Respuesta :
The total evaporation loss of water is 87.873 ×[tex]10^{-5}[/tex] lbm /day.
Explanation:
Assume A is the water and B is air.
A is diffusing to non diffusing B.
[tex]N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)[/tex]
By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is [tex]0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}[/tex].
Total pressure P = 1 atm = 101.325 KPa
[tex]P_{A_{1} }[/tex] = 23.76 mm Hg
[tex]P_{A_{1} }[/tex] = [tex]\frac{23.76}{760}[/tex]
[tex]P_{A_{1} }[/tex] = 0.03126 atm
[tex]P_{A_{1} }[/tex] = 3.167 K Pa
When air surrounded is dry air, then [tex]PA_{2}[/tex] = 0 mm Hg
R = 8.314 [tex]\frac{K P a \cdot m^{3}}{mol \cdot k}[/tex]
[tex]P_{B/M}[/tex] = [tex]\frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}[/tex]
[tex]P_{B_{1}}=P_{T}-P_{A1_{}}[/tex]
= 101.325 - 3.167
[tex]P_{B_{1} }[/tex] = 98.158 K Pa
[tex]P_{B_{2}}=P_{T}-P_{A_{2}}[/tex]
[tex]P_{B_{2} }[/tex] = 101.325 - 0
[tex]P_{B_{2} }[/tex] = 101.325 K Pa
[tex]P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}[/tex]
[tex]P_{B/M}[/tex] = 99.733 K Pa
Z = 1 ft = 0.3048 m
T = 298 K
[tex]N_{A}[/tex] = [tex]\frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }[/tex]
[tex]N_{A}[/tex] = 0.11077 × [tex]10^{-6}[/tex] mol/ [tex]m^{2}[/tex].s
[tex]N_{A}[/tex] = 0.11077 × [tex]10^{-6}[/tex] × 18 × (60×60×24)
[tex]N_{A}[/tex] = 0.1723 [tex]lb_{m}[/tex] / [tex]m^{2}[/tex].day
[tex]\tilde{N}_{A}=N_{A} \times Area[/tex]
Area of individual pipe is
[tex]A=\frac{\pi}{4}(0.0254)^{2}[/tex]
A = 0.00051 [tex]m^{2}[/tex]
[tex]\bar{N}_{\boldsymbol{A}}[/tex] = 0.1723 × 0.00051
[tex]\bar{N}_{\boldsymbol{A}}[/tex] = 0.000087873 lbm/day
In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is
= 10 × 0.000087873 = 0.00087873 lbm /day
The total evaporation loss of water is 87.873 ×[tex]10^{-5}[/tex] lbm /day.
