Answer:
The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)
That is the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Explanation:
The diffusion through a stagnant layer is given by
Where
= Diffusion coefficient or diffusivity
z = Thickness in layer of transfer
R = universal gas constant
= Pressure at first boundary
= Pressure at the destination boundary
T = System temperature
= System pressure
Where = 101.3 kPa 
, 
, 
0.5×101.3 = 50.65 kPa
Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m
R = 
T = 298 K and = 1.18 
= 1.8×10⁻⁵
= 5.153×10⁻⁴
Hence the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Answer:
One inlet stream to the mixer flows at 100.0 kg/hr and is 35wt% species-A and 65wt% species-B.
Explanation:
Answer:
i) 0.610 m or 610 mm
ii) 0.4 m or 400 mm
Explanation:
The pressure difference between the pipes is
a) Air
Pa + πha +Ha = Pb + πhb +Hb
Pa - Pb = π(hb-ha) + Hb-Ha
Relative density of air = 1.2754 kg /m3
Pa - Pb = 1.2754 * 0.4 + (0.3-0.2) = 0.610 m or 610 mm
b) paraffin of relative density of 0.75
Pa - Pb = π(hb-ha) + Hb-Ha
Pa - Pb = 0.75 * 0.4 + (0.3-0.2) = 0.4 m or 400 mm
Answer:
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