Determination of dynamic pressure in the duct. Calculation of air ducts Calculation of pressure losses due to friction in air ducts
Ventilation calculation this is the calculation of air ducts and ventilation ducts in systems supply and exhaust ventilation . Ventilation is used to supply and remove air with temperatures up to 80°C. The calculation is made according to the method of specific pressure losses. The total pressure loss, kgf/m², in the duct network for standard air (t = 20°C and γ = 1.2 kg/m³) is determined by the formula:
p =∑(Rl+Z),
where R is the pressure loss due to friction in the calculated segment kgf / m² per 1 m; l is the length of the duct section, m; Z - pressure loss due to local resistances in the calculated segment, kgf / m².
Friction pressure loss R, kgf/m² per 1 m in round air ducts is determined by the formula R= λd v²γ2g, where λ is the coefficient of friction resistance; d is the duct diameter, m; v is the speed of air movement in the duct, m/s; γ - volumetric mass of air moving through the duct, kgf/m³; v²γ / 2g - speed (dynamic) pressure, kgf / m².
The drag coefficient is adopted according to the Altshul formula:
where Δe is the absolute equivalent surface roughness of the air duct made of sheet steel, equal to 0.1 mm; d – duct diameter, mm; Re is the Reynolds number.
For air ducts made of other materials with an absolute equivalent roughness Ke≥0.1 mm, the values of R are taken with a correction factor n for friction pressure losses.
Δe value for other materials:
- Sheet steel - 0.1mm
- Viniplast - 0.1mm
- Asbestos-cement pipes - 0.11mm
- Brick - 4mm
- Plaster on the grid - 10mm
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m/s |
n at Δe, mm |
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Recommended speed of air movement in air ducts with mechanical stimulation. Industrial buildings main air ducts - up to 12 m/s, branch air ducts - 6 m/s. Public buildings main air ducts - up to 8 m/s, branch air ducts - 5 m/s.
In air ducts rectangular section the calculated value d is taken to be the equivalent diameter dev, at which the pressure loss in a round duct at the same air velocity is equal to the loss in a rectangular duct. The values of equivalent diameters, m, are determined by the formula
where A and B are the dimensions of the sides of the rectangular duct. It should be borne in mind that with equal air speed, a rectangular duct and a similar round duct have miscellaneous expenses air. The value of velocity (dynamic) pressure and specific friction pressure losses for round air ducts.
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v2γ2g |
m/s |
Amount of passing air m³/h |
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Friction pressure loss kgf/m² |
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Pressure loss Z, kgf / m², due to local resistances is determined by the formula
Z = ∑ζ(v²γ/2g),
where ∑ζ is the sum of the coefficients of local resistances on the estimated section of the duct. If the temperature of the transported air is not equal to 20°C for pressure losses calculated by the formula p =∑(Rl+Z), it is required to enter correction factors K1 - friction, K2 - local resistance.
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t °C |
t °C |
t °C |
t °C |
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If discrepancies in pressure losses along the branches of the air ducts are within 10%, iris dampers should be installed.
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:
Ptr \u003d (x * l / d) * (v * v * y) / 2g,
z = Q* (v*v*y)/2g,
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.
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:
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).
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
management 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 site: determine the diameter D (m) of the round or the area F (m 2) cross section rectangular duct:
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. Silencer hydraulic resistance 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 |
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20 | 0,62 | - | - | Tab. 25.1 | 0,09 |
| Withdrawal |
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90 | - | - | - | Tab. 25.11 | 0,19 | |
| Tee-pass |
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- | - | 0,3 | 0,8 | App. 25.8 | 0,2 | |
| ∑ = | 0,48 | |||||||
| 2 | Tee-pass |
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- | - | 0,48 | 0,63 | App. 25.8 | 0,4 |
| 3 | branch tee |
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- | 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 |
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- | - | 0,49 | 0,64 | Tab. 25.8 | 0,4 | |
| ∑ = | 1,44 | |||||||
| 5 | Tee-pass |
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- | - | 0,34 | 0,83 | App. 25.8 | 0,2 |
| 6 | Diffuser after fan |
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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 |
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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.,
When the parameters of the air ducts are known (their length, cross section, air friction coefficient 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
Note: the air flow rate 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 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, according to 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
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 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 cells of the table contain equivalent duct diameters in mm.
