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

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 25^{\circ}C 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 0.3  m^{2} . Both the body and the air were kept at 25^{\circ}C during the study. The vapor pressure of naphthalene at 25^{\circ}C is 11 Pa and the mass diffusivity of naphthalene in air at 25^{\circ}C is D_{AB} = 0.61 \times 10^{-5}  m^{2}/s . Determine the heat transfer coefficient under the same flow conditions over the same geometry.

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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 25^{\circ}C and 1 atm, at which \rho = 1.184  kg/m^{3},  C_{p} = 1007  J/kg \cdot K, and \alpha = 2.141 \times 10^{-5}  m^{2}/s (Table A–15).

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

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

 

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

\dot{m}_{evap} = \frac{m}{\Delta t} = \frac{0.012  kg}{\left ( 15 \times 60  s \right )} = 1.33 \times 10^{-5}  kg/s

 

Then the mass convection coefficient becomes

h_{mass} = \frac{\dot{m}}{\rho A_{s} \left ( w_{A,s} – w_{A,\infty } \right )}

 

                     = \frac{1.33 \times 10^{-5}  kg/s}{\left ( 1.184  kg/m^{3} \right ) \left ( 0.3 m^{2} \right )\left ( 4.8 \times 10^{-4}  –  0 \right )} = 0.0780  m/s

 

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

h_{heat} = \rho  C_{p} h_{mass} \left ( \frac{\alpha }{D_{AB}} \right )^{2/3} =

 

=  \left ( 1.184  kg/m^{3} \right )\left ( 1007  J/kg \cdot  ^{\circ}C \right )\left ( 0.0776  m/s \right ) \left ( \frac{2.141 \times 10^{-5}  m^{2}/s}{ 0.61 \times 10^{-5}  m^{2}/s} \right )^{2/3}

 

= 215  W/m^{2} \cdot  ^{\circ}C

 

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