Solve thisThis is from chapter mechanical properties of fluids

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