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4 years ago

What measures can be taken to mitigate or prevent the formation of stress corrosion cracks?

Engineering

1 answer:

sasho [114]4 years ago

7 0

Answer:

a) Selection and control of material

b) Control of stress

c) Control of the environment

Explanation:

The measures that can be taken to control and prevent the formation of strss corrosion cracks are:

a) Selection and control of material

It deals with the selection of the best quality material that can resist the stress corrosion cracks to a high extent.

b) Control of stress

It is to control or reduce the intensity of the major stress that causes the stress corrosion cracks.

c) Control of the environment

It includes the prevention of the material from the corrosive environment by removing the component from the region or controlling the environment condition around the object.

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A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container

kykrilka [37]

Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

$delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}$

In an isobaric process heat (Q) is defined as:

$Q= m*Cp*dT$

Replacing in the equation for entropy

$delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}$

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:

$delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}$

Solving the integral we get the expression to estimate the entropy change in the system

$delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})$

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is $0.85\frac{kJ}{Kg*K}$

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

$Q= m*Cp*dT$

With $dt=T_{2}-T_{1}$ clearing for T2 we get:

$T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K$

Now we can estimate the entropy change in the system

$delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}$

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.

b) see picture.

3 0

3 years ago

Technician A says that the use of methanol in internal combustion engines has declined over the years. Technician B says that th

enyata [817]

Answer:

Both Technician A and B are correct

Explanation:

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3 years ago

A logic chip used in a computer dissipates 3 W of power in an environment at 120°F, and has a heat transfer surface area of 0.08

UkoKoshka [18]

Answer:

attached below

Explanation: