Coefficient of local resistance in ventilation transitions. Calculation of aerodynamic resistances. Local resistance coefficients
The aerodynamic calculation of air ducts begins with drawing an axonometric diagram M 1:100, putting down the numbers of sections, their loads b m / h, and lengths 1, m. The direction of the aerodynamic calculation is determined - from the most remote and loaded section to the fan. When in doubt, when determining the direction, all possible options are calculated.
The calculation starts from a remote area, its diameter is calculated D, m, or flat
spad cross section rectangular air duct Р, m:
The beginning of the system at the fan
Administrative buildings 4-5 m/s 8-12 m/s
Industrial buildings 5-6 m/s 10-16 m/s,
Increasing as you get closer to the fan.
Using Appendix 21, we accept the nearest standard values \u200b\u200bof Dst or (a x b)st
Then we calculate the actual speed:
Or———————— ———— - , m/s.
FACT 3660 * (a * 6) st
For further calculations, we determine the hydraulic radius of rectangular ducts:
£>1 =--,m. a + b
To avoid using tables and interpolating the values of specific friction losses, we use a direct solution of the problem:
We define the Reynolds criterion:
Re = 64 100 * Rest * Ufact (for rectangular Rest = Ob) (14.6)
And the coefficient of hydraulic friction:
0.3164*Rae 0 25 at Rae< 60 ООО (14.7)
0.1266 * 0167 for R e > 60,000. (14.8)
The pressure loss in the calculated section will be:
Where KMS is the sum of the coefficients local resistance in the duct area.
Local resistances lying on the border of two sections (tees, crosses) should be attributed to the section with a lower flow rate.
The local resistance coefficients are given in the appendices.
Initial data:
Air duct material - galvanized sheet steel, thickness and dimensions in accordance with App. 21 .
The material of the air intake shaft is brick. Adjustable gratings of the PP type with possible sections are used as air distributors:
100 x 200; 200 x 200; 400 x 200 and 600 x 200 mm, shade factor 0.8 and maximum outlet air velocity up to 3 m/s.
The resistance of the insulated intake valve with fully open blades is 10 Pa. The hydraulic resistance of the air heater installation is 132 Pa (according to a separate calculation). Filter resistance 0-4 250 Pa. The hydraulic resistance of the muffler is 36 Pa (according to acoustic calculation). Based on the architectural requirements, the air ducts are designed with a rectangular section.
|
Supply L, m3/h |
Length 1, m |
Section a * b, m |
Losses in the section p, Pa |
|||||||
|
PP grating at the exit |
||||||||||
|
250×250 b =1030 |
|
|
|
With this material, the editors of the journal “Climate World” continue to publish chapters from the book “Ventilation and air conditioning systems. Design recommendations for
water and public buildings”. Author Krasnov Yu.S.
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 remote and loaded section to the fan. When in doubt, when determining the direction, all possible options are calculated.
The calculation starts from a remote section: the diameter D (m) of a round or the area F (m 2) of the cross section of a rectangular duct is determined:
The speed increases as you get closer to the fan.
According to Appendix H, the nearest standard values are taken from: D CT or (a x b) st (m).
Hydraulic radius of rectangular ducts (m):
 |
where - the sum of the local resistance coefficients in the duct section.
Local resistances at the border of two sections (tees, crosses) are attributed to the section with a lower flow rate.
Local resistance coefficients are given in the appendices.
Scheme of the supply ventilation system serving the 3-storey administrative building
Calculation example
Initial data:
| No. of plots | supply 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 section Δр, pa |
| outlet grating pp | 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 dimensions of which correspond to app. N from . The material of the air intake shaft is brick. Adjustable gratings of the PP type with possible sections are used as air distributors: 100 x 200; 200 x 200; 400 x 200 and 600 x 200 mm, shade factor 0.8 and maximum outlet air velocity up to 3 m/s.
The resistance of the insulated intake valve with fully open blades is 10 Pa. The hydraulic resistance of the air heater installation 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 calculation). Based on architectural requirements, rectangular ducts are designed.
Cross-sections of brick channels are taken according to Table. 22.7.
