Arun
Last Activity: 5 Years ago
Let the speed of the flow be v.
The diameter of the tap = d = 1.25 cm = 1.25 ×10-2 m
Density of water = ρ = 103 kg m–3
Viscosity = η = 10-3 Pa s
The volume of the water flowing out per second is
Q = v × πd2/4
v = 4Q/d2π
The Reynolds number is given by
R = ρvd/η
= 4ρQ/πdη
= 4 ×103 × Q/(3.14 × 1.25 ×10-2 × 10-3)
= 1.019 × 108 Q
Initially, Q = 0.48 L / min = 0.48 × 10-3 / 60 = 8 × 10-6
R = 815
Since this is below 1000, the flow is steady.
After some time, Q = 3 L / min = 3 × 10-3 / 60 = 5 × 10-5
R = 5095
The flow will be turbulent