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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:

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To calculate the mobility of electron and mobility of hole for gallium antimonide we have,

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Where

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$8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)$

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$2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))$

$0.1436 = 7.6*10^{-3}\mu_e+\mu_h$

Then, solving the equation:

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3 years ago

1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per

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Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

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To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

s1=7.234KJ/kg

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

h2=Enthalpy(Water;s=s1;P=P2)

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T2=330.8°C

5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

h3=2750KJ/kg

T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation

W=m1(h1-h2)+m3(h2-h3)

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