3. Block diagram of the system
To analyze the functioning of the system in case of failure of some of its elements should be based on the concept to develop a block diagram based on serial or parallel connection of elements.
Connection type elements in the block diagram is affected by failures of individual elements on the performance of the whole system.
Serial communication is done then, if component failure leads to a malfunction of the entire system.
Parallel connection element is satisfied if, when the failure of one of the elements does not lead to malfunction of the entire system, as parallel connected element starts to act as the element that refused (principle of redundancy).
Since modern functional systems are a complex set of interconnected units, it is advisable to pre-system divided into functional blocks. Grouping units (elements) in blocks should be carried out in accordance with their functional connections, and then a block diagram detailing the elements.
|
I |
|
II |
|
|
Figure 3 Structural block diagram of the fuel system
1I 1II 2II
I II
Figure 4 Structural (detailed) scheme of fuel system
Block I: 1I fuel tank;.
Block II: 1II - booster pump. 2II - check valve.
4. Tree of events for r2 situation
The difficult situation R2 (Figure 5)in the air can occur in the event of failure of one of three engines engine G2П G2С , G2Л. In turn, the engine failure may occur at the termination of the fuel supply pump (Z1), failure (Z2), damage to the pipeline feeding fuel (Z3) or joint failure of the filter element (Z4) and bypass valve (Z5).
Figure 5. tree of events for R2 occurrence
Using the tree events that make up the minimum cross-sections for the difficult situation of flight R2. The first, second and third minimum cross section in accordance with the tree of events includes a filter element (Z4) and the bypass valve (Z5), causing the event of failure of all three engines filter G2.1. In the rest of the minimum cross-section composed of one order of elements Z1, Z2, Z3 for three engines.
At a minimum cross sections build computational model for determining the probability of a functional failure of the system, resulting in a difficult situation R2.
QR2(t)=QZ4П* QZ5П + QZ4С* QZ5С + QZ4Л* QZ5Л+ QZ1П + QZ2П + QZ3П + QZ1С + QZ2С + QZ3С + QZ1Л + QZ2Л + QZ3Л = 0,86*10-3 * 1,44*10-3 * 3+3*0,86*10-4+3*1,44*10-4+3*1,1*10-4=1,06*10-3
5. Calculation of non failure operation indices of system and its elements
Failure rate for the first element in the block is λ1 = 4.3˖10-4, for the second element is λ2 = 7.2 ˖10-4, for the third element equals λ3 = 4.3 ˖10-5.
Probabilities of non-failure work during time t = 4h taking into account exponential distribution, for 4 elements obtain:
P1=
;
P2=
P3=
;
Q1=1-P1=1-
;
Q2=1-P2=1-
Q3=1-P3=1-
.
PI(1,2,3)=P1*P2*P3=
*
*
=0.999522
PII(
=Q1*P2*P3=0.00017985*
*
=0.179795*10-3
PIII(1,
,3)=P1*Q2*P3=0.999828015*0.000287959*0.9999828=0.287904*10-3
PIV(1,2,
)=P1*P2*Q3=0.999828015*0.999712041*0.0000172=0.171920*10-4
PV(
,
,3)=
Q1*
Q2*
P3=0.17985*10-3*0.287959*10-3*0.9999828=0.0517*10-6
PVI(
,2,
)=
Q1*
P2*
Q3=0.17985*10-3*0.999712041*0.0172*10-3=0.00309*10-6
PVII(1,
,
)=
P1*
Q2*
Q3=0.999828015*0.287959*10-3*0.0172*10-3=0.00495*10-6
PVIII(
,
,
)=
Q1*
Q2*
Q3=0.17985*10-3*0.287959*10-30.0172*10-3=0.00089*10-9