξ = 0.17
b) tee for straight passage:
ξ = 0,25
Section 3-3’:
a) branch tee
Section 3-4:
a) 90° bend: ξ = 0.17
b) tee for straight passage:
Section 4-4’:
a) branch tee
Section 4-5:
a) tee for straight passage:
Section 5-5’:
a) branch tee
Section 5-6:
a) 90° bend: ξ = 0.17
b) tee for straight passage:
Section 6-6’:
a) branch tee
Section 6-7:
a) tee for straight passage:
ξ = 0,15
Section 7-8:
a) tee for straight passage:
ξ = 0,25
Section 8-9:
a) 2 bends 90°: ξ = 0.17
b) tee for straight passage:
Section 10-11:
a) 90° bend: ξ = 0.17
b) to the output: ξ = 1.4
Section 12-13:
a) to the output: ξ = 1.4
b) 90° bend: ξ = 0.17
c) tee for straight passage:
Section 13-13’
a) branch tee
Section 7-13:
a) 90° bend: ξ = 0.17
b) tee for straight passage:
ξ = 0,25
c) branch tee:
ξ = 0,8
Section 14-15:
a) to the output: ξ = 1.4
b) 90° bend: ξ = 0.17
c) tee for straight passage:
Section 15-15’:
a) branch tee
Section 15-16:
a) 2 bends 90°: ξ = 0.17
b) tee for straight passage:
ξ = 0,25
Section 16-16’:
a) branch tee
Section 8-16:
a) tee for straight passage:
ξ = 0,25
b) branch tee:
Aerodynamic calculation of the supply system P1
|
Flow, L, m³/h |
Length, l, m |
Duct dimensions |
Air speed V, m/s |
Losses per 1 m of section length R, Pa |
Coeff. roughness m |
Friction losses Rlm, Pa |
Amount of KMS, Σξ |
Dynamic pressure Рд, Pa |
Local resistance losses, Z |
Pressure loss in the area, ΔР, Pa |
||||||||||||
|
Sectional area F, m² |
Equivalent diameter |
|||||||||||||||||||||
Let us make a discrepancy in the supply system P1, which should be no more than 10%.
Since the discrepancy exceeds the permissible 10%, it is necessary to install a diaphragm.
I install the diaphragm in the area 7-13, V = 8.1 m/s, R C = 20.58 Pa
Therefore, for an air duct with a diameter of 450, I install a diaphragm with a diameter of 309.
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 with a remote section: determine the diameter D (m) of the round or the area F (m 2) of the cross section of the rectangular air duct:
The speed increases as you approach the fan.
According to Appendix H, the nearest standard values are taken: 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.
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). Air ducts are designed based on architectural requirements rectangular section.
The cross-sections of brick channels are taken according to table. 22.7.
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.,
1. Friction losses:
Ptr = (x*l/d) * (v*v*y)/2g,
z = Q* (v*v*y)/2g,
Note: air flow speed in the table is given in meters per second
Using rectangular ducts
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:
Table of equivalent duct diameters
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 losses pressure (in kg/sq.m.) are calculated using the formula:
where R is pressure loss due to friction per 1 linear meter air duct, l - length of the air duct in meters, z - 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 determine suitable diameter(or 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
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 
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.
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
agricultural 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 with a remote section: determine the diameter D (m) of the round or the area F (m 2) of the cross section of the rectangular air duct:
The speed increases as you approach the fan.
According to Appendix H, the nearest standard values are taken: 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.
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. The hydraulic resistance of the muffler is 36 Pa (according to acoustic calculations). Based on architectural requirements, rectangular air ducts are designed.
The cross-sections of brick channels are taken according to table. 22.7.
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.,
„Ventilation and air conditioning systems. Design recommendations for industrial and public buildings”, chapter 15. “Thermocool”
|
UDC 697.9 Determination of local resistance coefficients of tees in ventilation systems O. D. Samarin, Ph.D., Associate Professor (National Research University MGSU) The current situation with determining the values of local resistance coefficients (KMR) of ventilation network elements during their aerodynamic calculation is considered. An analysis of some modern theoretical and experimental works in the area under consideration is given and the shortcomings of the existing reference literature regarding the ease of using its data for carrying out engineering calculations using MS Excel spreadsheets are identified. The main results of approximation of the available tables for the KMS of unified tees on the branch during discharge and suction in ventilation and air conditioning systems are presented in the form of corresponding engineering formulas. An assessment of the accuracy of the obtained dependencies and the permissible range of their applicability is given, and recommendations for their use in the practice of mass design are presented. The presentation is illustrated with numerical and graphic examples. Keywords:local resistance coefficient, tee, branch, discharge, suction. |
UDC 697.9 Determination of local resistance coefficients of teeth in ventilating systems O. D. Samarin, PhD, Assistant Professor, National Research Moscow State University of Civil Engineering (NR MSUCE) The current situation is reviewed with the definition of values of coefficients of local resistance (CLR) of elements of the ventilation systems at their aerodynamic calculation. The analysis of some contemporary theoretical and experimental works in this fi eld is given and defi ciencies are identified in the existing reference literature for the usability of its data to perform engineering calculations using MS Excel spreadsheets. The main results of approximation of the existing tables to the CLR for the uniform teas on the branch of the injection and the suction in the ventilating and air-conditioning systems are presented in the appropriate engineering formulas. The estimation of accuracy of the obtained dependencies and valid range of their applicability are given, as well as recommendations for their use in practice mass design. The presentation is illustrated by numerical and graphical examples. Keywords:coefficient of local resistance, tee, branch, injection, suction. |
When air flow moves in air ducts and channels of ventilation and air conditioning systems (V and AC), in addition to pressure losses due to friction, a significant role is played by losses on local resistances - shaped parts of air ducts, air distributors and network equipment.
