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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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The diffusion coefficients for species A in metal B are given at two temperatures:
Kruka [31]

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

D = D_0e^{\frac{-Q_d}{RT} }

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

R = 8.314\ J/mol*K

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}

and

ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}

You might notice that these equations have the form of  

d=y-ax

You can solve this equation system easily using calculator, and you will eventually get

D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol

After you got those 2 parameters, the rest is easy, you can just plug them all   including the given temperature of 1180°C into the Arrhenius equation

6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}

And you should get D = 2.76*10^-16 m^/s as an answer for c)

5 0
3 years ago
Conditions of special concern: i. Suggest two reasons each why distillation columns are run a.) above or b.) below ambient press
lutik1710 [3]

Solution :

Methods for selling pressure of a distillation column :

a). Set, \text{based on the pressure required to condensed} the overhead stream using cooling water.

  (minimum of approximate 45°C condenser temperature)

b). Set, \text{based on highest temperature} of bottom product that avoids decomposition or reaction.

c). Set, \text{based on available highest } not utility for reboiler.

Running the distillation column above the ambient pressure because :

The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.

Run the reactor at an evaluated temperature because :

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Run the reaction at an evaluated pressure because :

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2 years ago
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Answer:

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4 0
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Answer:

the third statement is true

Explanation:

given data

Lenovos cost more than Dells

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solution

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and

2nd statement that is express as

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