CN100405375C - Sampling pipe network design authentication and optimization method of air suction type smoke sensity fire detecting system - Google Patents

Abstract

The present invention relates to a design authentication and optimization method of a sampling pipe network of an air suction type smoke sense fire detecting system, which belongs to a fire alarm matching technology. The present invention relates to a matching design method for a super-early air suction smoke sense fire detecting system. The method of the present invention carries out the analog simulation of a sampling pipe network, and analyzes the sampling pipe network of the fire detector in the angle of fluid motion so as to establish a mathematical model of the sampling pipe network according to related hydromechanical principle; the length of sampling pipes. The distance of sampling holes, and the number of the sampling holes are defined according to the requirements of engineering application. Initial sample hole diameter is defined according to a set value. The present invention has the requirements that sampling points have uniform detection sensitivity, namely inspiratory capacity, and detection corresponding time which is not more than 120 seconds. When design requirements can not be realized, the design requirements can be preferentially realized by adjusting the hole diameter of each sampling hole.

Description

Sampling pipe network design verification of air suction type smoke detection system and optimization method

Technical field

The invention belongs to the fire protection warning supporting technology, particularly a kind of air suction type smoke fire detector engineering design of pipe networks checking and optimization method are the supporting method for designing of super early stage air suction type smoke fire detector.

Background technology

The air suction type smoke fire detector adopts the working method of active inspiration, because it can have higher detection sensitivity to smog, therefore, has obtained using widely in some special application scenarios.Air suction type smoke detection system mainly is made up of gas sampling pipe network and detection main frame two parts:

A. the effect of gas sampling pipe network is the air sample of gathering in the protected area.Thieff hatch is set on the gas sample tube, owing to survey the effect of asepwirator pump in the main frame, produces negative pressure in the pipe network, form a stable air-flow, the air sample in the protected area is drawn into thieff hatch, enters sonde body by sampling pipe network.

B. survey main frame the concentration of the smoke particle in the air sample that collects is analyzed comparison, report to the police when reaching the response threshold time stage.

The element of sampling pipe network is a sampling pipe.Sampling pipe has battle of some along duct wall, but the Direct Sampling air is thieff hatch, also can draw tubule, stretches into the position of desiring sampled air, is kapillary.System is delivered to air sampling on the detector by thieff hatch or kapillary, and the other end of sampling conduit is distal end cap (perforate), with the smog sensitivity of the different thieff hatchs of balance.As shown in Figure 1.

The inventive method is by carrying out analog simulation to sampling pipe network, according to hydromechanical relative theory, the fire detector sampling pipe network analyzed from the fluid motion angle set up mathematical model, by the engineering application requirements, determine the sampling pipe range, thieff hatch spacing and thieff hatch quantity, determine initial sampled aperture according to setting value, require all sampled points to have balanced detection sensitivity (inspiratory capacity) and be not more than 120 seconds corresponding time of detection (GB requirement), as do not reach designing requirement, preferentially realize by adjusting each thieff hatch aperture.

Current domestic do not have corresponding disposal system and method, external similar disposal system also can only be carried out simulation calculation for itself supporting equipment, and can not be applied to other products.

Air suction type smoke fire detector engineering design of pipe networks system and method provided by the invention can be realized following function:

1) shows the design of sampling pipe network in the mode of 3-D view;

2) can import the dxf file layout drawing of room or whole building;

3), and can make amendment to sampling pipe, thieff hatch, diameter capillaceous by graphic interface design detector, sample lines, thieff hatch, kapillary;

4) draw input variables such as bend pipe, pipe range, caliber, aperture, quantity capillaceous and length by figure, according to mathematics

Model can calculate each sampled point gas sample transmission time, throughput ratio and carry out optimal design;

5) graphic file, result of calculation can show, stores, print;

6) can realize the design on single tube road and multitube road;

7) provide detailed help file.

Summary of the invention

At the deficiencies in the prior art, the invention provides a kind of air suction type smoke fire detector engineering design of pipe networks checking and optimization method, supporting super early stage air suction type smoke fire detector uses.Because the perforate of air suction type detector sampling pipe network should keep the inspiratory capacity equalization, detection to whole space could keep optimum condition like this, should guarantee that simultaneously the gas sample transmission time is in 120 seconds, therefore need verify calculating to every data of sampling pipe network, if do not reach above-mentioned requirements, in the methods of the invention, then can realize the purpose of optimal design by adjusting the aperture.

The inventive method specific implementation step is as follows:

Step 1, input data;

Input is carried out the needed data of emulation to sampling pipe network, and promptly the intrinsic data of engineering are reappeared former design proposal in system, comprise following content:

1, the bending situation of sampling pipe, the length of every section sampling pipe;

2, the quantity of thieff hatch is positioned at the position of sampling pipe, the aperture of each thieff hatch;

3, quantity capillaceous is positioned at the position of sampling pipe, each aperture capillaceous and length;

4, the aperture of distal end cap.

Step 2, the intrinsic rationality of checking engineering, calculating comprises the inspiratory capacity of all sampled points of distal end cap, percent inspiration and inspiratory duration;

The needed ultimate principle of checking rationality is as follows:

1, the foundation of pipeline hydrodynamic analysis and mathematical model

If the kinematic parameter of fluid during through its occupied space each point do not change in time, so mobile steady flow that is called; Otherwise,, then be called unsteady fluid flow if kinematic parameter changes in time.Native system is because the gettering ability of detector asepwirator pump is limited, and gas flow rate is not too high in the pipeline, this gas can be considered as incompressible fluid, sampling pipe network is in case determine that fluid parameter is just definite in the pipeline, change in time simultaneously, be steady flow, meet the applicable elements of Bernoulli equation.

