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4 years ago

For each of the following stacking sequences found in FCC metals, cite the type of planar defect that exists:

Engineering

1 answer:

lana [24]4 years ago

6 0

Answer:

a) The planar defect that exists is twin boundary defect.

b) The planar defect that exists is the stacking fault.

Explanation:

I am using bold and underline instead of a vertical line.

a. A B C A B C B A C B A

In this stacking sequence, the planar defect that occurs is twin boundary defect because the stacking sequence at one side of the bold and underlined part of the sequence is the mirror image or reflection of the stacking sequence on the other side. This shows twinning. Hence it is the twin boundary inter facial defect.

b. A B C A B C B C A B C

In this stacking sequence the planar defect that occurs is which occurs is stacking fault defect. This underlined region is HCP like sequence. Here BC is the extra plane hence resulting in the stacking fault defect. The fcc stacking sequence with no defects should be A B C A B C A B C A B C. So in the above stacking sequence we can see that A is missing in the sequence. Instead BC is the defect or extra plane. So this disordering of the sequence results in stacking fault defect.

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An insulated mixing chamber receives 0.5 kg/s of steam at 3 MPa and 300°C through one inlet, and saturated liquid water at 3 MPa

maxonik [38]

Answer:

$\dot m_{2} = 0.199\,\frac{kg}{s}$

Explanation:

The mixing chamber can be modelled by applying the First Law of Thermodynamics:

$\dot W_{in}+\dot m_{1}\cdot h_{1} +\dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0$

Since that mass flow rate of water at inlet 1 is the only known variable, the expression has to be simplified like this:

$\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - z\cdot h_{3} = 0$

Besides, the following expression derived from the Principle of Mass Conservation is presented below:

$1 + y = z$

Then, the expression is simplified afterwards:

$\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - (1+y)\cdot h_{3} = 0$

$\frac{\dot W_{in}}{\dot m_{1}} +h_{1} - h_{3} + y\cdot (h_{2}-h_{3}) = 0$

Specific enthalpies are obtained from steam tables and described as follows:

State 1 (Superheated vapor)

$h = 2994.3\,\frac{kJ}{kg}$

State 2 (Saturated liquid)

$h = 1008.3\,\frac{kJ}{kg}$

State 3 (Liquid-Vapor mixture)

$h = 2444.22\,\frac{kJ}{kg}$

The ratio of the stream at state 2 to the stream at state 1 is:

$y = \frac{\frac{\dot W_{in}}{\dot m_{1}}+h_{1}-h_{3}}{h_{3}-h_{2}}$

$y = \frac{\frac{10\,kW}{0.5\,\frac{kg}{s} }+2994.3\,\frac{kJ}{kg}-2444.22\,\frac{kJ}{kg} }{2444.22\,\frac{kJ}{kg}-1008.3\,\frac{kJ}{kg} }$

$y = 0.397$

The mass flow rate of the saturated liquid is:

$\dot m_{2} = y\cdot \dot m_{1}$

$\dot m_{2} = 0.397\cdot (0.5\,\frac{kg}{s} )$

$\dot m_{2} = 0.199\,\frac{kg}{s}$

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3 years ago

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Vinil7 [7]

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3 years ago

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3 years ago

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Answer:

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3 years ago

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rewona [7]

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