Local resistance coefficients
Section 1. RR grating at the exit with a section of 200 × 400 mm (calculated separately):
| No. of plots | Type of local resistance | Sketch | Angle α, deg. | Attitude | Rationale | KMS | ||
| F0/F1 | L 0 /L st | f pass / f st | ||||||
| 1 | Diffuser |
 |
20 | 0,62 | — | — | Tab. 25.1 | 0,09 |
| Withdrawal |
 |
90 | — | — | — | Tab. 25.11 | 0,19 | |
| Tee-pass |
 |
— | — | 0,3 | 0,8 | App. 25.8 | 0,2 | |
| ∑ = | 0,48 | |||||||
| 2 | Tee-pass |
 |
— | — | 0,48 | 0,63 | App. 25.8 | 0,4 |
| 3 | branch tee |
 |
— | 0,63 | 0,61 | — | App. 25.9 | 0,48 |
| 4 | 2 outlets | 250×400 | 90 | — | — | — | App. 25.11 | |
| Withdrawal | 400×250 | 90 | — | — | — | App. 25.11 | 0,22 | |
| Tee-pass |
 |
— | — | 0,49 | 0,64 | Tab. 25.8 | 0,4 | |
| ∑ = | 1,44 | |||||||
| 5 | Tee-pass |
 |
— | — | 0,34 | 0,83 | App. 25.8 | 0,2 |
| 6 | Diffuser after fan |
 |
h=0.6 | 1,53 | — | — | App. 25.13 | 0,14 |
| Withdrawal | 600×500 | 90 | — | — | — | App. 25.11 | 0,5 | |
| ∑= | 0,64 | |||||||
| 6a | Confuser in front of the fan |
 |
D g \u003d 0.42 m | Tab. 25.12 | 0 | |||
| 7 | Knee | 90 | — | — | — | Tab. 25.1 | 1,2 | |
| Louvre grille | Tab. 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”
- Refrigeration machines and refrigeration units. Refrigeration center design example
- “Calculation of heat balance, moisture intake, air exchange, construction of J-d diagrams. Multi zone air conditioning. Solution examples»
- Designer. Materials of the journal "Climate World"
- Basic air parameters, filter classes, heater power calculation, standards and regulations, table of physical quantities
- Separate technical solutions, equipment What is an elliptical plug and why is it needed
Impact of Current Temperature Regulations on Data Center Power Consumption New Methods for Improving the Energy Efficiency of Data Center Air Conditioning Systems Increasing the efficiency of a solid fuel fireplace Heat recovery systems in refrigeration plants The microclimate of wine storages and equipment for its creation Selection of equipment for specialized outdoor air supply systems (DOAS) Tunnel ventilation system. TLT-TURBO GmbH equipment Application of Wesper equipment in the complex for deep oil processing of the enterprise "KIRISHINEFTEORGSINTEZ" Air exchange control in laboratory rooms Integrated use of underfloor air distribution systems (UFAD) in combination with chilled beams Tunnel ventilation system. Choosing a ventilation scheme Calculation of air-thermal curtains based on a new type of presentation of experimental data on heat and mass losses Experience in creating a decentralized ventilation system during the reconstruction of a building Cold beams for laboratories. Use of dual energy recovery Ensuring reliability at the design stage Utilization of heat released during the operation of the refrigeration plant of an industrial enterprise - Method of aerodynamic calculation of air ducts Methodology for selecting a split system from DAICHI Vibration characteristics of fans The new standard for thermal insulation design Applied issues of classification of premises according to climatic parameters Optimization of control and structure of ventilation systems Variators and drainage pumps from EDC New reference book from ABOK A new approach to the construction and operation of refrigeration systems for air-conditioned buildings
|
Purpose |
Basic requirement | ||||
| Noiselessness | Min. head loss | ||||
| Main channels | main channels | Branches | |||
| tributary | Hood | tributary | Hood | ||
| Living spaces | 3 | 5 | 4 | 3 | 3 |
| Hotels | 5 | 7.5 | 6.5 | 6 | 5 |
| Institutions | 6 | 8 | 6.5 | 6 | 5 |
| Restaurants | 7 | 9 | 7 | 7 | 6 |
| The shops | 8 | 9 | 7 | 7 | 6 |
Based on these values, the linear parameters of the air ducts should be calculated.
