Answer:
The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.
Explanation:
Here the first think you have to consider is the definition of the Reynolds number ([tex]Re[/tex]) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:
[tex]Re=\frac{\rho v D}{\mu}[/tex]
where [tex]\rho[/tex]is the density of the fluid, [tex]\mu[/tex] is the viscosity, D is the pipe diameter and v is the velocity of the fluid.
Now, we know that Re=2100. So the velocity is:
[tex]v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s[/tex]
For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:
[tex]D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm[/tex]
The velocity in pipe be "5.158 m/sec" and pipe diameter should be "0.0066 m".
According to the question,
Diameter, D = 10.0 mm or,
= 0.01 m
Oil density, ρ = 855 kg/m³
Viscosity, μ = 2.1 × 10⁻² pa.sec
Flow, Re = 2100
We know that,
Reynolds number, Re = [tex]\frac{DV \rho}{\mu}[/tex]
By substituting the values,
2100 = [tex]\frac{0.01\times v\times 855}{2.1\times 10^{-2}}[/tex]
v = 5.158 m/sec
Now,
Flowrate, Q = Velocity × Area
or,
= v × [tex]\frac{\pi}{4}[/tex] × D²
= 5.158 × [tex]\frac{3.14}{4}[/tex] × (0.01)²
= 4.05 × 10⁻⁴ m³/sec
hence,
The diameter of pipe be:
→ Re = [tex]\frac{Dv \rho}{\mu}[/tex]
2100 = [tex]\frac{D\times 5.158\times 925}{1.5\times 10^{-2}}[/tex]
D = 6.602 × 10⁻³ or,
= 0.0066 m
Thus the above response is correct.
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