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Paul [167]
2 years ago
10

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture tough

ness of and a yield strength of 1400 MPa (205,000 psi). The flaw size resolution limit of the flaw detection apparatus is 4.0 mm (0.16 in.). If the design stress is one half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.
Engineering
1 answer:
sp2606 [1]2 years ago
5 0

Answer:

Not subject to detection

Explanation:

Assuming the value of strain fracture toughness is 77 Mpa \sqrt m

The design stress is half hence \sigma=0.5\times 1400=700 Mpa

Critical flaw size, a_c=\frac {1}{\pi}(\frac {K_{1c}}{Y/sigma})^{2}

Where Y is dimensionless parameter, \sigma is applied stress, K_{1c} is plane strain fracture toughness, a_c is critical length of surface crack

a_c=\frac {1}{\pi}(\frac {77}{1*700})^{2}= 0.0038515496\approx 0.00385m

The critical length of surface crack is therefore 3.85 mm, which is less than detection apparatus size given as 4 mm

Since the critical flaw size is less than the resolution limit of flaw detection apparatus, the critical flaw for this plate is not subjected to detection.

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Answer:

Re=\dfrac{\rho\ v\ l}{\mu }

Explanation:

Reynolds number:

  Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is  low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

Re=\dfrac{F_i}{F_v}

For plate can be given as

Re=\dfrac{\rho\ v\ l}{\mu }

Where  ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.

Reynolds\ number\is \

7 0
3 years ago
Air within a piston cylinder assembly executes a Carnot refrigeration cycle between hot and cold reservoirs at TH=600 K and TC=3
Nataly [62]

Answer:

See explaination

Explanation:

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Kindly check attachment for the step by step solution of the given problem.

5 0
3 years ago
The combustion chamber has different shapes depending on the make and model of the engine. True or false
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5 0
2 years ago
Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1
Mariana [72]

Answer:

Koch's adjusted basis in machine 2 after the exchange is $60,000

Explanation:

given data

fair market value = $60,000

originally purchased Machine 1 = $76,900

Machine 1 adjusted basis = $40,950

Machine 2 seller purchase = $64,050

Machine 2 adjusted basis = $55,950

solution

As he exchanged machine for another at $60,000

and this exchanged in fair market

so adjusted basis =  $50,000

Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell

Adjusted basis increases when a person deducts expenses from factor taxes and operating statements

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3 0
3 years ago
Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin
Vladimir79 [104]

Answer:

a)

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

b)

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

c)

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

a)

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 =  1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 =  1.013 μm

b)

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 =  0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.243 μm

c)

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 =  0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.608 μm

6 0
3 years ago
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