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.
(a) At steady state, calculate Na in Kmol/m-s
(b) Calculate NB. Explain.
1 answer:
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Answer:
distance = 22.57 ft
superelevation rate = 2%
Explanation:
given data
radius = 2,300-ft
lanes width = 12-ft
no of lane = 2
design speed = 65-mph
solution
we get here sufficient sight distance SSD that is express as
SSD = 1.47 ut + ..............1
here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here
so put here value and we get
SSD = 1.47 × 65 ×2.5 +
solve it we get
SSD = 644 ft
so here minimum distance clear from the inside edge of the inside lane is
Ms = Rv ( 1 - ) .....................2
here Rv is = R - one lane width
Rv = 2300 - 6 = 2294 ft
put value in equation 2 we get
Ms = 2294 ( 1 - )
solve it we get
Ms = 22.57 ft
and
superelevation rate for the curve will be here as
R = ..................3
here f is coefficient of friction that is 0.10
put here value and we get e
2300 =
solve it we get
e = 2%