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Vitek1552 [10]
3 years ago
7

Give the approximate temperature (in K) at which creep deformation becomes an important consideration for each of the following

metals: nickel, copper, iron, tungsten, lead, and aluminum.
Engineering
1 answer:
andrezito [222]3 years ago
5 0

Answer:

691K, 543K, 725K, 1473K, 240K, 373K

Explanation:

Creep deformation of any metal is the transformational tendency of a metal to distort rapidly or slowly when attacked by any form of mechanical stress. The temperature significant for a metal to deform is gotten by the division of the actual temperature of the metal by its melting point. This is termed homologous temperature which is 0.4 or higher. It is calculated by the equation:

0.4Tm

Therefore for the listed metals...

For Nickel, 0.4Tm = 0.4 ×(1455 + 273) = 691 K

For Copper, 0.4Tm = 0.4 ×(1085 + 273) = 543 K

For Iron, 0.4Tm = 0.4 ×(1538 + 273) = 725 K

For Tungsten, 0.4Tm = 0.4 ×(3410 + 273) = 1473 K

For Lead, 0.4Tm = 0.4 × (327 + 273) = 240 K

For Aluminium, 0.4Tm = 0.4 ×(660 + 273) = 373 K

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Answer:

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Explanation:

Given data :

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surface temperature of a person throughout the year = 32⁰c = 305 k

coefficient of heat transfer by natural convection (h)= 2 w /m^2 k

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An explanation to the condition of feeling chilled during the winter and comfortable during summer  can be explained with the calculation below

The heat transfer from the surface body of a person the the room is carried out by convection and this can be calculated as

q = hΔt = 2 * ( 305  -  293) = 2 * 12 = 24 w /m^2

also calculate heat transfer through radiation using this formula

q_{rad} = εσ [ (Temperature of body)^4- (temperature of room at each season)^4 ]

ε = 0.90,(emissivity)    σ = 5.67 *10^-8 w/^2 . k ( Boltzmann's constant)

during summer :

q_{rad} = (0.90)*(5.67*10^-8)* ( 305^4 - 300^4 ) = 28.253 W/m^2

therefore the net heat transfer during the summer = 24 w/m^3 + 28.253 W/m^3 = 52.253 W/m^2

During winter :

q_{rad} = (0.90)*(5.67*10^-8) * ( 305^4 - 287^4 ) = 95.375 W/m^3

therefore the net heat transfer during the winter

= 24 w/m^3 + 95.375w/m^3 = 119.375 w/m^2

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Answer:

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low temperature reservoir T_{L} =400k

thermal efficiency n_{1}= n_{2}

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n_{1} =1-\frac{T}{T_{H} }

The thermal efficiency of second heat engine can be written as

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The temperature of intermediate reservoir can be defined as  

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La probabilidad de que un nuevo producto tenga éxito es de 0.85. Si se eligen 10 personas al azar y se les pregunta si compraría
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Answer:

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Sabemos que la probabilidad de que un nuevo producto tenga éxito es de 0.85. Sabemos también que se eligen 10 personas al azar y se les pregunta si comprarían el nuevo producto. Para responder a la pregunta, primero definiremos la siguiente variable aleatoria :

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Ahora bien, si suponemos que la probabilidad de que el nuevo producto tenga éxito se mantiene constante (p=0.85) y además suponemos que hay independencia entre cada una de las personas al azar a las que se les preguntó ⇒ Podemos modelar a X como una variable aleatoria Binomial. Esto se escribe :

X ~ Bi(n,p) en donde ''n'' es el número de personas entrevistadas y ''p'' es la probabilidad de éxito (una persona adquiriendo el producto) en cada caso.

Utilizando los datos ⇒ X ~ Bi(10,0.85)

La función de probabilidad de la variable aleatoria binomial es :

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Si reemplazamos los datos de la pregunta en la función de probabilidad obtenemos :

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Nos piden la probabilidad de que por lo menos 8 personas adquieran el nuevo producto, esto es :

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