Algorithm for calculating air pressure losses
The calculation must begin with drawing up a diagram of the ventilation system with the obligatory indication of the spatial location of the air ducts, the length of each section, ventilation grilles, additional equipment for air purification, technical fittings and fans. Losses are determined first for each individual line, and then summed up. For a separate technological section, the losses are determined using the formula P = L × R + Z, where P is the air pressure loss in the calculated section, R is the loss per linear meter of the section, L is the total length of the air ducts in the section, Z is the loss in the additional fittings of the system ventilation.
To calculate the pressure loss in a circular duct, the formula Ptr is used. = (L/d×X) × (Y×V)/2g. X is the tabular coefficient of air friction, depends on the material of manufacture of the air duct, L is the length of the calculated section, d is the diameter of the air duct, V is the required air flow rate, Y is the air density, taking into account temperature, g is the acceleration of fall (free). If the ventilation system has square air ducts, then table No. 2 should be used to convert round values to square ones.
Tab. No. 2. Equivalent diameters of round ducts for square
| 150 | 200 | 250 | 300 | 350 | 400 | 450 | 500 | |
| 250 | 210 | 245 | 275 | |||||
| 300 | 230 | 265 | 300 | 330 | ||||
| 350 | 245 | 285 | 325 | 355 | 380 | |||
| 400 | 260 | 305 | 345 | 370 | 410 | 440 | ||
| 450 | 275 | 320 | 365 | 400 | 435 | 465 | 490 | |
| 500 | 290 | 340 | 380 | 425 | 455 | 490 | 520 | 545 |
| 550 | 300 | 350 | 400 | 440 | 475 | 515 | 545 | 575 |
| 600 | 310 | 365 | 415 | 460 | 495 | 535 | 565 | 600 |
| 650 | 320 | 380 | 430 | 475 | 515 | 555 | 590 | 625 |
| 700 | 390 | 445 | 490 | 535 | 575 | 610 | 645 | |
| 750 | 400 | 455 | 505 | 550 | 590 | 630 | 665 | |
| 800 | 415 | 470 | 520 | 565 | 610 | 650 | 685 | |
| 850 | 480 | 535 | 580 | 625 | 670 | 710 | ||
| 900 | 495 | 550 | 600 | 645 | 685 | 725 | ||
| 950 | 505 | 560 | 615 | 660 | 705 | 745 | ||
| 1000 | 520 | 575 | 625 | 675 | 720 | 760 | ||
| 1200 | 620 | 680 | 730 | 780 | 830 | |||
| 1400 | 725 | 780 | 835 | 880 | ||||
| 1600 | 830 | 885 | 940 | |||||
| 1800 | 870 | 935 | 990 |
The horizontal is the height of the square duct, and the vertical is the width. Equivalent value round section is at the intersection of the lines.
Air pressure losses in bends are taken from table No. 3.
Tab. No. 3. Loss of pressure on bends
To determine the pressure loss in the diffusers, the data from Table No. 4 are used.
Tab. No. 4. Pressure loss in diffusers
Table No. 5 gives a general diagram of losses in a straight section.
Tab. No. 5. Diagram of air pressure losses in straight air ducts
All individual losses in a given section of the duct are summarized and corrected with Table No. 6. Tab. No. 6. Calculation of the flow pressure drop in ventilation systems
During design and calculations, existing regulations recommend that the difference in pressure loss between individual sections should not exceed 10%. The fan should be installed in the section of the ventilation system with the highest resistance, the most distant air ducts should have the minimum resistance. If these conditions are not met, then it is necessary to change the layout of air ducts and additional equipment, taking into account the requirements of the regulations.
You can also use the approximate formula:
0.195 v 1.8
R f . (10) d 100 1 , 2
Its error does not exceed 3–5%, which is sufficient for engineering calculations.
The total friction pressure loss for the entire section is obtained by multiplying the specific losses R by the length of the section l, Rl, Pa. If air ducts or channels from other materials are used, it is necessary to introduce a correction for roughness βsh according to Table. 2. It depends on the absolute equivalent roughness of the duct material K e (Table 3) and the value of v f .
table 2 |
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Correction values βsh |
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v f , m/s |
βsh at K e , mm |
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Table 3 Absolute equivalent roughness of duct material
Plasterer- |
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ka on the grid |
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K e , mm |
For steel air ducts βsh = 1. More detailed values of βsh can be found in Table. 22.12. With this correction in mind, the adjusted friction pressure loss Rl βsh , Pa, is obtained by multiplying Rl by the value βsh . Then determine the dynamic pressure on the participants
under standard conditions ρw = 1.2 kg/m3.
