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Content
1. Basic theoretical information………………………………………………………… 2
2. Supplemential information………………………………………………………….... 3
3. Design information…………………………………………………………………… 4
4. Algorithm of finding a malfunction…………………………………………………... 5
5. Conclusion……………………………………………………………………………...7
6. Literature……………………………………………………………………………….8
Basic theoretical information
Hydraulic system of Tu-154 consists of main HS and emergency HS. Functions of the main HS are:
- landing gear extension and retraction
- flaps extension and retraction
- landing gear brake
- nose landing gear guidance
- opening and closing of landing gear doors
- cargo door controlling
Functions of the emergency HS are:
- emergency flaps extension
- emergency brake
- emergency door opening
- emetgency cargo door controlling
Nominal pressure of main HS is 120…155 kgf/cm2, of emergency HS – 160…175 kgf/cm2. The pressure source of main HS is 2-а 623АНМ with engine actuator, for emergency HS – HC-14.
Hydro turbo pumps drive pumping and supplying fuel to the engine is a hydraulic turbine. The working fluid for the turbine is a high-pressure fuel supplied to the turbine nozzle assembly of the two drive centrifugal pumps SDS-80, installed one on each remote box engine units. Fuel supplied to the pump inlet SDS-80 and going on to drive hydro turbo pumps, taken from the fuel supply line to the engines.
The urgent requirements of engineering practice, therefore, soon gave rise to a new science of fluid motion, hydraulics, in which researchers took to the second path, that of extensive experimenting and accumulation of factual data for application to engineering problems. At its origin, hydraulics was a purely empirical science. Today, though, whenever necessary it employs the methods of theoretical hydromechanics for the solution of various problems; and, conversely, in theoretical hydromechanics experiments are widely used to verify the validity of its conclusions. Thus, the iifference in the methods employed by either science is gradually lisappearing, and with it the boundary line between them.
Supplemental information
Table 1. supplemental information
№ | Units | Typical failures and malfunctions of units (diagnosis ) | Typical features of failures and malfunctions () |
Hydraulic pumps | Destruction of plungers and gears. Seal failure of pistons. Spring failure or shrink. Unbalance. | Response time deviation of actuating units. Pressure drop of hydraulic system. Outside noise. Pump overheat. Frequent hydraulic pump off-loading. Metal chips appearance in filters. | |
Hydraulic pump off-loading device | Slide valve coupling jamming. Spring failure or shrink. Seal failure. Jamming of control and check valves. | Deviation of working parameters of actuating units. Frequent hydraulic pump off-loading. Pressure drop of hydraulic system. Hydraulic fluid leakage. | |
Check valves | Spring failure. Valves jamming in opened or closed position. Seal failure. | Rapid pressure drop of hydraulic system. Frequent hydraulic pump off-loading. Overfilling of hydraulic tanks. Deviation of working parameters of actuating units. | |
Filters
| Filters clogging. Filter screen rupture. Seal failure. Wiring malfunction. Spring failure or shrink. Wear of friction pair. | Actuating units clogging. Pressure drop of hydraulic system. Signal lamp actuation. Deviation of working parameters of actuating units. Frequent hydraulic pump off-loading. | |
Slide valve distributor
| Unbalance. Seal failure. Wear of seals. Slide valve jamming. | Metal chip appearance on filters. Increasing of hydraulic fluid flowing. Hydraulic fluid leakage. | |
Hydraulic cylinders | Wear of seals. Complication of actuating of friction pairs. Spring failure. Valve jamming. | Failure of actuating units. Deviation of working parameters of actuating units. | |
Hydraulic accumulators | Seal failure of gas chamber. Seal failure of membranes. Seal failure of fluid chamber. Cup wearing. | Pressure drop of hydraulic system. Frequent hydraulic pump off-loading. Hydraulic fluid leakage. Low pressure of nitrogen. Deviation of working parameters of actuating units. |
Let’s mark the following features:
- frequent hydraulic pump off-loading
- deviation of working parameters of actuating units
- pressure drop of hydraulic system
- actuating units clogging
- metal chip appearance on the filters
- failure of actuating units
- hydraulic fluid leakage.
