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Fittoniya [83]
2 years ago
6

Plastically deforming a metal specimen near room temperature generally leads to which of the following property changes? 5. (a)

An increased tensile strength and a decreased ductility (b) A decreased tensile strength and an increased ductility c) An increased tensile strength and an increased ductility (d) A decreased tensile strength and a decreased ductility
Engineering
2 answers:
algol132 years ago
8 0

Answer:

The answer is (a) An increased tensile strength and a decreased ductility.

Explanation:

When a metal specimen is plastically deformed near room temperature, it implies that the metal specimen is being cold worked (it is undergoing cold working), or it’s shape is being changed without applying heat. Cold working generally leads to an increase in tensile strength and yield strength, and a decrease in ductility.

andrezito [222]2 years ago
4 0

Answer:

Option a) An increased tensile strength and decreased ductility.

Explanation:

Plastic deformation is when the material once reshaped under applied stress is not able to regain its original shape and is deformed permanently even after the removal of the applied force on a metal specimen at room temperature.

This process is irreversible and is done at microscopic level by the use of dislocation defects. At room temperatures instead of being removed, these accumulate and acts as pin points thus enhancing the tensile strength of the metal followed by a reduction in its ductility.

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A system consisting of 3 lb of water vapor in a piston–cylinder assembly, initially at 350°F and a volume of 71.7 ft3, is expand
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Answer:

isobaric expansion = 281.09 Btu

isothermal compression= 72 Btu

Explanation:

The first law of thermodynamics is:

Q_{AB}=W_{AB}+deltaU_{AB}

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W= work

U=internal energy  

W_{AB}=P*(V_{B}-V_{A})

U_{AB}=n*C_{v}(T_{2}-T_{1})

P=pressure, V= volume, T= temperature, n =  moles, Cv= specific heat at constant volume.

In a isobaric process heat transferred is:

Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})

For an isothermal process (T2-T1 = 0) so

Q=P*(V_{B}-V_{A})= W_{AB}

From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.

For the first process

Q=P*(V_{B}-V_{A})+n*C_{v}(T_{2}-T_{1})

we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)

specific-volume=\frac{71.7 ft^{3}}{3Lb}=23.9\frac{ft^{3}}{Lb}

we use this information in the appropriate unit to find the pressure in thermodynamic tables.

T1= 176°C

v1= 1.49 m^3/kg

P=1.37 bar

in the second state we have

P=1.37 bar =137000Pa

v_{2}=\frac{85.38ft^{3}}{3Lb}= 28.46\frac{ft^{3}}{Lb}

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T2= 255°C

n=mass/Mw = 3Lb*\frac {1kg}{2.2Lb}*\frac{1000gr}{1kg}*\frac{1mol}{18gr}=75.75 mol

we usually find Cp on tables for water but from the Mayer relation we have:

C_{v}=C_{p}+R

Cp for water vapor is: 33.12 J/mol*K

R=8.314 J/mol*K

Cv= 41.434 J/mol*K

replacing in the equation for Q

Q=137000 Pa*(2.41m^{3}-2.030m^{3})+75.75mol*41.434\frac{ J}{mol*K}*(528.15-449.81 K)=296569J

296569J =281.09 Btu

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