Calculation of aerodynamic drag. Determination of the coefficients of local resistances of tees in ventilation systems KMS of the ventilation grill

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With this material the editors of the magazine "Climate World" continue to publish chapters from the book "Ventilation and Air Conditioning Systems. Design guidelines for production
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 area to the fan. If in doubt, when determining the direction, all possible options are calculated.

The calculation starts from a remote area: the diameter D (m) of a round or area F (m 2) of the cross-section of a rectangular duct is determined:

The speed increases as you approach the fan.

According to Appendix H, the nearest standard values \u200b\u200bare 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 at the duct section.

Local resistances at the border of two sections (tees, crosses) are referred to a section with a lower flow rate.

Local resistance coefficients are given in the appendices.

Diagram of the supply ventilation system serving a 3-storey administrative building

Calculation example

Initial data:

No. of sites	feed 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 Δp, pa
PP grating 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 dimensions of which correspond to the app. H from. The material of the air intake shaft is brick. As air distributors used adjustable grilles of the PP type with possible cross-sections: 100 x 200; 200 x 200; 400 x 200 and 600 x 200 mm, shading coefficient 0.8 and maximum outlet air speed up to 3 m / s.

The resistance of the intake insulated valve with fully open blades is 10 Pa. The hydraulic resistance of the heating installation is 100 Pa (according to a separate calculation). Resistance of the G-4 filter 250 Pa. The hydraulic resistance of the muffler is 36 Pa (according to the acoustic calculation). Rectangular air ducts are designed based on architectural requirements.

The sections of the brick channels are taken according to the table. 22.7.

Local resistance coefficients

Section 1. PP grating at the outlet with a section of 200 × 400 mm (calculated separately):

No. of sites	Local resistance type	Sketch	Angle α, deg.	Attitude			Justification	CCM
No. of sites	Local resistance type	Sketch	Angle α, deg.	F 0 / F 1	L 0 / L st	f pass / f st	Justification	CCM
1	Diffuser		20	0,62	—	—	Tab. 25.1	0,09
	Diversion		90	—	—	—	Tab. 25.11	0,19
	Tee-passage		—	—	0,3	0,8	Adj. 25.8	0,2
							∑ =	0,48
2	Tee-passage		—	—	0,48	0,63	Adj. 25.8	0,4
3	Branch tee		—	0,63	0,61	—	Adj. 25.9	0,48
4	2 branches	250 × 400	90	—	—	—	Adj. 25.11
	Diversion	400 × 250	90	—	—	—	Adj. 25.11	0,22
	Tee-passage		—	—	0,49	0,64	Tab. 25.8	0,4
							∑ =	1,44
5	Tee-passage		—	—	0,34	0,83	Adj. 25.8	0,2
6	Diffuser after fan		h \u003d 0.6	1,53	—	—	Adj. 25.13	0,14
	Diversion	600 × 500	90	—	—	—	Adj. 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
	Louvered grille						Tab. 25.1	1,3
							∑ =	1,44
Table 2. Determination of local resistances

Yu.S. Krasnov,

„Ventilation and air conditioning systems. Design guidelines for industrial and public buildings ", chapter 15." Thermocool "

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Aerodynamic calculation of air ducts begins with drawing an axonometric scheme M 1: 100, setting the numbers of sections, their loads L m / h, and lengths 1, m. The direction of the aerodynamic calculation is determined - from the most remote and loaded section to the fan. If in doubt, all possible options are calculated when determining the direction.

The calculation starts from a remote area, its diameter is calculated D, m, or

Square cross-section of a rectangular duct P, m:

System start at fan

Administrative buildings 4-5 m / s 8-12 m / s

Industrial buildings 5-6 m / s 10-16 m / s,

Increasing as you approach the fan.

Using Appendix 21, we take the nearest standard values \u200b\u200bof Dst or (a x b) st

Then we calculate the actual speed:

2830 * d;

Or ———————— ———— -, m / s.

FACT 3660 * (a * 6) st

For further calculations, we determine the hydraulic radius of rectangular ducts:

£\u003e 1 \u003d -, m. a + b

To avoid using tables and interpolating the values \u200b\u200bof specific friction losses, we use a direct solution to the problem:

We define the Reynolds criterion:

Re \u003d 64 100 * Ost * Ufact (for rectangular Ost \u003d Ob) (14.6)

And the coefficient of hydraulic friction:

0.3164 * Re 0 25 at Re< 60 ООО (14.7)

0.1266 * Ne 0167 for Re\u003e 60,000. (14.8)

The pressure loss in the calculated area will be:

Where KMS is the sum of the coefficients of local resistances in the duct section.

Local resistances lying on the border of two sections (tees, crosses) should be attributed to a section with a lower flow rate.

Local resistance coefficients are given in the appendices.

Initial data:

Duct material - galvanized sheet steel, thickness and dimensions in accordance with App. 21.

The material of the air intake shaft is brick. As air diffusers, PP type adjustable grilles with possible cross-sections are used:

100 x 200; 200 x 200; 400 x 200 and 600 x 200 mm, shading coefficient 0.8 and maximum outlet air speed up to 3 m / s.

The resistance of the receiving insulated valve with fully open blades is 10 Pa. The hydraulic resistance of the air heater 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 the acoustic calculation). Based on architectural requirements, air ducts are designed with rectangular cross-sections.

	Delivery L, m3 / h	Length 1, m	Section a * b, m	Losses in the section p, Pa

PP grille at the outlet

250 × 250 b \u003d 1030

500 × 500 \u003d Lc \u003d 6850

L_ 0.5 * 0.5 / s 0.6 * 0.5

Appointment	Basic requirement
	Noiselessness	Min. head loss
	Trunk channels	Main channels		Branches
	Trunk channels	Inflow	Hood	Inflow	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 ducts should be calculated.

Algorithm for calculating the loss of air pressure

The calculation must begin with drawing up a diagram of the ventilation system with the obligatory indication of the spatial arrangement of air ducts, the length of each section, ventilation grilles, additional equipment for air purification, technical fittings and fans. Losses are determined first for each separate line, and then they are summed up. For a separate technological section, the losses are determined using the formula P \u003d L × R + Z, where P is the air pressure loss in the calculated section, R is the losses per running meter of the section, L is the total length of the air ducts in the section, Z is the losses in the additional system fittings ventilation.

To calculate the pressure loss in a circular duct, the formula Ptr is used. \u003d (L / d × X) × (Y × V) / 2g. X is the tabular coefficient of air friction, depends on the material 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 the temperature, g is the acceleration of falling (free). If the ventilation system has square ducts, then table No. 2 should be used to convert round values \u200b\u200bto 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.

The air pressure losses in the bends are taken from table 3.

Tab. No. 3. Pressure loss at bends

To determine the pressure loss in the diffusers, the data from Table 4 are used.

Tab. No. 4. Pressure loss in diffusers

Table 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 this section of the duct are summed up and corrected with table No. 6. Tab. No. 6. Calculation of the decrease in flow pressure in ventilation systems

During design and calculations, existing regulations recommend that the difference in the magnitude of pressure losses between individual sections does not exceed 10%. The fan should be installed in the section of the ventilation system with the highest resistance, the farthest air ducts should have the lowest resistance. If these conditions are not met, then it is necessary to change the layout of the air ducts and additional equipment, taking into account the requirements of the provisions.

The calculation starts from a remote area: the diameter D (m) of a round or area F (m 2) of the cross-section of a rectangular duct is determined:

The speed increases as you approach the fan.

According to Appendix H, the nearest standard values \u200b\u200bare 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 at the duct section.

Local resistances at the border of two sections (tees, crosses) are referred to a section with a lower flow rate.

Local resistance coefficients are given in the appendices.

Diagram of the supply ventilation system serving a 3-storey administrative building

Calculation example

Initial data:

No. of sites	feed 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 Δp, pa
PP grating 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 sections of the brick channels are taken according to the table. 22.7.

