5. 5. Momentum & Heat Transfer Analogy

Example 5:Momentum & Heat Transfer Analogy Air flows through a smooth tube, 2.5 cm diameter and 10 m long, at 37oC. If the pressure drop through the tube is 10000 Pa, estimate

  • the air velocity through the tube and the friction factor

  • the heat transfer coefficient using Colburn Analogy [\(j_H = (\text{St})(\text{Pr})^{0.67}\)], where \(\text{St}\) is the Stanton Number and \(\text{Pr}\) is the Prandtl Number.

Gas constant, \(R\) = 82.06 cm3.atm/mol.K. Darcy friction factor \(= 0.184/\text{Re}^{0.2}\). Other relevant properties of air under the given conditions: viscosity = \(1.8\times10^{-5}\) kg/m.s, density = 1.134 kg/m3, specific heat capacity, \(C_p\) = 1.046 kJ/kg.oC, thermal conductivity = 0.028 W/m..oC (GATE-2002)

Solution: Pressure drop due to friction is related to velocity as \[\Delta P = \frac{2fL\rho v^2}{D} \label{pr1}\] Given: \(f=\text{ Darcy friction factor } = 0.184/\text{Re}^{0.2}\). \[\text{Darcy friction factor } = 4 \times \text{ Fanning friction factor}\] In Eqn.[1], \(f\) denotes Fanning friction factor. Therefore, \[f = 0.25 \times 0.184/\text{Re}^{0.2} = 0.046/\text{Re}^{0.2}\] Expanding, \[f = \frac{0.046}{(Dv\rho/\mu)^0.2} = \frac{0.046\mu^{0.2}}{(Dv\rho)^{0.2}} \label{pr2}\]


Substituting this in Eqn.[1], \[\begin{aligned} \Delta P &=& \frac{2\times0.046\times\mu^{0.2}L\rho v^2}{D^{1.2}v^{0.2}\rho^{0.2}}\\ &=& \frac{0.092\mu^{0.2}L\rho^{0.8}v^{1.8}}{D^{1.2}} \end{aligned}\] Substituting for the known quantities, \[10000 = \frac{0.092\times(1.8\times10^{-5})^{0.2}\times10\times(1.134)^{0.8}\times v^{1.8}}{(2.5\times10^{-2})^{1.2}}\] Solving, \(v\) = air velocity through the tube = 47.6 m/s.

From Eqn.[2], \[f = \frac{0.046\times(1.8\times10^{-5})^{0.2}}{(2.5\times10^{-2}\times22.02\times1.134)^{0.2}} = 0.0049\]

By Colburn analogy, \[j_H = (\text{St})(\text{Pr})^{0.67} = \frac{f}{2}\] where

\(\text{St}\) = Stanton number = \(\text{Nu}/(\text{Re}\cdot\text{Pr}) = h/(\rho C_pv)\)
\(\text{Pr}\) = Prandtl number = \(C_p\mu/k\)
\(f\) = Fanning friction factor

Therefore, \[\frac{h}{1.134\times1046\times47.6}\times\left(\frac{1046\times1.8\times10^{-5}}{0.028}\right)^{0.67} = \frac{0.0049}{2}\] Solving, \(h\) = heat transfer coefficient = 180.5 W/m2.K