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3 years ago

6

A flat, circular hydrostatic air bearing has an outer diameter of 160 mm and a 5-mm-deep recess from the 50-mm diameter to the b

earing center. The air pressure in the recess is 2 MPa, and the pressure around the outside of the bearing is 0.1 MPa.
1. Determine the pressure distribution in the bearing if 0.3 kg of air is pumped through the bearing per second. The air density is 1 kg/m^3 at atmospheric pressure and the air viscosity is 18.2 x 10^-6 Pa-s at 22°C.
2. Also, calculate the air film thickness and the friction torque if the bearing rotates at 8000 rpm.

1 answer:

OLEGan [10]3 years ago

7 0

Answer:

W=15510.06

Air film thickness =1.8255×10^-4m

Torque=8.2l where l is length of bearing

Explanation:

Please find attached file for complete answer solution and explanation.

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The answer is both B and D

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Consider a turbojet powered airplane flying at a standard altitude of 30,000ft at a velocity of 500 mph. The turbojet engine its

grandymaker [24]

Answer:

T = 20.42 N

Explanation:

given data

standard altitude = 30,000 ft

velocity Ca = 500 mph = 0.4 m/s

inlet areas Aa = 7 ft² = 0.65 m²

exit areas Aj = 4.5 ft² = 0.42 m²

velocity at exit Cj = 1600 ft/s = 487.68 m/s

pressure exit $\rho$ j = 640 lb/ft² = 0.3 bar

solution

we get here thrust of the turbojet that is express as

thrust of the turbojet T = Mg × Cj - Ma × Ca + ( $\rho$ j Aj - $\rho$ a Ag ) .............1

here Ma = Mg

Ma = $\rho$ a × Ca Aa = 0.042 kg/s

put value in equation 1 we get

T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )

T = 20.42 N

5 0

3 years ago

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of

Orlov [11]

Answer: 133.88 MPa approximately 134 MPa

Explanation:

Given

Plane strains fracture toughness, k = 26 MPa

Stress at which fracture occurs, σ = 112 MPa

Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m

Critical internal crack length, l' = 6 mm = 6*10^-3 m

We know that

σ = K/(Y.√πa), where

112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]

112 MPa = 26 MPa / Y.√(3.142 * 0.043)

112 = 26 / Y.√1.35*10^-2

112 = 26 / Y * 0.116

Y = 26 / 112 * 0.116

Y = 26 / 13

Y = 2

σ = K/(Y.√πa), using l'instead of l and, using Y as 2

σ = 26 / 2 * [√3.142 * (6*10^-3/2)]

σ = 26 / 2 * √(3.142 *3*10^-3)

σ = 26 / 2 * √0.009426

σ = 26 / 2 * 0.0971

σ = 26 / 0.1942

σ = 133.88 MPa

8 0

3 years ago

The heat flux through a 1-mm thick layer of skin is 1.05 x 104 W/m2. The temperature at the inside surface is 37°C and the tempe

miss Akunina [59]

Answer:

a) Thermal conductivity of skin: $k_{skin}=1.5W/mK$

b) Temperature of interface: $T_{interface}=35.6\°C$

Heat flux through skin: $\frac{Q}{A}=2100W/m^2$

Explanation:

a)

$k=\frac{QL}{A(T_{2}-T_{1})}$

Where: $k$ is thermal conductivity of a material, $\frac{Q}{A}$ is heat flux through a material, $L$ is the thickness of the material, $T_{1}$ is the temperature on the first side and $T_{2}$ is the temperature on the second side

$k_{skin}=\frac{QL}{A(T_{2}-T_{1})}$

$k_{skin}=\frac{Q}{A}*\frac{L}{(T_{2}-T_{1})}$

$k_{skin}=1.05*10^{4}*\frac{1*10^{-3}}{(37-30)}$

$k_{skin}=1.5W/mK$

b)

$k_{insulation}=\frac{k_{skin}}{2}$

$k_{insulation}=\frac{1.5}{2}$

$k_{insulation}=0.75W/mK$

The heat flux between both surfaces is constant, assuming the temperature is maintained at each surface.

$\frac{Q}{A}=\frac{k(T_{2}-T_{1})}{L}$

$\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}=\frac{k_{insulation}(T_{interface}-T_{insulation})}{L_{insulation}}$

$\frac{1.5*(37-T_{interface})}{0.001}=\frac{0.75*(T_{interface}-30)}{0.002}$

$55500-1500T_{interface}=375T_{interface}-11250$

$1875T_{interface}=66750$

$T_{interface}=35.6\°C$

$\frac{Q}{A}=\frac{k_{skin}(T_{skin}-T_{interface})}{L_{skin}}$

$\frac{Q}{A}=\frac{1.5*(37-35.6)}{0.001}$

$\frac{Q}{A}=2100W/m^2$

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3 years ago

An incremental encoder is rotating at 15 rpm. On the wheel there are 40 holes. How many degrees of rotation would 1 pulse be?

elena-s [515]

Answer:

1 pulse rotate = 9 degree

Explanation:

given data

incremental encoder rotating = 15 rpm

wheel holes = 40

solution

we get here first 1 revolution time

as 15 revolution take = 60 second

so 1 revolution take = $\frac{60}{15}$

1 revolution take = 4 seconds

and

40 pulse are there for 1 revolution

40 pulse for 360 degree

so 1 pulse rotate is = $\frac{360}{40}$

1 pulse rotate = 9 degree

3 0

3 years ago