Answer:
In both cases, reynolds number is greater than 2000,so the flows can't be laminar.
Explanation:
A) For flow in a tube of uniform diameter, the reynolds number is defined as;
Re = 2ρvr/η
where;
ρ is the fluid density,
v its speed,
η is viscosity
r is the tube radius.
In this question,
We are given;
r = 0.25cm = 0.25 × 10^(-2) m
η of water has a standard value of 1.005 × 10^(-3)
ρ of water has a standard value of 1000 kg/m³
In the reynolds equation, we don't know the velocity. So let's calculate it from;
Q' = vA
Where; Q' is flow rate = 0.5 L/s = 0.0005 m³/s
Area = πr² = π × (0.25 × 10^(-2))²
Area = 1.963 × 10^(-5) m²
So, v = Q/A = 0.0005/(1.963 × 10^(-5)) = 25.5 m/s
So, Re = 2ρvr/η = (2*1000*25.5*0.25 × 10^(-2))/(1.005 × 10^(-3))
Re = 126865.67
Re > 2000 and so the flow is not laminar
B) Now,
radius = 0.9cm = 0.9 × 10^(-2) m
So, A = πr² = π × (0.9 × 10^(-2))²
A = 2.5447 × 10^(-4) m²
v = Q/A = 0.0005/(2.5447 × 10^(-4))
v = 1.965 m/s
Re = 2ρvr/η = (2*1000*1.965*0.9 × 10^(-2))/(1.005 × 10^(-3))
Re = 35194.03
Re > 2000. So flow is not laminar.
