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3 years ago

15

When removing the balance shaft assembly: Technician A inspects the bearings for unusual wear or damage. Technician B smoothens

a damaged camshaft or balance shaft journal by lightly sanding it. Who is correct

1 answer:

liraira [26]3 years ago

4 0

Answer:

Technicians A

Explanation:

The technician A would also tap the the teeth with a blunt end to break it loose. He also remove the underlying O ring in the front case groove. Technicians A also remove the driven gear bolt that secures the oil pump driven gear to the left balance shaft.

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You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the

Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, $\sigma$ /Strain, ∈

initial Strain, $\epsilon_i = \frac{\sigma}{E}$

$\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}$

$\epsilon_i = 0.002$

creep rate in the steady state

$\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )$

$\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )$

but Tinitial = 0

$\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001$

$\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}$

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0

3 years ago

Please help i give brainliest

Mazyrski [523]

Answer:

A mock-up

Explanation:

It is made of cheap and easy to access parts.

5 0

3 years ago

A body whose velocity is constant has a. positive acceleration b. negative acceleration g. zero acceleration d. all of the above

adoni [48]

Answer:

option (c) is the correct answer which is zero acceleration.

Explanation:

It is given in the question that the velocity is constant.

Now,

the options are provided in relation to the acceleration.

We know,

acceleration is rate of change of velocity per unit time i.e

acceleration = $\frac{dV}{dt}$

since, the change in velocity is given to be zero,

thus, dV/dt = 0

hence,

acceleration = 0

therefore, option (c) is the correct answer which is zero acceleration.

4 0

3 years ago

P1.30 shows a gas contained in a vertical piston– cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is a

iogann1982 [59]

Answer:

1.737 kJ

Explanation:

Thinking process:

Step 1

Data:

Area of the shaft = 0.8 cm²

Combined mass of shaft and piston = (24.5 + 0.5) kg

= 25 kg

Piston diameter = 0.1 m

External atmospheric pressure = 1 bar = 101.3 kPa

Pressure inside the gas cylinder = 3 bar = 3 × 101.3 kPa

g = 9.81 m/s²

Step 2

Draw a free body diagram

Step 3: calculations

area of the piston = 0.0314 m²

Change in the elevation of the piston, $\deltaz$

$\delta$ z = $\frac{PE}{mg}$

= $\frac{0.2*10x^{3} }{25*9.81}$

= 0.82 m

Next, we evaluate the work done by the shaft:

$W_{s} = F_{s} Z$

= (1668) ( 0.082)

= 1. 37 kJ

Net area for work done = A (piston) - Area of shaft

= $\pi*(0.1)^{2} - 0.8 cm^{2}$

= 77.7 cm²

= 0.007774 m²

Work done in overcoming atmospheric pressure:

Wₐ = PAZ

=101.3 kPa * 0.007774 * 0.82

= 0.637 kJ

total work = work done by shaft + work to overcome atmospheric pressure = 0.367 + 1.37

= 1.737 kJ Ans

6 0

3 years ago

The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is

slava [35]

Question:

The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Eₛₜ = 200 GPa and vₛₜ = 0.3.

Answer:

See explanation below

Explanation:

Given:

d = 2m = 2*10³ = 2000

thickness, t = 10 mm

Length of strain guage = 20 mm

i) Let's calculate d/t

$\frac{d}{t} = \frac{2000}{10} = 200$

Since $\frac{d}{t}$ is greater than length of strain guage, the pressure vessel is thin.

For the minimum normal stress, we have:

$\sigma max= \frac{pd}{4t}$

$\sigma max= \frac{2000p}{4 * 20}$

= 50p

For the minimum normal strain due to pressure, we have:

$E_max= \frac{change in L}{L_g}$

$= \frac{0.012}{20} = 0.60*10^-^3$

The minimum normal stress for a thin pressure vessel is 0.

$\sigma _min = 0$

i) Let's use Hookes law to calculate the pressure causing this deformation.

$E_max = \frac{1}{E} [\sigma _max - V(\sigma _initial + \sigma _min)]$

Substituting figures, we have:

$0.60*10^-^3 = \frac{1}{200*10^9} [50p - 0.3 (50p + 0)]$

$120 * 10^6 = 35p$

$p = \frac{120*10^6}{35}$

$p = 3.429 * 10^6$

p = 3.4 MPa

ii) Calculating the maximum in-plane shear stress, we have:

$\frac{\sigma _max - \sigma _int}{2}$

$= \frac{50p - 50p}{2} = 0$

Max in plane shear stress = 0

iii) To find the absolute maximum shear stress at a point on the outer surface of the vessel, we have:

$\frac{\sigma _max - \sigma _min}{2}$

$= \frac{50p - 0}{2} = 25p$

since p = 3.429 MPa

25p = 25 * 3.4 MPa

= 85.71 ≈ 85.7 MPa

The absolute maximum shear stress at a point on the outer surface of the vessel is 85.7 MPa

6 0

3 years ago