Next, local resistances are detected on the site, local resistance coefficients (LMR) ξ are determined and the sum of LMR in this section (Σξ) is calculated. All local resistances are entered into the statement in the following form.
STATEMENT KMS VENTILATION SYSTEMS
Etc.
AT the column “local resistances” records the names of the resistances (bend, tee, cross, elbow, grate, air distributor, umbrella, etc.) available in this area. In addition, their number and characteristics are noted, according to which the CMR values are determined for these elements. For example, for a round bend, this is the angle of rotation and the ratio of the radius of rotation to the diameter of the duct r / d , for a rectangular outlet - the angle of rotation and dimensions of the sides of the duct a and b . For side openings in an air duct or duct (for example, at the installation site of an air intake grille) - the ratio of the opening area to the cross section of the air duct
f resp / f about . For tees and crosses on the passage, the ratio of the cross-sectional area of the passage and the trunk f p / f s and the flow rate in the branch and in the trunk L o / L s is taken into account, for tees and crosses on the branch - the ratio of the cross-sectional area of the branch and the trunk f p / f s and again, the value of L about /L with. It should be borne in mind that each tee or cross connects two adjacent sections, but they refer to one of these sections, in which the air flow L is less. The difference between tees and crosses on a run and on a branch has to do with how the design direction runs. This is shown in fig. 11. Here, the calculated direction is shown by a thick line, and the directions of air flows are shown by thin arrows. In addition, it is signed exactly where in each option the trunk, passage and exit are located.
branch tee for right choice relations fп / fс , fо /fс and L о /L с . Note that in supply ventilation systems, the calculation is usually carried out against the movement of air, and in exhaust systems, along this movement. The sections to which the considered tees belong are indicated by checkmarks. The same applies to crosses. As a rule, although not always, tees and crosses on the passage appear when calculating the main direction, and on the branch they appear when aerodynamic linking of secondary sections (see below). In this case, the same tee in the main direction can be considered as a tee per passage, and in the secondary
– as a branch with a different coefficient. KMS for crosses
accepted in the same size as for the corresponding tees.
Rice. 11. Tee calculation scheme
Approximate values of ξ for common resistances are given in Table. 4.
Table 4 |
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Values ξ of some local resistances |
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Name |
Name |
|||
resistance |
resistance |
|||
Elbow round 90o, |
The grate is not adjustable |
|||
r/d = 1 |
may RS-G (exhaust or |
|||
Rectangular elbow 90o |
air intake) |
|||
Tee in the passage (on- |
sudden expansion |
|||
oppression) |
||||
Branch tee |
sudden constriction |
|||
Tee in the passage (all- |
First side hole |
|||
stie (entrance to the air |
||||
Branch tee |
–0.5* … |
boron mine) |
||
Plafond (anemostat) ST-KR, |
Rectangular elbow |
|||
90o |
||||
Grille adjustable RS- |
Umbrella over exhaust |
|||
VG (supply) |
||||
*) negative CMR can occur at low Lo /Lc due to air ejection (suction) from the branch by the main flow.
More detailed data for the KMS are given in Table. 22.16 - 22.43. For the most common local resistances -
tees in the passage - KMR can also be approximately calculated using the following formulas:
0.41f "25L" 0.24 |
0.25 at |
0.7 and |
f "0.5 (11) |
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- for tees during injection (supply); |
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at L" |
0.4 you can use the simplified formula |
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prox int 0. 425 0. 25 f p "; |
||||||||||
0.2 1.7f" |
0.35 0.25f" |
2.4L" |
0. 2 2 |
|||||||
– for suction tees (exhaust).
Here L" |
f about |
and f" |
f p |
|||||||
f c |
||||||||||
After determining the value of Σξ, the pressure loss at local resistances Z P d, Pa, and the total pressure loss are calculated
on the section Rl βsh + Z , Pa.
The results of the calculations are entered in the table in the following form.
