煤矿安全外文翻译文献

煤矿安全外文翻译文献

away from node i; EQij =-1 if branch j is connected to node i and the air ow goes into node i; EQij =0 if branch j is not connected to node i.

Let us assume that the mine ventilation network employs one main fan that is connected with the ambient outside of the mine. Also let node 1 be connected to the fan branch. Then the airow in the fan branch can be expressed as

n?eQmjQj?Qm,

j?1(4)

or

eQmQ=Qm, (5)

where Qm is airow quantity through fan (main) branch, eQm =[eQm1;:::;eQmn]is1×n matrix, includes the values of eQmj;j =1;:::;n are defined as follows: eQmj =1 if branch j is connected to node 1 and the air flow goes away from node 1;eQmj =-1 if branch j is connected to node 1 and the air ow goes into node 1;eQmj =0 if branch j is not connected to node 1.

Similarly, a mine ventilation network also satisfies Kirchho’s voltage law, i.e., the sum of the pressure drops around any loop in the networkmust be equal to zero, or mathematically,

?EHijHj?0j?1n, i=1,…,l-k

(6)

or

EHH=0, (7)

where Hj is the pressure drop of the branch j; l is a number of the links in the network, l=n-nc +1; H is a vector of pressure drops, EH is (l - k) × n fundamental mesh matrix, in which each mesh is formed by a link and a unique chain in the tree connecting two endpoints of the link, k is a number of meshes, containing fan branch, it is equal to the number of links, connected to the fan branch at its end. EH =[EHij], the elements of

煤矿安全外文翻译文献

EHij are defined as follows: EHij = 1 if branch j is contained in mesh i and has the same direction, EHij =-1 if branch j is contained in mesh i and has the opposite direction, EHij =0 if branch j is not contained in mesh i. Considering meshes, containing the fan branch, express the pressure drop in the fan branch as

?eHmjHj??Hmj?1n,i=1,…,k,

(8)

or

eHmH=-Hm, (9)

where Hm is the pressure drop of the fan branch, eHm is k×n matrix, includes the values of eHmij;j=1,…,n which defined as follows: eHmij =1 if branch j is contained in mesh i and has the same direction, eHmij =-1 if branch j is contained in mesh i and has the opposite direction, eHmij = 0 if branch j is not contained in mesh i. The dynamics of the fan branch can be expressed as

Hm (10)

where d denotes the equivalent pressure drop generated by fan, and Rm is the resistance coefficient in the fan branch. 2.2. Non-minimal model of the network

In order to establish the state equation, one has to find independent variables as states of the system. By virtue of the concepts of a tree and a link, they can easily be found. So the first step is to describe the tree of the mine ventilation network such that the fan branch is contained in it, and take the airflow quantities of link branches as state variables. For convenience of analysis, we label the air flow quantities of link branches from 1 to N -nc +1, where N =n+1. Define

=d-RmQm;

煤矿安全外文翻译文献

Q=

?Q1???????Qc??QN?nc?1???Q???Q?a??N?nc?2??????Q??n??, H=

?H1???????Hc??HN?nc?1???H???H?a??N?nc?2??????H??n??,

(11)

so that Qc and Hc matrices describe airflow quantity and pressure drop, respectively, in the links, and Qa and Ha matrices describe them in the tree branches, excluding the fan branch. With the notation

?Kc0? Q2D?(Q1Q1,...,QnQn),K?(K1,...,Kn)????0Ka? (1) can be rewritten as

???KQ2R?KHQD(13)

.

Proposition 2.1. There exist matrices A; B; C; YRQ;YQ and Yd of appropriate dimensions so that the full order model of mine ventilation network can be expressed as

??AQ2R?BQ?CdQD(14)

,

??YQ2R?YQ?YdQRQDQd(15)

where Q is the state, R and d are the inputs, and H is the output of the system. Proof. The matrices EH;EQ;eHm and eQm can be represented in the form:

EH=[EHcEHa], (16)

eHm (17)

=[eHmceHma],eQm

,

EQ=[EQcEQa],

=[eQmceQma],

煤矿安全外文翻译文献

where

?EHc??e??Il?l?Hmc?(18)

EQa?I(N?l?1)?(N?l?1),

, eQma =0.

(19)

Let us now express the tree airflow quantities through link airflows. From (3), (11) and (16), we have

?E(20)

QcEQa??Qc??Q??0?a? ,

With (19)

?1Qa??EQaEQcQc??EQcQc.

(21)

Now express the link pressure drops through the fan branch pressure drop and tree pressure drops. From (7), (9) and (17), we can get

?EH??EHc??EHa??0?H?H?H??e??e?c?e?a?1?Hm???Hm??Hmc??Hma?,

(22)

From this equation, using (18) one can find Hc as

?EHa??0?Hc????Ha??1?Hme???Hma?(23)

Using (10), rewrite (23)as

Hc?SHaHa?RmSQQ?Sdd,

,

(24) where