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matrenka [14]
3 years ago
5

Give the approximate temperature at which creep deformation becomes an important consideration for each of the following metals:

Engineering
1 answer:
IrinaVladis [17]3 years ago
8 0

Answer:

The creep temperature for different materials is as follows.

A. <u>Nickel :- </u>T_{c} = 690.4 K

B. <u>Copper :-  </u>T_{c} = 542.8 K

C.<u> Iron :-</u>T_{c} = 723.6 K

D.<u> Tungsten :-</u>T_{c} = 1469.2 K

E. <u>Lead :- </u>T_{c} = 240.2 K

F. <u>Aluminium :- </u>T_{c} = 373.2 K

Explanation:

Creep is the slow plastic deformation occur at high elevated temperature.

The approximate temperature for different metals is as follows at which creep becomes considerable.

A. <u>Nickel :-</u>  The relation between melting temperature & creep temperature is given by the formula.

T_{c}  = 0.4 T_{m}

The melting temperature of nickel = 1726 K

Creep temperature T_{c} = 0.4 × 1726

T_{c} = 690.4 K

B. <u>Copper :- </u>The relation between melting temperature & creep temperature for copper is given by the formula.

T_{c}  = 0.4 T_{m}

The melting temperature of copper = 1357 K

Creep temperature T_{c} = 0.4  × 1357

T_{c} = 542.8 K

C.<u> Iron :- </u>The relation between melting temperature & creep temperature for iron is given by the formula

T_{c}  = 0.4 T_{m}

The melting temperature of iron = 1809 K

Creep temperature T_{c} = 0.4  × 1809

T_{c} = 723.6 K

D.<u> Tungsten :- </u>The relation between melting temperature & creep temperature for tungsten is given by the formula

T_{c}  = 0.4 T_{m}

The melting temperature of iron = 3673 K

Creep temperature T_{c} = 0.4  × 3673

T_{c} = 1469.2 K

E. <u>Lead :-</u>The relation between melting temperature & creep temperature for lead is given by the formula

T_{c}  = 0.4 T_{m}

The melting temperature of lead = 600.5 K

Creep temperature T_{c} = 0.4  × 600.5

T_{c} = 240.2 K

F. <u>Aluminium :- </u>The relation between melting temperature & creep temperature for aluminium is given by the formula

T_{c}  = 0.4 T_{m}

The melting temperature of lead = 933 K

Creep temperature T_{c} = 0.4  × 933

T_{c} = 373.2 K

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A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he
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Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

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Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

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Now by putting the values in the above equation we get

{T_2}=300\times \dfrac{368}{301}\ K

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