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kirill115 [55]
3 years ago
5

A thin plastic membrane is used to separate helium from a gas stream. Under steady-state conditions the concentration of helium

in the membrane is known to be 0.02 and 0.005 kmol/m^3 at the inner and outer surfaces, respectively. If the member is 1mm thick and the binary diffusion coefficient of helium with respect to the plastic is 10^-9 m^2/s, what is the diffusive flux?
Engineering
1 answer:
Juli2301 [7.4K]3 years ago
5 0

Answer:

N_A=1.5*10^-8 kmol/s.m^2

Explanation:

<u>KNOWN: </u>

Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.  

<u>FIND:</u>

Molar diffusion flux.  

<u>ASSUMPTIONS:</u>

(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.  

<u>ANALYSIS:</u> The molar flux may be obtained from

 N_A=D_AB/L(C_A,1-C_A,2)

       =10^-9 m^2/s/0.001 m(0.02-0.005)kmol/m^3

N_A=1.5*10^-8 kmol/s.m^2

<u>COMMENTS:</u> The mass flux is:

n_A,x=M_a*N_A,x

n_A,x=6*10^-8 kg/s m^2

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Answer:

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Explanation:

given data

mass = 120 kg

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displacement = 5 mm

to find out

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solution

we will apply here frequency formula that is

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here  ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99    .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})

put here value

0.005 = 120/k   ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})  

k ×∈ × √(1-2∈²) = 1200       ......................3

so by equation 3 and 2

\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}

\varepsilon^{2} - \varepsilon^{4}  = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

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Here, the C lagoon= 20 mg/L

Q in= Q out= 8640 m³/d

C in= 100 mg/L

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An induced-draft cooling tower cools 90,000 gallons per minute of water from 84 to 68oF. Air at 14.61 psia, 70oF dry bulb and 60
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Qw = 6607.33[28.89 - 20] * 4.18

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oulet condition,

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check the attached file for complete solution

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The volume of 1.5 kg of helium in a frictionless piston-cylinder device is initially 6 m3. Now, helium is compressed to 2 m3 whi
coldgirl [10]

Answer:

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Final volume, v₂ = 2 m³

Pressure, P = 200 kPa

As we know,

Work, W=p(v_{2}-v_{1})

On putting the estimated values, we get

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Now,

Gas ideal equation will be:

⇒  pv_{1}=mRT_{1}

On putting the values. we get

⇒  200000\times 6=1.5\times 2077\times T_{1}

⇒  T_{1}=\frac{1200000}{3115.5}

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⇒  pv_{2}=mRT_{2}

On putting the values, we get

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⇒  T_{2}=\frac{400000}{3115.5}

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3 0
3 years ago
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