AERODYNAMIC CALCULATION OF THE VENTILATION SYSTEM
Estimated |
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Duct dimensions |
pressure |
||||||||||||||
on friction |
Rlβ w |
Rd , |
|||||||||||||
βw |
|||||||||||||||
d or |
f op, |
ff , |
Vf , |
d eq |
|||||||||||
l , m |
a×b |
||||||||||||||
When the calculation of all sections of the main direction is completed, the values of Rl βsh + Z for them are summarized and the total resistance is determined.
ventilation network resistance P network = Σ(Rl βw + Z ).
After calculating the main direction, one or two branches are linked. If the system serves several floors, you can select floor branches on intermediate floors for linking. If the system serves one floor, link branches from the main that are not included in the main direction (see the example in paragraph 4.3). The calculation of the linked sections is carried out in the same sequence as for the main direction, and recorded in a table in the same form. Linkage is considered completed if the amount
pressure loss Σ(Rl βsh + Z ) along the linked sections deviates from the sum Σ(Rl βsh + Z ) along parallel connected sections of the main direction by no more than 10%. Sections along the main and linked directions from the point of their branching to the end air distributors are considered to be connected in parallel. If the circuit looks like the one shown in Fig. 12 (the main direction is marked with a thick line), then direction 2 alignment requires that the value of Rl βsh + Z for section 2 be equal to Rl βsh + Z for section 1, obtained from the calculation of the main direction, with an accuracy of 10%. Linkage is achieved by selecting the diameters of round or cross-sectional dimensions of rectangular air ducts in the linked sections, and if this is not possible, by installing throttle valves or diaphragms on the branches.
The selection of a fan should be carried out according to the manufacturer's catalogs or according to the data. The fan pressure is equal to the sum of pressure losses in the ventilation network in the main direction, determined in the aerodynamic calculation of the ventilation system, and the sum of pressure losses in the elements of the ventilation unit ( air valve, filter, air heater, silencer, etc.).
Rice. 12. A fragment of the scheme of the ventilation system with the choice of a branch for linking
Finally, it is possible to choose a fan only after an acoustic calculation, when the issue of installing a silencer is decided. Acoustic calculation can be performed only after preliminary selection of the fan, since the initial data for it are the sound power levels emitted by the fan into the air ducts. Acoustic calculation is carried out, guided by the instructions of chapter 12. If necessary, calculate and determine the size of the silencer , , then finally select the fan.
4.3. Calculation example supply system ventilation
The supply ventilation system for the dining room is considered. The application of air ducts and air distributors to the plan is given in clause 3.1 in the first variant ( typical scheme for halls).
System Diagram
1000х400 5 8310 m3/h
2772 m3/h2 |
|||||||
More details on the calculation methodology and the necessary initial data can be found at,. The corresponding terminology is given in .
STATEMENT OF KMS SYSTEM P1
local resistance |
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924 m3/h |
||||||
1. Elbow round 90о r /d =1 |
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2. Tee in the passage (pressure) |
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fp / fc |
Lo/Lc |
|||||
fp / fc |
Lo/Lc |
|||||
1. Tee in the passage (pressure) |
||||||
fp / fc |
Lo/Lc |
|||||
1. Tee in the passage (pressure) |
||||||
fp / fc |
Lo/Lc |
|||||
1. Rectangular elbow 1000×400 90o 4 pcs |
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1. Air intake shaft with umbrella |
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(first side hole) |
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1. Air intake louvre |
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STATEMENT OF KMS OF THE P1 SYSTEM (Branch No. 1) |
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local resistance |
||||||
1. Air distributor PRM3 at flow rate |
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924 m3/h |
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1. Elbow round 90о r /d =1 |
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2. Branch tee (injection) |
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fo / fc |
Lo/Lc |
|||||
APPENDIX Characteristics of ventilation grilles and shades
I. Living sections, m2, supply and exhaust louvered gratings RS-VG and RS-G
Length, mm |
Height, mm |
|||||
Speed coefficient m = 6.3, temperature coefficient n = 5.1.
II. Characteristics of ceiling lamps ST-KR and ST-KV
Name |
Dimensions, mm |
f fact, m 2 |
||
Dimensional |
Interior |
|||
Plafond ST-KR |
||||
(round) |
||||
Plafond ST-KV |
||||
(square) |
||||
Speed coefficient m = 2.5, temperature coefficient n = 3.
