Question

Water flows through a 10 mm diameter and 250 m long smooth pipe at an average velocity of 0.1 m/s. The density and the viscosity of water are 997 kg/m³ and 855 × 10^-6 N.s/m², respectively. Assuming fully-developed flow, the pressure drop (in Pa) in the pipe is _______

Answer 1

Concept:

• Reynold’s number: It is the ratio of inertia force and viscous force.

          \({R_e} = \frac{{\rho VD}}{\mu }\) (for a pipe)

          where,

          ρ = density of the fluid, V = Velocity of fluid, D = diameter of pipe, μ = dynamic viscosity of fluid

• For a pipe flow:

          If Reynold’s number (Re) < 2000 → flow is laminar

          If Reynold’s number (Re) > 4000 → flow is turbulent

          If 2000 < Re < 4000 → flow is transitional

• Pressure drop in a pipe is given by:

          ΔP = \(\frac{{32\mu VL}}{{{D^2}}}\)

          where,

          μ = Dynamic viscosity of fluid (Ns/m²)

          V = Velocity of fluid (m/s)

          D = Diameter of pipe (m)

Calculation:

Given:

D = 10 mm = 0.01 m

L = 250 m

\(\begin{array}{l} \rho = 997\;kg/{m^3}\\\mu = 855 \times {10^{ - 6}}Ns/{m^2} \end{array}\)

V = 0.1 m/s

Reynold's number

\({R_e} = \frac{{\rho VD}}{\mu } = \frac{{997 \times 0.1 \times 0.01}}{{855 \times {{10}^{ - 6}}}}\)

⇒ R_e = 1166.08 < 2000

∴ flow is laminar.

Pressure drop in the pipe,

\(\begin{array}{l} {\rm{\Delta }}p = \frac{{32\mu VL}}{{{D^2}}}\\ \Rightarrow \Delta p= \frac{{32 \times 0.1 \times 855 \times {{10}^{ - 6}} \times 250}}{{{{\left( {0.01} \right)}^2}}} = 6840\;Pa \end{array}\)​