Question

Water flows through a tube of diameter 25 mm at an average velocity of 1.0m/s. The properties of water ρ = 1000 kg/m³, μ = 7.25 × 10^-4 N.s/m², k = 0.625 W/mK, Pr = 4.85. Using Nu = 0.023 Re^0.8Pr^0.4, the convective heat transfer coefficient (in W/m².K) is ______

Answer 1

Concept:

Reynold’s number for a pipe is given by:

\(Re = \frac{{\rho VD}}{\mu}\)

where, ρ = density of water, V = velocity of water, D = diameter of pipe, μ = dynamic viscosity

Nusselt number for a pipe is given by:

\(Nu = \frac{{hD}}{k}\)

where, h = heat transfer coefficient, D = diameter of the pipe, k = thermal conductivity

Calculation:

Given:

ρ = 1000 kg/m³, μ = 7.25 × 10^-4 N.s/m², k = 0.625 W/mK, Pr = 4.85, V = 1 m/s

D = 25 mm = 0.025 m

Reynold’s number

\(Re = \frac{{\rho VD}}{\mu } = \frac{{1000 \times 1 \times 0.025}}{{7.25 \times {{10}^{ - 4}}}} = 34482.758\)

Now,

 Nu = 0.023 Re^0.8Pr^0.4

Nu = 0.023 × (34482.758)^0.8 × (4.85)^0.4

Nu = 184.5466

\(\frac{{hD}}{k} = 184.5466\)

\(h = \frac{{184.5466 \times 0.625}}{{0.025}} = 4613.66\frac{W}{{{m^2}K}}\)