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2 years ago

The best approach to keeping your car in safe working order is to

Engineering

2 answers:

Soloha48 [4]2 years ago

3 0

Answer:

Explanation:

Your car's owner's manual will provide a maintenance schedule designed to keep your brakes in good condition. Following it is the easiest way to avoid expense repairs and the potential for catastrophic brake failure. But at the very least, you should have your brakes inspected every year.

Helga [31]2 years ago

3 0

Answer:

1. Perform regular based-on-time or based-on-use inspections and adjustments.

2. Perform regular good quality replacements according to use or/and environment conditions

Explanation:

The best approach to keeping a car in a safe working order is to perform preventive maintenance activities to avoid unexpected failures that could put at risk people in a car. These preventive maintenance activities could be categorized into two groups:

Regular based-on-time or based-on-use inspections and adjustments.
Regular good quality replacements.

Those maintenance activities based-on-time (generally, based on days) or based-on-use (according to miles or kilometers) inspections and adjustments include those performed, at least, on main components fluids levels and working conditions of:

Motor oil level and oil quality conditions.
Motor coolant system level.
Braking system liquid level.
Steering system liquid level.
Tires pneumatic pressure.
Oil leakage presence below motor, gears box (manual transmission) or automatic transmission or in front or/and rear axles.
Condition of shock absorbers (suspension system).

Those maintenance activities based on regular good quality replacements depend on the evaluation of the previous ones and include main components such as:

Motor oil.
Spark plugs.
Oil/Air/Gas filters.
Headlights and stop lights.
Battery.
Tires.
Brake's pads and brake rotor rectification.

Good quality means replacements that meet quality norms and well-performed maintenance activities.

Again, these are activities that are performed in main and critical components, that is, a failure on them may have a high impact with high consequences (including loss of life) for those inside the car.

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Steam enters a turbine operating at steady state at 1 MPa, 200 °C and exits at 40 °C with a quality of 83%. Stray heat transfer

Andrei [34K]

Answer:

(a) Work out put=692.83 $\frac{KJ}{Kg}$

(b) Change in specific entropy=0.0044 $\frac{KJ}{Kg-K}$

Explanation:

Properties of steam at 1 MPa and 200°C

$h_1=2827.4\frac{KJ}{Kg},s_1=6.69\frac{KJ}{Kg-K}$

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of turbine is 0.83 and temperature T=40°C.So from steam table we can find pressure corresponding to saturation temperature 40°C.

Properties of saturated steam at 40°C

$h_f= 167.5\frac{KJ}{Kg} ,h_g= 2537.4\frac{KJ}{Kg}$

$s_f= 0.57\frac{KJ}{Kg-K} ,s_g= 8.25\frac{KJ}{Kg-K}$

So the enthalpy of steam at the exit of turbine

$h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}$

$h_2=167.5+0.83(2537.4-167.5)\frac{KJ}{Kg}$

$h_2=2134.57\frac{KJ}{Kg}$

$s_2=s_f+x(s_g-s_f)\frac{KJ}{Kg-K}$

$s_2=0.57+0.83(8.25-0.57)\frac{KJ}{Kg-K}$

$s_2=6.6944\frac{KJ}{Kg-K}$

(a)

Work out put = $h_1-h_2$

=2827.4-2134.57 $\frac{KJ}{Kg}$

Work out put =692.83 $\frac{KJ}{Kg}$

(b) Change in specific entropy

$s_2-s_1=6.6944-6.69\frac{KJ}{Kg-K}$

Change in specific entropy =0.0044 $\frac{KJ}{Kg-K}$

3 0

2 years ago

A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t

Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

$\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa$

The design factor for metals is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}$

The design factor for plastic and ceramics is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}$

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17$

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09$

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125$

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12$

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039$

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26$

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

$DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5$