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Rus_ich [418]
3 years ago
15

An aluminum oxide component must not fail when a tensile stress of 12.5 MPa is applied. Determine the maximum allowable surface

crack length if the surface energy of aluminum oxide is 0.90 J/m2. The modulus of elasticity of this material is 393 GPa.
Engineering
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

C=\frac {2E\gamma}{\pi \sigma_c^{2}} where E is the modulus  of elasticity, \gamma is surface energy and \sigma_c is tensile stress

Substituting the given values

C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm

The maximum allowable surface crack is 1.44 mm

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I have a stream with three components, A, B, and C, coming from another process. The stream is 50 % A, and the balance is equal
tigry1 [53]

Answer:

X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}

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Explanation:

A + 2B + 4C ⇒ 2X + 3Y

Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis

Say stream is 10 moles, this give

A = 5moles

B = 2.5mole

C = 2.5moles

from the balanced equation above,

1mole of A ⇒ 4moles of C

∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C

also;

2mole of B ⇒ 4moles of C

∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C

so clearly from above reactant C is the limiting reactant.

<em>Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end</em>

<em></em>

Formula to obtain conversion is:

Conversion = (Amount of A used up)/(Amount of A fed into the system)

X_{A} = \frac{N_{Ao}-N_{A}}{N_{Ao}}

where, Nₐ₀-Nₐ = is the amount in moles of A used up

            Nₐ₀ = amount in moles of A fed into the system

The next question is what mole of reactant C will give 0.1mole fraction of Y

Recall our basis = 10moles

<em>from conservation of mass law</em>, 10mole of product must come out which 0.1 moles fraction is Y

therefore amount Y in the product is = 0.1x10 = 1mole

if  3moles of Y ⇒ 4mole of C

∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C

calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y

Nₐ₀-Nₐ = 1.33

Nₐ₀ = 2.5

Conversion X = 1.33/2.5 = <u>0.533</u>

5 0
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Answer:

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Explanation:

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A) -kdT(r1)/dr = h[T∞ - T(r1)]

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Answer:

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