Therefore, the distance below the tap outlet where turbulent flow commences is 56.5 mm.
To determine this
The Reynolds number can be calculated using the stream diameter (d = 5 mm = 0.005 m) and the kinematic viscosity of water at 5°C (v = 1.004 10-5 m2/s):
Re = ρvd/μ
where:
ρ is the density of water (ρ = 1000 kg/m³)
v is the velocity of the water (v = ?)
μ is the dynamic viscosity of water (μ = νρ = 1.004 × 10^-3 kg m^-1 s^-1)
Step 2: Determine the maximum velocity for laminar flow
To find the maximum velocity for laminar flow, we set Re = 2100 and solve for v:
2100 = (1000 kg/m³)(v)(0.005 m)/(1.004 × 10^-3 kg m^-1 s^-1)
Solving for v, we get:
v = 0.42 m/s
Therefore, the maximum velocity for the water to flow out of the tap in a laminar manner is 0.42 m/s.
Step 3: Calculate the distance for the onset of turbulent flow
If the initial velocity of the water is 0.42 m/s, the critical Reynolds number (Re_cr = 2300) can be used to calculate the distance below the tap outlet where turbulent flow begins. The crucial Reynolds number is the point at which a laminar flow becomes turbulent.
Assuming the flow is constant and incompressible, the Hagen-Poiseuille equation can be used to connect the pressure difference (P) between the tap outlet and the point where turbulent flow begins to the flow rate (Q) and fluid viscosity ():
ΔP = 8μLQ/πr⁴
where:
L is the distance below the tap outlet (L = ?)
r is the radius of the stream (r = 0.0025 m)
We can also relate the flow rate (Q) to the velocity of the water (v) and the cross-sectional area of the stream (A):
Q = Av
where:
A = πr²
Substituting the second equation into the first equation and solving for L, we get:
L = (4μvρ)/(πΔr³)
Using the initial velocity (v = 0.42 m/s), the density of water (ρ = 1000 kg/m³), the viscosity of water (μ = 1.004 × 10^-3 kg m^-1 s^-1), and the radius of the stream (r = 0.0025 m), we can calculate the distance (L):
L = (4(1.004 × 10^-3 kg m^-1 s^-1)(0.42 m/s)(1000 kg/m³))/(π(0.42 m/s)(0.0025 m)³
Solving for L, we get:
L = 0.0565 m = 56.5 mm