Bernoulli equation (Bernoulli) claim the fluid energy equation again, reflected that hydrodynamic is strong, the relation between flow velocity and position height, is a most important equation in the fluid dynamics.The Bernoulli equation of steady air flow is as follows:

${\mathrm{<figure-callout\; id="1"\; label="p"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">p</figure-callout>}}_1+\frac{\mathrm{\&rho;}{v_1}^2}{2}+g({\mathrm{\&rho;}}_a-\mathrm{\&rho;})(z_2-z_1)={\mathrm{<figure-callout\; id="2"\; label="p"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">p</figure-callout>}}_2+\frac{\mathrm{\&rho;}{v_2}^2}{2}+p_{l1-2}$

P in the formula ₁, P ₂---the relative pressure of two sections is called static pressure.It is not the pressure of stationary fluid, but the dynamic pressure that causes with the speed a kind of habitual appellation of analogy mutually.

G (ρ _a-ρ) (z ₂-z ₁)---the product of difference in height and difference of elevation is called the position and presses.The position is pressed and can just can be born.ρ _aBe atmospheric density, ρ is a gas density in the pipeline.

---dynamic pressure.

p _L1-2---the pressure loss between two sections.

Density difference is very little when between gas in the pipe and air, perhaps difference in height very hour, the position is pressed and can be ignored; The sampling pipe network of native system is located in the same horizontal plane substantially, and gas and atmospheric density are more or less the same in the pipe, can not consider that the position of fluid is pressed.Equation can be reduced to:

${\mathrm{<figure-callout\; id="1"\; label="p"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">p</figure-callout>}}_1+\frac{\mathrm{\&rho;}{v_1}^2}{2}={\mathrm{<figure-callout\; id="2"\; label="p"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">p</figure-callout>}}_2+\frac{\mathrm{\&rho;}{v_2}^2}{2}+p_{l1-2}---\left(1\right)$

In above-mentioned Bernoulli equation application conditions, require flow constant along journey, i.e. fluid free inflow or outflow between selected useful area, but in the application of sampling pipe network, need conflux by thieff hatch, in this case, can only list the Bernoulli equation of total stream by gross energy conservation and conversion rule.

By above analysis, the inventive method can be set up the mathematical model of the gas transmission of sampling pipe network by having the Bernoulli equation that confluxes.

1.1 the mathematical model of standard sample pipeline gas transmission

The calculation diagram of sampling pipe network as shown in Figure 2, sampling pipe is a prismatic air intake duct, its diameter is dd, sectional area is AA.End is opened a point 0, and the some footpath is d (0), area A (0); Sidewall is opened (n-1) individual point, and the diameter of point is respectively d (i), and the area of corresponding point is A (i).By getter device (drawing fan or asepwirator pump) join at sampling pipe and getter device (section n) locate to produce a vacuum tightness P _Jn

According to equation (1), be benchmark with atmosphere liquid level o '-o ', for the tube section at thieff hatch place (1,2 ... n-1), can list system of equations, suc as formula (2):

$\left.\begin{array}{c}{P}_j\left(0\right)+P_y\left(0\right)+P_d\left(0\right)=P_j\left(1\right)\\ ................................\\ P_j\left(0\right)+\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}{P}_y\left(j\right)+P_d(i-1)=P_j\left(i\right)\\ P_j\left(0\right)+\underset{j=0}{\overset{n-2}{\mathrm{\&Sigma;}}}{P}_y\left(j\right)+P_d(n-2)=P_j(n-1)\end{array}\right\}---\left(2\right)$

P in the formula _j---air is by the local pressure loss of thieff hatch, Pa;

P _d---the dynamic pressure on certain section sampling pipe, Pa;

P _y---along the pressure loss on the pipeline section, Pa.

And,

$P_d\left(i\right)=\frac{v^2\left(i\right)\&CenterDot;\mathrm{\&rho;}}{2}=\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_s\left(i\right)}{\mathrm{AA}}\right)}^2=\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{\underset{j=0}{\overset{i}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2---\left(3\right)$

ν in the formula---the air velocity on the sampling pipeline section i, m/s;

The density of ρ---air, Kg/m ³

Qs (i)---the air mass flow at sampling pipeline section i place, m ³/ s;

Q (i)---the inspiratory capacity of sampled point i, m ³/ s.

$P_j\left(i\right)=\mathrm{\&mu;}\&CenterDot;\frac{{v_r}^2\left(i\right)\&CenterDot;\mathrm{\&rho;}}{2}=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q\left(i\right)}{A\left(i\right)}\right)}^2---\left(4\right)$

μ in the formula---the coefficient of shock resistance of thieff hatch inlet,

V _r(i)---the air velocity of thieff hatch inlet, m/s.

$P_y\left(i\right)=(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(i\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{v^2\left(i\right)\&CenterDot;\mathrm{\&rho;}}{2}=(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(i\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{\underset{j=0}{\overset{i}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2---\left(5\right)$

The coefficient of friction resistance of λ in the formula---pipeline;

The coefficient of shock resistance of ξ---pipeline;

L (i)---cross section i is to the length in cross section (i+1), m.

So, can obtain about Q (0), Q (1) ... n-1 the standard sample hole equation of Q (n-1), as follows:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^2\left(0\right)}{2}+\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(j\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2+{\frac{\mathrm{\&rho;}}{2}\&CenterDot;\left(\frac{\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^2\left(i\right)}{2}---\left(6\right)$

1.2 the mathematical model of kapillary sample lines gas transmission

Kapillary sampling calculation diagram such as Fig. 3, the kapillary sampling is to connect the kapillary that diameter is d1 on the sidewall of the sampling pipe identical with standard sample, the kapillary sectional area is A1.Length capillaceous is L1 (i), and end capillaceous connects the plug that diameter is d (i).

Equally, according to the fluid mechanics energy conservation theory, be benchmark with atmosphere liquid level o '-o ', the tube section i for kapillary thieff hatch place can list equation, as shown in the formula 7:

$P_j\left(0\right)+\underset{j=0}{\overset{1}{\mathrm{\&Sigma;}}}{P}_y\left(j\right)+P_d\left(i\right)={P_j}^{,}\left(i\right)+{P_y}^{,}\left(i\right)+{P_d}^{,}\left(i\right)---\left(7\right)$

P in the formula _j' (i)---the local pressure loss at place, capillary inlet, Pa.