Such losses are proportional to the dynamic pressure R d = ρ v²/2, where ρ is the air density, approximately equal to 1.2 kg/m³ at a temperature of about +20 °C; v— its speed [m/s], determined, as a rule, in the cross section of the channel behind the resistance.
Proportionality coefficients ξ, called local resistance coefficients (KMC), for various elements systems B and HF are usually determined from tables available, in particular, in a number of other sources. Greatest difficulty in this case, it most often causes a search for KMS for tees or branch nodes. The fact is that in this case it is necessary to take into account the type of tee (pass or branch) and the mode of air movement (discharge or suction), as well as the ratio of air flow in the branch to the flow in the trunk L´ o = L o /L c and the cross-sectional area of the passage to the cross-sectional area of the trunk F´ p = F p /F s.
For tees during suction, it is also necessary to take into account the ratio of the cross-sectional area of the branch to the cross-sectional area of the trunk F´ o = F o /F s. In the manual, the relevant data is given in table. 22.36-22.40. However, when carrying out calculations using Excel spreadsheets, which is now quite common due to the widespread use of various standard software and the convenience of formatting the calculation results, it is desirable to have analytical formulas for the CMS, at least in the most common ranges of changes in the characteristics of tees .
In addition, this would be advisable in the educational process to reduce technical work students and transferring the main load to development constructive solutions systems
Similar formulas are available in such a fairly fundamental source as, but there they are presented in a very generalized form, without taking into account the design features of specific elements of existing ventilation systems, and also use a significant number of additional parameters and in some cases require access to certain tables. On the other hand, appeared in Lately programs for automated aerodynamic calculations of V and HF systems use some algorithms for definitions of CMS, but, as a rule, they are unknown to the user and may therefore raise doubts about their validity and correctness.
Also, at present, some works are appearing, the authors of which continue research to refine the calculation of the CMR or expand the range of parameters of the corresponding element of the system for which the results obtained will be valid. These publications appear both in our country and abroad, although in general their number is not very large, and are based primarily on numerical modeling of turbulent flows using a computer or on direct experimental studies. However, the data obtained by the authors is, as a rule, difficult to use in the practice of mass design, since they are not yet presented in engineering form.
In this regard, it seems appropriate to analyze the data contained in the tables and obtain on their basis approximation dependencies that would have the simplest and most convenient form for engineering practice and at the same time would adequately reflect the nature of the existing dependencies for CMC tees. For their most common varieties - tees on the passage (unified branch nodes), this problem was solved by the author in the work. At the same time, it is more difficult to find analytical relationships for tees on a branch, since the dependencies themselves look more complex here. General form approximation formulas, as always in such cases, are obtained based on the location of the calculated points on the correlation field, and the corresponding coefficients are selected using the least squares method in order to minimize the deviation of the constructed graph using Excel. Then for some of the most common ranges F p /F s, F o /F s and L o /L s you can get the expressions:
at L´ about= 0.20-0.75 and F´ about= 0.40-0.65 - for tees during discharge (supply);
at L´ about = 0,2-0,7, F´ about= 0.3-0.5 and F´ p= 0.6-0.8 - for suction (exhaust) tees.
The accuracy of dependencies (1) and (2) is demonstrated in Fig. 1 and 2, which show the results of processing the table. 22.36 and 22.37 for KMS standardized tees (branch nodes) on a branch round section when suctioned. In the case of a rectangular cross-section, the results will differ insignificantly.
It can be noted that the discrepancy here is greater than for tees per passage, and averages 10-15%, sometimes even up to 20%, but for engineering calculations this may be acceptable, especially taking into account the obvious initial error contained in the tables, and Simultaneously simplifying calculations when using Excel. At the same time, 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. First of all, this simplifies calculations when using Excel spreadsheets. At the same time, Fig. 1 and 2 make it possible to verify that the found analytical dependencies quite adequately reflect the nature of the influence of all the main factors on the CMC of tees and the physical essence of the processes occurring in them during the movement of air flow.