Using these features together with a 7 different diagnosis, which are marked on the table, taking the quantity of diagnosis occurrence and calculating the probability of diagnosis occurrence, we can build a table 1:
Design information
Table 2 Design information
Name of the unit | Diagnosis () | Malfunction features () | Quantity of diagnosis occurrence () | Probability of diagnosis occurrence () | ||||||
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Hydraulic pump | - destruction of plungers or gears | + |
| + |
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| 0.233 | |
Hydraulic pump off-loading device | - slide valve coupling jamming |
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| + |
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| + | 0.209 | |
Check valves | - valves jamming |
| + |
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| 0.186 | |
Filters | - filters clogging |
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| + |
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| 0.07 | |
Slide valve distributor | - slide valve jamming |
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| + |
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| 0.116 | |
Hydraulic cylinder | - wear of seals |
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| + |
| 0.14 | |
Hydraulic accumulators | - seal failure | + |
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| + | 0.047 | |
Decsription of diagnosis, being controlled. | Number of diagnosis, which are controlled by a feature number i | |||||||||
Diagnosis, being controlled |
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Probability of diagnosis occurrence is calculated by the formula:
, where - quantity of diagnosis occurrences
Quantity of information I can be found by the formula:
, where - total amount of diagnosis probability
Algorithm of finding a malfunction
Table 3. algorithm of finding a malfunction beginning
Number of row | Feature, being controlled | Diagnosis of system units | Total amount of diagnosis probability | Quantity of information | ||||||||||
0.233 | 0.233 | 0.783 | ||||||||||||
0.047 | 0.047 | 0.274 | ||||||||||||
0.233 | 0.047 | 0.28 | 0.855 | |||||||||||
0.186 | 0.186 | 0.693 | ||||||||||||
0.233 | 0.233 | 0.783 | ||||||||||||
| 0.209 | 0.209 |
| |||||||||||
7 | 0.233 | 0.209 | 0.442 | 0.99 | ||||||||||
8 | 0.07 | 0.07 | 0.366 | |||||||||||
0.116 | 0.116 | 0.518 | ||||||||||||
10 | 0.14 | 0.14 | 0.584 | |||||||||||
0.209 | 0.209 | 0.74 | ||||||||||||
0.047 | 0.047 | 0.274 | ||||||||||||
0.209 | 0.047 | 0.256 | 0.821 |
We can see, that row #7 gives the biggest information. So, we need to control feature X3. For this purpose we will analyze two separate situations – without a malfunction of this feature (R=0), excluding the diagnosis, which give us a malfunction. And with the malfunction (R=1), excluding the diagnosis, which don’t give us a malfunction. All parameters (probability of diagnosis occurrence gi, total amount of diagnosis probability Qn and quantity of information I)
must be recalculated respectively. Row #7 will be excluded for both.
Number of row | |||||||||
- | - | - | - | - | |||||
0.083 | 0.083 | 0.414 | |||||||
0.083 | 0.083 | 0.414 | |||||||
0.333 | 0.333 | 0.918 | |||||||
- | - | - | - | - | |||||
- | - | - | - | - | |||||
0.125 | 0.125 | 0.544 | |||||||
0.208 | 0.208 | 0.738 | |||||||
0.25 | 0.25 | 0.811 | |||||||
- | - | - | - | - | |||||
0.083 | 0.083 | 0.414 | |||||||
0.083 | 0.083 | 0.414 | |||||||
Number of row | ||||
0.526 | `0.526 | 0.998 | ||
- | - | |||
0.526 | 0.526 | 0.998 | ||
- | - | |||
0.526 | 0.526 | 0.998 | ||
0.474 | 0.474 | 0.998 | ||
- | - | |||
- | - | |||
- | - | |||
0.474 | 0.474 | 0.998 | ||
- | - | |||
0.474 | 0.474 | 0.998 |
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Номера строк | ||||||
- | - | - | - | |||
0.125 | 0.125 | 0.544 | ||||
0.125 | 0.125 | 0.544 | ||||
- | - | - | - | |||
- | - | - | - | |||
0.188 | 0.188 | 0.696 | ||||
0.313 | 0.313 | 0.896 | ||||
0.375 | 0.375 | 0.954 | ||||
- | - | - | - | |||
0.125 | 0.125 | 0.544 | ||||
0.125 | 0.125 | 0.544 |
Номера строк | |||||
- | - | - | |||
0.2 | 0.2 | 0.722 | |||
0.2 | 0.2 | 0.722 | |||
- | - | - | |||
- | - | - | |||
0.3 | 0.3 | 0.881 | |||
0.5 | 0.5 | 1 | |||
- | - | - | |||
0.2 | 0.2 | 0.772 | |||
0.2 | 0.2 | 0.772 |
Номера строк | ||||
- | - | |||
0.4 | 0.4 | 0.971 | ||
0.4 | 0.4 | 0.971 | ||
- | - | |||
- | - | |||
0.6 | 0.6 | 0.971 | ||
- | - | |||
0.4 | 0.4 | 0.971 | ||
0.4 | 0.4 | 0.971 |
It is a final stage of an algorithm.
Conclusion
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R=1
Which can be read as follows: first of all we need to make a check for pressure drop of hydraulic system (X3), if this feature occurs, it can be a destruction of plungers(D1) or slide valve jamming(D2) with the same probability. If there is no pressure drop, we need make a check for deviation of working parameters for actuating units (X2), if this feature occurs, we have a valve jamming (D3), if not, we need to make a check for a failure of actuating units (X6). If this feature occurs, we have a wear of seals (D6), if not, we need to make a check for a metal chip on a filters (X5). If this feature occurs, we have a slide valve jamming (D5), if not, it can be a filter clogging (D4) or a seal failure (D7) with a same probability.
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