Air at 52°C, 101.3 kPa, and 20 percent relative humidity enters a 5-cm-diameter tube with an average velocity of 6 m/s. The tube inner surface is wetted uniformly with water, whose vapor pressure at 52°C is 13.6 kPa. While the temperature and pressure of air remain constant, the partial pressure of vapor in the outlet air is increased to 10 kPa. Detemine (a) the average mass transfer coefficient in m/s, (b) the log-mean driving force for mass transfer in molar concentration units, (c) the water evaporation rate in kg/h, and (d) the length of the tube.

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The liquid layer on the inner surface of a circular pipe is dried by blowing air through it. The average mass transfer coefficient, log-mean driving force for mass transfer (in molar concentration units, the evaporation rate, and the tube length are to be determined.

Assumptions 1 The low mass flux model and thus the analogy between heat and mass transfer is applicable since the mass fraction of vapor in the air is low (about 2 percent for saturated air at 325 K). 2 The flow is fully developed.

Properties Because of low mass flux conditions, we can use dry air properties for the mixture at the specified temperature of 52?C and 1 atm, for which (Table A-15). The mass diffusivity of water vapor in air at 52+273 = 325 K is determined from Eq. 14-15 to be



Analysis (a) The Reynolds number of the flow is



which is greater than 10,000 and thus the flow is turbulent. The Schmidt number in this case is



Therefore, the Sherwood number in this case is determined from Table 14-13 to be



Using the definition of Sherwood number, the mass transfer coefficient is determined to be



(b) The log-mean driving force for mass transfer (in molar concentration units) is determined as follows



(c) The evaporation rate is determined from



(d) The tube length is determined from

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