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4 years ago

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Diffusion of Ammonia in an Aqueous Solution Ammonia (A)-water (B) solution ta 278 K and 4 mm thick is in contact with an organic

liquid at this interface. The concentration of ammonia in the organic phase is held constant and is such that the equilibrium concentration of ammonia at the interface is 2.0 wt% ammonia (density of aqueous solution is 991.7 Kg/m) The concentration of ammonia in water 4 mm away is 10 wt% (density of solution is 961.7 Kg/m). Water and the organic phase are insoluble in each other. The diffusion coefficient of NH3 in water is 1.24 × 10-9 m2/s.
(a) At steady state, calculate Na in Kmol/m-s

(b) Calculate NB. Explain.

1 answer:

Tom [10]4 years ago

4 0

Answer:

Explanation:

The pictures below shows the whole explanation for the problem

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Ammonia contained in a piston-cylinder assembly, initially saturated vapor at 0o F, undergoes an isothermal process during which

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$-P_{2(a)} =15.6lbf/in^2\\-P_{2(b)} =30.146lbf/in^2\\ T_{2(a)} =0^oF\\T_{2(b)} =0^oF\\x_{2(b)} =49.87percent$

Explanation:

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<u>AMMONIA </u>

state(1)

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Isothermal process $T=C$

a)

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b)

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To find the final state by giving the quality in lbf/in we assume the friction is neglected and the system is in equilibrium.

state(1)

using PVT data for saturated ammonia

$-P_{1} =30.416 lbf/in^2\\-v_{1} =v_{g} =9.11ft^3/lb$

then the state exists in the supper heated region.

a) from standard data

$-v_{1(a)} =2v_{1} =18.22ft^3/lb\\-T_{1} =0^oF$

$at\\P_{x} =14lbf/in^2\\-v_{x} =20.289 ft^3/kg$

$at\\P_{y} =16 lbf/in^2\\-v_{y} =17.701ft^3/kg$

assume linear interpolation

$\frac{P_{x}-P_{2(b)} }{P_{x}- P_{y} } =\frac{v_{x}-v_{1(a)} }{v_{x}-v_{y} }$

$P_{1(b)}=P_{x} -(P_{x} -P_{y} )*\frac{v_{x}- v_{1(b)} }{v_{x}-v_{y} }\\ \\P_{1(b)} =14-(14-16)*\frac{20.289-18.22}{20.289-17.701} =15.6lbf/in^2$

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$-v_{2(a)} =2v_{1} =4.555ft^3/lb\\v_{g}$

from standard data

$-v_{f} =0.02419ft^3/kg\\-v_{g} =9.11ft^3/kg\\v_{f}$

then the state exist in the wet zone

$-P_{s} =30.146lbf/in^2\\v_{2(a)} =v_{f} +x(v_{g} -v_{f} )$

$x=\frac{v_{2(a)-v_{f} } }{v_{g} -v_{f} } \\x=\frac{4.555-0.02419}{9.11-0.02419} =49.87%$

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Answer:

$\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}$

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