Answer:
1.8 mm
Explanation:
given data
thick = 2.5 mm
flux = 1 × kg/m²
high pressure surface is 2 kg/m³
solution
we use fick first law for steady state diffusion
J = D × ..........1
we take here Ca to point which concentration of nitrogen is 2 kg/m³
so we solve Xb
Xb = Xa + D ×
assume Xa = 0 at surface
Xb = 0 + ( 12 × ) ×
Xb = 1.8 ×
Xb = 1.8 mm
Answer:
0.0406 m/s
Explanation:
Given:
Diameter of the tube, D = 25 mm = 0.025 m
cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²
Mass flow rate = 0.01 kg/s
Now,
the mass flow rate is given as:
mass flow rate = ρAV
where,
ρ is the density of the water = 1000 kg/m³
A is the area of cross-section of the pipe
V is the average velocity through the pipe
thus,
0.01 = 1000 × 4.9 × 10⁻⁴ × V
or
V = 0.0203 m/s
also,
Reynold's number, Re =
where,
ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s
thus,
Re =
or
Re = 611.39 < 2000
thus,
the flow is laminar
hence,
the maximum velocity = 2 × average velocity = 2 × 0.0203 m/s
or
maximum velocity = 0.0406 m/s