Table of equivalent duct diameters
The pressure distribution in the ventilation system must be known when setting up and regulating the system, when determining the flow rates in individual sections of the system, and when solving many other ventilation problems.
Distribution of pressures in ventilation systems with mechanical induction of air movement. Consider an air duct with a fan (Fig. XI.3). In section 1-/, the static pressure is zero (i.e., equal to the air pressure at the level of the duct). The total pressure in this section is equal to the dynamic pressure рді determined by the formula (XI.1). In section II-II, the static pressure pstіі>0 (numerically equal to the pressure loss due to friction between sections II-II and I-/). With a constant cross section of the duct, the static pressure line is straight. The total pressure line is also straight,
Parallel line rst. The vertical distance between these lines determines the dynamic pressure рДі.
In the diffuser, located between sections II-II and III-III, there is a change in the flow rate. The dynamic pressure decreases along the air flow. In this regard, the static pressure changes and may even increase, as shown in the figure (pstіі>pstііі).
The total pressure in section III-III, created by the fan, is lost to friction Drtr and in local resistance(diffuser Lrdif, at the exit Arnyh). The total pressure loss on the discharge side is:
The static pressure outside the duct on the suction side is zero. In the immediate vicinity of the opening within the suction plume, the air flow already has kinetic energy. The vacuum within the suction jet is negligible.
At the inlet to the duct, the flow velocity increases, which means that the kinetic energy of the flow also increases. Therefore, according to the law of conservation of energy, the potential energy of the flow must decrease. Taking into account pressure losses L/?POt in any section on the suction side
Per \u003d 0 - rd - Drpot - (XI. 24)
In the suction duct, as well as on the discharge side, the total pressure is equal to the pressure difference at the beginning of the duct and the pressure loss up to the considered section:
Pp \u003d 0-DrpOt. (XI.25)
From formulas (XI.24) and (XI.25) it follows that in each section of the air duct on the suction side, the values of p0m and pn are less than zero. By absolute value the static pressure is greater than the total pressure, but formula (XI.2) is also valid for this case.
The static pressure line goes below the full pressure line. The sharp decrease in the static pressure line after section VI-VI is explained by the narrowing of the flow at the duct inlet due to the formation of a vortex zone. Between sections V-V and IV-IV, the diagram shows a confuser with a turn. The decrease in the static pressure line between these sections occurs due to an increase in both the flow velocity in the confuser and pressure losses. Plots of static pressure in fig. XI.3 are shaded.
At point B, the lowest total pressure in the duct system is observed. Numerically, it is equal to the pressure loss on the suction side:
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A - full and static in the discharge duct; b - the same, in the suction duct; c - dynamic in the discharge duct; g - dynamic in the suction duct
The fan creates a pressure drop equal to the difference between the maximum and minimum values of the total pressure (rll - Rpb)> increasing the energy of 1 m3 of air passing through it by the value
The pressure created by the fan is used to overcome the resistance to air movement through the ducts:
Rveit \u003d DRvs + Drnagn. (XI.27)
Professor P. N. Kamenev suggested plotting pressure diagrams on the suction duct from absolute zero pressure (absolute vacuum). At the same time, the construction of lines pst. abs and rp. abs fully corresponds to the case of injection.
The pressure in the ducts is measured with a micromanometer. To measure the static pressure, the hose from the micromanometer is connected to a fitting attached to the duct wall, and to measure the total pressure, to a pneumometric Pitot tube, the opening of which is directed towards the flow (Fig. XI.4, a, b).