REFERENCES
1. Samarin O.D. Selection of supply air equipment ventilation units(air conditioners) type KCKP. Guidelines for the implementation of course and diploma projects for students of the specialty 270109 "Heat and gas supply and ventilation". – M.: MGSU, 2009. – 32 p.
2. Belova E.M. Central air conditioning systems in buildings. - M.: Euroclimate, 2006. - 640 p.
3. SNiP 41-01-2003 "Heating, ventilation and air conditioning". - M.: GUP TsPP, 2004.
4. Catalog of equipment "Arktos".
5. sanitary devices. Part 3. Ventilation and air conditioning. Book 2. / Ed. N.N. Pavlov and Yu.I. Schiller. – M.: Stroyizdat, 1992. – 416 p.
6. GOST 21.602-2003. System of design documents for construction. Rules for the implementation of working documentation for heating, ventilation and air conditioning. - M.: GUP TsPP, 2004.
7. Samarin O.D. On the regime of air movement in steel air ducts.
// SOK, 2006, No. 7, p. 90-91.
8. Designer's Handbook. Internal sanitary devices. Part 3. Ventilation and air conditioning. Book 1. / Ed. N.N. Pavlov and Yu.I. Schiller. – M.: Stroyizdat, 1992. – 320 p.
9. Kamenev P.N., Tertichnik E.I. Ventilation. - M.: ASV, 2006. - 616 p.
10. Krupnov B.A. Terminology for building thermophysics, heating, ventilation and air conditioning: guidelines for students of the specialty "Heat and gas supply and ventilation".
With this material, the editors of the journal “Climate World” continue to publish chapters from the book “Ventilation and air conditioning systems. Design recommendations for
water and public buildings.” Author Krasnov Yu.S.
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 remote and loaded section to the fan. When in doubt, when determining the direction, all possible options are calculated.
The calculation starts from a remote section: the diameter D (m) of a round or the area F (m 2) of the cross section of a rectangular duct is determined:
The speed increases as you get closer to the fan.
According to Appendix H, the nearest standard values are taken from: 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 duct section.
Local resistances at the border of two sections (tees, crosses) are attributed to the section with a lower flow rate.
Local resistance coefficients are given in the appendices.
Scheme of the supply ventilation system serving the 3-storey administrative building
Calculation example
Initial data:
| No. of plots | supply 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 section Δр, pa |
| outlet grating pp | 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 dimensions of which correspond to app. N out. The material of the air intake shaft is brick. Adjustable gratings of the PP type with possible sections are used as air distributors: 100 x 200; 200 x 200; 400 x 200 and 600 x 200 mm, shade factor 0.8 and maximum outlet air velocity up to 3 m/s.
The resistance of the insulated intake valve with fully open blades is 10 Pa. The hydraulic resistance of the air heater installation 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 calculation). Based on architectural requirements, rectangular ducts are designed.
Cross-sections of brick channels are taken according to Table. 22.7.
Local resistance coefficients
Section 1. RR grating at the exit with a section of 200 × 400 mm (calculated separately):
| No. of plots | Type of local resistance | Sketch | Angle α, deg. | Attitude | Rationale | KMS | ||
| F0/F1 | L 0 /L st | f pass / f st | ||||||
| 1 | Diffuser |
 |
20 | 0,62 | - | - | Tab. 25.1 | 0,09 |
| Withdrawal |
 |
90 | - | - | - | Tab. 25.11 | 0,19 | |
| Tee-pass |
 |
- | - | 0,3 | 0,8 | App. 25.8 | 0,2 | |
| ∑ = | 0,48 | |||||||
| 2 | Tee-pass |
 |
- | - | 0,48 | 0,63 | App. 25.8 | 0,4 |
| 3 | branch tee |
 |
- | 0,63 | 0,61 | - | App. 25.9 | 0,48 |
| 4 | 2 outlets | 250×400 | 90 | - | - | - | App. 25.11 | |
| Withdrawal | 400×250 | 90 | - | - | - | App. 25.11 | 0,22 | |
| Tee-pass |
 |
- | - | 0,49 | 0,64 | Tab. 25.8 | 0,4 | |
| ∑ = | 1,44 | |||||||
| 5 | Tee-pass |
 |
- | - | 0,34 | 0,83 | App. 25.8 | 0,2 |
| 6 | Diffuser after fan |
 |
h=0.6 | 1,53 | - | - | App. 25.13 | 0,14 |
| Withdrawal | 600×500 | 90 | - | - | - | App. 25.11 | 0,5 | |
| ∑= | 0,64 | |||||||
| 6a | Confuser in front of the fan |
 |
D g \u003d 0.42 m | Tab. 25.12 | 0 | |||
| 7 | Knee | 90 | - | - | - | Tab. 25.1 | 1,2 | |
| Louvre grille | Tab. 25.1 | 1,3 | ||||||
| ∑ = | 1,44 | |||||||
| Table 2. Determination of local resistances | ||||||||
Krasnov Yu.S.,
1. Friction loss:
Ptr \u003d (x * l / d) * (v * v * y) / 2g,
z = Q* (v*v*y)/2g,
Permissible speed method
Note: the air flow rate 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, find their equivalent diameters using the table below.