Local resistance coefficients

Section 1. PP grating at the outlet with a section of 200 × 400 mm (calculated separately):

No. of sites	Local resistance type	Sketch	Angle α, deg.	Attitude			Justification	CCM
No. of sites	Local resistance type	Sketch	Angle α, deg.	F 0 / F 1	L 0 / L st	f pass / f st	Justification	CCM
1	Diffuser		20	0,62	-	-	Tab. 25.1	0,09
	Diversion		90	-	-	-	Tab. 25.11	0,19
	Tee-passage		-	-	0,3	0,8	Adj. 25.8	0,2
							∑ =	0,48
2	Tee-passage		-	-	0,48	0,63	Adj. 25.8	0,4
3	Branch tee		-	0,63	0,61	-	Adj. 25.9	0,48
4	2 branches	250 × 400	90	-	-	-	Adj. 25.11
	Diversion	400 × 250	90	-	-	-	Adj. 25.11	0,22
	Tee-passage		-	-	0,49	0,64	Tab. 25.8	0,4
							∑ =	1,44
5	Tee-passage		-	-	0,34	0,83	Adj. 25.8	0,2
6	Diffuser after fan		h \u003d 0.6	1,53	-	-	Adj. 25.13	0,14
	Diversion	600 × 500	90	-	-	-	Adj. 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
	Louvered grille						Tab. 25.1	1,3
							∑ =	1,44
Table 2. Determination of local resistances

Yu.S. Krasnov,

1. Friction loss:

Ptr \u003d (x * l / d) * (v * v * y) / 2g,

z \u003d Q * (v * v * y) / 2g,

Method of admissible speeds

Note: the air flow rate in the table is in meters per second.

Using rectangular ducts

The head loss diagram shows the diameters of the circular 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), rectangular ducts are chosen. As a rule, the width of the duct is 2 times the height).

Table of equivalent duct diameters

When the parameters of the air ducts are known (their length, cross-section, coefficient of air friction against the surface), it is possible to calculate the pressure loss in the system at the projected air flow rate.

The total pressure loss (in kg / m2) is calculated using the formula:

where R is the pressure loss due to friction per 1 running 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 cross-section).

1. Friction loss:

In a round duct, the friction pressure loss P tr is calculated as follows:

Ptr \u003d (x * l / d) * (v * v * y) / 2g,

where x is the coefficient of frictional drag, 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 acceleration of gravity (9 , 8 m / s2).

Note: If the duct has a rectangular, rather than circular cross section, the equivalent diameter must be substituted into the formula, which for a duct with sides A and B is equal to:

2. Losses for local resistance:

Pressure losses on local resistances are calculated by the formula:

z \u003d Q * (v * v * y) / 2g,

where Q is the sum of the local resistance coefficients in the duct section 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 acceleration of gravity (9.8 m / s2 ). The Q values \u200b\u200bare tabulated.

Method of admissible speeds

When calculating a network of air ducts using the method of permissible speeds, the optimal air speed is taken as the initial data (see table). Then the required section of the duct and the pressure loss in it are considered.

The procedure for aerodynamic calculation of air ducts using 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 farthest and most loaded areas from the fan.

Knowing the optimal air speed for a given room and the volume of air passing through the duct in 1 hour, we determine the appropriate diameter (or section) of the 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 for 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 consistently link all the 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 sections of air ducts, increasing the resistance.

Maximum air speed table depending on duct requirements

Constant head loss method

This method assumes a constant head loss per 1 running meter of the duct. The dimensions of the duct network are determined based on this. 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, the speed on the main section of the air duct is selected.

According to the speed defined in clause 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 done in the diagram below.

The most loaded branch is determined and its length is taken as the equivalent length of the air distribution system. This is most often the distance to the farthest diffuser.

Multiply the equivalent system length by the head loss from item 2. The pressure loss on the diffusers is added to the value obtained.

Now, according to the diagram below, the diameter of the initial air duct coming from the fan is determined, and then the diameters of the remaining sections of the network according to the corresponding air flow rates. In this case, a constant initial head loss is taken.

Diagram for determining the head loss and diameter of air ducts

The head loss diagram shows the diameters of the circular 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 duct width is 2 times the height).

In the table, the horizontal indicates the duct height in mm, the vertical indicates its width, and the table cells contain the equivalent duct diameters in mm.