9 A manufacturer wishes to know the reliability of a skid protection system to b

Question

9 A manufacturer wishes to know the reliability of a skid protection system to be used on military tractor trailers. The system consists of: (a) Two battery or generator powered sensors per wheel (b) One logic unit per sensor to predict wheel skid. (c) A command unit, which operates an electric or an engine vacuum solenoid. (d) The solenoids in (c) operate an actuator that controls the pressure to the brake. The system block diagram (not reliability diagram) is shown in Figure 3.18. Logic unit Vacuumm solenoid Sensor Actuator Sensor Logic unit Electric solenoid Battery Identical to above for four wheels Command unit Identical to above for four wheels Generator Figure 3.18 Block diagram for a skid control system. For the following component reliabilities calculate the system reliability. COMPONENT Battery Generator Sensor Logic unit Command unit Vacuum solenoid Electric solenoid Actuator RELIABILITY 0.90 0.99 0.98 0.97 0.99 0.97 0.98 0.99

Explanation / Answer

Here battery and generator in parallel

So, reliability of that portion R(Battery ll parallel) = 1 - (1 - Rbattery) * (1 - RGenerator) = 1 - ( 1- 0.90) * (1 - 0.99) = 0.999

Here now for each wheel, sensor and logic unit are in series and then this thing in repeat is in parellat to the other sensor and logic unit.

Reliability (One wheel) = 1 - (1 - Sensor ll Logic Unit) * (1 - Sensor ll Logic Unit) = 1- (1 - 0.98 * 0.97) * (1 - 0.98 * 0.97) = 0.99756

Now all four wheels are in parallel so

Reliability (all four wheels) = 1 - (1 - 0.99756)4 = 0.999999999999 or say 1

Now we have third system of Command unit which is independent.

As fourth system, we have Actuator in series with Vacuum solenoid and Electric solenoid in parallel

R(one wheel) = R(actuator) * [1 - (1 - RVaccum Solenoid) * (1 - RElectric Solenoid)]

= 0.99 * [ 1 - (1 - 0.97) *(1 - 0.98) ] = 0.99 * 0.9994 = 0.9894

Now for 4 wheels which are in parallel.

R(4 wheels) = 1 - (1 - 0.9894)4= 0.999999 or say 1

so now there are 4 system reliability we got. All these system are in series.

Now,

R(System) = 0.999 * 1 * 0.99 * 1 = 0.989

so system reliability is 0.989

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