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Contoh Soal Geankoplis

This document provides an example calculation to determine the rate of evaporation of water diffusing through air in a metal tube. Given information includes the temperatures and pressures of the air and water, as well as the diffusion path length. The example uses equations from Geankoplis to calculate the rate of evaporation in both English and SI units. Key steps involve determining vapor pressures, converting values using appropriate conversion factors, and substituting into the relevant equations to obtain the final results.

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Alfhine Takamoto
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258 views11 pages

Contoh Soal Geankoplis

This document provides an example calculation to determine the rate of evaporation of water diffusing through air in a metal tube. Given information includes the temperatures and pressures of the air and water, as well as the diffusion path length. The example uses equations from Geankoplis to calculate the rate of evaporation in both English and SI units. Key steps involve determining vapor pressures, converting values using appropriate conversion factors, and substituting into the relevant equations to obtain the final results.

Uploaded by

Alfhine Takamoto
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Geankoplis
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Example 6.2-2 Geankoplis

Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K.

The total pressure of air (assumed dry) is 1.01325 x 105 Pa (1.0 atm) and the

temperature is 293 K (20 °C). Water evaporates and diffuses through the air in the tube,

and the diffusion path z2-z1 is 0.1524 m (0.5 ft) long. The diagram is similar to Fig. 6.2-

2a. Calculate the rate of evaporation at steady state in lb mol/h.ft2 and kg mol/s.m2. The

diffusivity of water vapor at 293 K and 1 am pressure is 0.250 x 10-4 m2/s. Assume that

the system is isothermal. Use SI and English units.


Solution
• The diffusivity is converted to ft2/h by using the conversion factor:

• From Appendix A.2 the vapor pressure of water at 20 °C is 17.54 mm, or pA1 =
17.54/760 = 0.0231 atm = 0.0231(1.01325 x 105) = 2.341 x 103 Pa, pA2 = 0 (pure air).
Since the temperature is 20 °C (68 °F), T = (460 + 68) °R = 293 K. From Appendix
A.1, R = 0.730 ft3.atm/lb mol.°R . To calculate the value of pBM from Eq. (6.2-21)
• Since pB1 is close to pB2, the linear mean (pB1+pB2)/2 could be used and would be
very close to pBM.
• Substituting into Eq. (6.2-22) with z2-z1 = 0.5 ft (0.1524m),

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