P _y' (i)---the pressure loss of kapillary pipeline, Pa.

P _d' (i)---by the dynamic pressure that kapillary calculates, Pa.

And,

${P_j}^{\&prime;}\left(i\right)=\mathrm{\&mu;}\&CenterDot;\frac{v^{\&prime;\&prime;2}\left(i\right)\&CenterDot;\mathrm{\&rho;}}{2}=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q\left(i\right)}{A\left(i\right)}\right)}^2---\left(8\right)$

ν in the formula "---the air velocity of kapillary i inlet, m/s.

${P_y}^{\&prime;}\left(i\right)=(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(i\right)}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">d</figure-callout>}1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^{\&prime;2}\left(i\right)}{2}=(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(i\right)}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">d</figure-callout>}1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q\left(i\right)}{A1}\right)}^2---\left(9\right)$

Air velocity in ν in the formula---the kapillary i, m/s.

${P_d}^{\&prime;}\left(i\right)=\frac{\mathrm{\&rho;}\&CenterDot;v^{\&prime;2}\left(i\right)}{2}=\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q\left(i\right)}{A1}\right)}^2---\left(10\right)$

All the other physical quantitys as mentioned above.

Can obtain equally about Q (0), Q (1) ... the kapillary equation of Q (n-1), as follows:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^2\left(0\right)}{2}+\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(j\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2$

$=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^{\&prime;\&prime;2}\left(i\right)}{2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(i\right)}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">d</figure-callout>}1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^{\&prime;2}\left(i\right)}{2}+\frac{v^{\&prime;2}\left(i\right)\&CenterDot;\mathrm{\&rho;}}{2}---\left(11\right)$

1.3 the mathematical model of the pressure loss in the pipeline

When gas flows in pipeline, be subjected to and the reciprocal fluid resistance of moving phase, consumed energy becomes the pressure loss.The pressure loss is divided into prolongs two kinds of stroke pressure loss and local pressure losses.

1.3.1 prolong the stroke pressure loss

Air prolongs stroke pressure loss owing to overcoming the pressure loss that viscous force causes, being called on whole flow process, suc as formula (12):

$P_m=\mathrm{\&lambda;}\&CenterDot;\frac{L}{d}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="V"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">V</figure-callout>}}^2}{2}---\left(12\right)$

λ in the formula---the coefficient of friction resistance also claims to prolong the journey resistance coefficient;

L---pipeline prolong Cheng Changdu;

D---internal diameter of the pipeline;

The mean flow rate of gas in V---the pipeline;

The density of ρ---air, Kg/m ³

1.3.2 shock resistance

When the local excessively pipe fitting of gas stream (as elbow etc.), and the size of gas flow rate or direction or both are all changed, cause the exchange and the vortex of local gas generation momentum, and consumed energy falls thereby produce local pressure.The calculating of general local pressure loss can be according to formula (13):

$P_z=\mathrm{\&xi;}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;{\mathrm{<figure-callout\; id="2"\; label="v"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">v</figure-callout>}}^2}{2}---\left(13\right)$

ξ in the formula---coefficient of shock resistance.

Coefficient of shock resistance ξ is relevant with shape, the reynolds number Re of pipeline, is obtained by experimental formula mostly.There are following several coefficient of shock resistance for this operating mode.

1) bend pipe

The coefficient of shock resistance of bend pipe is relevant with the ratio of the radius-of-curvature of bend pipe and caliber, can be according to the table 5-4 " value of elbow k " of " Hydro-mechanics ", the coefficient of shock resistance of bend pipe ${\mathrm{\&xi;}}_1=k\frac{{\mathrm{\&theta;}}^{*}}{90},$ Because the bend pipe that relates among the present invention all adopts 90 degree, so ξ ₁With k value together, check in local loss coefficient of bend in pipeline ξ ₁

2) section sudden enlargement

As shown in Figure 4, the coefficient of shock resistance of section sudden enlargement is relevant with the ratio of the area of two sections, can check in the coefficient of shock resistance ξ of section sudden enlargement according to table " coefficient of shock resistance of circular air channel conic diffuse " in " heating ventilator design manual " ₂

3) T shape interflow threeway

As shown in Figure 5, the coefficient of shock resistance of T shape interflow threeway and the length of trunk line and branch pipe(tube) are than relevant, this engineering is got the equal diameters of trunk line and branch pipe(tube), look into the table of " heating ventilator design manual " according to physical length ratio and " justify air channel T shape interflow threeway ", then the coefficient of shock resistance ξ of T shape interflow threeway ₃As shown in Table 1.

The coefficient of shock resistance ξ of table one T shape interflow threeway ₃

L3/L1	0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.0
												Branch pipe(tube) ξ 3′	-0.9	-0.52	-0.24	-0.08	0.32	0.42	0.57	0.72	0.86	0.99	1.1
Trunk line ξ 3″	0	0.16	0.27	0.38	0.46	0.53	0.57	0.59	0.6	0.59	0.55

4) conduit entrance

As shown in Figure 6, the main conduit entrance form that relates among the present invention has 3 kinds, is respectively distal end cap import, the import of sidewall thieff hatch, capillary end import.The coefficient of shock resistance ξ of distal end cap import, the import of sidewall thieff hatch, capillary end import ₄', ξ ₄", ξ ₄In " ', the actual engineering involved in the present invention, should meet the resistance coefficient of sharp-edged import, value 0.5.

1.4 detector gettering ability (static pressure) mathematical model

Air suction type detector and pipe network join, and by the fan generation static pressure of bleeding, detector is in case design is finished, and its gettering ability determines, can use curve representation at link to each other with the detector static pressure and the wind speed of place generation of pipe network end.The performance curve of selected detector as shown in Figure 7.

