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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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155fts

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A soil has a bulk unit density of 1.91 Mg/m3 and a water content of 9.5%. The value of Gs = 2.7. Calculate the void ratio and de
hichkok12 [17]

Answer:

2.0978 g/cm^3

Explanation:

Given data:

p=1.91 g/cm^{3}

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G_{s}= 2.70

e =?

S_{r} =?

solution:

w=\frac{mass of water}{mass of solid} =\frac{m_{w} }{m_{s} }

e=\frac{volume of solid}{volume of solid} =\frac{V_{s} }{V_{s} }

assume total volume V_{total}=1 cm^{3}=V_{w} +V_{s} +V_{air}

p=\frac{m_{total} }{V_{total} }

m_{total} =1.91 g\\m_{total} =m_{w} +m_{s}\\w=\frac{m_{w} }{m_{s}}\\ 0.095.m_{s}=m_{w}\\1.91=m_{s}+0.095.m_{s}\\m_{s}=1.744 g\\m_{w}=0.166 g\\

p_{w} =\frac{m_{w} }{V_{w} } ==> V\\G_{S}=\frac{m_{s} }{V_{s}.p_{w}}\\2.70=\frac{1.744}{V_{s}.1 } \\\V_{s}=0.646 cm^{3} \\V_{v}=V_{T}-V_{S}=0.354 cm^{3} \\e=\frac{V_{v}}{V_{S}} =\frac{0.354}{0.646}=0.55=55 percent\\

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now find p

p=\frac{m_{t}(=m_{w} +m_{s} )}{V_{t} }

p=2.0978 g/cm^3

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4 years ago
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