Analogy between Heat and Mass Transfer Heat transfer coefficients in complex geometries with complicated boundary conditions can be determined by mass transfer measurements on similar geometries under similar flow conditions using volatile solids such as naphthalene and dichlorobenzene and utilizing

Question

Analogy between Heat and Mass Transfer

Heat transfer coefficients in complex geometries with complicated boundary conditions can be determined by mass transfer measurements on similar geometries under similar flow conditions using volatile solids such as naphthalene and dichlorobenzene and utilizing the Chilton–Colburn analogy between heat and mass transfer at low mass flux conditions. The amount of mass transfer during a specified time period is determined by weighing the model or measuring the surface recession. During a certain experiment involving the flow of dry air at [latex] 25^{\circ}C [/latex] and 1 atm at a free stream velocity of 2 m/s over a body covered with a layer of naphthalene, it is observed that 12 g of naphthalene has sublimated in 15 min (Fig. 14–50). The surface area of the body is [latex] 0.3 m^{2} [/latex]. Both the body and the air were kept at [latex] 25^{\circ}C [/latex] during the study. The vapor pressure of naphthalene at [latex] 25^{\circ}C [/latex] is 11 Pa and the mass diffusivity of naphthalene in air at [latex] 25^{\circ}C [/latex] is [latex] D_{AB} = 0.61 imes 10^{-5} m^{2}/s [/latex]. Determine the heat transfer coefficient under the same flow conditions over the same geometry.

Accepted Answer

SOLUTION Air is blown over a body covered with a layer of naphthalene, and the rate of sublimation is measured. The heat transfer coefficient under the same flow conditions over the same geometry is to be determined.

Assumptions 1 The low mass flux conditions exist so that the Chilton–Colburn analogy between heat and mass transfer is applicable (will be verified). 2 Both air and naphthalene vapor are ideal gases.

Properties The molar mass of naphthalene is 128.2 kg/kmol. Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of [latex] 25^{\circ}C [/latex] and 1 atm, at which [latex] ho = 1.184 kg/m^{3}, C_{p} = 1007 J/kg \cdot K, [/latex] and [latex] \alpha = 2.141 imes 10^{-5} m^{2}/s [/latex] (Table A–15).

Analysis The incoming air is free of naphthalene, and thus the mass fraction of naphthalene at free stream conditions is zero, [latex] w_{A,\infty } = 0 [/latex]. Noting that the vapor pressure of naphthalene at the surface is 11 Pa, its mass fraction at the surface is determined to be

[latex] w_{A,s} = \frac{P_{A,s}}{P} \left ( \frac{M_{A}}{M_{air}} ight ) = \frac{11 Pa}{101,325 Pa} \left ( \frac{128.2 kg/kmol}{29 kg/kmol} ight ) = 4.8 imes 10^{-4} [/latex]

which confirms that the low mass flux approximation is valid. The rate of evaporation of naphthalene in this case is

[latex] \dot{m}_{evap} = \frac{m}{\Delta t} = \frac{0.012 kg}{\left ( 15 imes 60 s ight )} = 1.33 imes 10^{-5} kg/s [/latex]

Then the mass convection coefficient becomes

[latex] h_{mass} = \frac{\dot{m}}{ ho A_{s} \left ( w_{A,s} - w_{A,\infty } ight )} [/latex]

[latex] = \frac{1.33 imes 10^{-5} kg/s}{\left ( 1.184 kg/m^{3} ight ) \left ( 0.3 m^{2} ight )\left ( 4.8 imes 10^{-4} - 0 ight )} = 0.0780 m/s [/latex]

Using the analogy between heat and mass transfer, the average heat transfer coefficient is determined from Eq. to be

[latex] h_{heat} = ho C_{p} h_{mass} \left ( \frac{\alpha }{D_{AB}} ight )^{2/3} = [/latex]

[latex]= \left ( 1.184 kg/m^{3} ight )\left ( 1007 J/kg \cdot ^{\circ}C ight )\left ( 0.0776 m/s ight ) \left ( \frac{2.141 imes 10^{-5} m^{2}/s}{ 0.61 imes 10^{-5} m^{2}/s} ight )^{2/3} [/latex]

[latex] = 215 W/m^{2} \cdot ^{\circ}C [/latex]

Discussion Because of the convenience it offers, naphthalene has been used in numerous heat transfer studies to determine convection heat transfer coefficients.

Question : Analogy between Heat and Mass Transfer Heat transfer coeffic...

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