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3 years ago

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. Nickel :- $T_{c}$ = 690.4 K

B. Copper :- $T_{c}$ = 542.8 K

C. Iron :- $T_{c}$ = 723.6 K

D. Tungsten :- $T_{c}$ = 1469.2 K

E. Lead :- $T_{c}$ = 240.2 K

F. Aluminium :- $T_{c}$ = 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. Nickel :- 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. Copper :- 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. Iron :- 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. Tungsten :- 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. Lead :-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. Aluminium :- 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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Answer:

As the temperature of a thermal radiator is increased

Group of answer choices

the object appears redder.
the object appears bluer.
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<h3>How temperature affects thermal radiation.</h3>

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3 years ago

The wall has dimensions of 100 ft x 10 ft including window and door areas. The wall has a total window area of 80 ft2 and a door

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Answer:

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Explanation:

According to the question, the total area includes the areas of the wall, window and door.

Total area, A = 100 * 10 = 1000 ft²

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Area of the door, $A_{d} = 8 * 2.5 = 20 ft^2$

Area of the wall, $A_{wall} = A - (A_{d} + A_{wind} )$

$A_{wall} = 1000 - ( 80 + 20)\\A_{wall} = 900 ft^2$

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For door, $h_{d} = 0.1Btu/hr-ft^2-F$

To get the overall heat - transfer coefficient for the combination:

$AU = A_{wall} h_{wall} + A_{wind} h_{wind} + A_{d} h_{d}\\1000U = (900*0.2) + (80*0.1) + (20*0.1)\\1000U = 190\\U = 190/1000\\U = 0.19 Btu/hr-ft^2 -F$

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