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kherson [118]
3 years ago
5

Answer the following questions, and very briefly explain your answer:

Engineering
1 answer:
alexandr1967 [171]3 years ago
5 0

Answer:

A) micro defects are left behind on the surface of metal components during the manufacturing process. These defects, in the form of micro-cracks or pits, becomes initiation sites for crack propagation or corrosion. Removing these imperfections on the surface of metal parts by electroplating greatly improves the life of metal components.

B) it will reduce fatigue crack growth.

Dispersion hardening involves the inclusion of small, hard particles in the metal, thus restricting the movement of dislocations, and thereby raising the strength properties. In dispersion hardening it is assumed that the precipitates do not deform with the matrix and that a moving dislocation bypasses the obstacles (precipitates) by moving in the clean pieces of crystal between the precipitated particles.

C) stress concentrations such as changes in section with sharp corners caused yielding, which will typically occur first at a stress concentration. For ductile materials localised plastic deformation can cause a redistribution of stress, enabling the component to continue to carry load. Brittle materials will typically fail at the stress concentration. Repeated loading may cause a fatigue crack to initiate and slowly grow at a stress concentration leading to the failure of even ductile materials. Fatigue cracks always start at stress raisers, so removing such defects increases the fatigue strength.

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3 years ago
How many atom are in Caci2 • 2H2O
Taya2010 [7]

Answer:

Nine

Explanation:

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3 0
3 years ago
Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally
vichka [17]

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Technician A says that TSBs are typically updates to the owner's manual. Technician B says that TSBs are generally

updated information on model changes that do not affect the technician. Who is correct? the answer is c

4 0
3 years ago
Refrigerant-134a at 400 psia has a specific volume of 0.1144 ft3/lbm. Determine the temperature of the refrigerant based on (a)
vekshin1

Answer:

a) Using Ideal gas Equation, T = 434.98°R = 435°R

b) Using Van Der Waal's Equation, T = 637.32°R = 637°R

c) T obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is T = 559.67°R = 560°R

Explanation:

a) Ideal gas Equation

PV = mRT

T = PV/mR

P = pressure = 400 psia

V/m = specific volume = 0.1144 ft³/lbm

R = gas constant = 0.1052 psia.ft³/lbm.°R

T = 400 × 0.1144/0.1052 = 434.98 °R

b) Van Der Waal's Equation

T = (1/R) (P + (a/v²)) (v - b)

a = Van Der Waal's constant = (27R²(T꜀ᵣ)²)/(64P꜀ᵣ)

R = 0.1052 psia.ft³/lbm.°R

T꜀ᵣ = critical temperature for refrigerant-134a (from the refrigerant tables) = 673.6°R

P꜀ᵣ = critical pressure for refrigerant-134a (from the refrigerant tables) = 588.7 psia

a = (27 × 0.1052² × 673.6²)/(64 × 588.7)

a = 3.596 ft⁶.psia/lbm²

b = (RT꜀ᵣ)/8P꜀ᵣ

b = (0.1052 × 673.6)/(8 × 588.7) = 0.01504 ft³/lbm

T = (1/0.1052) (400 + (3.596/0.1144²) (0.1144 - 0.01504) = 637.32°R

c) The temperature for the refrigerant-134a as obtained from the refrigerant tables at P = 400 psia and v = 0.1144 ft³/lbm is

T = 100°F = 559.67°R

7 0
3 years ago
What kind of situations breaks the causal link?
34kurt

Answer:A novus actus

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