Notes:
- If there is not enough space (for example, during reconstruction), choose rectangular ducts. As a rule, the width of the duct is 2 times the height).
Table of equivalent duct diameters
When the parameters of air ducts are known (their length, cross-section, coefficient of friction of air on the surface), it is possible to calculate the pressure loss in the system at the projected air flow.
The 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 duct, l is the length of the duct in meters, z is the pressure loss due to local resistances (with a variable section).
1. Friction loss:
In a round duct, friction pressure losses P tr are calculated as follows:
Ptr \u003d (x * l / d) * (v * v * y) / 2g,
where x is the coefficient of friction resistance, l is the length of the duct in meters, d is the diameter of the duct in meters, v is the air flow velocity in m/s, y is the air density in kg/m3, g is the free fall acceleration (9 .8 m/s2).
Note: If the air duct has not a round, but a rectangular cross section, the equivalent diameter must be substituted into the formula, which for an air duct with sides A and B is equal to: dequiv = 2AB/(A + B)
2. Losses due to local resistance:
Pressure losses due to local resistances are calculated according to the formula:
z = Q* (v*v*y)/2g,
where Q is the sum of the coefficients of local resistances in the section of the duct for which the calculation is made, v is the air flow velocity in m/s, y is the air density in kg/m3, g is the free fall acceleration (9.8 m/s2 ). The Q values are contained in tabular form.
Permissible speed method
When calculating the air duct network using the method of permissible speeds, the optimal air speed is taken as the initial data (see table). Then, the required cross-section of the duct and the pressure loss in it are considered.
The procedure for the aerodynamic calculation of air ducts according to the method of permissible speeds:
Draw a diagram of the air distribution system. For each section of the duct, indicate the length and amount of air passing in 1 hour.
We start the calculation from the most distant from the fan and the most loaded sections.
Knowing the optimal air velocity for a given room and the volume of air passing through the duct in 1 hour, we determine the appropriate diameter (or cross section) of the duct.
We calculate the pressure loss due to friction P tr.
According to 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 next branches of the air distribution network is determined as the sum of the pressure losses in the sections located before this branch.
In the process of calculation, it is necessary to sequentially link all the branches of the network, equating the resistance of each branch to the resistance of the most loaded branch. This is done with diaphragms. They are installed on lightly loaded sections of air ducts, increasing resistance.
Table of maximum air speed depending on duct requirements
Constant Head Loss Method
This method assumes a constant pressure loss per 1 linear meter of the duct. Based on this, the dimensions of the duct network are determined. The method of constant head loss is quite simple and is used at the stage of the feasibility study of ventilation systems:
Depending on the purpose of the room, according to the table of permissible air velocities, the speed on the main section of the duct is selected.
Based on the speed determined in paragraph 1 and on the basis of the design air flow, the initial pressure loss is found (per 1 m of the duct length). This is the diagram below.
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 system length by the head loss from step 2. The head loss at the diffusers is added to the value obtained.
Now, using the diagram below, determine the diameter of the initial 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 head loss and duct diameter 
The head loss diagram shows the diameters of round ducts. If rectangular ducts are used instead, find their equivalent diameters using the table below.
Notes:
If space permits, it is better to choose round or square ducts;
If there is not enough space (for example, during reconstruction), rectangular ducts are chosen. As a rule, the width of the duct is 2 times the height).
The table shows the height of the duct in mm horizontally, the vertical width, and the table cells contain equivalent duct diameters in mm.