The difference between the total and static pressures is equal to the value of the dynamic pressure. This difference can be measured directly with a micromanometer, as shown in Fig. XI.4, c, d. The value of rd determines the speed, m / s:
V = V2prfp, (XI.28)
According to which the air flow in the duct is calculated, m3 / h:
L = 3600y/. (XI.29)
Pressure distribution in ventilation systems with natural induction of air movement. The features of such systems are the vertical arrangement of their channels in the building, low values of available pressures and, consequently, low speeds. The operation of systems with natural induction of air movement depends on the design features of the system and the building, the difference in density between the outdoor and indoor air, wind speed and direction. However, when choosing the design dimensions individual elements ventilation systems (sections of channels and shafts, areas of louvered grilles) it is enough to carry out a calculation for the case when the building does not affect the work.
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A - diagrams of absolute aerostatic pressures in the channel closed with plugs 1 - inside the channel; 2 - outside the channel; b - excess pressure diagram in the same channel; c - diagrams of overpressures for the movement of air through the channel; d - diagrams of excess pressures in the mine and in the "wide channel" attached to it; d-diagrams of excess pressures in the channel and shaft in the presence of a branch; e - excess pressure diagrams with natural induction of air movement in the ventilation system of a multi-storey building; g - diagrams of excess pressures with mechanical induction of air movement; (pst> Rp ~ lines, respectively, of static and total pressure inside the channel and shaft; Pn - line of static pressure outside the channel and shaft)
Let us consider the simplest case, when a vertical channel of height Yak filled with warm air with temperature tB, closed at the top and bottom with plugs. The channel is surrounded by outside air with temperature ta.
Let us assume that the pressure inside and outside the channel at the level of its top is equal to pa (to ensure this condition, it is enough to leave a small hole in the top plug). Then, in accordance with Pascal's law, the absolute pressure at any level (at a distance h from the top of the channel) is equal to: outside pst n=pa4-^pp£, and inside pstk=pa4--hpBg. The distribution of absolute pressures inside the channel (line 1) and outside it (line 2) is shown in Fig. XI.5, a.
In the "channel - ambient air" system, one can use conditional values of excess pressures, i.e., conditionally take the aerostatic pressure inside the channel at any level as zero. The diagram of these pressures outside the channel has the shape of a triangle (Fig. XI.5,6J. The base of the triangle
Drk = Hk Drg
Is the available pressure, Pa, which determines the movement of air through the channel.
When air moves through the channel (Fig. XI.5, c), pressure losses are the sum of losses at the inlet, friction and outlet. On fig. XI.5, c shows the distribution of total and static pressures (in excess pressure relative to conventional zero). The dynamic pressure pd is equal to the difference between pp and pst. The static pressure (its diagram is shaded in the figure) along the entire length of the channel is less than the excess aerostatic pressure outside the channel pH. In some cases, ZONES with Pst > pH can be observed in the channel. For example, in the channel before the constriction (Fig. XI.5, d), under certain conditions, the static pressure may exceed the pressure pH. Polluted air will leak through leaks in this area of the channel.
If vertical ventilation duct combines two (Fig. XI, 5, (3) or more (Fig. XI.5, e) branches, it is recommended to connect them not at the level of the air inlet to the branch, but somewhat higher (one, two floors or more). This recommendation is given in the light of accumulated operating experience.When connecting a branch at the level of point A instead of the level of point B, the available pressure Drotv increases (see Fig. XI.5, e), therefore, the resistance of the channel and the stability of the system also increase.
On fig. XI.5, e, f static pressure diagrams are shaded. The total pressure decreases in height to the value of the outlet loss, and the dynamic pressure at a constant channel cross section increases in height, since after the branch is connected, the flow rate in the channel increases.
AT recent times ventilation systems with vertical channels and mechanical induction of air movement are being introduced. In these systems, the air moves under the action of a fan and gravitational forces. The construction of the pressure distribution in such systems is similar to that considered above. The peculiarity lies in the fact that the static pressure in front of the fan is determined by the vacuum created by the fan (see the diagram in Fig. XI.5,g). In this case, the available pressure for air movement in the system
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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, one should calculate linear parameters air ducts.
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 arrangement 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 design section, R is the loss in running meter plot, L - total length air ducts on the site, Z - losses in the additional fittings of the ventilation system.
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 duct, L is the length of the calculated section, d is the diameter of the 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. The equivalent value of the circular 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 It is recommended 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.