This performance curve is carried out the single order match.Obtain

$P=a\&CenterDot;V_s+b=a\&CenterDot;\frac{Q_s}{\mathrm{AA}}+b.---\left(14\right)$

P is a static pressure in the formula, V _sBe pipe network and detector junction wind speed, Q _sBe the pipe network total flow, AA is for being responsible for sectional area, and a, b represent fitting parameter in the formula, and available Matlab obtains a=-9.9968 to curve fit, b=82.5425.

The corresponding specific performance properties curve of the distribution of each pipe network, he and the intersection point of detector actual performance curve are exactly the detector working point of this kind of correspondence pipe network, the performance curve formula (15) of simultaneous pipe network and the performance curve formula (14) of detector,

P＝S·Q _s ² (15)

P is the static pressure in the sampling pipe in the formula;

Can obtain:

$S\&CenterDot;{Q_s}^2-a\&CenterDot;\frac{Q_s}{\mathrm{AA}}-b=0---\left(16\right)$

This is about Q _sThe Second-order Nonlinear Equations formula.Can use the flat-sawn method to find the solution Q to it _s

Use above-mentioned principle, as follows to the concrete proof procedure of air suction type smoke detection system sampling pipe network design rationality:

1, determines resistance coefficient.

The coefficient of friction resistance λ of match entrance, generally speaking, coefficient of friction resistance λ is reynolds number Re and relative roughness Function, under the operating mode of reality, because inhaled air contains the solid particle of part, and because cleaning period, pipeline is simplified to coarse pipe, λ gets between 0.025～0.05.

If certain section sampling pipe has bending, the coefficient of shock resistance ξ of this section ₁Be 0.131, if sampling pipe is a straight tube, coefficient of shock resistance ξ ₁Be 0.

According to 1.3.1 4) match entrance coefficient of shock resistance ξ ₄', determine on the density p=1.204Kg/m3, kapillary of air because the coefficient of shock resistance ξ that the caliber sudden change produces ₂=0.87, sampling pipe diameter d d=21mm, the calculating sampling tube section amasss AA=3.14dd ²/ 4 and area A (the i)=3.14d (i) of each sampled point ²/ 4.

2, the inspiratory capacity of calculating each sampled point accounts for the number percent QQ (i) and the pipe network impedance S of getter capacity.

The inspiratory capacity of supposing distal end cap is 1, and at this moment the flow of the 0th section sampling pipe promptly is the inspiratory capacity 1 of distal end cap.Inspiratory capacity based on distal end cap is 1, and for common thieff hatch, according to formula (6), for kapillary, the kapillary sampling is to connect diameter d 1 to be the kapillary of 8.5mm on the sidewall of the sampling pipe identical with standard sample, and the kapillary sectional area is A1.The general L1 of length capillaceous (i) is 1～6m, and it is the plug of 3～8mm that end capillaceous connects diameter d (i), according to formula 11, and the inspiratory capacity of other thieff hatchs of cycle calculations successively.Can obtain based on the distal end cap inspiratory capacity like this is 1, and the inspiratory capacity of each thieff hatch is Q (0), Q (1) ..., Q (n-1), total inspiratory capacity Q _s=Q (0)+Q (1)+... + Q (n-1), then the inspiratory capacity of each sampled point accounts for number percent QQ (i)=Q (the i)/Q of getter capacity _s

The detector performance curve determines that when the pipe network data were determined, calculated amount at this moment also should be a point on the detector performance curve.The Q that calculates when being 1 according to the inspiratory capacity of supposing the pipe network distal end cap _sAnd P, can calculate impedance S according to formula (15).

3, calculate the inspiratory duration of total inspiratory capacity Qs and each sampled point, verify whether each sampled point inspiratory capacity is impartial, and whether the sampling time meets the requirements.

According to formula (16) and above the impedance S that calculated, a, b value fits by Matlab and obtains, application flat-sawn method is found the solution Qs.

The basis of flat-sawn method is the interpolation principle, also is to make the linearizing a kind of method of nonlinear equation.

The computing formula of iteration function is as follows:

${Q_s}^{(k+1)}\left(i\right)={Q_s}^{\left(k\right)}\left(i\right)-\frac{f_i\left({Q_s}^{\left(k\right)}\left(i\right)\right)}{f_i\left({Q_s}^{\left(k\right)}\left(i\right)\right)-f_i\left({Q_s}^{(k-1)}\left(i\right)\right)}({Q_s}^{\left(k\right)}\left(i\right)-{Q_s}^{(k-1)}\left(i\right))---\left(17\right)$

During calculating, at first any two groups of initial value Q _s ⁰And Q _s ¹(process of iteration can be chosen arbitrarily) is updated to formula (17), thereby obtains iterative formula.Control iteration error＜10 ⁽¹³⁾Finally can obtain the inspiratory capacity Qs of detector, i.e. total inspiratory capacity of sampling pipe network.According to the percent inspiration of before having tried to achieve, trying to achieve from the inspiratory capacity of every of distal end cap is Q (i)=Qs * QQ (i).

Can calculate the inspiratory duration of each sampled point then, promptly each sampled point at first can be calculated the gas transmission time t of every section sampling pipe to the gas transmission time of detector, t=L (i)/v=L (i)/(Q _g(i)/AA), the gas flow Q in every section pipe wherein _g(i) for enter the flow sum of all sampled points of sampling pipe from its front, in the gas transmission time that can calculate every section sampling pipe successively,, can calculate the gas transmission time of distal end cap at last every period superposition successively.

Can verify design result by the percent inspiration of each sampled point and the gas transmission time of distal end cap then, whether the inspiratory capacity of judging each thieff hatch balance, can deduct minimum value with the percent inspiration maximal value of sampled point, if value is greater than 5%, can think air-breathing imbalance, should be optimized by the optimum gas mobile equilibrium, the gas transmission time by judging distal end cap whether within required time (as 80 seconds, 60 seconds etc., set by concrete engineering), determine whether and need be optimized by the delivery time.

