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andrey2020 [161]
3 years ago
13

You have been assigned to design an open cylindrical storage tank 4 meters tall with a diameter of 8 meters to be made out of A-

36 steel. The tank will hold a fluid with density of rho = 1.2 g/cm^3 (or 1200 kg/m^3 ). Using a safety factor of 4, what wall thickness is required?
Engineering
1 answer:
Katen [24]3 years ago
8 0

Answer:

The required wall thickness is 1.506 \times 10^{-3} m

Explanation:

Given:

Fluid density \rho = 1200 \frac{kg}{m^{3} }

Diameter of tank d = 8 m

Length of tank l = 4 m

F.S = 4

For A-36 steel yield stress \sigma = 250 MPa,

Allowable stress \sigma _{allow} = \frac{\sigma}{F.S}

 \sigma _{allow} = \frac{250}{4} = 62.5 MPa

Pressure force is given by,

 P = \rho gh

 P = 1200 \times 9.8 \times 4

P = 47088 Pa

Now for a vertical pipe,

\sigma _{allow} = \frac{Pd}{4t}

Where t = required thickness

 t = \frac{Pd}{4 \sigma _{allow} }

 t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }

t = 1.506 \times 10^{-3} m

Therefore, the required wall thickness is 1.506 \times 10^{-3} m

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El tiempo hasta que falle un sistema informático sigue una distribución Exponencial con media de 600hs. (Utilice 3 decimales par
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X: '' El tiempo (en horas) hasta que falle un sistema informático ''

La variable aleatoria X será entonces una variable aleatoria continua.

Sabemos que sigue una distribución exponencial con una media de 600 hs.

Esto se escribe :

X ~ ε ( λ ) (I)

En donde λ es igual a la inversa de la media. Esto se escribe :

λ =\frac{1}{E(X)}

En donde E(X) es la media de la variable. Por ende, si reemplazamos los datos del ejercicio obtenemos ⇒

λ =\frac{1}{E(X)} ⇒ λ =\frac{1}{600}

Si reemplazamos el valor de λ en (I) obtenemos :

X ~ ε (\frac{1}{600})

La función de distribución de X (por ser una variable aleatoria exponencial) es :

F_{X}(x)=P(X\leq x)=  1 - e ^ ( - λx) con x > 0 y F_{X}(x)=0 en caso contrario.

Si reemplazamos el valor de λ en la función de distribución de X obtenemos :

F_{X}(x)=P(X\leq x)=1-e^{-\frac{x}{600}}  

Dado que la variable aleatoria X se distribuye de manera exponencial, el hecho de saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos aporta ninguna información sobre lo que ocurrirá después. Esta característica se conoce como propiedad de perdida de memoria de la variable aleatoria exponencial. Entonces, la probabilidad pedida se reduce a calcular :

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Dado que saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos dice nada sobre lo que ocurrirá instantes posteriores a esas 400 hs.

Calculamos entonces la probabilidad pedida :

P(X>200)=1-P(X\leq 200)=1-F_{X}(200)=1-(1-e^{-\frac{200}{600}})=e^{-\frac{1}{3}}=0.717

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