At the same time, the formulas given in this work are very simple, clear and easily accessible for engineering calculations, especially in 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 the local resistance coefficients of tees on a branch in a very wide range of cross-sectional ratios and air flow rates in the trunk and branches.
This is quite sufficient for the design of ventilation and air conditioning systems in most residential and public buildings.
The calculation begins with a remote section: determine the diameter D (m) of the round or the area F (m 2) of the cross section of the rectangular air duct: 
Table. Required hourly consumption fresh air, m 3 /h (cfm)
According to Appendix H, the nearest standard values are taken: D st or (a x b) st (m).
Actual speed (m/s): 
or 
Hydraulic radius of rectangular ducts (m): 
Reynolds criterion: Re = 64100 x D st x U fact (for rectangular ducts D st = D L).
Hydraulic friction coefficient: λ = 0.3164 x Re - 0.25 at Re ≤ 60000, λ = 0.1266 x Re - 0.167 at Re Pressure loss in the design area (Pa): 
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.
Table 1. Aerodynamic calculation
| No. of plots | flow L, m 3 / h | length L, m | U re k, m/s | section a x b, m | U f, m/s | D l , m | Re | λ | Kmc | losses on the site?р, pa |
| PP grid at outlet | 0.2 x 0.4 | 3,1 | - | - | - | 1,8 | 10,4 | |||
| 1 | 720 | 4,2 | 4 | 0.2 x 0.25 | 4,0 | 0,222 | 56900 | 0,0205 | 0,48 | 8,4 |
| 2 | 1030 | 3,0 | 5 | 0.25 x 0.25 | 4,6 | 0,25 | 73700 | 0,0195 | 0,4 | 8,1 |
| 3 | 2130 | 2,7 | 6 | 0.4 x 0.25 | 5,92 | 0,308 | 116900 | 0,0180 | 0,48 | 13,4 |
| 4 | 3480 | 14,8 | 7 | 0.4 x 0.4 | 6,04 | 0,40 | 154900 | 0,0172 | 1,44 | 45,5 |
| 5 | 6830 | 1,2 | 8 | 0.5 x 0.5 | 7,6 | 0,50 | 234000 | 0,0159 | 0,2 | 8,3 |
| 6 | 10420 | 6,4 | 10 | 0.6 x 0.5 | 9,65 | 0,545 | 337000 | 0,0151 | 0,64 | 45,7 |
| 6a | 10420 | 0,8 | Yu. | ø 0.64 | 8,99 | 0,64 | 369000 | 0,0149 | 0 | 0,9 |
| 7 | 10420 | 3,2 | 5 | 0.53 x 1.06 | 5,15 | 0,707 | 234000 | 0.0312 x n | 2,5 | 44,2 |
| Total losses: 185 Note. For brick channels with an absolute roughness of 4 mm and U f = 6.15 m/s, correction factor n = 1.94 (Table 22.12.). |
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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. The hydraulic resistance of the muffler is 36 Pa (according to acoustic calculations). Based on architectural requirements, rectangular air ducts are designed.
The cross-sections of brick channels are taken according to table. 22.7.
Section 1. PP grid at the outlet with a cross section of 200 x 400 mm (calculated separately): 
Dynamic pressure: 
Lattice KMC (Appendix 25.1) = 1.8.
Pressure drop in the grid: Δр - рД x KMC = 5.8 x 1.8 = 10.4 Pa.
Design fan pressure p: Δр vent = 1.1 (Δр air + Δр valve + Δр filter + Δр cal + Δр muffler) = 1.1 (185 + 10 + 250 + 100 + 36) = 639 Pa.
Fan flow: L fan = 1.1 x Lsyst = 1.1 x 10420 = 11460 m 3 /h.
Selected radial fan VTs4-75 No. 6.3, version 1: L = 11500 m 3 /h; Δр ven = 640 Pa (fan unit E6.3.090 - 2a), rotor diameter 0.9 x D pom, rotation speed 1435 min-1, electric motor 4A10054; N = 3 kW installed on the same axis as the fan. Unit weight 176 kg.
Checking fan motor power (kW): 
According to the aerodynamic characteristics of the fan, n fan = 0.75. 
Table 2. Determination of local resistances
| 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 | |
| Σ | 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 x 400 | 90 | - | - | - | Adj. 25.11 | |
| Retraction | 400 x 250 | 90 | - | - | - | Adj. 25.11 | 0,22 | |
| Tee-pass |  |
- | - | 0,49 | 0,64 | Table 25.8 | 0,4 | |
| Σ | 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 x 500 | 90 | - | - | - | Adj. 25.11 | 0,5 | |
| Σ | 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 | ||||||
| Σ | 1,44 | |||||||