Step 3, design is optimized;

(-) optimizes by the optimum gas mobile equilibrium:

1, design being optimized is to be based upon the calculating master routine to calculate on the basis of the total inspiratory capacity of sampling pipe network.The total inspiratory capacity of sampling pipe network is known, make the inspiratory capacity balance of each thieff hatch, inspiratory capacity Q that can the mean allocation thieff hatch _p=Qs/n.

2, make the inspiratory capacity of each thieff hatch as much as possible near Q _p, and rearrange the aperture of thieff hatch.Return the aperture d (i) of the thieff hatch after the optimization.

The specific implementation method is as follows:

Equating owing to set the inspiratory capacity of each sampled point, all is Q _pSo, can obtain inspiratory capacity Q (the 0)=Q (1) of each thieff hatch=...=Q (n-1)=Q _p, the flow of every section sampling pipe is Q respectively _s(0)=and Qp, Q _s(i)=(i+1) * Q _p, Q _s(n-1)=n * Q _p

By beginning from the nearest thieff hatch of detector, calculated hole diameters d (n-1), if the computing formula of common thieff hatch A (n-1) is:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{(n-1)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L(n-1)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{(n\&CenterDot;Q_p)}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{(n\&CenterDot;Q_p)}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}=P---\left(18\right)$

Can solve $d(n-1)=\sqrt{\frac{4\&CenterDot;A(n-1)}{\mathrm{\&pi;}}},$ P is the static pressure at thieff hatch place in the formula;

If kapillary then the computing formula of A (n-1) be:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{Q_p^2}{A{(n-1)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1(n-1)}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">d</figure-callout>}1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+(\frac{\mathrm{\&lambda;}\&CenterDot;L(n-1)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{(n\&CenterDot;Q_p)}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{(n\&CenterDot;Q_p)}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}$

$=P---\left(19\right)$

Can solve $d(n-1)=\sqrt{\frac{4\&CenterDot;A(n-1)}{\mathrm{\&pi;}}},$ P is the static pressure at kapillary place in the formula;

Then d (n-1) is integrated.The scope of common thieff hatch is an interval every 0.5mm from 2.5mm to 6mm; Diameter range capillaceous is an interval every 0.5mm from 3mm to 8mm.

Calculate the some footpath of other sampled points, at first calculate A (i), if i be common thieff hatch then computing formula as 20:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(i\right)}^2}+\frac{\mathrm{\&lambda;}\&CenterDot;L\left(i\right)}{\mathrm{dd}}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{((i+1)\&CenterDot;Q_p)}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{((i+1)\&CenterDot;Q_p)}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}=P_j(i+1)---\left(20\right)$

If i be kapillary then computing formula as 21:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{Q_p^2}{A{\left(i\right)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(i\right)}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">d</figure-callout>}1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(i\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{((i+1)\&CenterDot;Q_p)}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{((i+1)\&CenterDot;Q_p)}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}$

$=P_j(i+1)---\left(21\right)$

Known P in the formula _j(i+1) determine according to the concrete condition of i+1 sampled point, if i+1 sampled point is common thieff hatch, then:

$P_j(i+1)=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{(i+1)}^2}---\left(22\right)$

If i+1 sampled point is kapillary, then:

$P_j(i+1)=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{(i+1)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1(i+1)}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">d</figure-callout>}1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2---\left(23\right)$

So can solve $d\left(i\right)=\sqrt{\frac{4\&CenterDot;A\left(i\right)}{\mathrm{\&pi;}}}.$ Then d (i) is integrated.The scope of common thieff hatch is an interval every 0.5mm from 2.5mm to 6mm; Diameter range capillaceous is an interval every 0.5mm from 3mm to 8mm.

Calculate the some footpath of distal end cap, at first calculate A (0), if the 1st be common thieff hatch then computing formula as 24:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(0\right)}^2}+\frac{\mathrm{\&lambda;}\&CenterDot;L\left(0\right)}{\mathrm{dd}}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{{\mathrm{<figure-callout\; id="2"\; label="Q\; p"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">Q</figure-callout>}}_p}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{{\mathrm{<figure-callout\; id="2"\; label="Q\; p"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">Q</figure-callout>}}_p}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(1\right)}^2}---\left(24\right)$

If the 1st be kapillary then computing formula as 25:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(0\right)}^2}+\frac{\mathrm{\&lambda;}\&CenterDot;L\left(0\right)}{\mathrm{dd}}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{{\mathrm{<figure-callout\; id="2"\; label="Q\; p"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">Q</figure-callout>}}_p}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{{\mathrm{<figure-callout\; id="2"\; label="Q\; p"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">Q</figure-callout>}}_p}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(1\right)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(1\right)}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">d</figure-callout>}1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2---\left(25\right)$

So can solve $d\left(0\right)\sqrt{\frac{4\&CenterDot;A\left(0\right)}{\mathrm{\&pi;}}}\&CenterDot;$ Then d (0) is integrated, scope is an interval every 0.5mm from 2.5mm to 6mm.

3, calculate the inspiratory capacity of all sampled points that comprise distal end cap once more according to new d (i), percent inspiration and inspiratory duration, and the gas transmission time of percent inspiration by each sampled point and distal end cap verify optimizing the result.

(2) optimize by the delivery time:

If the maximum sampling time that the calculating master routine calculates surpasses allowed band, carry out time-optimized.

