Question

Fully developed conditions are known to exist for water flowing through a 25-mm-diameter tube at 0.01 kg/s and 27 C. What is the maximum velocity of the water in the tube

Answer 1

Answer:

0.0406 m/s

Explanation:

Given:

Diameter of the tube, D = 25 mm = 0.025 m

cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²

Mass flow rate = 0.01 kg/s

Now,

the mass flow rate is given as:

mass flow rate = ρAV

where,

ρ is the density of the water = 1000 kg/m³

A is the area of cross-section of the pipe

V is the average velocity through the pipe

thus,

0.01 = 1000 × 4.9 × 10⁻⁴ × V

or

V = 0.0203 m/s

also,

Reynold's number, Re = $\frac{VD}{\nu}$

where,

ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s

thus,

Re = $\frac{0.0203\times0.025}{0.833\times10^{-6}}$

or

Re = 611.39 < 2000

thus,

the flow is laminar

hence,

the maximum velocity =  2 × average velocity = 2 × 0.0203 m/s

or

maximum velocity = 0.0406 m/s