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# Question: the images might not show up but its just a...

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The images might not show up, but its just a demostration of what is written, dont think you need them to answer the question...

A light aircraft landing system uses air shock absorber to absorb landing motion of the aircraft (Figure 1).

Figure 1 – Closed end air cylinder as shock absorber (Note: aircraft in the figure is just an example for illustration only.  It is not a "light" aircraft)

The landing system has a rigid bracket structure where the wheels are connected to an air cylinder.  A light aircraft landing system consists of a closed end cylinder as shown in Figure 2.

Figure 2 – Closed end air cylinder as shock absorber

The shock absorber is mounted on an aircraft wheel structure.  The mass of aircraft is supported (floated) on the piston of the air cylinder.  The system parameters are diagrammatically shown in Figure 3.

Figure 3 – System parameters of closed end air cylinder

When the aircraft lands, the piston is pushed into the closed end chamber with a downward velocity $\frac{dy}{dt}$ (Note: Canvas can't display y-dot so the differential is written in full) where y is measured positive downwards starting from the zero point y(0) = 0.  The shock absorber system should resist and bring the velocity of the aircraft mass M to zero, i.e. $\frac{dy}{dt}$=0.  You can assume the piston is a perfect seal so no air in the lower chamber can escape.

Air pressure at the bottom chamber of the cylinder is denoted by P2.  The bottom air chamber of the air cylinder can be injected with compressed air to strengthen its ability to resist downward motion.  An initial air pressure at t = 0, i.e. P2(0) = Pi = 6 bars is injected to the air cylinder.

The upper cylinder chamber has an air vent to allow atmospheric air to go in or out while the piston moves down or up.  This reduces resistance of air pressure in top chamber.  Two frictional forces are opposing motion of the piston:

• Static friction due to tight sealing between piston and cylinder wall is a constant force acting against direction of piston movement, F = 500 N.
• Viscous friction due to lubricating oil acting between piston and cylinder wall and is given by ${F}_{r}=c\frac{dy}{dt}$ N, where c is the coefficient of viscosity.

The value of c depends on choice of lubricating oil:

• Grade A oil has a viscosity of 2000 Ns/m.
• Grade B oil has a viscosity of 1000 Ns/m.
• Grade C oil has a viscosity of 5000 Ns/m.

The following engineering data are available:

• Aircraft mass M = 3000 kg (portion of mass shared by this shock absorber)
• Landing velocity $\frac{dy}{dt}$ = 3 m/s downwards

Decision on the set of system parameters is based on several performance outcomes:

• The shock absorber cylinder should avoid hard piston to cylinder impact during landing.
• The rebounce due to shock pulse vibration should be minimised (to avoid piston bouncing back to hit the air vent end).

Which set of system parameters would you recommend?

Formulae applicable to this question:

The energy equation related to the work done (by the downward aircraft motion) and the change of internal energy in chamber 2 is given by:

${P}_{2}\frac{d{V}_{2}}{dt}=\frac{d\left({C}_{v}{\rho }_{2}{V}_{2}{T}_{2}\right)}{dt}$

where

• P2 = pressure in chamber 2
• V2 = volume of air in chamber 2
• Cv = specific heat of gas at constant volume (J/kg)
• ${\rho }_{2}$ = density of air inside cylinder chamber 2 (kg/m3)

Other engineering constants relevant to this system are:

• Atmospheric pressure P0 = 101325 N/m2
• Density of air at atmospheric pressure r0 = 1.225 kg/m3
• Ratio of specific heat of air = $\gamma$ = 1.4
• Universal gas constant = R = 286 J/kg/K
• Cp = specific heat of gas at constant pressure (J/kg)

Relationship of constants of air:

$\frac{R}{{C}_{v}}=\frac{{C}_{p}-{C}_{v}}{{C}_{v}}=\gamma -1$

D = diameter of air cylinder = 0.2 m

Grade A lubricating oil, c = 2000 Ns/m

y0 = length of downwards movement = 1 m

D = diameter of air cylinder = 0.1 m

Grade C lubricating oil, c = 5000 Ns/m

y0 = length of downwards movement = 0.5 m

D = diameter of air cylinder = 0.3 m

Grade B lubricating oil, c = 1000 Ns/m

y0 = length of downwards movement = 1 m

D = diameter of air cylinder = 0.1 m

Grade A lubricating oil, c = 2000 Ns/m

y0 = length of downwards movement = 2 m

D = diameter of air cylinder = 0.2 m

Grade B lubricating oil, c = 1000 Ns/m

y0 = length of downwards movement = 0.5 m

D = diameter of air cylinder = 0.4 m

Grade B lubricating oil, c = 1000 Ns/m

y0 = length of downwards movement = 1 m