1, from distal end cap, make the sampling time of distal end cap the shortest, maximal value will be got in the aperture of thieff hatch, maximal value is got 6mm, at this moment the inspiratory capacity of distal end cap is assumed to be 1, for guaranteeing the inspiratory capacity balance of each thieff hatch, other thieff hatch inspiratory capacity hypothesis still is 1, and the diameter of other thieff hatchs is calculated.The aperture A (i) that calculates thieff hatch sees formula 26:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{1}{{A\left(0\right)}^2}+\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(j\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{(j+1)}^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{i^2}{{\mathrm{<figure-callout\; id="2"\; label="AA"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">AA</figure-callout>}}^2}=P_j\left(i\right)---\left(26\right)$

If i be common thieff hatch then computing formula as 27:

$P_j\left(i\right)=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{1}{A{\left(i\right)}^2}---\left(27\right)$

If i be kapillary then computing formula as 28:

$P_j\left(i\right)=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{1}{{A\left(i\right)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(i\right)}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="C20051004691400202.png"\; state="\{\{state\}\}">d</figure-callout>}1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{1}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{1}{A1}\right)}^2---\left(28\right)$

So can solve $d\left(i\right)=\sqrt{\frac{4\&CenterDot;A\left(i\right)}{\mathrm{\&pi;}}}\&CenterDot;$ Then d (i) is integrated, the scope of common thieff hatch is an interval every 0.5mm from 2.5mm to 6mm; Diameter range capillaceous is an interval every 0.5mm from 3mm to 8mm.

2, calculate the inspiratory capacity of all sampled points that comprise distal end cap once more according to new d (i), percent inspiration and inspiratory duration, and the gas transmission time of percent inspiration by each sampled point and distal end cap verify optimizing the result.

Step 4, carry out actual engineering operation by the data after optimizing;

Carry out actual engineering construction by the data after optimizing, can make air suction type smoke detection system be operated in optimum condition, whole design space is protected.

Description of drawings

Fig. 1 is the standard sample synoptic diagram;

Fig. 2 calculates synoptic diagram for sampling pipe;

Fig. 3 calculates synoptic diagram for the kapillary sampling;

Fig. 4 enlarges synoptic diagram for section;

Fig. 5 is T shape interflow threeway synoptic diagram;

Fig. 6 is the conduit entrance synoptic diagram;

Fig. 7 is a detector family curve synoptic diagram.

Embodiment

The specific embodiment of the present invention is as follows:

Input variable is total pipe range 50m, at 8 meters of distance distal end cap one bending is arranged, totally 10 thieff hatchs, and thieff hatch is 5m at interval, and the distal end cap diameter is 4mm, all the other thieff hatch diameter 3mm.

By formula 6,11,16,17 result of calculations are as follows:

Every hole inspiratory capacity (L/m): Q (0)=4.489

Q(1)＝2.53

Q(2)＝2.55

Q(3)＝2.586

Q(4)＝2.643

Q(5)＝2.726

Q(6)＝2.837

Q(7)＝2.979

Q(8)＝3.157

Q(9)＝3.372

Percent inspiration: QQ (0)=15.03%

QQ(1)＝8.47％

QQ(2)＝8.54％

QQ(3)＝8.66％

QQ(4)＝8.85％

QQ(5)＝9.13％

QQ(6)＝9.5％

QQ(7)＝9.98％

QQ(8)＝10.57％

QQ(9)＝11.29％

Gas transmission time: T (0)=87.22

T(1)＝64.09

T(2)＝49.29

T(3)＝38.44

T(4)＝29.89

T(5)＝22.87

T(6)＝16.95

T(7)＝11.85

T(8)＝7.4

T(9)＝3.48

Total inspiratory capacity: Qs=29.869L/m

The percent inspiration maximal value of result of calculation sampled point and the difference of minimum value are 6.56%, surpass 5%, can think to be optimized the air-flow imbalance by the optimum gas mobile equilibrium, if the regulation gas transmission time can not surpass 70 seconds, also should be optimized by the shortest inspiratory duration.

Optimize the back by the optimum gas mobile equilibrium by formula 18～25:

The aperture is 3mm.

Every hole inspiratory capacity (L/m): Q (0)=2.815

Q(1)＝2.818

Q(2)＝2.83

Q(3)＝2.856

Q(4)＝2.901

Q(5)＝2.971

Q(6)＝3.069

Q(7)＝3.201

Q(83)＝.369

Q(9)＝3.576

Percent inspiration: QQ (0)=9.26%

QQ(1)＝9.27％

QQ(2)＝9.31％

QQ(3)＝9.39％

QQ(4)＝9.54％

QQ(5)＝9.77％

QQ(6)＝10.09％

QQ(7)＝10.53％

QQ(8)＝11.08％

QQ(9)＝11.76％

Gas transmission time: T (0)=106.96

T(1)＝70.07

T(2)＝51.63

T(3)＝39.36

T(4)＝30.19

T(5)＝22.88

T(6)＝16.84

T(7)＝11.71

T(8)＝7.29

T(9)＝3.42

Total inspiratory capacity: Qs=30.405L/m

The percent inspiration maximal value of result of calculation sampled point and the difference of minimum value are 2.5%, can think air balance.Optimize the back by formula 26～28 by the time:

The aperture is respectively

d(0)＝6

d(1)＝6

d(2)＝6

d(3)＝5.5

d(4)＝5

d(5)＝4.5

d(6)＝4

d(7)＝4

d(8)＝3.5

d(9)＝3.5

Every hole inspiratory capacity (L/m): Q (0)=4.803

Q(1)＝4.847

Q(2)＝5.015

Q(3)＝4.513

Q(4)＝4.11

Q(5)＝3.732

Q(6)＝3.325

Q(7)＝3.743

Q(8)＝3.225

Q(9)＝3.606

Percent inspiration: QQ (0)=11.74%

QQ(1)＝11.85％

QQ(2)＝12.25％

QQ(3)＝11.03％

QQ(4)＝10.04％

QQ(5)＝9.12％

QQ(6)＝8.13％

QQ(7)＝9.15％

QQ(8)＝7.88％

QQ(9)＝8.81％

Gas transmission time: T (0)=64.97

T(1)＝43.35

T(2)＝32.59

T(3)＝25.51

T(4)＝20.09

T(5)＝15.63

T(6)＝11.79

T(7)＝8.37

T(8)＝5.32

T(9)＝2.54

Total inspiratory capacity: Qs=40.921L/m

The percent inspiration maximal value of result of calculation sampled point and the difference of minimum value are 4.37%, can think that air balance, maximum gas transmission time are 64.97 seconds, less than 70 seconds, meet the requirements.

Claims (1)

1. air suction type smoke detection system sampling pipe network design verification and optimization method is characterized in that the inventive method may further comprise the steps:

Step 1, input data;

Input is carried out the needed data of emulation to sampling pipe network, and promptly the intrinsic data of engineering are reappeared former design proposal in system, comprise following content:

1) the bending situation of sampling pipe, the length of every section sampling pipe;

2) quantity of thieff hatch is positioned at the position of sampling pipe, the aperture of each thieff hatch;

3) quantity capillaceous is positioned at the position of sampling pipe, each aperture capillaceous and length;

4) aperture of distal end cap;

Step 2, the intrinsic rationality of checking engineering, calculating comprises the inspiratory capacity of all sampled points of distal end cap, percent inspiration and inspiratory duration;

1) determines resistance coefficient;

2) inspiratory capacity of calculating each sampled point accounts for the number percent QQ (i) and the pipe network impedance S of getter capacity;

Sampling pipe is a prismatic air intake duct, and its diameter is dd, and sectional area is AA; End is opened a point 0, and the some footpath is d (0), area A (0); Sidewall is opened (n-1) individual point, and the diameter of point is respectively d (i), and the area of corresponding point is A (i);

The inspiratory capacity of supposing distal end cap is 1, and at this moment the flow of the 0th section sampling pipe promptly is the inspiratory capacity 1 of distal end cap, is 1 based on the inspiratory capacity of distal end cap, for common thieff hatch, according to formula

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^2\left(0\right)}{2}+\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(j\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^2\left(i\right)}{2}---\left(6\right)$

μ in the formula---the coefficient of shock resistance of thieff hatch inlet;

V---the air velocity on the sampling pipeline section i, m/s;

The density of ρ---air, Kg/m ³

Q (i)---the inspiratory capacity of sampled point i, m ³/ s;

The coefficient of friction resistance of λ---pipeline;

The coefficient of shock resistance of ξ---pipeline;

L (i)---cross section i arrives the length in cross section (i+1), m,

For kapillary, the kapillary sampling is to connect the kapillary that diameter is d1 on the sidewall of the sampling pipe identical with standard sample, and the kapillary sectional area is A1; Length capillaceous is L1 (i), and end capillaceous connects the plug that diameter is d (i); According to formula

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^2\left(0\right)}{2}+\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(j\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}Q\left(j\right)}{\mathrm{AA}}\right)}^2$

$=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^{\&prime;\&prime;2}\left(i\right)}{2}+\left(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(i\right)}{d1}\text{+\&xi;}\right)\&CenterDot;\frac{\mathrm{\&rho;}\&CenterDot;v^{\&prime;2}\left(i\right)}{2}+\frac{v^{\&prime;2}\left(i\right)\&CenterDot;\mathrm{\&rho;}}{2}---\left(11\right),$

V in the formula "---the air velocity of kapillary i inlet, m/s;

Air velocity in V '---the kapillary i, m/s;

λ---the coefficient of friction resistance also claims to prolong the journey resistance coefficient;

Qs (i)---the air mass flow at sampling pipeline section i place, m ³/ s,

The inspiratory capacity of each thieff hatch of cycle calculations successively; Be 1 based on the distal end cap inspiratory capacity then, the inspiratory capacity of each thieff hatch is Q (0), Q (1), Q (n-1), total inspiratory capacity Qs=Q (0)+Q (1)+... + Q (n-1), then the inspiratory capacity of each sampled point accounts for number percent QQ (i)=Q (the i)/Qs of getter capacity;

The detector performance curve determines that when the pipe network data were determined, calculated amount at this moment also should be a point on the detector performance curve, then according to formula

P=SQ _s ²(15) meter

Calculate impedance S; P is the static pressure in the sampling pipe in the formula;

3) inspiratory duration of the total inspiratory capacity Qs of calculating and each sampled point verifies whether each sampled point inspiratory capacity is impartial, and whether the sampling time meets the requirements;

The computing formula of iteration function is as follows:

${Q_s}^{(k+1)}\left(i\right)={Q_s}^{\left(k\right)}\left(i\right)-\frac{f_i\left({Q_s}^{\left(k\right)}\left(i\right)\right)}{f_i\left({Q_s}^{\left(k\right)}\left(i\right)\right)-f_i\left({Q_s}^{(k-1)}\left(i\right)\right)}({Q_s}^{\left(k\right)}\left(i\right)-{Q_s}^{(k-1)}\left(i\right))---\left(17\right)$

At first choose any two groups of initial value Q _s ⁰And Q _s ¹Be updated to formula (17), thereby obtain iterative formula, control iteration error＜10 ⁽¹³⁾, finally obtain the inspiratory capacity Qs of detector, i.e. total inspiratory capacity of sampling pipe network; According to the percent inspiration of before having tried to achieve, trying to achieve from the inspiratory capacity of every of distal end cap is Q (i)=Qs * QQ (i);

Calculate the inspiratory duration of each sampled point then, promptly each sampled point is at first calculated the gas transmission time t of every section sampling pipe to the gas transmission time of detector, t=L (i)/v=L (i)/(Q _g(i)/AA), the gas flow Q in every section pipe wherein _g(i) for enter the flow sum of all sampled points of sampling pipe from its front, in the gas transmission time of calculating every section sampling pipe successively,, calculate the gas transmission time of distal end cap every period superposition successively;

By the percent inspiration of each sampled point and the gas transmission time of distal end cap design result is verified at last, whether the inspiratory capacity of judging each thieff hatch balance, percent inspiration maximal value with sampled point deducts minimum value, if value is greater than 5%, think air-breathing imbalance, should be optimized by the optimum gas mobile equilibrium, whether the gas transmission time by judging distal end cap within the engineering required time, determines whether and need be optimized by the delivery time;

Step 3, design is optimized;

(1) optimize by the optimum gas mobile equilibrium:

1) the total inspiratory capacity of sampling pipe network is known, make the inspiratory capacity balance of each thieff hatch, the inspiratory capacity Q of mean allocation thieff hatch _p=Qs/n;

2) make the inspiratory capacity of each thieff hatch near Q _p, and rearrange the aperture of thieff hatch, return the aperture d (i) of the thieff hatch after the optimization;

By begin calculated hole diameters d (n-1) from the nearest thieff hatch of detector:

If the computing formula of common thieff hatch A (n-1) is:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{(n-1)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L(n-1)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{\left(n{\&CenterDot;Q}_p\right)}^2}{A{A}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{(n\&CenterDot;Q_p)}^2}{A{A}^2}=P---\left(18\right)$

Then $d(n-1)=\sqrt{\frac{4\&CenterDot;A(n-1)}{\mathrm{\&pi;}}},$ P is the static pressure at thieff hatch place in the formula;

If kapillary then the computing formula of A (n-1) be:

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{(n-1)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1(n-1)}{d1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+(\frac{\mathrm{\&lambda;}\&CenterDot;L(n-1)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{(n\&CenterDot;Q_p)}^2}{A{A}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{(n\&CenterDot;Q_p)}^2}{A{A}^2}$

$=P---\left(19\right)$

Then $d(n-1)=\sqrt{\frac{4\&CenterDot;A(n-1)}{\mathrm{\&pi;}}},$ Then d (n-1) is integrated; P is the static pressure at kapillary place in the formula;

Calculate the some footpath of other sampled points, at first calculate A (i):

If i be common thieff hatch then computing formula as (20),

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(i\right)}^2}+\frac{\mathrm{\&lambda;}\&CenterDot;L\left(i\right)}{\mathrm{dd}}\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{((i+1)\&CenterDot;Q_p)}^2}{A{A}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{((i+1)\&CenterDot;Q_p)}^2}{A{A}^2}=P_j(i+1)---\left(20\right)$

P in the formula _J---air is by the local pressure loss of thieff hatch, Pa, if i be kapillary then computing formula as (21),

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(i\right)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(i\right)}{d1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(i\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{((i+1)\&CenterDot;Q_p)}^2}{A{A}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{((i+1)\&CenterDot;Q_p)}^2}{A{A}^2}$

$=P_j(i+1)---\left(21\right)$

Known P in the formula _J(i+1) determine according to the concrete condition of i+1 sampled point, if i+1 sampled point is common thieff hatch, then:

$P_j(i+1)=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{(i+1)}^2}---\left(22\right)$

If i+1 sampled point is kapillary, then:

$P_j(i+1)=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{(i+1)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1(i+1)}{d1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2---\left(23\right)$

Then solve $d\left(i\right)=\sqrt{\frac{4\&CenterDot;A\left(i\right)}{\mathrm{\&pi;}}},$ Then d (i) is integrated;

Calculate the some footpath of distal end cap, at first calculate A (0):

If the 1st be common thieff hatch then computing formula as (24):

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(0\right)}^2}+\frac{\mathrm{\&lambda;}\&CenterDot;L\left(0\right)}{\mathrm{dd}}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{A}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{A}^2}=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(1\right)}^2}---\left(24\right)$

If the 1st be kapillary then computing formula as (25):

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(0\right)}^2}+\frac{\mathrm{\&lambda;}\&CenterDot;L\left(0\right)}{\mathrm{dd}}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{A}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{A}^2}=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{Q_p}^2}{A{\left(1\right)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(1\right)}{d1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{Q_p}{A1}\right)}^2---\left(25\right)$

Then solve $d\left(0\right)=\sqrt{\frac{4\&CenterDot;A\left(0\right)}{\mathrm{\&pi;}}},$ Then d (0) is integrated;

3) calculate inspiratory capacity, percent inspiration and the inspiratory duration of all sampled points that comprise distal end cap once more according to new d (i), and verify once more optimizing the result by the percent inspiration of each sampled point and the gas transmission time of distal end cap;

(2) optimize by the delivery time:, carry out time-optimized if the maximum sampling time surpasses allowed band;

1) from distal end cap, make the sampling time of distal end cap the shortest, maximal value will be got in the aperture of thieff hatch, at this moment the inspiratory capacity of distal end cap is assumed to be 1, for guaranteeing the inspiratory capacity balance of each thieff hatch, other thieff hatch inspiratory capacity hypothesis still is 1, and the diameter of other thieff hatchs is calculated, and the aperture A (i) that calculates thieff hatch sees formula (26):

$\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{1}{A{\left(0\right)}^2}+\underset{j=0}{\overset{i-1}{\mathrm{\&Sigma;}}}(\frac{\mathrm{\&lambda;}\&CenterDot;L\left(j\right)}{\mathrm{dd}}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{{(j+1)}^2}{A{A}^2}+\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{i^2}{A{A}^2}=P_j\left(i\right)---\left(26\right)$

If i be common thieff hatch then computing formula as (27):

$P_j\left(i\right)=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{1}{A{\left(i\right)}^2}---\left(27\right)$

If i be kapillary then computing formula as (28):

$P_j\left(i\right)=\mathrm{\&mu;}\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;\frac{1}{A{\left(i\right)}^2}+(\frac{\mathrm{\&lambda;}\&CenterDot;L1\left(i\right)}{d1}+\mathrm{\&xi;})\&CenterDot;\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{1}{A1}\right)}^2+\frac{\mathrm{\&rho;}}{2}\&CenterDot;{\left(\frac{1}{A1}\right)}^2---\left(28\right)$

Then solve $d\left(i\right)=\sqrt{\frac{4\&CenterDot;A\left(i\right)}{\mathrm{\&pi;}}},$ Then d (i) is integrated;

2) calculate the inspiratory capacity of all sampled points that comprise distal end cap once more according to new d (i), percent inspiration and inspiratory duration, and verify optimizing the result by the percent inspiration of each sampled point and the gas transmission time of distal end cap;

Step 4, carry out actual engineering operation by the data after optimizing;

Carry out actual engineering construction by the data after optimizing, make air suction type smoke detection system be operated in optimum condition.

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